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5682bf2e-6055-4b76-b576-97a1edcdcc28 | # ANATOMY OF FLOWERING PLANTS
# CHAPTER 6
# 6.1 The Tissue System
You can very easily see the structural similarities and variations in the external morphology of the larger living organism, both plants and animals. Similarly, if we were to study the internal structure, one also finds several similarities as well as differences. This chapter introduces you to the internal structure and functional organisation of higher plants. Study of internal structure of plants is called anatomy. Plants have cells as the basic unit, cells are organised into tissues and in turn the tissues are organised into organs. Different organs in a plant show differences in their internal structure. Within angiosperms, the monocots and dicots are also seen to be anatomically different. Internal structures also show adaptations to diverse environments.
# 6.1 THE TISSUE SYSTEM
We were discussing types of tissues based on the types of cells present. Let us now consider how tissues vary depending on their location in the plant body. Their structure and function would also be dependent on location. On the basis of their structure and location, there are three types of tissue systems. These are the epidermal tissue system, the ground or fundamental tissue system and the vascular or conducting tissue system.
# 6.1.1 Epidermal Tissue System
The epidermal tissue system forms the outer-most covering of the whole plant body and comprises epidermal cells, stomata and the epidermal appendages – the trichomes and hairs. The epidermis is the outermost layer of the primary plant body. It is made up of elongated, compactly | 0 | 11 | Biology | 06 |
3a2958c7-de76-4f2f-9517-69f30901e718 | # BIOLOGY
arranged cells, which form a continuous layer. Epidermis is usually single-layered. Epidermal cells are parenchymatous with a small amount of cytoplasm lining the cell wall and a large vacuole. The outside of the epidermis is often covered with a waxy thick layer called the cuticle which prevents the loss of water. Cuticle is absent in roots. Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two bean-shaped cells known as guard cells which enclose stomatal pore. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells (away from the stomatal pore) are thin and the inner walls (towards the stomatal pore) are highly thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus (Figure 6.1).
# 6.1.1 Epidermal Cells
- Epidermal cells
- Subsidiary cells
- Chloroplast
- Guard cells
- Stomatal pore
Figure 6.1 Diagrammatic representation: (a) stomata with bean-shaped guard cells (b) stomata with dumb-bell shaped guard cell
The cells of epidermis bear a number of hairs. The root hairs are unicellular elongations of the epidermal cells and help absorb water and minerals from the soil. On the stem the epidermal hairs are called trichomes. The trichomes in the shoot system are usually multicellular. They may be branched or unbranched and soft or stiff. They may even be secretory. The trichomes help in preventing water loss due to transpiration.
# 6.1.2 The Ground Tissue System
All tissues except epidermis and vascular bundles constitute the ground tissue. It consists of simple tissues such as parenchyma, collenchyma and sclerenchyma. Parenchymatous cells are usually present in cortex, pericycle, pith and medullary rays, in the primary stems and roots. In leaves, the ground tissue consists of thin-walled chloroplast containing cells and is called mesophyll.
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c06d56d6-f476-4561-a69d-26c9150127ca | # ANATOMY OF FLOWERING PLANTS
# 6.1.3 The Vascular Tissue System
The vascular system consists of complex tissues, the phloem and the xylem. The xylem and phloem together constitute vascular bundles (Figure 6.2). In dicotyledonous stems, cambium is present between phloem and xylem. Such vascular bundles because of the presence of cambium possess the ability to form secondary xylem and phloem tissues, and hence are called open vascular bundles. In the monocotyledons, the vascular bundles have no cambium present in them. Hence, since they do not form secondary tissues they are referred to as closed. When xylem and phloem within a vascular bundle are arranged in an alternate manner along the different radii, the arrangement is called radial such as in roots. In conjoint type of vascular bundles, the xylem and phloem are jointly situated along the same radius of vascular bundles. Such vascular bundles are common in stems and leaves. The conjoint vascular bundles usually have the phloem located only on the outer side of xylem.
# 6.2 ANATOMY OF DICOTYLEDONOUS AND MONOCOTYLEDONOUS PLANTS
For a better understanding of tissue organisation of roots, stems and leaves, it is convenient to study the transverse sections of the mature zones of these organs.
# Figure 6.2 Various types of vascular bundles:
(a) radial
(b) conjoint closed
(c) conjoint open
# 6.2.1 Dicotyledonous Root
Look at Figure 6.3 (a), it shows the transverse section of the sunflower root. The internal tissue organisation is as follows: The outermost layer is epiblema. Many of the cells of epiblema protrude in the form of unicellular root hairs. The cortex consists of several layers of thin-walled parenchyma cells. | 2 | 11 | Biology | 06 |
5d35dcb8-9f1b-43b6-ab2c-90f7a9f7c8e1 | # 74
# BIOLOGY
with intercellular spaces. The innermost layer of the cortex is called endodermis. It comprises a single layer of barrel-shaped cells without any intercellular spaces. The tangential as well as radial walls of the endodermal cells have a deposition of water-impermeable, waxy material suberin in the form of casparian strips. Next to endodermis lies a few layers of thick-walled parenchyomatous cells referred to as pericycle. Initiation of lateral roots and vascular cambium during the secondary growth takes place in these cells. The pith is small or inconspicuous.
(a) The parenchymatous cells which lie between the xylem and the phloem are called conjunctive tissue. There are usually two to four xylem and phloem patches. Later, a cambium ring develops between the xylem and phloem. All tissues on the innerside of the endodermis such as pericycle, vascular bundles and pith constitute the stele.
# 6.2.2 Monocotyledonous Root
The anatomy of the monocot root is similar to the dicot root in many respects (Figure 6.3 b). It has epidermis, cortex, endodermis, pericycle, vascular bundles and pith. As compared to the dicot root which have fewer xylem bundles, there are usually more than six (polyarch) xylem bundles in the monocot root. Pith is large and well developed. Monocotyledonous roots do not undergo any secondary growth.
# 6.2.3 Dicotyledonous Stem
The transverse section of a typical young dicotyledonous stem shows that the epidermis is the outermost protective layer of the stem.
# Figure 6.3
T.S.: (a) Dicot root (Primary) (b) Monocot root
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3ef1b079-bfe3-431b-a457-40442f049504 | # ANATOMY OF FLOWERING PLANTS
(Figure 6.4 a). Covered with a thin layer of cuticle, it may bear trichomes and a few stomata. The cells arranged in multiple layers between epidermis and pericycle constitute the cortex. It consists of three sub-zones. The outer hypodermis consists of a few layers of collenchymatous cells just below the epidermis, which provide mechanical strength to the young stem. Cortical layers below hypodermis consist of rounded thin walled parenchymatous cells with conspicuous intercellular spaces. The innermost layer of the cortex is called the endodermis. The cells of the endodermis are rich in starch grains and the layer is also referred to as the starch sheath. Pericycle is
# Figure 6.4 T.S. of stem : (a) Dicot (b) Monocot
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adfae95d-8ebf-4057-a3fe-a50ca5b3f059 | # 76
# BIOLOGY
present on the inner side of the endodermis and above the phloem in the form of semi-lunar patches of sclerenchyma. In between the vascular bundles there are a few layers of radially placed parenchymatous cells, which constitute medullary rays. A large number of vascular bundles are arranged in a ring; the ‘ring’ arrangement of vascular bundles is a characteristic of dicot stem. Each vascular bundle is conjoint, open, and with endarch protoxylem. A large number of rounded, parenchymatous cells with large intercellular spaces which occupy the central portion of the stem constitute the pith.
# 6.2.4 Monocotyledonous Stem
The monocot stem has a sclerenchymatous hypodermis, a large number of scattered vascular bundles, each surrounded by a sclerenchymatous bundle sheath, and a large, conspicuous parenchymatous ground tissue (Figure 6.4 b). Vascular bundles are conjoint and closed. Peripheral vascular bundles are generally smaller than the centrally located ones. The phloem parenchyma is absent, and water-containing cavities are present within the vascular bundles.
# 6.2.5 Dorsiventral (Dicotyledonous) Leaf
The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, epidermis, mesophyll and vascular system. The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata. The tissue between the upper and the lower epidermis is called the mesophyll. Mesophyll, which possesses chloroplasts and carry out photosynthesis, is made up of parenchyma. It has two types of cells – the palisade parenchyma and the spongy parenchyma. The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged.
Figure 6.5 T.S. of leaf : (a) Dicot (b) Monocot
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04dade93-54f7-4eda-9db0-16eb6cac54b4 | # ANATOMY OF FLOWERING PLANTS
vertically and parallel to each other. The oval or spongy parenchyma is situated below the palisade cells and extends to the lower epidermis. There are numerous large spaces and air cavities between these cells. The vascular system includes vascular bundles, which can be seen in the veins and the midrib. The size of the vascular bundles are dependent on the size of the veins. The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick walled bundle sheath cells. Look at Figure 6.5 (a) and find the position of xylem in the vascular bundle.
# 6.2.6 Isobilateral (Monocotyledonous) Leaf
The anatomy of isobilateral leaf is similar to that of the dorsiventral leaf in many ways. It shows the following characteristic differences. In an isobilateral leaf, the stomata are present on both the surfaces of the epidermis; and the mesophyll is not differentiated into palisade and spongy parenchyma (Figure 6.5 b).
In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells. These are called bulliform cells. When the bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards to minimise water loss.
The parallel venation in monocot leaves is reflected in the near similar sizes of vascular bundles (except in main veins) as seen in vertical sections of the leaves.
# SUMMARY
Anatomically, a plant is made of different kinds of tissues. The plant tissues are broadly classified into meristematic (apical, lateral and intercalary) and permanent (simple and complex). Assimilation of food and its storage, transportation of water, minerals and photosynthates, and mechanical support are the main functions of tissues. There are three types of tissue systems – epidermal, ground and vascular. The epidermal tissue systems are made of epidermal cells, stomata and the epidermal appendages. The ground tissue system forms the main bulk of the plant. It is divided into three zones – cortex, pericycle and pith. The vascular tissue system is formed by the xylem and phloem. On the basis of presence of cambium, location of xylem and phloem, the vascular bundles are of different types. The vascular bundles form the conducting tissue and translocate water, minerals and food material.
Monocotyledonous and dicotyledonous plants show marked variation in their internal structures. They differ in type, number and location of vascular bundles. The secondary growth occurs in most of the dicotyledonous roots and stems.
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f9a2f85c-e719-4b49-84b7-81d93f21f4b3 | # BIOLOGY
# EXERCISES
1. Draw illustrations to bring out the anatomical difference between
- (a) Monocot root and Dicot root
- (b) Monocot stem and Dicot stem
2. Cut a transverse section of young stem of a plant from your school garden and
observe it under the microscope. How would you ascertain whether it is a
monocot stem or a dicot stem? Give reasons.
3. The transverse section of a plant material shows the following anatomical
features -
- (a) the vascular bundles are conjoint, scattered and surrounded by a sclerenchymatous bundle sheaths.
- (b) phloem parenchyma is absent. What will you identify it as?
4. What is stomatal apparatus? Explain the structure of stomata with a labelled diagram.
5. Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
6. How is the study of plant anatomy useful to us?
7. Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.
# 2024-25 | 7 | 11 | Biology | 06 |
0d2a54cc-7d75-46ee-b718-877a09a35eff | # CHAPTER 17
# LOCOMOTION AND MOVEMENT
# 17.1 Types of Movement
Movement is one of the significant features of living beings. Animals and plants exhibit a wide range of movements. Streaming of protoplasm in the unicellular organisms like Amoeba is a simple form of movement. Movement of cilia, flagella and tentacles are shown by many organisms.
# 17.2 Muscle
Human beings can move limbs, jaws, eyelids, tongue, etc. Some of the movements result in a change of place or location. Such voluntary movements are called locomotion. Walking, running, climbing, flying, swimming are all some forms of locomotory movements. Locomotory structures need not be different from those affecting other types of movements. For example, in Paramoecium, cilia helps in the movement of food through cytopharynx and in locomotion as well. Hydra can use its tentacles for capturing its prey and also use them for locomotion. We use limbs for changes in body postures and locomotion as well. The above observations suggest that movements and locomotion cannot be studied separately. The two may be linked by stating that all locomotions are movements but all movements are not locomotions.
# 17.3 Skeletal System
Methods of locomotion performed by animals vary with their habitats and the demand of the situation. However, locomotion is generally for search of food, shelter, mate, suitable breeding grounds, favourable climatic conditions or to escape from enemies/predators.
# 17.4 Joints
Cells of the human body exhibit three main types of movements, namely, amoeboid, ciliary and muscular.
# 17.5 Disorders of Muscular and Skeletal System
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1a57d946-30b7-4cd9-a2ca-a5108623024d | # BIOLOGY
Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.
Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.
Movement of our limbs, jaws, tongue, etc, require muscular movement. The contractile property of muscles are effectively used for locomotion and other movements by human beings and majority of multicellular organisms. Locomotion requires a perfect coordinated activity of muscular, skeletal and neural systems. In this chapter, you will learn about the types of muscles, their structure, mechanism of their contraction and important aspects of the skeletal system.
# 17.2 MUSCLE
You have studied in Chapter 8 that the cilia and flagella are the outgrowths of the cell membrane. Flagellar movement helps in the swimming of spermatozoa, maintenance of water current in the canal system of sponges and in locomotion of Protozoans like Euglena. Muscle is a specialised tissue of mesodermal origin. About 40-50 per cent of the body weight of a human adult is contributed by muscles. They have special properties like excitability, contractility, extensibility and elasticity. Muscles have been classified using different criteria, namely location, appearance and nature of regulation of their activities. Based on their location, three types of muscles are identified: (i) Skeletal (ii) Visceral and (iii) Cardiac.
Skeletal muscles are closely associated with the skeletal components of the body. They have a striped appearance under the microscope and hence are called striated muscles. As their activities are under the voluntary control of the nervous system, they are known as voluntary muscles too. They are primarily involved in locomotory actions and changes of body postures.
Visceral muscles are located in the inner walls of hollow visceral organs of the body like the alimentary canal, reproductive tract, etc. They do not exhibit any striation and are smooth in appearance. Hence, they are called smooth muscles (nonstriated muscle). Their activities are not under the voluntary control of the nervous system and are therefore known as involuntary muscles. They assist, for example, in the transportation of food through the digestive tract and gametes through the genital tract. | 1 | 11 | Biology | 17 |
32cbfa28-4212-4297-b690-bc956cc2620a | # LOCOMOTION AND MOVEMENT
As the name suggests, Cardiac muscles are the muscles of heart. Many cardiac muscle cells assemble in a branching pattern to form a cardiac muscle. Based on appearance, cardiac muscles are striated. They are involuntary in nature as the nervous system does not control their activities directly.
Let us examine a skeletal muscle in detail to understand the structure and mechanism of contraction. Each organised skeletal muscle in our body is made of a number of muscle bundles or fascicles held together by a common collagenous connective tissue layer called fascia. Each muscle bundle contains a number of muscle fibres (Figure 17.1).
Fascicle (muscle bundle)
Muscle fibre (muscle cell)
Sarcolemma
Blood capillary
Figure 17.1 Diagrammatic cross sectional view of a muscle showing muscle bundles and muscle fibres
Each muscle fibre is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm. Muscle fibre is a syncitium as the sarcoplasm contains many nuclei. The endoplasmic reticulum, i.e., sarcoplasmic reticulum of the muscle fibres is the store house of calcium ions. A characteristic feature of the muscle fibre is the presence of a large number of parallelly arranged filaments in the sarcoplasm called myofilaments or myofibrils. Each myofibril has alternate dark and light bands on it. A detailed study of the myofibril has established that the striated appearance is due to the distribution pattern of two important proteins – Actin and Myosin. The light bands contain actin and is called I-band or Isotropic band, whereas the dark band called ‘A’ or Anisotropic band contains | 2 | 11 | Biology | 17 |
4313b38d-eca3-4002-8481-822b1867222f | # 220
# BIOLOGY
myosin. Both the proteins are arranged as rod-like structures, parallel to each other and also to the longitudinal axis of the myofibrils. Actin filaments are thinner as compared to the myosin filaments, hence are commonly called thin and thick filaments respectively. In the centre of each ‘I’ band is an elastic fibre called ‘Z’ line which bisects it. The thin filaments are firmly attached to the ‘Z’ line. The thick filaments in the ‘A’ band are also held together in the middle of this band by a thin fibrous membrane called ‘M’ line. The ‘A’ and ‘I’ bands are arranged alternately throughout the length of the myofibrils. The portion of the myofibril between two successive ‘Z’ lines is considered as the functional unit of contraction and is called a sarcomere (Figure 17.2). In a resting state, the edges of thin filaments on either side of the thick filaments partially overlap the free ends of the thick filaments leaving the central part of the thick filaments. This central part of thick filament, not overlapped by thin filaments is called the ‘H’ zone.
# (a)
# (b)
Figure 17.2 Diagrammatic representation of (a) anatomy of a muscle fibre showing a sarcomere (b) a sarcomere
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f27b9de1-dba2-4669-975f-c190b9872ce4 | # LOCOMOTION AND MOVEMENT
# 17.2.1 Structure of Contractile Proteins
Each actin (thin) filament is made of two ‘F’ (filamentous) actins helically wound to each other. Each ‘F’ actin is a polymer of monomeric ‘G’ (Globular) actins. Two filaments of another protein, tropomyosin also run close to the ‘F’ actins throughout its length. A complex protein Troponin is distributed at regular intervals on the tropomyosin. In the resting state a subunit of troponin masks the active binding sites for myosin on the actin filaments (Figure 17.3a).
Each myosin (thick) filament is also a polymerised protein. Many monomeric proteins called Meromyosins (Figure 17.3b) constitute one thick filament. Each meromyosin has two important parts, a globular head with a short arm and a tail, the former being called the heavy meromyosin (HMM) and the latter component, i.e.; the head and short arm projects outwards at regular distance and angle from each other from the surface of a polymerised myosin filament and is known as cross arm. The globular head is an active ATPase enzyme and has binding sites for ATP and active sites for actin.
(a)
Actin binding sites Head ATP binding sites
(b)
Figure 17.3 (a) An actin (thin) filament (b) Myosin monomer (Meromyosin)
# 17.2.2 Mechanism of Muscle Contraction
Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.
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e762b418-2b19-4985-89c9-05774d8055bc | # BIOLOGY
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron along with the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (Acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge (Figure 17.4).
Figure 17.4 Stages in cross bridge formation, rotation of head and breaking of cross bridge
This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. It is clear from the above steps, that during shortening of the muscle, i.e., contraction, the ‘I’ bands get reduced, whereas the ‘A’ bands retain the length (Figure 17.5). The myosin, releasing the ADP and P goes back to its relaxed state. A new ATP binds and the cross-bridge is broken (Figure 17.4). The ATP is again hydrolysed by the myosin head and the cycle of cross bridge formation. | 5 | 11 | Biology | 17 |
b2bebe10-fc48-49ba-8839-42c2e47adb93 | # LOCOMOTION AND MOVEMENT
# 223
# Figure 17.5
Sliding-filament theory of muscle contraction (movement of the thin filaments and the relative size of the I band and H zones)
and breakage is repeated causing further sliding. The process continues till the Ca++ ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments. This causes the return of ‘Z’ lines back to their original position, i.e., relaxation. The reaction time of the fibres can vary in different muscles. Repeated activation of the muscles can lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen in them, causing fatigue. Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red fibres. These muscles also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production. These muscles, therefore, can also be called aerobic muscles. On the other hand, some of the muscles possess very less quantity of myoglobin and therefore, appear pale or whitish. These are the White fibres. Number of mitochondria are also few in them, but the amount of sarcoplasmic reticulum is high. They depend on anaerobic process for energy.
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1a2c95b2-05aa-4fd4-93b6-5a1d3b90c0a5 | # 17.3 SKELETAL SYSTEM
Skeletal system consists of a framework of bones and a few cartilages. This system has a significant role in movement shown by the body. Imagine chewing food without jaw bones and walking around without the limb bones. Bone and cartilage are specialised connective tissues. The former has a very hard matrix due to calcium salts in it and the latter has slightly pliable matrix due to chondroitin salts. In human beings, this system is made up of 206 bones and a few cartilages. It is grouped into two principal divisions – the axial and the appendicular skeleton.
Axial skeleton comprises 80 bones distributed along the main axis of the body. The skull, vertebral column, sternum and ribs constitute axial skeleton. The skull (Figure 17.6) is composed of two sets of bones – cranial and facial, that totals to 22 bones. Cranial bones are 8 in number. They form the hard protective outer covering, cranium for the brain. The facial region is made up of 14 skeletal elements which form the front part of the skull. A single U-shaped bone called hyoid is present at the base of the buccal cavity and it is also included in the skull. Each middle ear contains three tiny bones – Malleus, Incus and Stapes, collectively called Ear Ossicles. The skull region articulates with the superior region of the
# Figure 17.6 Diagrammatic view of human skull
|Frontal bone|Parietal bone|Sphenoid bone|
|---|---|---|
|Ethmoid bone|Lacrimal bone|Nasal bone|
|Temporal bone|Zygomatic bone|Occipital bone|
|Maxilla|Occipital condyle|Mandible|
|Hyoid bone|Hyoid bone|Hyoid bone| | 7 | 11 | Biology | 17 |
2c0a36a7-b9d5-4bbb-815e-59810eabb8d4 | # LOCOMOTION AND MOVEMENT
# 225
Our vertebral column (Figure 17.7) is formed by 26 serially arranged units called vertebrae and is dorsally placed. It extends from the base of the skull and constitutes the main framework of the trunk. Each vertebra has a central hollow portion (neural canal) through which the spinal cord passes. First vertebra is the atlas and it articulates with the occipital condyles. The vertebral column is differentiated into cervical (7), thoracic (12), lumbar (5), sacral (1-fused) and coccygeal (1-fused) regions starting from the skull. The number of cervical vertebrae are seven in almost all mammals including human beings. The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back. Sternum is a flat bone on the ventral midline of thorax.
There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic. First seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage. The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. These are called vertebrochondral (false) ribs. Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are therefore, called floating ribs. Thoracic vertebrae, ribs and sternum together form the rib cage (Figure 17.8).
The bones of the limbs along with their girdles constitute the appendicular skeleton. Each limb is made of 30 bones. The bones of the hand (fore limb) are
# Figure 17.7 Vertebral column (right lateral view)
# Figure 17.8 Ribs and rib cage | 8 | 11 | Biology | 17 |
1f8fd354-59db-4fa1-98e1-26f2a757ac31 | # BIOLOGY
# 17.4 JOINTS
Joints are essential for all types of movements involving the bony parts of the body. Locomotory movements are no exception to this.
# Figure 17.9
Right pectoral girdle and upper arm. (frontal view)
Humerus, radius and ulna, carpals (wrist bones – 8 in number), metacarpals (palm bones – 5 in number) and phalanges (digits – 14 in number) (Figure 17.9). Femur (thigh bone – the longest bone), tibia and fibula, tarsals (ankle bones – 7 in number), metatarsals (5 in number) and phalanges (digits – 14 in number) are the bones of the legs (hind limb) (Figure 17.10). A cup shaped bone called patella cover the knee ventrally (knee cap).
Pectoral and Pelvic girdle bones help in the articulation of the upper and the lower limbs respectively with the axial skeleton. Each girdle is formed of two halves. Each half of pectoral girdle consists of a clavicle and a scapula (Figure 17.9). Scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs. The dorsal, flat, triangular body of scapula has a slightly elevated ridge called the spine which projects as a flat, expanded process called the acromion. The clavicle articulates with this. Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint. Each clavicle is a long slender bone with two curvatures. This bone is commonly called the collar bone.
Pelvic girdle consists of two coxal bones (Figure 17.10). Each coxal bone is formed by the fusion of three bones – ilium, ischium and pubis. At the point of fusion of the above bones is a cavity called acetabulum to which the thigh bone articulates. The two halves of the pelvic girdle meet ventrally to form the pubic symphysis containing fibrous cartilage.
# Figure 17.10
Right pelvic girdle and lower limb bones (frontal view) | 9 | 11 | Biology | 17 |
d0a6f37f-4ce4-4f2a-837d-0a4e24daff6c | # LOCOMOTION AND MOVEMENT
This. Joints are points of contact between bones, or between bones and cartilages. Force generated by the muscles is used to carry out movement through joints, where the joint acts as a fulcrum. The movability at these joints vary depending on different factors. Joints have been classified into three major structural forms, namely, fibrous, cartilaginous and synovial.
Fibrous joints do not allow any movement. This type of joint is shown by the flat skull bones which fuse end-to-end with the help of dense fibrous connective tissues in the form of sutures, to form the cranium.
In cartilaginous joints, the bones involved are joined together with the help of cartilages. The joint between the adjacent vertebrae in the vertebral column is of this pattern and it permits limited movements.
Synovial joints are characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones. Such an arrangement allows considerable movement. These joints help in locomotion and many other movements. Ball and socket joint (between humerus and pectoral girdle), hinge joint (knee joint), pivot joint (between atlas and axis), gliding joint (between the carpals) and saddle joint (between carpal and metacarpal of thumb) are some examples.
# 17.5 DISORDERS OF MUSCULAR AND SKELETAL SYSTEM
- Myasthenia gravis: Auto immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.
- Muscular dystrophy: Progressive degeneration of skeletal muscle mostly due to genetic disorder.
- Tetany: Rapid spasms (wild contractions) in muscle due to low Ca++ in body fluid.
- Arthritis: Inflammation of joints.
- Osteoporosis: Age-related disorder characterised by decreased bone mass and increased chances of fractures. Decreased levels of estrogen is a common cause.
- Gout: Inflammation of joints due to accumulation of uric acid crystals.
# SUMMARY
Movement is an essential feature of all living beings. Protoplasmic streaming, ciliary movements, movements of fins, limbs, wings, etc., are some forms exhibited by animals. A voluntary movement which causes the animal to change its place, is | 10 | 11 | Biology | 17 |
a581edc2-60cb-4c21-9102-b8e60d2aa1f0 | # BIOLOGY
called locomotion. Animals move generally in search of food, shelter, mate, breeding ground, better climate or to protect themselves.
The cells of the human body exhibit amoeboid, ciliary and muscular movements. Locomotion and many other movements require coordinated muscular activities. Three types of muscles are present in our body. Skeletal muscles are attached to skeletal elements. They appear striated and are voluntary in nature. Visceral muscles, present in the inner walls of visceral organs are nonstriated and involuntary. Cardiac muscles are the muscles of the heart. They are striated, branched and involuntary. Muscles possess excitability, contractility, extensibility and elasticity.
Muscle fibre is the anatomical unit of muscle. Each muscle fibre has many parallelly arranged myofibrils. Each myofibril contains many serially arranged units called sarcomere which are the functional units. Each sarcomere has a central ‘A’ band made of thick myosin filaments, and two half ‘I’ bands made of thin actin filaments on either side of it marked by ‘Z’ lines. Actin and myosin are polymerised proteins with contractility. The active sites for myosin on resting actin filament are masked by a protein-troponin. Myosin head contains ATPase and has ATP binding sites and active sites for actin. A motor neuron carries signal to the muscle fibre which generates an action potential in it. This causes the release of Ca++ from sarcoplasmic reticulum. Ca++ activates actin which binds to the myosin head to form a cross bridge. These cross bridges pull the actin filaments causing them to slide over the myosin filaments and thereby causing contraction. Ca++ are then returned to sarcoplasmic reticulum which inactivate the actin. Cross bridges are broken and the muscles relax.
Repeated stimulation of muscles leads to fatigue. Muscles are classified as Red and White fibres based primarily on the amount of red coloured myoglobin pigment in them.
Bones and cartilages constitute our skeletal system. The skeletal system is divisible into axial and appendicular. Skull, vertebral column, ribs and sternum constitute the axial skeleton. Limb bones and girdles form the appendicular skeleton. Three types of joints are formed between bones or between bone and cartilage – fibrous, cartilaginous and synovial. Synovial joints allow considerable movements and therefore, play a significant role in locomotion.
# EXERCISES
1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.
2. Define sliding filament theory of muscle contraction.
3. Describe the important steps in muscle contraction.
2024-25 | 11 | 11 | Biology | 17 |
16a88be0-d2f6-4378-b063-d3dff595f4e6 | # LOCOMOTION AND MOVEMENT
# 229
# 4. Write true or false. If false change the statement so that it is true.
- (a) Actin is present in thin filament
- (b) H-zone of striated muscle fibre represents both thick and thin filaments.
- (c) Human skeleton has 206 bones.
- (d) There are 11 pairs of ribs in man.
- (e) Sternum is present on the ventral side of the body.
# 5. Write the difference between:
- (a) Actin and Myosin
- (b) Red and White muscles
- (c) Pectoral and Pelvic girdle
# 6. Match Column I with Column II:
|Column I|Column II|
|---|---|
|(a) Smooth muscle|(iv) Involuntary|
|(b) Tropomyosin|(ii) Thin filament|
|(c) Red muscle|(i) Myoglobin|
|(d) Skull|(iii) Sutures|
# 7. What are the different types of movements exhibited by the cells of human body?
# 8. How do you distinguish between a skeletal muscle and a cardiac muscle?
# 9. Name the type of joint between the following:
- (a) atlas/axis
- (b) carpal/metacarpal of thumb
- (c) between phalanges
- (d) femur/acetabulum
- (e) between cranial bones
- (f) between pubic bones in the pelvic girdle
# 10. Fill in the blank spaces:
- (a) All mammals (except a few) have __________ cervical vertebra.
- (b) The number of phalanges in each limb of human is __________
- (c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely __________ and __________.
- (d) In a muscle fibre Ca++ is stored in __________
- (e) __________ and __________ pairs of ribs are called floating ribs.
- (f) The human cranium is made of __________ bones.
# 2024-25 | 12 | 11 | Biology | 17 |
45db6140-00aa-4639-8a5f-61c54adfe10f | # CHAPTER 12
# RESPIRATION IN PLANTS
# 12.1 Do Plants Breathe?
All of us breathe to live, but why is breathing so essential to life? What happens when we breathe? Also, do all living organisms, including plants and microbes, breathe? If so, how?
# 12.2 Glycolysis
All living organisms need energy for carrying out daily life activities, be it absorption, transport, movement, reproduction or even breathing.
# 12.3 Fermentation
Where does all this energy come from? We know we eat food for energy – but how is this energy taken from food? How is this energy utilised? Do all foods give the same amount of energy? Do plants ‘eat’? Where do plants get their energy from? And micro-organisms – for their energy requirements, do they eat ‘food’?
# 12.4 Aerobic Respiration
You may wonder at the several questions raised above – they may seem to be very disconnected. But in reality, the process of breathing is very much connected to the process of release of energy from food. Let us try and understand how this happens.
# 12.5 The Respiratory Balance Sheet
All the energy required for ‘life’ processes is obtained by oxidation of some macromolecules that we call ‘food’. Only green plants and cyanobacteria can prepare their own food; by the process of photosynthesis they trap light energy and convert it into chemical energy that is stored in the bonds of carbohydrates like glucose, sucrose and starch.
# 12.6 Amphibolic Pathway
We must remember that in green plants too, not all cells, tissues and organs photosynthesise; only cells containing chloroplasts, that are most often located in the superficial layers, carry out photosynthesis. Hence, even in green plants all other organs, tissues and cells that are non-green, need food for oxidation. Hence, food has to be translocated to all non-green parts.
# 12.7 Respiratory Quotient
Animals are heterotrophic, i.e., they obtain food from plants. | 0 | 11 | Biology | 12 |
a97767ed-4b67-4567-8a1c-c692c23395e5 | # BIOLOGY
directly (herbivores) or indirectly (carnivores). Saprophytes like fungi are dependent on dead and decaying matter. What is important to recognise is that ultimately all the food that is respired for life processes comes from photosynthesis. This chapter deals with cellular respiration or the mechanism of breakdown of food materials within the cell to release energy, and the trapping of this energy for synthesis of ATP.
Photosynthesis, of course, takes place within the chloroplasts (in the eukaryotes), whereas the breakdown of complex molecules to yield energy takes place in the cytoplasm and in the mitochondria (also only in eukaryotes). The breaking of the C-C bonds of complex compounds through oxidation within the cells, leading to release of considerable amount of energy is called respiration. The compounds that are oxidised during this process are known as respiratory substrates. Usually carbohydrates are oxidised to release energy, but proteins, fats and even organic acids can be used as respiratory substances in some plants, under certain conditions. During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP. Hence, it is important to understand that the energy released by oxidation in respiration is not (or rather cannot be) used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell. This energy trapped in ATP is utilised in various energy-requiring processes of the organisms, and the carbon skeleton produced during respiration is used as precursors for biosynthesis of other molecules in the cell.
# 12.1 DO PLANTS BREATHE?
Well, the answer to this question is not quite so direct. Yes, plants require O2 for respiration to occur and they also give out CO2. Hence, plants have systems in place that ensure the availability of O2. Plants, unlike animals, have no specialised organs for gaseous exchange but they have stomata and lenticels for this purpose. There are several reasons why plants can get along without respiratory organs. First, each plant part takes care of its own gas-exchange needs. There is very little transport of gases from one plant part to another. Second, plants do not present great demands for gas exchange. Roots, stems and leaves respire at rates far lower than animals do. Only during photosynthesis are large volumes of gases exchanged and, each leaf is well adapted to take care of its own needs during these periods. When cells photosynthesise, availability of O2 is not a problem in these cells since O2 is released within the cell. | 1 | 11 | Biology | 12 |
715a6db5-dd59-48b9-8f7d-425149c9c848 | # RESPIRATION IN PLANTS
The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant. ‘This is true for leaves’, you may ask, ‘but what about thick, woody stems and roots?’ In stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.
The complete combustion of glucose, which produces CO2 and H2O as end products, yields energy most of which is given out as heat.
C6H12O6 + 6O2 ⟶ 6CO2 + 6H2O + Energy
If this energy is to be useful to the cell, it should be able to utilise it to synthesise other molecules that the cell requires. The strategy that the plant cell uses is to catabolise the glucose molecule in such a way that not all the liberated energy goes out as heat. The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough such that the energy released can be coupled to ATP synthesis. How this is done is, essentially, the story of respiration.
During the process of respiration, oxygen is utilised, and carbon dioxide, water and energy are released as products. The combustion reaction requires oxygen. But some cells live where oxygen may or may not be available. Can you think of such situations (and organisms) where O2 is not available? There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen. Even among present-day living organisms, we know of several that are adapted to anaerobic conditions. Some of these organisms are facultative anaerobes, while in others the requirement for anaerobic condition is obligate. In any case, all living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen. This breakdown of glucose to pyruvic acid is called glycolysis.
# 12.2 GLYCOLYSIS
The term glycolysis has originated from the Greek words, glycos for sugar, and lysis for splitting. The scheme of glycolysis was given by Gustav Embden, Otto Meyerhof, and J. Parnas, and is often referred to as the EMP pathway. In anaerobic organisms, it is the only process in respiration. Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants, this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage. | 2 | 11 | Biology | 12 |
0a439953-e9b4-4d28-a8e7-8572839f49b1 | # 156
# BIOLOGY
|Glucose|ATP (6C)|ADP|Glucose-6-phosphate (6C)|
|---|---|---|---|
|Fructose-6-phosphate|ATP (6C)|ADP|Fructose 1, 6-bisphosphate (6C)|
|Triose phosphate|Triose phosphate|(glyceraldehyde-3-phosphate)|(Dihydroxy acetone phosphate)|
|NAD+| |NADH+H+| |
|2 × Triose bisphosphate|(1,3 bisphosphoglyceric acid)| | |
|ADP|ATP|2 × Triose phosphate|(3-phosphoglyceric acid)|
|2 × 2-phosphoglycerate|HO|2 × phosphoenolpyruvate|ADP|
|ATP|2 × Pyruvic acid|(3C)| |
Figure 12.1 Steps of glycolysis
Sucrose is converted into glucose and fructose by the enzyme, invertase, and these two monosaccharides readily enter the glycolytic pathway. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are same. The various steps of glycolysis are depicted in Figure 12.1. In glycolysis, a chain of ten reactions, under the control of different enzymes, takes place to produce pyruvate from glucose. While studying the steps of glycolysis, please note the steps at which utilisation or synthesis of ATP or (in this case) NADH + H+ take place.
ATP is utilised at two steps: first in the conversion of glucose into glucose 6-phosphate and second in the conversion of fructose 6-phosphate to fructose 1, 6-bisphosphate. The fructose 1, 6-bisphosphate is split into dihydroxyacetone phosphate and 3-phosphoglyceraldehyde (PGAL). We find that there is one step where NADH + H+ is formed from NAD+; this is when 3-phosphoglyceraldehyde (PGAL) is converted to 1, 3-bisphosphoglycerate (BPGA). Two redox-equivalents are removed (in the form of two hydrogen atoms) from PGAL and transferred to a molecule of NAD+. PGAL is oxidised and with inorganic phosphate to get converted into BPGA. The conversion of BPGA to 3-phosphoglyceric acid (PGA), is also an energy yielding process; this energy is trapped by the formation of ATP. Another ATP is synthesised during the conversion of PEP to pyruvic acid.
Can you then calculate how many ATP molecules are directly synthesised in this pathway from one glucose molecule?
Pyruvic acid is then the key product of glycolysis. What is the metabolic fate of pyruvate? This depends on the cellular need. | 3 | 11 | Biology | 12 |
b39eb309-d3dc-4cd7-9e5b-01de0d96d2f1 | # RESPIRATION IN PLANTS
There are three major ways in which different cells handle pyruvic acid produced by glycolysis. These are lactic acid fermentation, alcoholic fermentation and aerobic respiration. Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to CO2 and H2O, however, organisms adopt Krebs’ cycle which is also called as aerobic respiration. This requires O2 supply.
# 12.3 FERMENTATION
In fermentation, say by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to CO2 and ethanol. The enzymes, pyruvic acid decarboxylase and alcohol dehydrogenase catalyse these reactions. Other organisms like some bacteria produce lactic acid from pyruvic acid. The steps involved are shown in Figure 12.2. In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid by lactate dehydrogenase. The reducing agent is NADH + H+ which is reoxidised to NAD+ in both the processes.
In both lactic acid and alcohol fermentation not much energy is released; less than seven per cent of the energy in glucose is released and not all of it is trapped as high energy bonds of ATP. Also, the processes are hazardous – either acid or alcohol is produced.
What is the net ATP that is synthesised (calculate how many ATP are synthesised and deduct the number of ATP utilised during glycolysis) when one molecule of glucose is fermented to alcohol or lactic acid? Yeasts poison themselves to death when the concentration of alcohol reaches about 13 per cent. What then would be the maximum concentration of alcohol in beverages that are naturally fermented? How do you think alcoholic beverages of alcohol content greater than this concentration are obtained?
What then is the process by which organisms can carry out complete oxidation of glucose and extract the energy stored to | 4 | 11 | Biology | 12 |
d6611084-928c-4218-a257-a38175d83ae5 | # BIOLOGY
synthesise a larger number of ATP molecules needed for cellular metabolism? In eukaryotes these steps take place within the mitochondria and this requires O. Aerobic respiration is the process that leads to a complete oxidation of organic substances in the presence of oxygen, and releases CO2, water and a large amount of energy present in the substrate. This type of respiration is most common in higher organisms. We will look at these processes in the next section.
# 12.4 AEROBIC RESPIRATION
For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria. The crucial events in aerobic respiration are:
- The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO2.
- The passing on of the electrons removed as part of the hydrogen atoms to molecular O with simultaneous synthesis of ATP.
What is interesting to note is that the first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria. Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including NAD and Coenzyme A.
|Pyruvic acid|CoA|NAD+|
|---|---|---|
|Pyruvate dehydrogenase → Mg2+|Pyruvate dehydrogenase → Mg2+|Pyruvate dehydrogenase → Mg2+|
|Acetyl CoA|CO2|NADH|
|H+|H+|H+|
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis). The acetyl CoA then enters a cyclic pathway, tricarboxylic acid cycle, more commonly called as Krebs’ cycle after the scientist Hans Krebs who first elucidated it.
# 12.4.1 Tricarboxylic Acid Cycle
The TCA cycle starts with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yield citric acid (Figure 12.3). The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released. Citrate is then isomerised to isocitrate. It is followed by two successive steps of decarboxylation, leading to the formation of α-ketoglutaric acid. | 5 | 11 | Biology | 12 |
8e8f6b82-292b-49c3-a48d-860f1860d006 | # RESPIRATION IN PLANTS
and then succinyl-CoA. In the remaining steps of citric acid cycle, succinyl-CoA is oxidised to Pyruvate (3C) allowing the cycle to continue. During the conversion of succinyl-CoA to succinic acid a molecule of GTP is synthesised. This is a substrate level phosphorylation. In a coupled reaction GTP is converted to GDP with the simultaneous synthesis of ATP from ADP.
Also there are three points in the cycle where NAD is reduced to NADH + H+ and one point where FAD+ is reduced to FADH2. The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid, the first member of the cycle. In addition it also requires regeneration of NAD+ and FAD+ from NADH and FADH2 respectively. The summary equation for this phase of respiration may be written as follows:
Pyruvic acid
+ 4NAD+ + FAD+ + 2 H2O + ADP + Pi
→ 3CO2 + 4NADH + 4H+ + FADH2 + ATP
We have till now seen that glucose has been broken down to release CO2 and eight molecules of NADH + H+; two of FADH2 have been synthesised besides just two molecules of ATP in TCA cycle. You may be wondering why we have been discussing respiration at all – neither O2 has come into the picture nor the promised large number of ATP has yet been synthesised. Also what is the role of the NADH + H+ and FADH2 that is synthesised? Let us now understand the role of O2 in respiration and how ATP is synthesised.
# 12.4.2 Electron Transport System (ETS) and Oxidative Phosphorylation
The following steps in the respiratory process are to release and utilise the energy stored in NADH + H+ and FADH2. This is accomplished when they are oxidised through the electron transport system and the electrons are passed on to O2 resulting in the formation of H2O. The metabolic pathway through which the electron passes from one carrier to another, is called the electron transport system (ETS) and it is present in the inner mitochondrial membrane. Electrons from NADH are passed through the electron transport system. | 6 | 11 | Biology | 12 |
c8546d49-f969-4f6d-8be7-5ae875e27ffc | # BIOLOGY
produced in the mitochondrial matrix during citric acid cycle are oxidised by an NADH dehydrogenase (complex I), and electrons are then transferred to ubiquinone located within the inner membrane. Ubiquinone also receives reducing equivalents via FADH (complex II) that is generated during oxidation of succinate in the citric acid cycle. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome b c1 complex (complex III). Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex III and IV.
|Inter-membrane space|Inner Mitochondrial membrane|Matrix|
|---|---|---|
|NADH + H+|(Fe-S) < FMN <- 2e-|NAD|
|Complex I|(NADH dehydrogenase)| |
|QH|Complex II|(Succinate dehydrogenase)|
|(Fe-S) < FAD <-|Fumarate| |
|Complex III|(Cytochrome bc1)| |
|Cyt c <~Fe-S<-Cyt b| | |
|UQH|Complex IV|(Cytochrome oxidase)|
|ADP + P|ATP synthase|ATP|
When the electrons pass from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate. The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH produces 2 molecules of ATP. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.
You have already studied about the mechanism of membrane-linked ATP synthesis as explained by chemiosmotic hypothesis in the earlier chapter. As mentioned earlier, the energy released during the electron
Figure 12.4 Electron Transport System (ETS) | 7 | 11 | Biology | 12 |
0c3acc69-29f3-4d2b-8617-c862cc763273 | # RESPIRATION IN PLANTS
transport system is utilised in synthesising ATP with the help of ATP synthase (complex V). This complex consists of two major components, F1 and F0 (Figure 12.5). The F1 headpiece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. F0 is an integral membrane protein complex that forms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F1 component for the production of ATP. For each ATP produced, 4H+ passes through F0 from the intermembrane space to the matrix down the electrochemical proton gradient.
Figure 12.5 Diagramatic presentation of ATP synthesis in mitochondria
# 12.5 THE RESPIRATORY BALANCE SHEET
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
- There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
- The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
- None of the intermediates in the pathway are utilised to synthesise any other compound.
- Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.
But this kind of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 38 ATP molecules during aerobic respiration of one molecule of glucose. | 8 | 11 | Biology | 12 |
bb5cd83d-92a6-4e4e-9f9c-3bc1fdc6f0e9 | # 12.6 AMPHIBOLIC PATHWAY
Glucose is the favoured substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired, as has been mentioned earlier then they do not enter the respiratory pathway at the first step. See Figure 12.6 to see the points of entry of different substrates in the respiratory pathway. Fats would need to be broken down into glycerol and fatty acids first. If fatty acids were to be respired they would first be degraded to acetyl CoA and enter the pathway. Glycerol would enter the pathway after being converted to PGAL. The proteins would be degraded by proteases and the individual amino acids (after deamination) depending on their structure would enter the pathway at some stage within the Krebs’ cycle or even as pyruvate or acetyl CoA.
Since respiration involves breakdown of substrates, the respiratory process has traditionally been considered a catabolic process and the respiratory pathway as a catabolic pathway. But is this understanding correct? We have discussed above, at which points in the respiratory pathway different substrates would enter if they were to be respired and used to derive energy. What is important to recognize is that it is these very compounds that would be withdrawn from the respiratory pathway for the synthesis of the said substrates. Hence, fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesize fatty acids, CoA would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of protein too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one. | 9 | 11 | Biology | 12 |
c1f4fe03-c609-4288-8a3b-5bad526d57cc | # RESPIRATION IN PLANTS
# 12.7 RESPIRATORY QUOTIENT
Let us now look at another aspect of respiration. As you know, during aerobic respiration, O2 is consumed and CO2 is released. The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.
RQ = volume of CO2 evolved / volume of O2 consumed
The respiratory quotient depends upon the type of respiratory substrate used during respiration.
When carbohydrates are used as substrate and are completely oxidised, the RQ will be 1, because equal amounts of CO2 and O2 are evolved and consumed, respectively, as shown in the equation below: | 10 | 11 | Biology | 12 |
908b8218-c5c8-421f-aca9-b26f2baa9107 | # BIOLOGY
C6H12 + 6O2 ⟶ 6CO2 + 6H2O + Energy
RQ = 6CO2 / 6O2 = 1.0
When fats are used in respiration, the RQ is less than 1. Calculations for a fatty acid, tripalmitin, if used as a substrate is shown:
2(C51H98O6) + 145O2 ⟶ 102CO2 + 98H2O + Energy
Tripalmitin
RQ = 102CO2 / 145O2 = 0.7
When proteins are respiratory substrates the ratio would be about 0.9.
What is important to recognise is that in living organisms respiratory substrates are often more than one; pure proteins or fats are never used as respiratory substrates.
# SUMMARY
Plants unlike animals have no special systems for breathing or gaseous exchange. Stomata and lenticels allow gaseous exchange by diffusion. Almost all living cells in a plant have their surfaces exposed to air.
The breaking of C-C bonds of complex organic molecules by oxidation cells leading to the release of a lot of energy is called cellular respiration. Glucose is the favoured substrate for respiration. Fats and proteins can also be broken down to yield energy. The initial stage of cellular respiration takes place in the cytoplasm. Each glucose molecule is broken through a series of enzyme catalysed reactions into two molecules of pyruvic acid. This process is called glycolysis.
The fate of the pyruvate depends on the availability of oxygen and the organism. Under anaerobic conditions either lactic acid fermentation or alcohol fermentation occurs. Fermentation takes place under anaerobic conditions in many prokaryotes, unicellular eukaryotes and in germinating seeds. In eukaryotic organisms aerobic respiration occurs in the presence of oxygen. Pyruvic acid is transported into the mitochondria where it is converted into acetyl CoA with the release of CO2. Acetyl CoA then enters the tricarboxylic acid pathway or Krebs’ cycle operating in the matrix of the mitochondria. NADH + H+ and FADH2 are generated in the Krebs’ cycle. The energy in these molecules as well as that in the NADH + H+ synthesised during glycolysis are used to synthesise ATP. This is accomplished through a | 11 | 11 | Biology | 12 |
aa2c9f1f-61a0-4c91-9add-2109f9980910 | # RESPIRATION IN PLANTS
system of electron carriers called electron transport system (ETS) located on the inner membrane of the mitochondria. The electrons, as they move through the system, release enough energy that are trapped to synthesise ATP. This is called oxidative phosphorylation. In this process O2 is the ultimate acceptor of electrons and it gets reduced to water.
The respiratory pathway is an amphibolic pathway as it involves both anabolism and catabolism. The respiratory quotient depends upon the type of respiratory substance used during respiration.
# EXERCISES
1. Differentiation between
1. Respiration and Combustion
2. Glycolysis and Krebs’ cycle
3. Aerobic respiration and Fermentation
2. What are respiratory substrates? Name the most common respiratory substrate.
3. Give the schematic representation of glycolysis?
4. What are the main steps in aerobic respiration? Where does it take place?
5. Give the schematic representation of an overall view of Krebs’ cycle.
6. Explain ETS.
7. Distinguish between the following:
1. Aerobic respiration and Anaerobic respiration
2. Glycolysis and Fermentation
3. Glycolysis and Citric acid Cycle
8. What are the assumptions made during the calculation of net gain of ATP?
9. Discuss “The respiratory pathway is an amphibolic pathway.”
10. Define RQ. What is its value for fats?
11. What is oxidative phosphorylation?
12. What is the significance of step-wise release of energy in respiration?
2024-25 | 12 | 11 | Biology | 12 |
0b9fa7d2-212a-4e7b-bbcb-ff8d6d96a29e | # Unit 3
# Classification of Elements and Periodicity in Properties
The Periodic Table is arguably the most important concept in chemistry, both in principle and in practice. It is the everyday support for students, it suggests new avenues of research to professionals, and it provides a succinct organization of the whole of chemistry. It is a remarkable demonstration of the fact that the chemical elements are not a random cluster of entities but instead display trends and lie together in families. An awareness of the Periodic Table is essential to anyone who wishes to disentangle the world and see how it is built up from the fundamental building blocks of the chemistry, the chemical elements.
After studying this Unit, you will be able to:
- appreciate how the concept of grouping elements in accordance to their properties led to the development of Periodic Table.
- understand the Periodic Law;
- understand the significance of atomic number and electronic configuration as the basis for periodic classification;
- name the elements with Z > 100 according to IUPAC nomenclature;
- classify elements into s, p, d, f blocks and learn their main characteristics;
- recognise the periodic trends in physical and chemical properties of elements;
- compare the reactivity of elements and correlate it with their occurrence in nature;
- explain the relationship between ionization enthalpy and metallic character;
- use scientific vocabulary appropriately to communicate ideas related to certain important properties of atoms e.g., atomic/ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valence of elements.
# 3.1 Why do We Need to Classify Elements?
We know by now that the elements are the basic units of all types of matter. In 1800, only 31 elements were known. By 1865, the number of identified elements had more than doubled to 63. At present, 114 elements are known. Of them, the recently discovered elements are man-made. Efforts to synthesise new elements are continuing. With such a large number of elements it is very difficult to study individually the chemistry of all these elements and their innumerable compounds individually. To ease out this problem, scientists searched for a systematic way to organise their knowledge by classifying the elements. Not only that it would rationalize known chemical facts about elements, but even predict new ones for undertaking further study. | 0 | 11 | Chemistry | 103 |
a8c5fe37-053a-4e7e-ba3f-152f9c1e6c72 | # 3.2 Genesis of Periodic Classification
The classification of elements into groups and development of Periodic Law and Periodic Table are the consequences of systematising the knowledge gained by a number of scientists through their observations and experiments. The German chemist, Johann Dobereiner in early 1800’s was the first to consider the idea of trends among properties of elements. By 1829 he noted a similarity among the physical and chemical properties of several groups of three elements (triads). In each case, he noticed that the middle element of each of the triads had an atomic weight about half way between the atomic weights of the other two (Table 3.1). Also the properties of the middle element were in between those of the other two members. Since Dobereiner’s relationship, referred to as the law of triads, seemed to work only for a few elements, it was dismissed as coincidence.
The next reported attempt to classify elements was made by a French geologist, A.E.B. de Chancourtois in 1862. He arranged the then known elements in order of increasing atomic weights and made a cylindrical table of elements to display the periodic recurrence of properties. This also did not attract much attention. The English chemist, John Alexander Newlands in 1865 propounded the law of octaves. He arranged the elements in increasing order of their atomic weights and noted that every eighth element had properties similar to the first element (Table 3.2). The relationship was just like every eighth note that resembles the first in octaves of music. Newlands’s Law of Octaves seemed to be true only for elements up to calcium. Although his idea was not widely accepted at that time, he, for his work, was later awarded Davy Medal in 1887 by the Royal Society, London.
The Periodic Law, as we know it today owes its development to the Russian chemist, Dmitri Mendeleev (1834-1907) and the German chemist, Lothar Meyer (1830-1895). Working independently, both the chemists in 1869 proposed that on arranging elements in the increasing order of their atomic weights, similarities appear in physical and chemical properties at regular intervals.
# Table 3.1 Dobereiner’s Triads
|Element|Atomic Weight|Element|Atomic Weight|Element|Atomic Weight|
|---|---|---|---|---|---|
|Li|7|Ca|40|Cl|35.5|
|Na|23|Sr|88|Br|80|
|K|39|Ba|137|I|127|
# Table 3.2 Newlands’ Octaves
|Element|Li|Be|B|C|N|O|F|
|---|---|---|---|---|---|---|---|
|At. wt.|7|9|11|12|14|16|19|
|Element|Na|Mg|Al|Si|P|S|Cl|
|At. wt.|23|24|27|29|31|32|35.5|
|Element|K|Ca| | | | | |
|At. wt.|39|40| | | | | | | 1 | 11 | Chemistry | 103 |
b236068b-02f4-42fe-a988-e79bd8c6b95e | # Chemistry
a periodically repeated pattern. Unlike classification if the order of atomic weight was strictly followed. He ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together. For example, iodine with lower atomic weight than that of tellurium (Group VI) was placed in Group VII along with fluorine, chlorine, bromine because of similarities in properties (Fig. 3.1). At the same time, keeping his primary aim of arranging the elements of similar properties in the same group, he proposed that some of the elements were still undiscovered and, therefore, left several gaps in the table. For example, both gallium and germanium were unknown at the time Mendeleev published his Periodic Table. Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. Mendeleev’s system of classifying elements was more elaborate than that of Lothar Meyer’s. He fully recognized the significance of periodicity and used broader range of physical and chemical properties to classify the elements. In particular, Mendeleev relied on the similarities in the empirical formulas and properties of the compounds formed by the elements. He realized that some of the elements did not fit in with his scheme of table 3.3.
# Table 3.3 Mendeleev’s Predictions for the Elements Eka-aluminium (Gallium) and Eka-silicon (Germanium)
|Property|Eka-aluminium (predicted)|Gallium (found)|Eka-silicon (predicted)|Germanium (found)|
|---|---|---|---|---|
|atomic weight|68|70|72|72.6|
|density/(g/cm³)|5.9|5.94|5.5|5.36|
|melting point/K|Low|302.93|High|1231|
|formula of oxide|E₂O₃|Ga₂O₃|EO₂|GeO₂|
|formula of chloride|E Cl₃|GaCl₃|ECl₄|GeCl₄| | 2 | 11 | Chemistry | 103 |
3e11aa85-2be2-467f-8440-eb312ead9ca7 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
# SERIES AND GROUPS IN THE PERIODIC SYSTEM OF ELEMENTS
# 2024-25
Fig. 3.1 Mendeleev’s Periodic Table published earlier
77 | 3 | 11 | Chemistry | 103 |
14a2607e-069a-408e-8a41-13ce3c4bcfb4 | # Chemistry
# 3.3 Modern Periodic Law and the Physical and Chemical Properties of Elements
We must bear in mind that when Mendeleev developed his Periodic table, chemists knew nothing about the internal structure of atom. However, the beginning of the 20th century witnessed profound developments in theories about sub-atomic particles. In 1913, the English physicist, Henry Moseley observed regularities in the characteristic X-ray spectra of the elements. A plot of (where is frequency of X-rays emitted) against atomic number (Z) gave a straight line and not the plot of vs atomic mass. He thereby showed that the atomic number is a more fundamental property of an element than its atomic mass. Mendeleev’s Periodic law was, therefore, accordingly modified. This is known as the modern Periodic law and can be stated as:
The physical and chemical properties of the elements are periodic functions of their atomic numbers.
The Periodic law revealed important analogies among the 94 naturally occurring elements (neptunium and plutonium like actinium and protoactinium are also found in pitch blende – an ore of uranium). It stimulated renewed interest in Inorganic Chemistry and has carried into the present with the creation of artificially produced short-lived elements.
You may recall that the atomic number is equal to the nuclear charge (i.e., number of protons) or the number of electrons in a neutral atom. It is then easy to visualize the significance of quantum numbers and electronic configurations in periodicity of elements. In fact, it is now recognized that the Periodic Law is essentially the consequence of the periodic variation in electronic configurations, which indeed determine the properties of the elements.
* Glenn T. Seaborg’s work in the middle of the 20th century starting with the discovery of plutonium in 1940, followed by those of all the transuranium elements from 94 to 102 led to reconfiguration of the periodic table placing the actinoids below the lanthanoids. In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been named Seaborgium (Sg) in his honour.
# 3.4 Nomenclature of Elements with Atomic Numbers > 100
The naming of the new elements had been traditionally the privilege of the discoverer (or discoverers) and the suggested name was ratified by the IUPAC. In recent years this has led to some controversy. The new elements with very high atomic numbers are so unstable that only minute quantities, sometimes only. | 4 | 11 | Chemistry | 103 |
84da2dfb-4d27-4be2-9f8b-3f1193de89fd | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
IA
IIA
III A
IV A
V A
VI A
VII A
VIII
IB
IIB
III B
IV B
V B
VI B
VII B
2024-25
Fig. 3.2 Long form of the Periodic Table of the Elements with their atomic numbers and ground state outer electronic configurations. The groups are numbered 1-18 in accordance with the 1984 IUPAC recommendations. This notation replaces the old numbering scheme of IA–VIIA, VIII, IB–VIIB and 0 for the elements. | 5 | 11 | Chemistry | 103 |
a8ddd5d7-81b6-4544-b769-117e266bc680 | # Chemistry
a few atoms of them are obtained. Their synthesis and characterisation, therefore, require highly sophisticated costly equipment and laboratory. Such work is carried out with competitive spirit only in some laboratories in the world. Scientists, before collecting the reliable data on the new element, at times get tempted to claim for its discovery. For example, both American and Soviet scientists claimed credit for discovering element 104. The Americans named it Rutherfordium whereas Soviets named it Kurchatovium. To avoid such problems, the IUPAC has made recommendation that until a new element’s discovery is proved, and its name is officially recognised, a systematic nomenclature be derived directly from the atomic number of the element using the numerical roots for 0 and numbers 1-9. These are shown in Table 3.4. The roots are put together in order of
# Table 3.4 Notation for IUPAC Nomenclature of Elements
|digit|name|abbreviation|
|---|---|---|
|0|nil|n|
|1|un|u|
|2|bi|b|
|3|tri|t|
|4|quad|q|
|5|pent|p|
|6|hex|h|
|7|sept|s|
|8|oct|o|
|9|enn|e|
# Table 3.5 Nomenclature of Elements with Atomic Number Above 100
|atomic number|name according to IUPAC nomenclature|symbol|IUPAC Official Name|symbol|
|---|---|---|---|---|
|101|Unnilunium|Unu|Mendelevium|Md|
|102|Unnilbium|Unb|Nobelium|No|
|103|Unniltrium|Unt|Lawrencium|Lr|
|104|Unnilquadium|Unq|Rutherfordium|Rf|
|105|Unnilpentium|Unp|Dubnium|Db|
|106|Unnilhexium|Unh|Seaborgium|Sg|
|107|Unnilseptium|Uns|Bohrium|Bh|
|108|Unniloctium|Uno|Hassium|Hs|
|109|Unnilennium|Une|Meitnerium|Mt|
|110|Ununnillium|Uun|Darmstadtium|Ds|
|111|Unununnium|Uuu|Rontgenium|Rg|
|112|Ununbium|Uub|Copernicium|Cn|
|113|Ununtrium|Uut|Nihonium|Nh|
|114|Ununquadium|Uuq|Flerovium|Fl|
|115|Ununpentium|Uup|Moscovium|Mc|
|116|Ununhexium|Uuh|Livermorium|Lv|
|117|Ununseptium|Uus|Tennessine|Ts|
|118|Ununoctium|Uuo|Oganesson|Og| | 6 | 11 | Chemistry | 103 |
69b73dd4-510a-45eb-9d48-42e5ac35502b | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
Thus, the new element first gets a temporary name, with symbol consisting of three letters. Later permanent name and symbol are given by a vote of IUPAC representatives from each country. The permanent name might reflect the country (or state of the country) in which the element was discovered, or pay tribute to a notable scientist. As of now, elements with atomic numbers up to 118 have been discovered. Official names of all elements have been announced by IUPAC.
# Problem 3.1
What would be the IUPAC name and symbol for the element with atomic number 120?
# Solution
From Table 3.4, the roots for 1, 2 and 0 are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn and unbinilium.
# 3.5 Electronic Configurations of Elements and the Periodic Table
In the preceding unit we have learnt that an electron in an atom is characterised by a set of four quantum numbers, and the principal quantum number (n) defines the main energy level known as shell. We have also studied about the filling of electrons into different subshells, also referred to as orbitals (s, p, d, f) in an atom. The distribution of electrons into orbitals of an atom is called its electronic configuration. An element’s location in the Periodic Table reflects the quantum numbers of the last orbital filled. In this section we will observe a direct connection between the electronic configurations of the elements and the long form of the Periodic Table.
# (a) Electronic Configurations in Periods
The period indicates the value of n for the outermost or valence shell. In other words, successive period in the Periodic Table is associated with the filling of the next higher principal energy level (n = 1, n = 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period (n = 1) starts with the filling of the lowest level (1s) and therefore has two elements — hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period (n = 2) starts with lithium and the third electron enters the 2s orbital. The next element, beryllium has four electrons and has the electronic configuration 1s²2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed at neon (2s²2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) begins at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals gives rise to the third period of 8 elements from sodium to argon. The fourth period (n = 4) starts at potassium, and the added electrons fill up the 4s orbital. Now you may note that before the 4p orbital is filled, filling up of 3d orbitals becomes energetically favourable and we come across the so-called 3d transition series of elements. This starts from scandium (Z = 21) which has the electronic configuration 3d¹4s². The 3d orbitals are filled at zinc (Z = 30) with electronic configuration 3d¹⁰4s². The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in this fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39). This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals, in the order — filling up of the 4f orbitals begins with cerium (Z = 58) and ends at lutetium (Z = 71) to give the 4f-inner transition series which is called the lanthanoid series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. | 7 | 11 | Chemistry | 103 |
2a6b0933-7259-4b0f-bb2a-2f401c247e6d | # Chemistry
Actinium (Z = 89) gives the 5f-inner transition series known as the actinoid series. The 4f- and 5f-inner transition series of elements are placed separately in the Periodic Table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column. This similarity arises because these elements have the same number and same distribution of electrons in their outermost orbitals.
# Problem 3.2
How would you justify the presence of 18 elements in the 5th period of the Periodic Table?
# Solution
When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5th period.
(b) Groupwise Electronic Configurations
Elements in the same vertical column or group have similar valence shell electronic configurations, the same number of electrons in the outer orbitals, and similar properties. For example, the Group 1 elements (alkali metals) all have ns1 valence shell electronic configuration as shown below.
|Atomic Number|Symbol|Electronic Configuration|
|---|---|---|
|3|Li|1s²2s¹ (or) [He]2s¹|
|11|Na|1s²2s²2p⁶3s¹ (or) [Ne]3s¹|
|19|K|1s²2s²2p⁶3s²3p⁶4s¹ (or) [Ar]4s¹|
|37|Rb|1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹ (or) [Kr]5s¹|
|55|Cs|1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶6s¹ (or) [Xe]6s¹|
|87|Fr|[Rn]7s¹|
Thus it can be seen that the properties of an element have periodic dependence upon its atomic number and not on relative atomic mass.
# 3.6 Electronic Configurations and Types of Elements:
# 3.6.1 The s-Block Elements
The elements of Group 1 (alkali metals) and Group 2 (alkaline earth metals) which have ns1 and ns2 outermost electronic configuration belong to the s-Block Elements. They are all. | 8 | 11 | Chemistry | 103 |
215ed1d3-d125-4d19-b688-44150cb681e6 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PRO
# Og
# PE
# TI
# Mc
# Nh
# 2024-25
# Fig. 3.3
The types of elements in the Periodic Table based on the orbitals that are being filled. Also shown is the broad division of elements into METALS, METALLOIDS, and NON-METALS.
83 | 9 | 11 | Chemistry | 103 |
6b7694a9-1608-40c1-8a77-dd30f1e6420a | # Chemistry
Reactive metals with low ionization enthalpies. They lose the outermost electron(s) readily to form 1+ ion (in the case of alkali metals) or 2+ ion (in the case of alkaline earth metals). The metallic character and the reactivity increase as we go down the group. Because of high reactivity they are never found pure in nature. The compounds of the s-block elements, with the exception of those of lithium and beryllium are predominantly ionic.
# 3.6.2 The p-Block Elements
The p-Block Elements comprise those belonging to Group 13 to 18 and these together with the s-Block Elements are called the Representative Elements or Main Group Elements. The outermost electronic configuration varies from ns²np¹ to ns²np⁶ in each period. At the end of each period is a noble gas element with a closed valence shell ns²np⁶ configuration. All the orbitals in the valence shell of the noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. The noble gases thus exhibit very low chemical reactivity. Preceding the noble gas family are two chemically important groups of non-metals. They are the halogens (Group 17) and the chalcogens (Group 16). These two groups of elements have highly negative electron gain enthalpies and readily add one or two electrons respectively to attain the stable noble gas configuration. The non-metallic character increases as we move from left to right across a period and metallic character increases as we go down the group.
# 3.6.3 The d-Block Elements (Transition Elements)
These are the elements of Group 3 to 12 in the centre of the Periodic Table. These are characterised by the filling of inner d orbitals by electrons and are therefore referred to as d-Block Elements. These elements have the general outer electronic configuration (n-1)d¹⁻¹⁰ns⁰⁻² except for Pd where its electronic configuration is 4d¹⁰5s0. They are all metals. They mostly form coloured ions, exhibit variable valence (oxidation states), paramagnetism and oftenly used as catalysts. However, Zn, Cd and Hg which have the electronic configuration, (n-1)d¹⁰ns² do not show most of the properties of transition elements.
# 3.6.4 The f-Block Elements (Inner-Transition Elements)
The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) – Lu(Z = 71) and Actinoids, Th(Z = 90) – Lr (Z = 103) are characterised by the outer electronic configuration (n-2)f¹⁻¹⁴ (n-1)d0–1ns². The last electron added to each element is filled in f-orbital. These two series of elements are hence called the Inner-Transition Elements (f-Block Elements). They are all metals. Within each series, the properties of the elements are quite similar. The chemistry of the early actinoids is more complicated than the corresponding lanthanoids, due to the large number of oxidation states possible for these actinoid elements. Actinoid elements are radioactive. Many of the actinoid elements have been made only in nanogram quantities or even less by nuclear reactions and their chemistry is not fully studied. The elements after uranium are called Transuranium Elements.
# Problem 3.3
The elements Z= 117 and 120 have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case.
# Solution
We see from Fig. 3.2, that element with Z = 117, would belong to the halogen family (Group 17) and the electronic configuration would be [Rn] 5f¹⁴6d¹⁰7s²7p⁵. The element with Z = 120, will be placed in Group 2 (alkaline earth metals), and will have the electronic configuration [Uuo]8s². | 10 | 11 | Chemistry | 103 |
e52084df-7a0a-484c-a9b1-ebffad90e062 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
# 3.6.5 metals, non-metals and metalloids
In addition to displaying the classification of elements into s-, p-, d-, and f-blocks, Fig. 3.3 shows another broad classification of elements based on their properties. The elements can be divided into metals and non-metals. Metals comprise more than 78% of all known elements and appear on the left side of the Periodic table. Metals are usually solids at room temperature (mercury is an exception; gallium and caesium also have very low melting points (303K and 302K, respectively)). Metals usually have high melting and boiling points. They are good conductors of heat and electricity. They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires). In contrast, non-metals are located at the top right hand side of the Periodic table. In fact, in a horizontal row, the property of elements change from metallic on the left to non-metallic on the right. Non-metals are usually solids or gases at room temperature with low melting and boiling points (boron and carbon are exceptions). They are poor conductors of heat and electricity. Most non-metallic solids are brittle and are neither malleable nor ductile. The elements become more metallic as we go down a group; the non-metallic character increases as one goes from left to right across the Periodic table. The change from metallic to non-metallic character is not abrupt as shown by the thick zig-zag line in Fig. 3.3. The elements (e.g., silicon, germanium, arsenic, antimony and tellurium) bordering this line and running diagonally across the Periodic table show properties that are characteristic of both metals and non-metals. These elements are called semi-metals or metalloids.
# 3.7 PERIODIC TRENDS IN PROPERTIES OF ELEMENTS
There are many observable patterns in the physical and chemical properties of elements as we descend in a group or move across a period in the Periodic Table. For example, within a period, chemical reactivity tends to be high in Group 1 metals, lower in elements towards the middle of the table, and increases to a maximum in the Group 17 non-metals. Likewise within a group of representative metals (say alkali metals) reactivity increases on moving down the group, whereas within a group of non-metals (say halogens), reactivity decreases down the group. But why do the properties of elements follow these trends? And how can we explain periodicity? To answer these questions, we must look into the theories of atomic structure and properties of the atom. In this section we shall discuss the periodic trends in certain physical and chemical properties and try to explain them in terms of number of electrons and energy levels.
# 3.7.1 trends in Physical Properties
There are numerous physical properties of elements such as melting and boiling points, heats of fusion and vaporization, energy of atomization, etc. which show periodic variations. However, we shall discuss the periodic trends with respect to atomic and ionic radii, ionization enthalpy, electron gain enthalpy and electronegativity.
# Problem 3.4
(a) Atomic Radius
You can very well imagine that finding the size of an atom is a lot more complicated than measuring the radius of a ball. Do you know why? Firstly, because the size of an atom (~ 1.2 Å i.e., 1.2 × 10–10 m in radius) is very | 11 | 11 | Chemistry | 103 |
cc788afc-1d2c-4fdc-beaf-adc56c7059a6 | # Chemistry
Small. Secondly, since the electron cloud surrounding the atom does not have a sharp boundary, the determination of the atomic size cannot be precise. In other words, there is no practical way by which the size of an individual atom can be measured. However, an estimate of the atomic size can be made by knowing the distance between the atoms in the combined state. One practical approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and from this value, the “Covalent radius” of the element can be calculated. For example, the bond distance in the chlorine molecule (Cl₂) is 198 pm and half this distance (99 pm) is taken as the atomic radius of chlorine. For metals, we define the term “metallic radius” which is taken as half the internuclear distance separating the metal cores in the metallic crystal. For example, the distance between two adjacent copper atoms in solid copper is 256 pm; hence the metallic radius of copper is assigned a value of 128 pm. For simplicity, in this book, we use the term atomic radius to refer to both covalent or metallic radius depending on whether the element is a non-metal or a metal. Atomic radii can be measured by X-ray or other spectroscopic methods.
# Table 3.6(a) Atomic Radii/pm Across the Periods
|Atom (Period II)|Li|Be|B|C|N|O|F|
|---|---|---|---|---|---|---|---|
|Atomic Radius|152|111|88|77|74|66|64|
# Table 3.6(b) Atomic Radii/pm Down a Family
|Atom (Group I)|Atomic Radius|Atom (Group 17)|Atomic Radius|
|---|---|---|---|
|Li|152|F|64|
|Na|186|Cl|99|
|K|231|Br|114|
|Rb|244|I|133|
|Cs|262|At|140| | 12 | 11 | Chemistry | 103 |
cee243b5-76c9-41e3-b3e4-63aece0cc1b1 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
# 87
Fig. 3.4 (a) Variation of atomic radius with atomic number across the second period
Fig. 3.4 (b) Variation of atomic radius with atomic number for alkali metals and halogens
# (b) Ionic Radius
The removal of an electron from an atom results in the formation of a cation, whereas gain of an electron leads to an anion. The ionic radii can be estimated by measuring the distances between cations and anions in ionic crystals. In general, the ionic radii of elements exhibit the same trend as the atomic radii. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge. For example, the ionic radius of fluoride ion (F–) is 136 pm whereas the atomic radius of fluorine is only 64 pm. On the other hand, the atomic radius of sodium is 186 pm compared to the ionic radius of 95 pm for Na⁺.
When we find some atoms and ions which contain the same number of electrons, we call them isoelectronic species*. For example, O2–, F–, Na⁺ and Mg²⁺ have the same number of electrons (10). Their radii would be different because of their different nuclear charges.
* Two or more species with same number of atoms, same number of valence electrons and same structure, regardless of the nature of elements involved.
# Problem 3.5
Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
# solution
Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius. Hence the largest species is Mg; the smallest one is Al³⁺.
# (c) Ionization Enthalpy
A quantitative measure of the tendency of an element to lose electron is given by its ionization Enthalpy. It represents the energy required to remove an electron from an isolated gaseous atom (X) in its ground state.
2024-25 | 13 | 11 | Chemistry | 103 |
f953bdc6-d473-4922-84c5-03aa77ae0ba0 | # Chemistry
In other words, the first ionization enthalpy for an element X is the enthalpy change (∆i H) for the reaction depicted in equation 3.1.
X(g) → X⁺(g) + e– (3.1)
The ionization enthalpy is expressed in units of kJ mol–1. We can define the second ionization enthalpy as the energy required to remove the second most loosely bound electron; it is the energy required to carry out the reaction shown in equation 3.2.
X⁺(g) → X²⁺(g) + e– (3.2)
Energy is always required to remove electrons from an atom and hence ionization enthalpies are always positive. The second ionization enthalpy will be higher than the first ionization enthalpy because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. In the same way, the third ionization enthalpy will be higher than the second and so on. The term “ionization enthalpy”, if not qualified, is taken as the first ionization enthalpy.
The first ionization enthalpies of elements having atomic numbers up to 60 are plotted in Fig. 3.5. The periodicity of the graph is quite striking. You will find maxima at the noble gases which have closed electron shells and very stable electron configurations. On the other hand, minima occur at the alkali metals and their low ionization enthalpies can be correlated with their high reactivity.
In addition, you will notice two trends: the first ionization enthalpy generally increases as we go across a period and decreases as we descend in a group. These trends are illustrated in Figs. 3.6(a) and 3.6(b) respectively for the elements of the second period and the first group of the periodic table. You will appreciate that the ionization enthalpy and atomic radius are closely related properties. To understand these trends, we have to consider two factors: (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other.
# 3.6 (a)
Fig. 3.6(a) First ionization enthalpies (∆iH) of elements of the second period as a function of atomic number (Z)
# 3.6 (b)
Fig. 3.6(b) ∆iH of alkali metals as a function of Z.
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3d06863c-10ae-4b77-9b09-7cad608bff8e | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
valence electron in an atom will be less than the 2s electrons of beryllium. Therefore, it is the actual charge on the nucleus because of “shielding” or “screening” of the valence electron from the nucleus by the intervening core electrons. For example, the 2s electron in lithium is shielded from the nucleus by the inner core of 1s electrons. As a result, the valence electron experiences a net positive charge which is less than the actual charge of +3. In general, shielding is effective when the orbitals in the inner shells are completely filled. This situation occurs in the case of alkali metals which have single outermost ns-electron preceded by a noble gas electronic configuration.
When we move from lithium to fluorine across the second period, successive electrons are added to orbitals in the same principal quantum level and the shielding of the nuclear charge by the inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus. Thus, across a period, increasing nuclear charge outweighs the shielding. Consequently, the outermost electrons are held more and more tightly and the ionization enthalpy increases across a period. As we go down a group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. In this case, increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.
# Problem 3.6
The first ionization enthalpy (∆i H) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol–1. Predict whether the first ∆i H value for Al will be more close to 575 or 760 kJ mol–1? Justify your answer.
# solution
It will be more close to 575 kJ mol–1. The value for Al should be lower than that of Mg because of effective shielding of 3p electrons from the nucleus by 3s-electrons.
# (d) Electron Gain Enthalpy
When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change accompanying the process is defined as the Electron Gain Enthalpy (∆egH). Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion as represented by equation 3.3.
X(g) + e– → X–(g) (3.3)
Depending on the element, the process of adding an electron to the atom can be either endothermic or exothermic. For many elements energy is released when an electron is added to the atom and the electron gain enthalpy is negative. For example, group 17 elements (the halogens) have very high. | 15 | 11 | Chemistry | 103 |
cf73fd94-bae5-42d7-9026-102885f19172 | # Chemistry
# Table 3.7 Electron Gain Enthalpies* / (kJ mol–1) of some main Group Elements
|Group 1|∆egH|Group 16|∆egH|Group 17|∆egH|Group 0|∆egH|
|---|---|---|---|---|---|---|---|
|H|– 73|O|– 141|F|– 328|He|+ 48|
|Li|– 60|S|– 200|Cl|– 349|Ne|+ 116|
|Na|– 53|Se|– 195|Br|– 325|Ar|+ 96|
|K|– 48|Te|– 190|I|– 295|Kr|+ 96|
|Rb|– 47|Po|– 174|At|– 270|Xe|+ 77|
|Cs|– 46| | | | |Rn|+ 68|
Negative electron gain enthalpies because they can attain stable noble gas electronic configurations by picking up an electron. On the other hand, noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. It may be noted that electron gain enthalpies have large negative values toward the upper right of the periodic table preceding the noble gases.
The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the positively charged nucleus. We should also expect electron gain enthalpy to become less negative as we go down a group because the size of the atom increases and the added electron would be farther from the nucleus. This is generally the case (Table 3.7). However, electron gain enthalpy of O or F is less negative than that of the succeeding element. This is because when an electron is added to O or F, the added electron goes to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. For the n = 3 quantum level (S or Cl), the added electron occupies a larger region of space and the electron-electron repulsion is much less.
# (e) Electronegativity
A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity. Unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity. However, a number of numerical scales of electronegativity of elements viz., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale have been developed. The one which is the most widely used is the Pauling scale. Linus Pauling, an American scientist, in 1922 assigned arbitrarily a value of 4.0 to fluorine, the element considered to have the greatest.
* In many books, the negative of the enthalpy change for the process depicted in equation 3.3 is defined as the ELECTRON AFFINITY (Ae) of the atom under consideration. If energy is released when an electron is added to an atom, the electron affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to add an electron to an atom, then the electron affinity of the atom is assigned a negative sign. However, electron affinity is defined as absolute zero and, therefore at any other temperature (T) heat capacities of the reactants and the products have to be taken into account in ∆egH = –Aₑ – 5/2 RT.
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9406c08d-1173-45ac-bda2-07a09ec3e5d9 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
Ability to attract electrons. Approximate values for the electronegativity of a few elements are given in Table 3.8(a). The electronegativity of any given element is not constant; it varies depending on the element to which it is bound. Though it is not a measurable quantity, it does provide a means of predicting the nature of force that holds a pair of atoms together – a relationship that you will explore later.
Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decreases down a group (say from fluorine to astatine) in the periodic table. How can these trends be explained? Can the electronegativity be related to atomic radii, which tend to decrease across each period from left to right, but increase down each group? The attraction between the outer (or valence) electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases.
# Fig. 3.7 The periodic trends of elements in the periodic table
# Table 3.8(a) Electronegativity Values (on Pauling scale) across the Periods
|atom (Period ii)|li|Be|B|C|N|O|F|
|---|---|---|---|---|---|---|---|
|Electronegativity|1.0|1.5|2.0|2.5|3.0|3.5|4.0|
# Table 3.8(b) Electronegativity Values (on Pauling scale) down a family
|atom (Group i)|Electronegativity Value|atom (Group 17)|Electronegativity Value|
|---|---|---|---|
|Li|1.0|F|4.0|
|Na|0.9|Cl|3.0|
|K|0.8|Br|2.8|
|Rb|0.8|I|2.5|
|Cs|0.7|At|2.2|
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b95b4dad-1c8a-4faa-ba72-7c218c51a620 | # CHEMISTRY
to gain electrons. Therefore, electronegativity is directly related to that non-metallic properties of elements. It can be further extended to say that the electronegativity is inversely related to the metallic properties of elements. Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or decrease in metallic properties) of elements. Similarly, the decrease in electronegativity down a group is accompanied by a decrease in non-metallic properties (or increase in metallic properties) of elements.
All these periodic trends are summarised in Figure 3.7.
# 3.7.2 Periodic trends in Chemical Properties
Most of the trends in chemical properties of elements, such as diagonal relationships, inert pair effect, effects of lanthanoid contraction etc. will be dealt with along the discussion of each group in later units. In this section we shall study the periodicity of the valence state shown by elements and the anomalous properties of the second period elements (from lithium to fluorine).
# Periodic States
The valence is the most characteristic property of the elements and can be understood in terms of their electronic configurations. The valence of representative elements is usually (though not necessarily) equal to the number of electrons in the outermost orbitals and/or equal to eight minus the number of outermost electrons as shown below. Nowadays the term oxidation state is frequently used for valence. Consider the two oxygen containing compounds: OF₂ and Na₂O. The order of electronegativity of the three elements involved in these compounds is F > O > Na. Each of the atoms of fluorine shares one electron with oxygen in the OF₂ molecule. Being the highest electronegative element, fluorine is given oxidation state –1. Since there are two fluorine atoms in this molecule, oxygen with outer electronic configuration 2s²2p⁴ shares two electrons with fluorine atoms and thereby exhibits oxidation state +2. In Na₂O, oxygen being more electronegative accepts two electrons, one from each of the two sodium atoms and, thus, shows oxidation state –2. On the other hand sodium with electronic configuration 3s¹ loses one electron to oxygen and is given oxidation state +1. Thus, the oxidation state of an element in a particular compound can be defined as the charge acquired by its atom on the basis of electronegative consideration from other atoms in the molecule.
# Problem 3.8
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;
- (a) silicon and bromine
- (b) aluminium and sulphur.
# Solution
(a) Silicon is a group 14 element with a valence of 4; bromine belongs to the halogen family with a valence of 1. Hence the formula of the compound formed would be SiBr₄.
(b) Aluminium belongs to group 13 with a valence of 3; sulphur belongs to group 16 elements with a valence of 2. Hence, the formula of the compound formed would be Al₂S₃.
# Periodic Trends
Some periodic trends observed in the valence of elements (hydrides and oxides) are shown in Table 3.9. Other such periodic trends which occur in the chemical behaviour of the elements are discussed elsewhere in the document.
|Group|1|2|13|14|15|16|17|18|
|---|---|---|---|---|---|---|---|---|
|number of valence electron|1|2|3|4|5|6|7|8|
|valence|1|2|3|4|3,5|2,6|1,7|0,8| | 18 | 11 | Chemistry | 103 |
56725e40-dd55-4755-91aa-1c45e1696b26 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
# Table 3.9
# Periodic trends in Valence of Elements as shown by the formulas of their Compounds
|Group|1|2|13|14|15|16|17|
|---|---|---|---|---|---|---|---|
|Formula of hydride|LiH|CaH₂|B₂H₆|CH₄|NH₃|H₂O|HF|
| |NaH| |AlH₃|SiH₄|PH₃|H₂S|HCl|
| |KH| | |GeH₄|AsH₃|H₂Se|HBr|
| | | | |SnH₄| |H₂Te|HI|
|Formula of oxide|Li₂O|MgO|B₂O₃|CO₂|N₂O₃, N₂O₅|-| |
| |Na₂O|CaO|Al₂O₃|SiO₂|P₄O₆, P₄O₁₀|SO₃|Cl₂O₇|
| |K₂O|SrO|Ga₂O₃|GeO₂|As₂O₃, As₂O₅|SeO₃|-|
| | |BaO|In₂O₃|SnO₂|Sb₂O₃, Sb₂O₅|TeO₃|-|
| | | |PbO₂|Bi₂O₃|-|-| |
This book. There are many elements which exhibit variable valence. This is particularly characteristic of transition elements and actinoids, which we shall study later.
# (b) Anomalous Properties of Second Period Elements
The first element of each of the groups 1 (lithium) and 2 (beryllium) and groups 13-17 (boron to fluorine) differs in many respects from the other members of their respective group. For example, lithium unlike other alkali metals, and beryllium unlike other alkaline earth metals, form compounds with pronounced covalent character; the other members of these groups predominantly form ionic compounds. In fact the behaviour of lithium and beryllium is more similar with the second element of the following group i.e., magnesium and aluminium, respectively. This sort of similarity is commonly referred to as diagonal relationship in the periodic properties.
What are the reasons for the different chemical behaviour of the first member of a group of elements in the s- and p-blocks compared to that of the subsequent members in the same group? The anomalous behaviour is attributed to their small size, large charge/radius ratio and high electronegativity of the elements. In addition, the first member of group has only four valence orbitals (2s and 2p) available for bonding, whereas the second member of the groups have nine valence orbitals (3s, 3p, 3d). As a consequence of this, the maximum covalency of the first member of each group is 4 (e.g., boron can only form [BF₄]⁻, whereas the other members of the groups can expand their valence shell to accommodate more than four pairs of electrons e.g., aluminium [AlF₆]³⁻ forms). Furthermore, the first member of p-block elements displays greater ability to form pπ – pπ multiple bonds to itself (e.g., C = C, C ≡ C, N = N, N ≡ N) and to other second period elements (e.g., C = O, C = N, C ≡ N, N = O) compared to subsequent members of the same group. | 19 | 11 | Chemistry | 103 |
b53fdc86-a5ab-4c46-8185-4af562cf95a0 | # Chemistry
# Problem 3.9
Are the oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ same?
Solution: No. The oxidation state of Al is +3 and the covalency is 6.
# 3.7.3 Periodic trends and Chemical reactivity
We have observed the periodic trends in certain fundamental properties such as atomic and ionic radii, ionization enthalpy, electron gain enthalpy and valence. We know by now that the periodicity is related to electronic configuration. That is, all chemical and physical properties are a manifestation of the electronic configuration of elements. We shall now try to explore relationships between these fundamental properties of elements with their chemical reactivity.
The atomic and ionic radii, as we know, generally decrease in a period from left to right. As a consequence, the ionization enthalpies generally increase (with some exceptions as outlined in section 3.7.1(a)) and electron gain enthalpies become more negative across a period. In other words, the ionization enthalpy of the extreme left element in a period is the least and the electron gain enthalpy of the element on the extreme right is the highest negative (note: noble gases having completely filled shells have rather positive electron gain enthalpy values). This results into high chemical reactivity at the two extremes and the lowest in the centre. Thus, the maximum chemical reactivity at the extreme left (among alkali metals) is exhibited by the loss of an electron leading to the formation of a cation and at the extreme right (among halogens) shown by the gain of an electron forming an anion. This property can be related with the reducing and oxidizing behaviour of the elements which you will learn later. However, among transition metals (3d series), the change in atomic radii is much smaller as compared to those of representative elements across the period. The change in atomic radii is still smaller among inner-transition metals (4f series). The ionization enthalpies are intermediate between those of s- and p-blocks. As a consequence, they are less electropositive than group 1 and 2 metals.
# Problem 3.10
Show by a chemical reaction with water that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Solution:
Na₂O with water forms a strong base whereas Cl₂O₇ forms strong acid.
Na₂O + H₂O → 2NaOH
Cl₂O₇ + H₂O → 2HClO₄
Their basic or acidic nature can be qualitatively tested with litmus paper. | 20 | 11 | Chemistry | 103 |
26315f36-1c20-42c4-905f-b224b97e57b4 | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
In a group, the increase in atomic and ionic radii with increase in atomic number generally results in a gradual decrease in ionization enthalpies and a regular decrease (with exception in some third period elements as shown in section 3.7.1(d)) in electron gain enthalpies in the case of main group elements. Thus, the metallic character increases down the group and non-metallic character decreases. This trend can be related with their reducing and oxidizing property which you will learn later. In the case of transition elements, however, a reverse trend is observed. This can be explained in terms of atomic size and ionization enthalpy.
# sUmmary
In this Unit, you have studied the development of the Periodic law and the Periodic table. Mendeleev’s Periodic table was based on atomic masses. Modern Periodic table arranges the elements in the order of their atomic numbers in seven horizontal rows (periods) and eighteen vertical columns (groups or families). Atomic numbers in a period are consecutive, whereas in a group they increase in a pattern. Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have incrementally increasing number of electrons from left to right, and, therefore, have different valencies.
Four types of elements can be recognized in the periodic table on the basis of their electronic configurations. These are s-block, p-block, d-block and f-block elements. Hydrogen with one electron in the 1s orbital occupies a unique position in the periodic table. Metals comprise more than seventy-eight percent of the known elements. Non-metals, which are located at the top of the periodic table, are less than twenty in number. Elements which lie at the border line between metals and non-metals (e.g., Si, Ge, As) are called metalloids or semi-metals. Metallic character increases with increasing atomic number in a group whereas decreases from left to right in a period. The physical and chemical properties of elements vary periodically with their atomic numbers.
Periodic trends are observed in atomic sizes, ionization enthalpies, electron gain enthalpies, electronegativity and valence. The atomic radii decrease while going from left to right in a period and increase with atomic number in a group. Ionization enthalpies generally increase across a period and decrease down a group. Electronegativity also shows a similar trend. Electron gain enthalpies, in general, become more negative across a period and less negative down a group. There is some periodicity in valence, for example, among representative elements, the valence is either equal to the number of electrons in the outermost orbitals or eight minus this number. Chemical reactivity is highest at the two extremes of a period and is lowest in the centre. The reactivity on the left extreme of a period is because of the ease of electron loss (or low ionization enthalpy). Highly reactive elements do not occur in nature in free state; they usually occur in the combined form. Oxides formed of the elements on the left are basic and of the elements on the right are acidic in nature. Oxides of elements in the centre are amphoteric or neutral.
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886472f4-f927-416b-abaf-8af2492b6276 | # Chemistry
# Exercises
1. What is the basic theme of organisation in the periodic table?
2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
5. In terms of period and group where would you locate the element with Z = 114?
6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
7. Which element do you think would have been named by
Lawrence Berkeley Laboratory
8. Seaborg’s group?
Why do elements in the same group have similar physical and chemical properties?
What does atomic radius and ionic radius really mean to you?
How do atomic radius vary in a period and in a group? How do you explain the variation?
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
F–
Ar
Mg²⁺
Rb⁺
Consider the following species:
- N³–
- O²–
- F–
- Na⁺
- Mg²⁺
- Al³⁺
1. What is common in them?
2. Arrange them in the order of increasing ionic radii.
Explain why cations are smaller and anions larger in radii than their parent atoms?
What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Hint: Requirements for comparison purposes.
Energy of an electron in the ground state of the hydrogen atom is –2.18×10⁻¹⁸ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol⁻¹.
Hint: Apply the idea of mole concept to derive the answer.
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why
Be has higher ∆i H than B
O has lower ∆i H than N and F?
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725a56df-2518-4e54-b308-cc45dcce995b | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
# 3.17
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
# 3.18
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
# 3.19
The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
|B|Al|Ga|In|Tl|
|---|---|---|---|---|
|801|577|579|558|589|
How would you explain this deviation from the general trend?
# 3.20
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl
# 3.21
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
# 3.22
What is the basic difference between the terms electron gain enthalpy and electronegativity?
# 3.23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
# 3.24
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
# 3.25
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
# 3.26
What are the major differences between metals and non-metals?
# 3.27
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
# 3.28
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
# 3.29
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
# 3.30
Assign the position of the element having outer electronic configuration
(i) ns²np⁴ for n=3
(ii) (n-1)d²ns² for n=4, and
(iii) (n-2)f⁷(n-1)d¹ns² for n=6, in the periodic table.
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9957c264-94b3-47a8-936a-a78dcd1399a2 | # Chemistry
# 3.31
The first (∆iH₁) and the second (∆iH₂) ionization enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
|Elements|∆H₁|∆H₂|∆egH|
|---|---|---|---|
|I|520|7300|–60|
|II|419|3051|–48|
|III|1681|3374|–328|
|IV|1008|1846|–295|
|V|2372|5251|+48|
|VI|738|1451|–40|
Which of the above elements is likely to be:
- (a) the least reactive element.
- (b) the most reactive metal.
- (c) the most reactive non-metal.
- (d) the least reactive non-metal.
- (e) the metal which can form a stable binary halide of the formula MX₂ (X=halogen).
- (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
# 3.32
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
- (a) Lithium and oxygen
- (b) Magnesium and nitrogen
- (c) Aluminium and iodine
- (d) Silicon and oxygen
- (e) Phosphorus and fluorine
- (f) Element 71 and fluorine
# 3.33
In the modern periodic table, the period indicates the value of:
- (a) atomic number
- (b) atomic mass
- (c) principal quantum number
- (d) azimuthal quantum number.
# 3.34
Which of the following statements related to the modern periodic table is incorrect?
- (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
- (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
- (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
- (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
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03d2b89c-ac6a-4a2a-b769-d6283de6ddba | # CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
- (a) Valence principal quantum number (n)
- (b) Nuclear charge (Z)
- (c) Nuclear mass
- (d) Number of core electrons.
3.36 The size of isoelectronic species — F–, Ne and Na⁺ is affected by
- (a) nuclear charge (Z)
- (b) valence principal quantum number (n)
- (c) electron-electron interaction in the outer orbitals
- (d) none of the factors because their size is the same.
3.37 Which one of the following statements is incorrect in relation to ionization enthalpy?
- (a) Ionization enthalpy increases for each successive electron.
- (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
- (c) End of valence electrons is marked by a big jump in ionization enthalpy.
- (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
3.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is:
- (a) B > Al > Mg > K
- (b) Al > Mg > B > K
- (c) Mg > Al > K > B
- (d) K > Mg > Al > B
3.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is:
- (a) B > C > Si > N > F
- (b) Si > C > B > N > F
- (c) F > N > C > B > Si
- (d) F > N > C > Si > B
3.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is:
- (a) F > Cl > O > N
- (b) F > O > Cl > N
- (c) Cl > F > O > N
- (d) O > F > N > Cl
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e4495b16-cb37-412e-bc0b-7d48f82c03b4 | # UNIT 1
# T1082CHOT
# SOME BASIC CONCEPTS OF CHEMISTRY
Chemistry is the science of molecules and their transformations. It is the science not so much of the one hundred elements but of the infinite variety of molecules that may be built from them.
After studying this unit, you will be able to:
- appreciate the contribution of India in the development of chemistry
- understand the role of chemistry in different spheres of life;
- explain the characteristics of three states of matter;
- classify different substances into elements, compounds and mixtures;
- use scientific notations and determine significant figures;
- differentiate between precision and accuracy;
- define SI base units and convert physical quantities from one system of units to another;
- explain various laws of chemical combination;
- appreciate significance of atomic mass, average atomic mass, molecular mass and formula mass;
- describe the terms – mole and molar mass;
- calculate the mass per cent of component elements constituting a compound;
- determine empirical formula and molecular formula for a compound from the given experimental data;
- perform the stoichiometric calculations.
# DEVELOPMENT OF CHEMISTRY
Chemistry, as we understand it today, is not a very old discipline. Chemistry was not studied for its own sake, rather it came up as a result of search for two interesting things:
1. Philosopher’s stone (Paras) which would convert all baser metals e.g., iron and copper into gold.
2. ‘Elixir of life’ which would grant immortality.
People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied that knowledge in various walks of life. Chemistry developed mainly in the form of Alchemy and Iatrochemistry during 1300-1600 CE. Modern chemistry took shape in the 18th century Europe, after a few centuries of alchemical traditions which were introduced in Europe by the Arabs. | 0 | 11 | Chemistry | 101 |
0e99ba4c-4e87-43f0-b673-470843db8e1e | # Chemistry
Other cultures – especially the Chinese and the Indian – had their own alchemical traditions. These included much knowledge of chemical processes and techniques. In ancient India, chemistry was called Rasayan Shastra, Rastantra, Ras Kriya or Rasvidya. It included metallurgy, medicine, manufacture of cosmetics, glass, dyes, etc. Systematic excavations at Mohenjodaro in Sindh and Harappa in Punjab prove that the story of development of chemistry in India is very old. Archaeological findings show that baked bricks were used in construction work. It shows the mass production of pottery, which can be regarded as the earliest chemical process, in which materials were mixed, moulded and subjected to heat by using fire to achieve desirable qualities.
Remains of glazed pottery have been found in Mohenjodaro. Gypsum cement has been used in the construction work. It contains lime, sand and traces of CaCO₃. Harappans made faience, a sort of glass which was used in ornaments. They melted and forged a variety of objects from metals, such as lead, silver, gold and copper. They improved the hardness of copper for making artefacts by using tin and arsenic. A number of glass objects were found in Maski in South India (1000–900 BCE), and Hastinapur and Taxila in North India (1000–200 BCE). Glass and glazes were coloured by addition of colouring agents like metal oxides.
Copper metallurgy in India dates back to the beginning of chalcolithic cultures in the subcontinent. There are much archeological evidences to support the view that technologies for extraction of copper and iron were developed indigenously. According to Rigveda, tanning of leather and dying of cotton were practised during 1000–400 BCE. The golden gloss of the black polished ware of northern India could not be replicated and is still a chemical mystery. These wares indicate the mastery with which kiln temperatures could be controlled. Kautilya’s Arthashastra describes the production of salt from sea.
A vast number of statements and material described in the ancient Vedic literature can be shown to agree with modern scientific findings. Copper utensils, iron, gold, silver ornaments and terracotta discs and painted grey pottery have been found in many archaeological sites in north India. Sushruta Samhita explains the importance of Alkalies. The Charaka Samhita mentions ancient Indians who knew how to prepare sulphuric acid, nitric acid and oxides of copper, tin and zinc; the sulphates of copper, zinc and iron and the carbonates of lead and iron.
Rasopanishada describes the preparation of gunpowder mixture. Tamil texts also describe the preparation of fireworks using sulphur, charcoal, saltpetre (i.e., potassium nitrate), mercury, camphor, etc. Nagarjuna was a great Indian scientist. He was a reputed chemist, an alchemist and a metallurgist. His work Rasratnakar deals with the formulation of mercury compounds. He has also discussed methods for the extraction of metals, like gold, silver, tin and copper. A book, Rsarnavam, appeared around 800 CE. It discusses the uses of various furnaces, ovens and crucibles for different purposes. It describes methods by which metals could be identified by flame colour.
Chakrapani discovered mercury sulphide. The credit for inventing soap also goes to him. He used mustard oil and some alkalies as ingredients for making soap. Indians began making soaps in the 18ᵗʰ century CE. Oil of Eranda and seeds of Mahua plant and calcium carbonate were used for making soap. The paintings found on the walls of Ajanta and Ellora, which look fresh even after ages, testify to a high level of science achieved in ancient India. Varähmihir’s Brihat Samhita is a sort of encyclopaedia, which was composed in the sixth century CE. It informs about the preparation of glutinous material to be applied on walls and roofs of houses and temples. It was prepared entirely from extracts of various plants, fruits, seeds and barks, which were concentrated by boiling, and then, treated with various resins. It will be interesting to test such materials scientifically and assess them for use. | 1 | 11 | Chemistry | 101 |
938625d2-5ca1-4d2a-978f-cbab3f2f245a | # SOME BASIC CONCEPTS OF CHEMISTRY
A number of classical texts, like Atharvaveda (1000 BCE) mention some dye stuff, the material used were turmeric, madder, sunflower, orpiment, cochineal and lac. Some other substances having tinting property were kamplcica, pattanga and jatuka.
Varähmihir’s Brihat Samhita references to perfumes and cosmetics. Recipes for hair dying were made from plants, like indigo and minerals like iron power, black iron or steel and acidic extracts of sour rice gruel. Gandhayukli describes recipes for making scents, mouth perfumes, bath powders, incense and talcum power.
Paper was known to India in the 17th century as account of Chinese traveller I-tsing describes. Excavations at Taxila indicate that ink was used in India from the fourth century. Colours of ink were made from chalk, red lead and minimum.
It seems that the process of fermentation was well-known to Indians. Vedas and Kautilya’s Arthashastra mention many types of liquors. Charaka Samhita also mentions ingredients, such as barks of plants, stem, flowers, leaves, woods, cereals, fruits and sugarcane for making Asavas.
The concept that matter is ultimately made of indivisible building blocks, appeared in India a few centuries BCE as a part of philosophical speculations. Acharya Kanda, born in 600 BCE, originally known by the name Kashyap, was the first proponent of the ‘atomic theory’. He formulated the theory of very small indivisible particles, which he named ‘Paramãnu’ (comparable to atoms). He authored the text Vaiseshika Sutras. According to him, all substances are aggregated form of smaller units called atoms (Paramãnu), which are eternal, indestructible, spherical, suprasensible and in motion in the original state.
He explained that this individual entity cannot be sensed through any human organ. Kanda added that there are varieties of atoms that are as different as the different classes of substances. He said these (Paramãnu) could form pairs or triplets, among other combinations and unseen values.
From the above discussion, you have learnt that chemistry deals with the composition, structure, properties and interaction of matter and is of much use to human beings in daily life. These aspects can be best described and understood in terms of basic constituents of matter that are atoms and molecules. That is why, chemistry is also called the science of atoms and molecules.
Can we see, weigh and perceive these entities (atoms and molecules)? Is it possible to count the number of atoms and molecules in a given mass of matter and have a quantitative relationship between the mass and the number of these particles? We will get the answer of some of these questions in this Unit. We will further describe how physical properties of matter can be quantitatively described using numerical values with suitable units.
2024-25 | 2 | 11 | Chemistry | 101 |
c66d2148-b145-46e4-8174-fdd019491936 | # 1.1 Importance of Chemistry
Chemistry plays a central role in science and is often intertwined with other branches of science. Principles of chemistry are applicable in diverse areas, such as weather patterns, functioning of brain and operation of a computer, production in chemical industries, manufacturing fertilisers, alkalis, acids, salts, dyes, polymers, drugs, soaps, detergents, metals, alloys, etc., including new material.
Chemistry contributes in a big way to the national economy. It also plays an important role in meeting human needs for food, healthcare products and other material aimed at improving the quality of life. This is exemplified by the large-scale production of a variety of fertilisers, improved variety of pesticides and insecticides. Chemistry provides methods for the isolation of life-saving drugs from natural sources and makes possible synthesis of such drugs. Some of these drugs are cisplatin and taxol, which are effective in cancer therapy. The drug AZT (Azidothymidine) is used for helping AIDS patients.
Chemistry contributes to a large extent in the development and growth of a nation. With a better understanding of chemical principles it has now become possible to design and synthesise new material having specific magnetic, electric and optical properties. This has lead to the production of superconducting ceramics, conducting polymers, optical fibres, etc. Chemistry has helped in establishing industries which manufacture utility goods, like acids, alkalies, dyes, polymers, metals, etc. These industries contribute in a big way to the economy of a nation and generate employment.
In recent years, chemistry has helped in dealing with some of the pressing aspects of environmental degradation with a fair degree of success. Safer alternatives to environmentally hazardous refrigerants, like CFCs (chlorofluorocarbons), responsible for ozone depletion in the stratosphere, have been successfully synthesised. However, many big environmental problems continue to be matters of grave concern to the chemists. One such problem is the management of the Green House gases, like methane, carbon dioxide, etc. Understanding of biochemical processes, use of enzymes for large-scale production of chemicals and synthesis of new exotic material are some of the intellectual challenges for the future generation of chemists. A developing country, like India, needs talented and creative chemists for accepting such challenges. To be a good chemist and to accept such challenges, one needs to understand the basic concepts of chemistry, which begin with the concept of matter. Let us start with the nature of matter.
# 1.2 Nature of Matter
You are already familiar with the term matter from your earlier classes. Anything which has mass and occupies space is called matter. Everything around us, for example, book, pen, pencil, water, air, all living beings, etc., are composed of matter. You know that they have mass and they occupy space. Let us recall the characteristics of the states of matter, which you learnt in your previous classes.
# 1.2.1 States of Matter
You are aware that matter can exist in three physical states viz. solid, liquid and gas. The constituent particles of matter in these three states can be represented as shown in Fig. 1.1. Particles are held very close to each other in solids in an orderly fashion and there is not much freedom of movement. In liquids, the particles are close to each other but they can move around. However, in gases, the particles are far apart as compared to those present in solid or liquid states and their movement is easy and fast. Because of such arrangement of particles, different states of matter exhibit the following characteristics:
- Solids have definite volume and definite shape.
- Liquids have definite volume but do not have definite shape. They take the shape of the container in which they are placed. | 3 | 11 | Chemistry | 101 |
87fe79cf-5999-40ff-889e-7bbccda23744 | # SOME BASIC CONCEPTS OF CHEMISTRY
Fig. 1.1 Arrangement of particles in solid, liquid and gaseous state
(iii) Gases have neither definite volume nor definite shape. They completely occupy the space in the container in which they are placed.
These three states of matter are interconvertible by changing the conditions of temperature and pressure.
Solid
Liquid
Gas
On heating, a solid usually changes to a liquid, and the liquid on further heating changes to gas (or vapour). In the reverse process, a gas on cooling liquifies to the liquid and the liquid on further cooling freezes to the solid.
# 1.2.2. Classification of Matter
In Class IX (Chapter 2), you have learnt that at the macroscopic or bulk level, matter can be classified as mixture or pure substance. These can be further sub-divided as shown in Fig. 1.2.
Fig. 1.2 Classification of matter
When all constituent particles of a substance are same in chemical nature, it is said to be a pure substance. A mixture contains many types of particles. A mixture contains particles of two or more pure substances which may be present in it in any ratio. Hence, their composition is variable. Pure substances forming mixture are called its components. Many of the substances present around you are mixtures. For example, sugar solution in water, air, tea, etc., are all mixtures.
A mixture may be homogeneous or heterogeneous. In a homogeneous mixture, the components completely mix with each other. This means particles of components of the mixture are uniformly distributed throughout the bulk of the mixture and its composition is uniform throughout. Sugar solution and air are the examples of homogeneous mixtures. In contrast to this, in a heterogeneous mixture, the composition is not uniform throughout and sometimes different components are visible. For example, mixtures of salt and sugar, grains and pulses along with some dirt (often stone pieces), are heterogeneous mixtures. You can think of many more examples of mixtures which you come across in daily life. It is worthwhile to mention here that the components of a mixture can be separated by using physical methods, such as simple hand-picking, filtration, crystallisation, distillation, etc.
Pure substances have characteristics different from mixtures. Constituent particles of pure substances have fixed composition. Copper, silver, gold, water and glucose are some examples of pure substances. Glucose contains carbon, hydrogen and oxygen in a fixed ratio and its particles are of same composition. Hence, like all other pure substances, glucose has a fixed composition. Also, its constituents—carbon, hydrogen and oxygen—cannot be separated by simple physical methods.
Pure substances can further be classified into elements and compounds. Particles of an element consist of only one type of atoms. These particles may exist as atoms or molecules. You may be familiar with atoms. | 4 | 11 | Chemistry | 101 |
8c353c87-ff29-4331-b312-a4b6304d50f7 | # Chemistry
and molecules from the previous classes; however, you will be studying about them in detail in Unit 2. Sodium, copper, silver, hydrogen, oxygen, etc., are some examples of elements. Their all atoms are of one type. However, the atoms of different elements are different in nature. Some elements, such as sodium or copper, contain atoms as their constituent particles, whereas, in some others, the constituent particles are molecules which are formed by two or more atoms. For example, hydrogen, nitrogen and oxygen gases consist of molecules, in which two atoms combine to give their respective molecules. This is illustrated in Fig. 1.3.
Water molecule (H2O) Carbon dioxide molecule (CO2)
Fig. 1.4 A depiction of molecules of water and carbon dioxide
Elements are present in a compound in a fixed and definite ratio and this ratio is characteristic of a particular compound. Also, the properties of a compound are different from those of its constituent elements. For example, hydrogen and oxygen are gases, whereas, the compound formed by their combination i.e., water is a liquid. It is interesting to note that hydrogen burns with a pop sound and oxygen is a supporter of combustion, but water is used as a fire extinguisher.
# 1.3 Properties of Matter and Their Measurement
# 1.3.1 Physical and Chemical Properties
Every substance has unique or characteristic properties. These properties can be classified into two categories — physical properties, such as colour, odour, melting point, boiling point, density, etc., and chemical properties, like composition, combustibility, reactivity with acids and bases, etc.
Physical properties can be measured or observed without changing the identity or the composition of the substance. The measurement or observation of chemical properties requires a chemical change to occur. Measurement of physical properties does not require occurrence of a chemical change. The examples of chemical properties are characteristic reactions of different substances; these include acidity or basicity, combustibility, etc. Chemists describe, interpret and predict the behaviour of substances on the basis of knowledge of their physical and chemical properties, which are determined by careful measurement and experimentation.
In the following section, we | 5 | 11 | Chemistry | 101 |
6ab3fb4a-6b4f-4d1a-8fae-6eebbbed9d2d | # SOME BASIC CONCEPTS OF CHEMISTRY
will learn about the measurement of physical properties.
# 1.3.2 measurement of physical properties
Quantitative measurement of properties is required for scientific investigation. Many properties of matter, such as length, area, volume, etc., are quantitative in nature. Any quantitative observation or measurement is represented by a number followed by units in which it is measured. For example, length of a room can be represented as 6 m; here, 6 is the number and m denotes metre, the unit in which the length is measured.
Earlier, two different systems of measurement, i.e., the English System and the Metric System were being used in different parts of the world. The metric system, which originated in France in late eighteenth century, was more convenient as it was based on the decimal system. Late, need of a common standard system was felt by the scientific community. Such a system was established in 1960 and is discussed in detail below.
# 1.3.3 the international System of Units (Si)
The International System of Units (in the SI system has seven base units French Le Systeme International d’Unités — abbreviated as SI) was established by the 11ᵗʰ General Conference on Weights and Measures (CGPM from Conference Generale des Poids et Measures). The CGPM is an inter-governmental treaty organisation created by a diplomatic treaty known as Metre Convention, which was signed in Paris in 1875.
|Base physical Quantity|Symbol for Si Unit|Name of Si Unit|Symbol for Si Unit|
|---|---|---|---|
|Length|l|metre|m|
|Mass|m|kilogram|kg|
|Time|t|second|s|
|Electric current|I|ampere|A|
|Thermodynamic temperature|T|kelvin|K|
|Amount of substance|n|mole|mol|
|Luminous intensity|Iv|candela|cd| | 6 | 11 | Chemistry | 101 |
08523ec7-1688-4763-a465-d86bb4d00515 | # Chemistry
The definitions of the SI base units are given in Table 1.2. The SI system allows the use of prefixes to indicate the multiples or submultiples of a unit. These prefixes are listed in Table 1.3. Let us now quickly go through some of the quantities which you will be often using in this book.
|Unit|Symbol|Definition|
|---|---|---|
|Length|metre|It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299792458 when expressed in the unit ms–1, where the second is defined in terms of the caesium frequency VCs.|
|Mass|kilogram|It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10–34 when expressed in the unit Js, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and VCs.|
|Time|second|It is defined by taking the fixed numerical value of the caesium frequency VCs, the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, to be 9192631770 when expressed in the unit Hz, which is equal to s–1.|
|Electric Current|ampere|It is defined by taking the fixed numerical value of the elementary charge e to be 1.602176634×10–19 when expressed in the unit C, which is equal to As, where the second is defined in terms of VCs.|
|Thermodynamic Temperature|kelvin|It is defined by taking the fixed numerical value of the Boltzmann constant k to be 1.380649×10–23 when expressed in the unit JK–1, which is equal to kg m2 s–2 K–1 where the kilogram, metre and second are defined in terms of h, c and VCs.|
|Amount of Substance|mole|One mole contains exactly 6.02214076×1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol–1 and is called the Avogadro number. The amount of substance, symbol n, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles.|
|Luminous Intensity|candela|It is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency 540×1012 Hz, Kcd, to be 683 when expressed in the unit lm·W–1, which is equal to cd·sr·W–1, or cd sr kg–1 m–2 s3, where the kilogram, metre and second are defined in terms of h, c and VCs.| | 7 | 11 | Chemistry | 101 |
9a414b15-ff61-4b86-a12f-a17445497d3c | # Some Basic Concepts of Chemistry
# Table 1.3 Prefixes used in the SI System
|multiple|Prefix|Symbol|
|---|---|---|
|10–24|yocto|y|
|10–21|zepto|z|
|10–18|atto|a|
|10–15|femto|f|
|10–12|pico|p|
|10–9|nano|n|
|10–6|micro|μ|
|10–3|milli|m|
|10–2|centi|c|
|10–1|deci|d|
|10|deca|da|
|102|hecto|h|
|103|kilo|k|
|106|mega|M|
|109|giga|G|
|1012|tera|T|
|1015|peta|P|
|1018|exa|E|
|1021|zeta|Z|
|1024|yotta|Y|
SI system, volume has units of m³. But again, in chemistry laboratories, smaller volumes are used. Hence, volume is often denoted in cm³ or dm³ units.
A common unit, litre (L) which is not an SI unit, is used for measurement of volume of liquids.
1 L = 1000 mL, 1000 cm³ = 1 dm³
# 1.3.4 Mass and Weight
The mass of a substance is the amount of matter present in it, while weight is the force exerted by gravity on an object. The mass of a substance is constant, whereas, its weight may vary from one place to another due to change in gravity. You should be careful in using these terms.
The mass of a substance can be determined accurately in the laboratory by using an analytical balance (Fig. 1.5).
The SI unit of mass as given in Table 1.1 is kilogram. However, its fraction named as gram (1 kg = 1000 g), is used in laboratories due to the smaller amounts of chemicals used in chemical reactions.
# 1.3.5 Volume
Volume is the amount of space occupied by a substance. It has the units of (length)³. So in
Fig. 1.6 Different units used to express volume | 8 | 11 | Chemistry | 101 |
59428f4b-eed7-4e16-a098-daeb495c52a5 | # Chemistry
In the laboratory, the volume of liquids or solutions can be measured by graduated cylinder, burette, pipette, etc. A volumetric flask is used to prepare a known volume of a solution. These measuring devices are shown in Fig. 1.7.
Generally, the thermometer with Celsius scale are calibrated from 0° to 100°, where these two temperatures are the freezing point and the boiling point of water, respectively. The Fahrenheit scale is represented between 32° to 212°.
The temperatures on two scales are related to each other by the following relationship:
°F = 9(°C) + 32
The Kelvin scale is related to Celsius scale as follows:
K = °C + 273.15
It is interesting to note that temperature below 0 °C (i.e., negative values) are possible in Celsius scale but in Kelvin scale, negative temperature is not possible.
Fig. 1.7 Some volume measuring devices
# 1.3.6 Density
The two properties — mass and volume discussed above are related as follows:
Density = Mass / Volume
Density of a substance is its amount of mass per unit volume. So, SI units of density can be obtained as follows:
SI unit of density = SI unit of mass / SI unit of volume = kg / m³
This unit is quite large and a chemist often expresses density in g cm–3, where mass is expressed in gram and volume is expressed in cm³. Density of a substance tells us about how closely its particles are packed. If density is more, it means particles are more closely packed.
Fig. 1.8 Thermometers using different temperature scales
# 1.4 Uncertainty in measurement
Many a time in the study of chemistry, one has to deal with experimental data as well as theoretical calculations. There are meaningful ways to handle the numbers conveniently and... | 9 | 11 | Chemistry | 101 |
a4824e23-890b-4254-a9cd-776d1e98ee5d | # SOME BASIC CONCEPTS OF CHEMISTRY
After defining a unit of measurement such as the kilogram or the metre, scientists agreed on reference standards that make it possible to calibrate all measuring devices. For getting reliable measurements, all devices such as metre sticks and analytical balances have been calibrated by their manufacturers to give correct readings. However, each of these devices is standardised or calibrated against some reference. The mass standard is the kilogram since 1889. It has been defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at International Bureau of Weights and Measures in Sevres, France. Pt-Ir was chosen for this standard because it is highly resistant to chemical attack and its mass will not change for an extremely long time.
Scientists are in search of a new standard for mass. This is being attempted through accurate determination of Avogadro constant. Work on this new standard focuses on ways to measure accurately the number of atoms in a well-defined mass of sample. One such method, which uses X-rays to determine the atomic density of a crystal of ultrapure silicon, has an accuracy of about 1 part in 10⁶ but has not yet been adopted to serve as a standard. There are other methods but none of them are presently adequate to replace the Pt-Ir cylinder. No doubt, changes are expected within this decade.
The metre was originally defined as the length between two marks on a Pt-Ir bar kept at a temperature of 0°C (273.15 K). In 1960 the length of the metre was defined as 1.65076373 ×10⁶ times the wavelength of light emitted by a krypton laser. Although this was a cumbersome number, it preserved the length of the metre at its agreed value. The metre was redefined in 1983 by CGPM as the length of path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. Similar to the length and the mass, there are reference standards for other physical quantities.
# 1.4.1 Scientific Notation
As chemistry is the study of atoms and molecules, which have extremely low masses and are present in extremely large numbers, a chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g mass of a H atom. It may look funny for a moment to write or count numbers involving so many zeros but it offers a real challenge to do simple mathematical operations of addition, subtraction, multiplication or division with such numbers. You can write any two numbers of the above type and try any one of the operations you like to accept as a challenge, and then, you will really appreciate the difficulty in handling such numbers.
This problem is solved by using scientific notation for such numbers, i.e., exponential notation in which any number can be represented in the form N ×10ⁿ, where n is an exponent having positive or negative values and N is a number (called digit term) which varies between 1.000... and 9.999.... Thus, we can write 232.508 as 2.32508 ×10² in scientific notation. Note that while writing it, the decimal had to be moved to the left by two places and same is the exponent (2) of 10 in the scientific notation. Similarly, 0.00016 can be written as 1.6 × 10–4. Here, the decimal has to be moved four places to the right and (–4) is the exponent in the scientific notation.
While performing mathematical operations on numbers expressed in scientific notations, the following points are to be kept in mind. | 10 | 11 | Chemistry | 101 |
95455141-34ba-4654-a51b-32cb3b2bcf33 | # Chemistry
# Multiplication and Division
These two operations follow the same rules which are there for exponential numbers, i.e.
(5.6 × 105) × (8 × 106) = (5.6 × 8) × 105+6
= 38.4 × 1011 = 3.84 × 1012
(9.8 × 10-2) × (2.5 × 10-6) = (9.8 × 2.5) × 10-2-6
= 24.50 × 10-8 = 2.450 × 10-7
(2.7 × 10-3) ÷ (5.5 × 10-4) = (2.7 ÷ 5.5) × 10-3-(-4)
= 0.4909 × 10-7 = 4.909 × 10-8
# Addition and Subtraction
For these two operations, first the numbers are written in such a way that they have the same exponent. After that, the coefficients (digit terms) are added or subtracted as the case may be.
Thus, for adding 6.65×104 and 8.95×103, exponent is made same for both the numbers.
Thus, we get (6.65×104) + (0.895×104)
Then, these numbers can be added as follows:
(6.65 + 0.895)×104 = 7.545×104
Similarly, the subtraction of two numbers can be done as shown below:
(2.5 × 10-2) – (4.8 × 10-3)
= (2.5 ×10-2) – (0.48 ×10-2)
= (2.5 – 0.48)×10-2 = 2.02 × 10-2
# 1.4.2 Significant Figures
Every experimental measurement has some amount of uncertainty associated with it because of limitation of measuring instrument and the skill of the person making the measurement.
For example, mass of an object is obtained using a platform balance and it comes out to be 9.4g. On measuring the mass of this object on an analytical balance, the mass obtained is 9.4213g.
The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures. Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit. Thus, if we write a result as 11.2 mL, we say the 11 is certain and 2 is uncertain and the uncertainty would be +1 in the last digit. Unless otherwise stated, an uncertainty of +1 in the last digit is always understood.
# Rules for Determining Significant Figures
1. All non-zero digits are significant. For example in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.
2. Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3. Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.
4. Zeros at the end or right of a number are significant, provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures. But, if otherwise, the terminal zeros are not significant if there is no decimal point. For example, 100 has only one significant figure, but 100 has three significant figures and 100.0 has four significant figures.
Such numbers are better represented in scientific notation. We can express the number 100 as 1×102 for one significant figure, 1.0×102 for two significant figures and 1.00×102 for three significant figures. | 11 | 11 | Chemistry | 101 |
88c800b7-d5ad-44c2-860d-b7f905e27e39 | # SOME BASIC CONCEPTS OF CHEMISTRY
# Counting the numbers of object
For example, 2 balls or 20 eggs, have infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000.
In numbers written in scientific notation, all digits are significant e.g., 4.01×10² has three significant figures, and 8.256×10–3 has four significant figures.
However, one would always like the results to be precise and accurate. Precision and accuracy are often referred to while we talk about the measurement.
Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result. For example, if the true value for a result is 2.00 g and student ‘A’ takes two measurements and reports the results as 1.95 g and 1.93 g. These values are precise as they are close to each other but are not accurate.
Another student ‘B’ repeats the experiment and obtains 1.94 g and 2.05 g as the results for two measurements. These observations are neither precise nor accurate. When the third student ‘C’ repeats these measurements and reports 2.01 g and 1.99 g as the result, these values are both precise and accurate.
This can be more clearly understood from the data given in Table 1.4.
**Table 1.4 Data to illustrate precision and accuracy**
|Measurements/g|1|2|Average (g)|
|---|---|---|---|
|Student A|1.95|1.93|1.940|
|Student B|1.94|2.05|1.995|
|Student C|2.01|1.99|2.000|
# Addition and Subtraction of Significant Figures
The result cannot have more digits to the right of the decimal point than either of the original numbers.
Example:
- 12.11
- 18.0
- 1.012
- 31.122
# Dimensional analysis
Often while calculating, there is a need to convert units from one system to the other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis. This is illustrated below.
Example:
A piece of metal is 3 inch (represented by in) long. What is its length in cm? | 12 | 11 | Chemistry | 101 |
f1623fc7-0464-43ce-bf29-c7256004bd5b | # Chemistry
# Solution
The above is multiplied by the unit factor. We know that 1 in = 2.54 cm.
From this equivalence, we can write:
1 in 2.54
2.54 = 1 = . cm
. cm = 1 in
Thus, 1 in equals 1 and .1 cm. Both of these are called unit factors. If some number is multiplied by these unit factors (i.e., 1), it will not be affected otherwise.
Say, the 3 in given above is multiplied by the unit factor. So,
2.54
3 in = 3 in × . cm = 3 × 2.54 cm = 7.62 cm
1 in
Now, the unit factor by which multiplication is to be done is that unit factor ( .1 cm in the above case) which gives the desired units i.e., the numerator should have that part which is required in the desired result.
It should also be noted in the above example that units can be handled just like other numerical part. It can be cancelled, divided, multiplied, squared, etc. Let us study one more example.
# Example
A jug contains 2 L of milk. Calculate the volume of the milk in m³.
# Solution
Since 1 L = 1000 cm³ and 1 m = 100 cm, which gives:
1 m = 1 = 100 cm
100 cm = 1 m
To get m³ from the above unit factors, the first unit factor is taken and it is cubed.
1 m³ = (1 m)³ = (100 cm)³
1 m³ = 1000000 cm³
Now 2 L = 2 × 1000 cm³.
# 1.5 Laws of Chemical Combinations
The combination of elements to form compounds is governed by the following five basic laws.
# 1.5.1 Law of Conservation of Mass
This law was put forth by Antoine Lavoisier in 1789. He performed careful experimental studies for combustion reactions and reached to the conclusion that in all physical and chemical changes, there is no net change in mass during the process. Hence, he reached to the conclusion that matter can neither be created nor destroyed. This is called ‘Law of Conservation of Mass’. This law formed the basis for several later developments in chemistry. In fact, this was the result of exact measurement of masses of reactants and products, and carefully planned experiments performed by Lavoisier. | 13 | 11 | Chemistry | 101 |
2f765b87-2e43-4bd8-9c0d-53cf2ec6d55a | # SOME BASIC CONCEPTS OF CHEMISTRY
# 1.5.2 Law of Definite Proportions
This law was given by a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. Proust worked with two samples of cupric carbonate — one of which was of natural origin and the other was synthetic. He found that the composition of elements present in it was the same for both the samples as shown below:
| |% of copper|% of carbon|% of oxygen|
|---|---|---|---|
|Natural Sample|51.35|9.74|38.91|
|Synthetic Sample|51.35|9.74|38.91|
Thus, he concluded that irrespective of the source, a given compound always contains the same elements combined together in the same proportion by mass. The validity of this law has been confirmed by various experiments. It is sometimes also referred to as law of Definite Composition.
# 1.5.3 Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
Hydrogen + Oxygen → Water
2g 16g 18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g 32g 34g
Here, the masses of oxygen (i.e., 16 g and 32 g), which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e., 16:32 or 1:2.
# 1.5.4 Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure. Thus, 100 mL of hydrogen combine with 50 mL of oxygen to give 100 mL of water vapour.
Hydrogen + Oxygen → Water
100 mL 50 mL 100 mL
Thus, the volumes of hydrogen and oxygen which combine (i.e., 100 mL and 50 mL) bear a simple ratio of 2:1. Gay Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportions by volume. The law of definite proportions, stated earlier, was with respect to mass. The Gay Lussac’s law was explained properly by the work of Avogadro in 1811.
# 1.5.5 Avogadro’s Law
In 1811, Avogadro proposed that equal volumes of all gases at the same temperature and pressure should contain equal number of molecules. Avogadro made a distinction between atoms and molecules which is quite understandable in present times. If we consider again the reaction of hydrogen and oxygen to produce water, we see that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water without leaving any unreacted oxygen.
Note that in the Fig. 1.9 (Page 16) each box contains equal number of molecules. In fact, Avogadro could explain the above result by considering the molecules to be polyatomic. If hydrogen and oxygen were considered as diatomic as recognised now, then the above results are easily understandable. However, Dalton and others believed at that time that atoms of the same kind. | 14 | 11 | Chemistry | 101 |
70312553-3fa7-43b4-8ed6-2921e5f80fa8 | # CHEMISTRY
Fig. 1.9 Two volumes of hydrogen react with one volume of oxygen to give two volumes of water vapour
cannot combine and molecules of oxygen or hydrogen containing two atoms did not exist. Dalton’s theory could explain the laws of chemical combination. However, it could not explain the laws of gaseous volumes. It could not provide the reason for combining of atoms, which was answered later by other scientists.
After about 50 years, in 1860, the first international conference on chemistry was held in Karlsruhe, Germany, to resolve various ideas. At the meeting, Stanislao Cannizaro presented a sketch of a course of chemical philosophy, which emphasised on the importance of Avogadro’s work.
# 1.6 Dalton’s atomic theory
Although the origin of the idea that matter is composed of small indivisible particles called ‘a-tomio’ (meaning, indivisible), dates back to the time of Democritus, a Greek Philosopher (460–370 BC), it again started emerging as a result of several experimental studies which led to the laws mentioned above.
In 1808, Dalton published John Dalton ‘A New System of Chemical Philosophy’, in which he proposed the following:
1. Matter consists of indivisible atoms.
2. All atoms of a given element have identical properties, including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.
# 1.7 atomic anD molecUlar maSSeS
After having some idea about the terms atoms and molecules, it is appropriate here to understand what do we mean by atomic and molecular masses.
# 1.7.1 atomic mass
The atomic mass or the mass of an atom is actually very-very small because atoms are extremely small. Today, we have sophisticated techniques e.g., mass spectrometry for determining the atomic masses fairly accurately. But in the nineteenth century, scientists could determine the mass of one atom relative to another by experimental means, as has been mentioned earlier.
Hydrogen, being the lightest atom was arbitrarily assigned a mass of 1 (without any units) and other elements were assigned masses relative to it. However, the present system of atomic masses is based on carbon-12 as the standard and has been agreed upon in 1961. Here, Carbon-12 is one of the isotopes of carbon and can be represented as 12C. In this system, 12C is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard. One atomic mass unit is defined as a mass exactly equal to one-twelfth of the mass of one carbon – 12 atom. | 15 | 11 | Chemistry | 101 |
350f8021-fc98-4f57-8bee-9c2b6ea4dff5 | # SOME BASIC CONCEPTS OF CHEMISTRY
# 1.7.3 molecular mass
Mass of an atom of hydrogen = 1.6736×10–24g
Thus, in terms of amu, the mass of hydrogen atom = 1.0078 amu = 1.0080 amu
Similarly, the mass of oxygen - 16 (¹⁶O) atom would be 15.995 amu.
‘u’ At present, ‘amu’ has been replaced by *u*, which is known as unified mass.
When we use atomic masses of elements in calculations, we actually use average atomic masses of elements, which are explained below.
# 1.7.2 average atomic mass
Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence), the average atomic mass of that element can be computed. For example, carbon has the following three isotopes with relative abundances and masses as shown against each of them.
|isotope|relative atomic mass|abundance (%)|
|---|---|---|
|12C|12|98.892|
|13C|13.00335|1.108|
|14C|14.00317|2 ×10–10|
From the above data, the average atomic mass of carbon will come out to be:
(0.98892) (12 u) + (0.01108) (13.00335 u) + (2 ×10–12) (14.00317 u) = 12.011 u
Similarly, average atomic masses for other elements can be calculated. In the periodic table of elements, the atomic masses mentioned for different elements actually represent their average atomic masses.
# 1.7.4 formula mass
Some substances, such as sodium chloride, do not contain discrete molecules as their constituent units. In such compounds, positive (sodium ion) and negative (chloride ion) entities are arranged in a three-dimensional structure, as shown in Fig. 1.10.
It may be noted that in sodium chloride, one Na+ ion is surrounded by six Cl– ions and vice-versa.
The formula, such as NaCl, is used to calculate the formula mass instead of molecular mass as in the solid state sodium chloride does not exist as a single entity.
Fig. 1.10 Packing of Na+ and Cl– ions in sodium chloride | 16 | 11 | Chemistry | 101 |
f208de6f-40c2-49a0-bd81-17584189510a | # Chemistry
Thus, the formula mass of sodium chloride is
atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u
# Problem 1.1
Calculate the molecular mass of glucose (C6H12O6) molecule.
# Solution
Molecular mass of glucose (C6H12O6)
= 6 (12.011 u) + 12 (1.008 u) + 6 (16.00 u)
= (72.066 u) + (12.096 u) + (96.00 u)
= 180.162 u
# 1.8 Mole Concept and Molar Masses
Atoms and molecules are extremely small in size and their numbers in even a small amount of any substance is really very large. To handle such large numbers, a unit of convenient magnitude is required.
Just as we denote one dozen for 12 items, score for 20 items, gross for 144 items, we use the idea of mole to count entities at the microscopic level (i.e., atoms, molecules, particles, electrons, ions, etc).
In SI system, mole (symbol, mol) was introduced as seventh base quantity for the amount of a substance.
The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol–1 and is called the Avogadro number.
The amount of substance, symbol n, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be.
In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to:
12 g/mol12C
1.992648 × 10–23 g/C atom
= 6.0221367 × 1023 atoms/mol
# Fig. 1.11 One mole of various substances | 17 | 11 | Chemistry | 101 |
7b8503f9-3e8d-40ee-8236-ff86af97e572 | # Some Basic Concepts of Chemistry
# 1.9.1 Empirical Formula for Molecular Formula
You would ask is: what is its formula or what are its constituents and in what ratio are they present in the given compound? For known compounds also, such information provides a check whether the given sample contains the same percentage of elements as present in a pure sample. In other words, one can check the purity of a given sample by analysing this data.
Let us understand it by taking the example of water (H₂O). Since water contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as follows:
Mass % of an element =
mass of that element in the compound × 100
molar mass of the compound
Molar mass of water = 18.02 g
Mass % of hydrogen =
11.18
Mass % of oxygen =
16.00 × 100 = 18.02
88.79
Let us take one more example. What is the percentage of carbon, hydrogen and oxygen in ethanol?
Molecular formula of ethanol is: C2H5OH
Molar mass of ethanol is:
(2 × 12.01 + 6 × 1.008 + 16.00) g = 46.068 g
Mass per cent of carbon =
24.02 g × 100 = 52.14%
46.068 g
Mass per cent of hydrogen =
6.048 g × 100 = 13.13%
46.068 g
Mass per cent of oxygen =
16.00 g × 100 = 34.73%
46.068 g
After understanding the calculation of per cent of mass, let us now see what information can be obtained from the per cent composition data. | 18 | 11 | Chemistry | 101 |
2b6b8381-3d45-42bc-8482-2f696995fb61 | # Chemistry
Step 3. Divide each of the mole values obtained above by the smallest number amongst them. Since 2.021 is the smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl.
In case the ratios are not whole numbers, then they may be converted into whole numbers by multiplying by the suitable coefficient.
Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements. CH2Cl is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula
1. Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. For CH2Cl, empirical formula mass is 12.01 + (2 × 1.008) + 35.453 = 49.48 g
2. Divide Molar mass by empirical formula mass
3. Multiply empirical formula by n obtained above to get the molecular formula. Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.
Thus, according to the above chemical reaction:
- One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g)
- One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g)
- 22.7 L of CH4(g) reacts with 45.4 L of O2(g) to give 22.7 L of CO2(g) and 45.4 L of H2O(g)
- 16 g of CH4(g) reacts with 2×32 g of O2(g) to give 44 g of CO2(g) and 2×18 g of H2O(g).
From these relationships, the given data can be interconverted as follows:
mass
Mass
=
Density
Volume
# 1.10 Stoichiometry and Stoichiometric Calculations
The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, element) and metron (meaning, measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction.
Before understanding how to calculate the amounts of reactants required or the products produced in a chemical reaction, let us study what information is available from the balanced chemical reaction.
# 1.10.1 Limiting Reagent
Many a time, reactions are carried out with the amounts of reactants that are different than the amounts as required by a balanced chemical reaction. In such situations, one reactant is in more amount than the amount required by the balanced chemical reaction. | 19 | 11 | Chemistry | 101 |
b17f2bd1-9bad-4042-8490-c86736828927 | # SOME BASIC CONCEPTS OF CHEMISTRY
reactant which is present in the least amount gets consumed after sometime and after that further reaction does not take place whatever be the amount of the other reactant. Hence, the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent.
In performing stoichiometric calculations, this aspect is also to be kept in mind.
# 1.10.2 reactions in Solutions
A majority of reactions in the laboratories are carried out in solutions. Therefore, it is important to understand as how the amount of substance is expressed when it is present in the solution. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways.
1. Mass per cent or weight per cent (w/w %)
2. Mole fraction
3. Molarity
4. Molality
Let us now study each one of them in detail.
# Balancing a chemical equation
According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides:
- 4 Fe(s) + 3O₂(g) → 2Fe₂O₃(s) (a) balanced equation
- 2 Mg(s) + O₂(g) → 2MgO(s) (b) balanced equation
- P₄(s) + O₂(g) → P₄O₁₀(s) (c) unbalanced equation
Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation.
P₄(s) + 5O₂(g) → P₄O₁₀(s) balanced equation
Now, let us take combustion of propane, C₃H₈. This equation can be balanced in steps.
1. Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products.
2. C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l) unbalanced equation
3. Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO₂ molecules are required on the right side.
4. C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)
5. Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side.
6. C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)
7. Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO₂ and 4×1= 4 in water). Therefore, five O₂ molecules are needed to supply the required 10 oxygen atoms.
8. C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
9. Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation.
2024-25 | 20 | 11 | Chemistry | 101 |
86c566c3-182c-4cc7-89d1-d6520ec68a58 | # Problem 1.3
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
# Solution
The balanced equation for the combustion of methane is:
CH₄ (g) + 2O₂(g) → CO₂ (g) + 2H₂O (g)
1. 16 g of CH₄ corresponds to one mole.
2. From the above equation, 1 mol of CH₄ (g) gives 2 mol of H₂O (g).
2 mol of water (H₂O) = 2 × (2 + 16) = 2 × 18 = 36 g
18 g H₂O
1 mol H₂O = 18 g H₂O ⇒ 2 mol H₂O = 1
Hence, 2 mol H₂O × 2 = 2 × 18 g H₂O = 36 g H₂O
# Problem 1.4
How many moles of methane are required to produce 22 g CO₂ (g) after combustion?
# Solution
According to the chemical equation,
CH₄ (g) + 2O₂(g) → CO₂ (g) + 2H₂O (g)
44 g CO₂ (g) is obtained from 16 g CH₄ (g).
[∴ 1 mol CO₂(g) is obtained from 1 mol of CH₄(g)]
Number of moles of CO₂ (g) = 22 g CO₂ (g) × 1 mol CO₂ (g) / 44 g CO₂ (g) = 0.5 mol CO₂ (g)
Hence, 0.5 mol CO₂ (g) would be obtained from 0.5 mol CH₄ (g) or 0.5 mol of CH₄ (g) would be required to produce 22 g CO₂ (g).
# Problem 1.5
50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₃ (g). Calculate the amount of NH₃ (g) formed. Identify the limiting reagent in the production of NH₃ in this situation.
# Solution
A balanced equation for the above reaction is written as follows:
N₂ (g) + 3H₂ (g) → 2NH₃ (g)
# Calculation of moles:
Number of moles of N₂:
50.0 kg N₂ × 1000 g N₂ / 1 kg N₂ × 1 mol N₂ / 28.0 g N₂ = 17.86 × 10² mol
Number of moles of H₂:
10.0 kg H₂ × 1000 g H₂ / 1 kg H₂ × 1 mol H₂ / 2.016 g H₂ = 4.96 × 10³ mol
According to the above equation, 1 mol N₂ (g) requires 3 mol H₂ (g), for the reaction. Hence, for 17.86 × 10² mol of N₂, the moles of H₂ (g) required would be:
17.86 × 10² mol N₂ × 3 mol H₂ (g) / 1 mol N₂ (g) = 5.36 × 10³ mol H₂
But we have only 4.96 × 10³ mol H₂. Hence, dihydrogen is the limiting reagent in this case. So, NH₃(g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 10³ mol.
Since 3 mol H₂ (g) gives 2 mol NH₃ (g):
4.96 × 10³ mol H₂ (g) × 2 mol NH₃ (g) / 3 mol H₂ (g) = 3.30 × 10³ mol NH₃ (g)
3.30 × 10³ mol NH₃ (g) is obtained.
If they are to be converted to grams, it is done as follows:
1 mol NH₃ (g) = 17.0 g NH₃ (g)
3.30 × 10³ mol NH₃ (g) × 17.0 g NH₃ (g) / 1 mol NH₃ (g) | 21 | 11 | Chemistry | 101 |
26551991-90ea-42b1-a70a-a45d1f509353 | # SOME BASIC CONCEPTS OF CHEMISTRY
# 3. Molarity
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution. Thus,
Molarity (M) = No. of moles of solute / Volume of solution in litres
Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it.
1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, we require 0.2 moles of NaOH dissolved in 1 litre solution.
Hence, for making 0.2M solution from 1M solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH and dilute the solution with water to 1 litre.
Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows:
If 1 mol is present in 1L or 1000 mL solution then, 0.2 mol is present in
1000 mL × 0.2 = 200 mL solution
Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.
In fact for such calculations, a general formula, M₁×V₁ = M₂× V₂ where M and V are molarity and volume, respectively, can be used.
In this case, M₁ is equal to 0.2M; V₁ = 1000 mL and, M₂ = 1.0M; V₂ is to be calculated.
Substituting the values in the formula:
0.2 M × 1000 mL = 1.0 M × V₂
Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution (in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration.
# 1. Mass per cent
It is obtained by using the following relation:
# Problem 1.6
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.
Solution
# 2. Mole Fraction
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB, respectively, then the mole fractions of A and B are given as: | 22 | 11 | Chemistry | 101 |
229bb647-f3e7-471e-8818-93e6a31986ed | # CHEMISTRY
# Problem 1.7
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
# Solution
Since molarity (M)
|Mass of NaCl in 1 L solution|= 3 × 58.5 = 175.5 g|
|---|---|
|Mass of 1L solution|= 1000 × 1.25 = 1250 g (since density = 1.25 g mL–1)|
|Mass of water in solution|= 1250 – 175.5 = 1074.5 g|
|Molality|= No. of moles of solute / Mass of solvent in kg|
| |= 3 mol / 1.0745 kg = 2.79 m|
Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.
# 4. Molality
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.
Thus, Molality (m) = No. of moles of solute / Mass of solvent in kg
# Summary
Chemistry, as we understand it today is not a very old discipline. People in ancient India already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life.
The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures.
When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English | 23 | 11 | Chemistry | 101 |
f381b36d-ebd4-4b80-b5a8-209f2a5a004c | # SOME BASIC CONCEPTS OF CHEMISTRY
and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units).
Since measurements involve recording of data, which are always associated with a certain amount of uncertainty, the proper handling of data obtained by measuring the quantities is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 1023. Hence, a convenient system of expressing the numbers in scientific notation is used. The uncertainty is taken care of by specifying the number of significant figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another.
The combination of different atoms is governed by basic laws of chemical combination — these being the law of conservation of mass, Law of Definite Proportions, Law of multiple proportions, Gay Lussac’s law of Gaseous volumes and Avogadro law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass.
The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities.
Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality.
# exerciSeS
1. Calculate the molar mass of the following:
1. H₂O
2. CO₂
3. CH₄
2. Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
4. Calculate the amount of carbon dioxide that could be produced when
1. 1 mole of carbon is burnt in air.
2. 1 mole of carbon is burnt in 16 g of dioxygen.
3. 2 moles of carbon are burnt in 16 g of dioxygen.
5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1. | 24 | 11 | Chemistry | 101 |
c41c06a7-8838-4bbb-9910-b93051eeccd3 | # Chemistry
# 1.6
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%.
# 1.7
How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
# 1.8
Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.
# 1.9
Calculate the atomic mass (average) of chlorine using the following data:
| | |% natural abundance|molar mass|
|---|---|---|---|
| |35Cl|75.77|34.9689|
| |37Cl|24.23|36.9659|
# 1.10
In three moles of ethane (C2H6), calculate the following:
- (i) Number of moles of carbon atoms.
- (ii) Number of moles of hydrogen atoms.
- (iii) Number of molecules of ethane.
# 1.11
What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?
# 1.12
If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?
# 1.13
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1 Pa = 1 N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.
# 1.14
What is the SI unit of mass? How is it defined?
# 1.15
Match the following prefixes with their multiples:
|Prefixes|Multiples|
|---|---|
|(i) micro|106|
|(ii) deca|109|
|(iii) mega|10–6|
|(iv) giga|10–15|
|(v) femto|10|
# 1.16
What do you mean by significant figures?
# 1.17
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
- (i) Express this in per cent by mass.
- (ii) Determine the molality of chloroform in the water sample.
# 1.18
Express the following in the scientific notation:
- (i) 0.0048
- (ii) 234,000
- (iii) 8008
- (iv) 500.0
- (v) 6.0012
# 1.19
How many significant figures are present in the following?
- (i) 0.0025
- (ii) 208
- (iii) 5005
2024-25 | 25 | 11 | Chemistry | 101 |
edab70c6-ff9f-4b21-a66a-e59d9220d18f | # SOME BASIC CONCEPTS OF CHEMISTRY
# 1.20
Round up the following upto three significant figures:
- (i) 34.216
- (ii) 10.4107
- (iii) 0.04597
- (iv) 2808
# 1.21
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
|mass of dinitrogen|mass of dioxygen|
|---|---|
|(i) 14 g|16 g|
|(ii) 14 g|32 g|
|(iii) 28 g|32 g|
|(iv) 28 g|80 g|
(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions:
- (i) 1 km = ...................... mm = ...................... pm
- (ii) 1 mg = ...................... kg = ...................... ng
- (iii) 1 mL = ...................... L = ...................... dm³
# 1.22
If the speed of light is 3.0 × 10⁸ m s–1, calculate the distance covered by light in 2.00 ns.
# 1.23
In a reaction
A + B2 AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
- (i) 300 atoms of A + 200 molecules of B
- (ii) 2 mol A + 3 mol B
- (iii) 100 atoms of A + 100 molecules of B
- (iv) 5 mol A + 2.5 mol B
- (v) 2.5 mol A + 5 mol B
# 1.24
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + H2 (g) 2NH3 (g)
- (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.
- (ii) Will any of the two reactants remain unreacted?
- (iii) If yes, which one and what would be its mass?
# 1.25
How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
# 1.26
If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
# 1.27
Convert the following into basic units:
- (i) 28.7 pm
- (ii) 15.15 pm
- (iii) 25365 mg
2024-25 | 26 | 11 | Chemistry | 101 |
0c83c7a2-ca33-435c-beab-eaeffc48db4d | # CHEMISTRY
# 1.28
Which one of the following will have the largest number of atoms?
- (i) 1 g Au (s)
- (ii) 1 g Na (s)
- (iii) 1 g Li (s)
- (iv) 1 g of Cl₂(g)
# 1.29
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
# 1.30
What will be the mass of one 12C atom in g?
# 1.31
How many significant figures should be present in the answer of the following calculations?
- (i) 0.02856 × 298.15 × 0.112
- (ii) 5 × 5.364
- (iii) 0.0125 + 0.7864 + 0.0215
# 1.32
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
|isotope|isotopic molar mass|abundance|
|---|---|---|
|36Ar|35.96755 g mol–1|0.337%|
|38Ar|37.96272 g mol–1|0.063%|
|40Ar|39.9624 g mol–1|99.600%|
# 1.33
Calculate the number of atoms in each of the following:
- (i) 52 moles of Ar
- (ii) 52 u of He
- (iii) 52 g of He
# 1.34
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate:
- (i) empirical formula
- (ii) molar mass of the gas
- (iii) molecular formula
# 1.35
Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction:
CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂(g) + H₂O(l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
# 1.36
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction:
4 HCl (aq) + MnO₂(s) → 2H₂O (l) + MnCl₂(aq) + Cl₂ (g)
How many grams of HCl react with 5.0 g of manganese dioxide? | 27 | 11 | Chemistry | 101 |
2aeb6c56-5b10-4b8c-a07d-648b74557d7c | # Unit 5
# Thermodynamics
It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown.
— Albert Einstein
After studying this Unit, you will be able to:
- explain the terms: system and surroundings;
- discriminate between close, open and isolated systems;
- explain internal energy, work and heat;
- state first law of thermodynamics and express it mathematically;
- calculate energy changes work and heat contributions in chemical systems;
- explain state functions: U, H;
- correlate ∆U and ∆H;
- measure experimentally ∆U and ∆H;
- define standard states for ∆H;
- calculate enthalpy changes for various types of reactions;
- state and apply Hess’s law of constant heat summation;
- differentiate between extensive and intensive properties;
- define spontaneous and non-spontaneous processes;
- explain entropy as a thermodynamic state function and apply it for spontaneity;
- explain Gibbs energy change (∆G); and establish relationship between ∆G and spontaneity, ∆G and equilibrium constant.
Chemical energy stored by molecules can be released as heat during chemical reactions when a fuel like methane, cooking gas or coal burns in air. The chemical energy may also be used to do mechanical work when a fuel burns as in an engine or to provide electrical energy through a galvanic cell like dry cell. Thus, various forms of energy are interrelated and under certain conditions, these may be transformed from one form into another. The study of these energy transformations forms the subject matter of thermodynamics. The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Thermodynamics is not concerned about how and at what rate these energy transformations are carried out, but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state.
In this unit, we would like to answer some of the important questions through thermodynamics, like:
- How do we determine the energy changes involved in a chemical reaction/process? Will it occur or not?
- What drives a chemical reaction/process?
- To what extent do the chemical reactions proceed?
2024-25 | 0 | 11 | Chemistry | 105 |
be41b481-40f8-4c94-bfd7-1855523625da | # THERMODYNAMICS
# 5.1 Thermodynamic Terms
We are interested in chemical reactions and the energy changes accompanying them. For this we need to know certain thermodynamic terms. These are discussed below.
# 5.1.1 The system and the surroundings
A system in thermodynamics refers to that part of universe in which observations are made and remaining universe constitutes the surroundings. The surroundings include everything other than the system. System and the surroundings together constitute the universe.
The universe = The system + The surroundings
However, the entire universe other than the system is not affected by the changes taking place in the system. Therefore, for all practical purposes, the surroundings are that portion of the remaining universe which can interact with the system. Usually, the region of space in the neighbourhood of the system constitutes its surroundings.
For example, if we are studying the reaction between two substances A and B kept in a beaker, the beaker containing the reaction mixture is the system and the room where the beaker is kept is the surroundings (Fig. 5.1).
# 5.1.2 Types of the system
We, further classify the systems according to the movements of matter and energy in or out of the system.
1. Open System
In an open system, there is exchange of energy and matter between system and surroundings [Fig. 5.2 (a)]. The presence of reactants in an open beaker is an example of an open system. Here the boundary is an imaginary surface enclosing the beaker and reactants.
2. Closed System
In a closed system, there is no exchange of matter, but exchange of energy is possible between system and the surroundings [Fig. 5.2 (b)]. The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.
Note that the system may be defined by physical boundaries, like beaker or test tube, or the system may simply be defined by a set of Cartesian coordinates specifying a particular volume in space. It is necessary to think of the system as separated from the surroundings by some sort of wall which may be real or imaginary. The wall that separates.
Fig. 5.1 System and the surroundings
Fig. 5.2 Open, closed and isolated systems.
* We could have chosen only the reactants as system then walls of the beakers will act as boundary. | 1 | 11 | Chemistry | 105 |
d50d3088-6c17-4bb1-8562-6916ad9221ed | # 3. Isolated System
In an isolated system, there is no exchange of energy or matter between the system and the surroundings [Fig. 5.2 (c)]. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of an isolated system.
# 5.1.3 The state of the system
The system must be described in order to make any useful calculations by specifying quantitatively each of the properties such as its pressure (p), volume (V), and temperature (T) as well as the composition of the system. We need to describe the system by specifying it before and after the change. You would recall from your Physics course that the state of a system in mechanics is completely specified at a given instant of time, by the position and velocity of each mass point of the system. In thermodynamics, a different and much simpler concept of the state of a system is introduced. It does not need detailed knowledge of motion of each particle because, we deal with average measurable properties of the system. We specify the state of the system by state functions or state variables.
The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. In order to completely define the state of a system it is not necessary to define all the properties of the system; as only a certain number of properties can be varied independently. This number depends on the nature of the system. Once these minimum number of macroscopic properties are fixed, others automatically have definite values.
The state of the surroundings can never be completely specified; fortunately it is not necessary to do so.
# 5.1.4 The internal energy as a state Function
When we talk about our chemical system losing or gaining energy, we need to introduce the change in the internal energy of the system by doing some work on it. Let us call the initial state of the system as state A and its temperature as TA. Let the internal energy of the system in state A be called UA. We can change the state of the system in two different ways. | 2 | 11 | Chemistry | 105 |
560d3441-5b72-4c32-83ef-b8beb214849e | # THERMODYNAMICS
One way: We do some mechanical work, say 1 kJ, by rotating a set of small paddles and thereby churning water. Let the new state be called B state and its temperature, as TB. It is found that TB > TA and the change in temperature, ∆T = TB–TA. Let the internal energy of the system in state B be UB and the change in internal energy, ∆U = UB– UA.
Second way: We now do an equal amount (i.e., 1kJ) electrical work with the help of an immersion rod and note down the temperature change. We find that the change in temperature is same as in the earlier case, say, TB– TA.
In fact, the experiments in the above manner were done by J. P. Joule between 1840–50 and he was able to show that a given amount of work done on the system, no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system.
So, it seems appropriate to define a quantity, the internal energy U, whose value is characteristic of the state of a system, whereby the adiabatic work, wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ∆U i.e.,
∆U = U₂ – U₁ = wad
Therefore, internal energy, U, of the system is a state function.
By conventions of IUPAC in chemical thermodynamics. The positive sign expresses that wad is positive when work is done on the system and the internal energy of system increases. Similarly, if the work is done by the system, wad will be negative because internal energy of the system decreases.
Can you name some other familiar state functions? Some of other familiar state functions are V, p, and T. For example, if we bring a change in temperature of the system from 25°C to 35°C, the change in temperature is 35°C–25°C = +10°C, whether we go straight up to 35°C or we cool the system for a few degrees, then take the system to the final temperature. Thus, T is a state function and the change in temperature is independent of the route taken.
By conventions of IUPAC in chemical thermodynamics. The q is positive, when heat is transferred from the surroundings to the system and the internal energy of the system increases and q is negative when heat is transferred from system to the surroundings resulting in decrease of the internal energy of the system.
* Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention.
2024-25 | 3 | 11 | Chemistry | 105 |
87e73f2b-54ce-482d-990a-ee2eac4d40cd | # Chemistry
# (c) The general case
Let us consider the general case in which a change of state is brought about both by doing work and by transfer of heat. We write change in internal energy for this case as:
∆U = q + w (5.1)
For a given change in state, q and w can vary depending on how the change is carried out. However, q + w = ∆U will depend only on initial and final state. It will be independent of the way the change is carried out. If there is no transfer of energy as heat or as work (isolated system) i.e., if w = 0 and q = 0, then ∆ U = 0.
The equation 5.1 i.e., ∆U = q + w is a mathematical statement of the first law of thermodynamics, which states that:
The energy of an isolated system is constant.
It is commonly stated as the law of conservation of energy i.e., energy can neither be created nor be destroyed.
Note: There is considerable difference between the character of the thermodynamic property energy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ∆U of the system.
# 5.2 Applications
Many chemical reactions involve the generation of gases capable of doing mechanical work or the generation of heat. It is important for us to quantify these changes and relate them to the changes in the internal energy. Let us see how!
# 5.2.1 Work
First of all, let us concentrate on the nature of work a system can do. We will consider only mechanical work i.e., pressure-volume work. For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If external pressure is pₑₓ which is greater than p, piston is moved inward till the pressure.
# Problem 5.1
Express the change in internal energy of a system when:
1. No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have?
2. No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
3. w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be?
Fig. 5.5 (a) Work done on an ideal gas in a cylinder when it is compressed by a constant external pressure, pₑₓ (in single step) is equal to the shaded area. | 4 | 11 | Chemistry | 105 |
73b42ba9-92e4-468b-a421-65aa4027db36 | # THERMODYNAMICS
inside becomes equal to pex. Let this change be achieved in a single step and the final volume be Vf. During this compression, suppose piston moves a distance, l and is cross-sectional area of the piston is A [Fig. 5.5(a)].
Then, volume change = l × A = ∆V = (Vf – Vi)
We also know, pressure = relation
Therefore, force on the piston = pex . A
If w is the work done on the system by movement of the piston then
w = force × distance = pex . A . l = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi) (5.2)
The negative sign of this expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system. Here (Vf – Vi) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive.
If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to – Σp ∆V [Fig. 5.5 (b)]
Fig.5.5 (c) pV-plot when pressure is not constant and changes in infinite steps (reversible conditions) during compression from initial volume, Vi to final volume, Vf. Work done on the gas is represented by the shaded area.
A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other.
2024-25 | 5 | 11 | Chemistry | 105 |
3a098ba1-f526-4869-92c1-8d2ca683b1e5 | 142
Processes other than reversible processes
are known as irreversible processes.
In chemistry, we face problems that can
be solved if we relate the work term to the
internal pressure of the system. We can
relate work to internal pressure of the system
under reversible conditions by writing
equation 5.3 as follows:
Vf Vf
w p dV ( p dp dV
rev Vi ex Vi in )
Since dp × dV is very small we can write
Vf
wrev pindV (5.4)
Vi
Now, the pressure of the gas (pinwhich we
can write as p now) can be expressed in terms
of its volume through gas equation. For n mol
of an ideal gas i.e., pV =nRT
p nRT
V
Therefore, at constant temperature
(isothermal process),
Vf dV V
wrev nRT V nRT ln Vᶠ
Vi i
= – 2.303 nRT log Vf (5.5)
Vi
Free expansion: Expansion of a gas in
vacuum (p = 0) is called free expansion.
ex
No work is done during free expansion of an
ideal gas whether the process is reversible or
irreversible (equation 5.2 and 5.3).
Now, we can write equation 5.1 in number
of ways depending on the type of processes.
Let us substitute w = – pₑₓ∆V (eq. 5.2) in
equation 5.1, and we get
U q pex V
If a process is carried out at constant volume
(∆V = 0), then ∆U = q
V
the subscript V in qV denotes that heat is
supplied at constant volume.
2024-25
chemIstry
Isothermal and free expansion of an
ideal gas
For isothermal (T = constant) expansion of
an ideal gas into vacuum; w = 0 since pₑₓ = 0.
Also, Joule determined experimentally that
q = 0; therefore, ∆U = 0
E q u a t i o n 5 . 1 , AU = q + W c a n b e
expressed for isothermal irreversible and
reversible changes as follows:
1. For isothermal irreversible change
2. q = – w = pₑₓ (Vf – Vi)
For isothermal reversible change
q = – w = nRT ln
V f
= 2.303 nRT log Vi
For adiabatic change, q = 0,
∆U = wad
Problem 5.2
Two litres of an ideal gas at a pressure of 10
atm expands isothermally at 25 °C into a
vacuum until its total volume is 10 litres.
How much heat is absorbed and how
much work is done in the expansion ?
solution
We have q = – w = pₑₓ (10 – 2) = 0(8) = 0
No work is done; no heat is absorbed.
Problem 5.3
Consider the same expansion, but
this time against a constant external
pressure of 1 atm.
solution
We have q = – w = pₑₓ (8) = 8 litre-atm
Problem 5.4
Consider the expansion given in problem
5.2, for 1 mol of an ideal gas conducted
reversibly.
solution V
We have q = – w = 2.303 nRT log Vᶠ
s
= 2.303 × 1 × 0.8206 × 298 × log 10
2 | 6 | 11 | Chemistry | 105 |
6b630448-7131-44ed-b132-12e2f1275d6d | # THERMODYNAMICS
= 2.303 x 0.8206 x 298 x log 5
= 2.303 x 0.8206 x 298 x 0.6990
= 393.66 L atm
# 5.2.2 enthalpy, H
# (a) A Useful New State Function
We know that the heat absorbed at constant volume is equal to change in the internal energy i.e., ∆U = qV. But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. We need to define another state function which may be suitable under these conditions.
We may write equation (5.1) as ∆U = qp – p∆V at constant pressure, where qp is heat absorbed by the system and –p∆V represent expansion work done by the system.
Let us represent the initial state by subscript 1 and final state by 2. We can rewrite the above equation as
U2–U1 = qp – p (V2 – V1)
On rearranging, we get qp = (U2 + pV2) – (U1 + pV1) (5.6)
Now we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as :
H = U + pV (5.7)
so, equation (5.6) becomes qp = H2 – H1 = ∆H
Although q is a path dependent function, H is a state function because it depends on U, p and V, all of which are state functions. Therefore, ∆H is independent of path. Hence, qp is also independent of path.
For finite changes at constant pressure, we can write equation 5.7 as ∆H = ∆U + ∆pV
Since p is constant, we can write ∆H = ∆U + p∆V (5.8)
It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.
Remember ∆H = qp, heat absorbed by the system at constant pressure.
∆H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions which absorb heat from the surroundings.
At constant volume (∆V = 0), ∆U = qV, therefore equation 5.8 becomes ∆H = ∆U = qV
The difference between ∆H and ∆U is not usually significant for systems consisting of only solids and / or liquids. Solids and liquids do not suffer any significant volume changes upon heating. The difference, however, becomes significant when gases are involved. Let us consider a reaction involving gases. If VA is the total volume of the gaseous reactants, VB is the total volume of the gaseous products, nA is the number of moles of gaseous reactants and nB is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,
pVA = nART
and pVB = nBRT
Thus, pVB – pVA = nBRT – nART = (nB – nA)RT
or p (VB – VA) = (nB – nA) RT
or p ∆V = ∆n RT (5.9)
Here, ∆ng refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
Substituting the value of p∆V from equation 5.9 in equation 5.8, we get ∆H = ∆U + ∆ngRT (5.10)
The equation 5.10 is useful for calculating ∆H from ∆U and vice versa.
# Problem 5.5
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100°C is 41 kJ mol–1. Calculate the internal energy change, when | 7 | 11 | Chemistry | 105 |
4b29c178-fce5-4b7e-847f-3a341b2c7292 | # Chemistry
1 mol of water is vapourised at 1 bar pressure and 100°C.
# Solution
(i) The change H₂O (l) → H₂O (g)
∆H = ∆U + ∆ngRT
or ∆U = ∆H – ∆ngRT, substituting the values, we get
∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K
= 41.00 kJ mol–1 – 3.096 kJ mol–1
= 37.904 kJ mol–1
Fig. 5.6(a) A gas at volume V and temperature T
Fig. 5.6(b) Partition, each part having half the volume of the gas
# (b) Extensive and Intensive Properties
In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive properties.
Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example, temperature, density, pressure etc. are intensive properties. A molar property, χm, is the value of an extensive property χ of the system for 1 mol of the substance. If n is the amount of matter, χm = χ/n is independent of the amount of matter. Other examples are molar volume, Vm and molar heat capacity, Cm.
Let us understand the distinction between extensive and intensive properties by considering a gas enclosed in a container of volume V and at temperature T [Fig. 5.6(a)]. Let us make a partition such that volume is halved, each part [Fig. 5.6(b)] now has one half of the original volume, V2, but the temperature will still remain the same i.e., T. It is clear that volume is an extensive property and temperature is an intensive property.
# (c) Heat Capacity
In this sub-section, let us see how to measure heat transferred to a system. This heat appears as a rise in temperature of the system in case of heat absorbed by the system. The increase of temperature is proportional to the heat transferred:
q = coeff × ΔT
The magnitude of the coefficient depends on the size, composition and nature of the system. We can also write it as q = C ΔT.
The coefficient, C is called the heat capacity.
Thus, we can measure the heat supplied by monitoring the temperature rise, provided we know the heat capacity.
When C is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature.
C is directly proportional to the amount of substance. The molar heat capacity of a substance, Cm = C/n, is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree Celsius (or one Kelvin). Specific heat, also called specific heat capacity, is the quantity. | 8 | 11 | Chemistry | 105 |
0bbeb145-48c2-4651-9728-f897a30c7e43 | # THERMODYNAMICS
of heat required to raise the temperature of one unit mass of a substance by one degree celsius (or one kelvin). For finding out the heat, q, required to raise the temperatures of a sample, we multiply the specific heat of the substance, c, by the mass m, and temperatures change, ∆T as
q = c × m × ∆T = C ∆T (5.11)
# (d) The Relationship between Cp and CV for an Ideal Gas
At constant volume, the heat capacity, CV is denoted by CV and at constant pressure, this is denoted by Cp. Let us find the relationship between the two.
We can write equation for heat, q at constant volume as qV = CV ∆T = ∆U at constant pressure as qp = Cp ∆T = ∆H
The difference between Cp and CV can be derived for an ideal gas as:
For a mole of an ideal gas, ∆H = ∆U + ∆(pV) = ∆U + ∆(RT) = ∆U + R∆T
∴ ∆H = ∆U + R ∆T (5.12)
On putting the values of ∆H and ∆U, we have
Cp ∆T = CV ∆T + R∆T
Cp = CV + R
Cp – CV = R (5.13)
# 5.3 Measurement of ∆U and ∆H: Calorimetry
We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry. In calorimetry, the process is carried out in a vessel called calorimeter, which is immersed in a known volume of a liquid. Knowing the heat capacity of the liquid in which calorimeter is immersed and the heat capacity of calorimeter, it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions:
For chemical reactions, heat absorbed at constant volume, is measured in a bomb calorimeter (Fig. 5.7). Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ∆V = 0. Temperature change of the calorimeter produced by the completed reaction is then converted to qV, by using the known heat capacity of the calorimeter with the help of equation 5.11. | 9 | 11 | Chemistry | 105 |
fec79a9f-f218-49b1-b49d-080bf15eb425 | # Chemistry
# (b) ∆H Measurements
Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter shown in Fig. 5.8. We know that ∆Η = qₚ (at constant p) and, therefore, heat absorbed or evolved, qₚ at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆ᵣH.
In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qₚ will be negative and ∆ᵣH will also be negative. Similarly in an endothermic reaction, heat is absorbed, qₚ is positive and ∆ᵣH will be positive.
Suppose q is the quantity of heat from the reaction mixture and CV is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
q = CV × ∆T
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
q = –CV× ∆T = – 20.7 kJ/K × (299 – 298) K = – 20.7 kJ
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK–1
For combustion of 1 mol of graphite,
12 g mol⁻¹ (–20.7 kJ) = – 2.48 × 10² kJ mol⁻¹, Since ∆n₍g₎ = 0, ∆H = ∆U = – 2.48 × 10² kJ mol⁻¹
# 5.4 Enthalpy change, ∆ᵣH
# Reaction – Reaction Enthalpy
In a chemical reaction, reactants are converted into products and is represented by,
Reactants → Products
The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol ∆ᵣH
∆ᵣH = (sum of enthalpies of products) – (sum of enthalpies of reactants)
∆ᵣH = ∑ aiHproducts – ∑ biHreactants (5.14)
Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric coefficients.
Fig. 5.8 Calorimeter for measuring heat changes at constant pressure (atmospheric pressure).
# Problem 5.6
1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation:
C (graphite) + O₂ (g) → CO₂ (g)
During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm? | 10 | 11 | Chemistry | 105 |
4fca14fa-89f1-48a0-8674-7a7fc0a41b5b | # THERMODYNAMICS
coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l)
∆ᵣH = ∑a H - ∑b H
= [Hₘ (CO₂, g) + 2Hₘ (H₂O, l)] – [Hₘ (CH₄, g) + 2Hₘ (O₂, g)]
where Hₘ is the molar enthalpy.
Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
# (a) Standard Enthalpy of Reactions
Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.
The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 1 bar; standard state of solid iron at 500 K is pure iron at 1 bar. Usually data are taken at 298 K.
Standard conditions are denoted by adding the superscript to the symbol ∆H, e.g., ∆H.
# (b) Enthalpy Changes during Phase Transformations
Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273 K).
H₂O(s) → H₂O(l); ∆fusH = 6.00 kJ mol–1
Here ∆fusH is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings.
The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, ∆fusH.
Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water
# Table 5.1 standard enthalpy changes of Fusion and Vaporisation
(Tf and Tb are melting and boiling points, respectively)
2024-25 | 11 | 11 | Chemistry | 105 |
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