Class 11 Chemistry

Charles Law

Two Frenchmen, Gay-Lussac and Jacques Charles, independently studied the thermal expansion of the gases, i.e., the variation of tire volume of a gas with temperature at a constant pressure. They observed that, at constant pressure, the gas expands when heated and contracts when cooled. Both of them arrived at the same conclusion, which is knorvn after their names as Gay-Lussac’s and Charles’s law more commonly as Charles’s law. It is stated as follows:

At constant pressure, the tire volume of a fixed amount of a gas is proportional to the temperature.

Formula For Volume In Chemistry

V ∝ T

or V/T = k at constant n and p,

where k is a constant, n is the number of moles of the gas and p is the pressure.

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Absolute Zero: Charles’s law raises an interesting question. If the volume of a gas decreases with decrease in temperature, will it become zero at any temperature? If so, at what temperature?

We could try to grasp this concept graphically as well. According to Charles’s law, the volume of a gas is a linear function of its temperature. Thus, a plot of the volume of a given amount of a gas (at constant pressure) against its temperature would be a straight line somewhat like those shown. Slopes of lines obtained at different pressures are different. If these straight line were to be extrapolated, as shown in the figure, they would touch the x-axis at -273.15°C, the temperature at which the volume would become zero.

Formula For Volume In Chemistry

For a particular value of temperature t and a fixed volume of gas V₀ at CTC, the following relation holds:

V_t = V₀ (1+ αt)

where V_t is the volume at temperature t and α is the cubic expansion coefficient. The value of α is 1/273.15

The volume of the gas at t°C may, therefore, be written as \(V_t=V_0\left(1+\frac{t}{273.15}\right)\)

It may be clearly seen from this expression that the volume changes by 1/273.15 of V₀ for every degree change in temperature Charles’s law can thus be formally stated as follows. The volume of a given amount of a gas at constant pressure increases (or decreases) by a constant fraction O273.15) volume at 0°C for each degree rise (or fall) in temperature.

The volume at -273.15°C is \(V_{-273.15}=V_0\left(1-\frac{273.15}{273.15}\right)=0\)

Thus, a direct implication of Charles’s law is that the volume of a gas becomes zero at -273.15 °C. Since one cannot have negative volumes, -273.15 °C should be the lowest possible temperature. This is why -273.15°C, the hypothetical temperature at which the volume of a gas should become zero, is referred to as the absolute zero of temperature. We use the word hypothetical because in reality, all gases liquefy before this temperature is reached. This temperature exists only in theory—no one has managed to reach the absolute zero of temperature.

Absolute Scale Of Temperature: Based on the above observations Lord Kelvin suggested the adoption of a new scale for the scientific measurement of temperature. The zero of this scale is at -273.15°C. This is taken as 0 K (not 0°K). It is now used by scientists the world over and is called the Kelvin scale. Temperatures expressed on the Kelvin scale and on the Celsius scale are related as follows.

TK = (t + 273.15)° C ≈ (t + 273)° C

Accepting the absolute scale of temperature for scientific measurements was not just an arbitrary choice. It can be justified by concepts you will learn about later in thermodynamics. In fact, the absolute scale or the kelvin scale is also called the thermodynamic scale. Also, many relations, such as the one between volume and temperature defined by Charles’s law, assume a simpler form if you use this scale. Let us try to express Charles’s law in terms of temperature on the Kelvin scale.

Formula For Volume In Chemistry

⇒ \(V_t=V_0\left(1+\frac{t}{273}\right)=V_0\left(\frac{273+t}{273}\right)\)

where V₀ = volume at 0°C and

V_t = volume at t°C.

But 273 + t = T, the temperature on the Kelvin scale which has -273.15°C as its zero.

Thus \(V_t=\frac{V_0}{273} T\)

On the Kelvin scale 273 is the temperature corresponding to V₀ (t = 0°C) and can be denoted as T₀. We can write the above equation as

⇒ \(\frac{V_t}{V_0}=\frac{T}{T_0}\)

or \(\frac{V}{T}\) = constant if T is the temperature on the Kelvin scale.

Thus, Charles’s law can be expressed as follows. The volume of a given mass of gas is directly proportional to its temperature on the Kelvin scale at a constant pressure. The following equation is convenient (for numerical problems)

⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2} \quad(n, p \text { are constant })\)

Remember that temperature is always expressed in kelvin, while dealing with pressure and volume in case of gases.

Application of Charles’s Law: According to Charles’s law the volume of a gas increases as its temperature rises. In other words, hot air is less dense than cold air, so if one fills a balloon with hot air and lets it go, it rises. In the early part of the twentieth century, hot-air balloons were used for meteorological observations.

Formula For Volume In Chemistry

Ballooning also became a sport. In the 1930s hydrogen balloons became more popular and were even used as a means of public transport (called airships). An accident, which caused a German airship to catch fire, brought an end to the use of airships for transport, but hot-air balloons and hydrogen balloons continued to be used for meteorological purposes and ballooning is still a sport. In fact, Charles was the first person to use hydrogen to inflate balloons.

Example 1. A certain mass of gas occupies a volume of 1 L at 2 7°C. Calculate the temperature at which the gas will occupy 500 cm³ if the pressure remains constant.
Solution:

V₁ = 1L = 1000 cm³  V₂ = 500 cm3

T₁ = 27°C = 300K  T₂ = ?

According to Charle’s law,

⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

∴ \(T_2=\frac{V_2 \times T_1}{V_1}=\frac{500 \times 300}{1000}=150 \mathrm{~K}\) = (150 – 273)°C = -123°C.

Example 2. Suppose you want to increase the volume of 500 cm³ of a gas by 25% while keeping the pressure constant. To what temperature would you have to heat the gas if its initial temperature is 25°C?
Solution:

V₁ = 500 cm³

V₂ = 500 + 25% of 500 = 500 + 125 = 625 cm

T₁ = 25° C = 298 K  T₂ = ?

According to Charle’s law

⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

∴ \(\quad T_2=\frac{V_2 \times T_1}{V_1}=\frac{625 \times 298}{500}=372.5 \mathrm{~K}\)

= \((372.5-273)^{\circ} \mathrm{C}=99.5^{\circ}\mathrm{C}\)

Pressure-temperature relationship The mathematical relationship between pressure and temperature was given by Gay-Lussac. It states that when the temperature of a fixed amount of a gas is changed keeping the volume constant, the pressure of the gas changes and is directly proportional to the temperature. As in the case of volume the pressure increases (or decreases) by 1/273 of its value at 0°C for each one degree rise (or fall) in temperature. Thus,

Formula For Volume In Chemistry

p ∝ T (n, V are constant)

or, \(\frac{p}{T}=\text { constant }\)

or \(\frac{p_1}{T_1}=\frac{p_2}{T_2}\).

Boyle’s Law

Robert Boyle, an Irish physicist, carried out a set of experiments to study the effect of pressure on the volume of a fixed mass of air. He devised a very simple apparatus, using a bent tube (J-tube), some mercury and a measuring scale.

Boyle performed his experiments in a room where the temperature remained fairly constant. He observed how the length of the air trapped above the mercury in the closed limb of the tube varied with the pressure applied on it and found a definite quantitative relationship between the two. He increased the pressure on the trapped air by adding more mercury in the open limb of the tube. These observations showed that the length of the trapped air (consequently, the volume of air as volume is proportional to the length) varied inversely with the pressure applied on it.

Boyle worked with air, but it was found later that all gases behave the way the air trapped in Boyle’s tube did (at constant temperature). This relationship between the volume and pressure of a fixed amount of gas at constant temperature can be expressed mathematically as follows.

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⇒ \(p \propto \frac{1}{V} or V \propto \frac{1}{p}(n, T constant)\)

V = \(\frac{k}{p}\)

or pV = k, where k is a constant.

Boyle’s Law Class 11

The value of the constant depends upon amount of gas (n) and its temperature (T).

This quantitative relationship is called Boyle’s law and can be stated as follows. For a fixed amount of gas, the pressure is inversely proportional to the volume if the temperature remains constant. In other words, the product of the pressure and volume of a given mass of gas is constant, provided the temperature remains constant.

Suppose the initial pressure and volume of a given mass of gas are p₁ and V₁ respectively at temperature T. If the pressure is then changed to p₂ while keeping the temperature constant and the volume becomes V₂ then according to Boyle’s law,

∴ \(p_1 V_1=p_2 V_2 \quad \text { or } \quad \frac{p_1}{p_2}=\frac{V_2}{V_1}\)

This implies that if the volume of the fixed amount of gas is doubled at constant temperature, its pressure will reduce to half. This way of expressing Boyle’s law is useful while solving numerical problems.

Experimental verification of Boyle’s law: Boyle’s law can be verified experimentally by measuring the volumes of a given mass of gas at different pressures, while keeping the temperature constant.

1. A plot of p against V at constant temperature is a hyperbola and it is called an isotherm. pV = constant, and the value of the constant depends upon n and T. Therefore, at every temperature for a fixed amount of gas, there is a separate pressure-volume curve shows such curves at different temperatures for a fixed amount of gas.
2. A plot of pV against p at constant temperature is a straight line parallel to the x-axis which indicates that pV remains constant with changing pressure.
3. A graph of V against 1/p gives a straight line through the origin, which shows that volume increases uniformly with increase in 1 /p or volume is inversely proportional to pressure.

Implications of Boyle’s law: The quantitative relationship discovered by Boyle between the pressure and volume of gas showed that a gas is compressible. When a given mass of a gas is compressed, the number of molecules it has does not change. They come closer and occupy less space. In other words, the gas becomes denser.

Boyle’s Law Class 11

This is why mountain air is rarer than the air at sea level. The air at sea level is denser because it is compressed by the mass of air above it. As one climbs higher and higher up a mountain, the pressure decreases and the density of air decreases. This is why mountaineers have to carry a supply of oxygen with them. For the same reason, the size of a weather balloon increases as it ascends to higher altitudes.

A relationship can be obtained between density and pressure by using Boyle’s law. You already know that density (d) = \(\frac{\text { mass }}{\text { volume }} \frac{(m)}{(V)}\)

But pV = k (from Boyle’s law).

∴ V = \(\frac{k}{p} \quad \text { or } \quad d=\frac{m}{k / p}=\left(\frac{m}{k}\right) p\)

This shows that the pressure of a gas is directly proportional to its density.

Example 1. 100 mL of CO₂ was collected at 27°C and 1 bar pressure. What would be the volume of the gas if the pressure changed to 0.96 bar at the same temperature?
Solution:

V₁ = 100 ml V₂ = ?

p₁ = 1 bar p₂ = 0.96 bar

From Boyle’s Law Class 11

p₁V₁ = p₂V₂ (at constant temperature)

∴ \(V_2=\frac{p_1 V_1}{p_2}=\frac{1 \times 100}{0.96}=105.5 \mathrm{~mL}\)

The volume of carbon dioxide = 105.5 mL.

Example 2. A gas occupies a volume of 2.0 L tif 745 mmHg pressure. Calculate the additional pressure required to decrease the volume of the gas to 1.5 L.
Solution:

Given p₁ = 745 mm p₂ = ?

V₁ = 2.0L V₂ = 1.5 L

From Boyle’s law, pV = pV

∴ p₂ = \(\frac{p_1 V_1}{V_2}=\frac{745 \times 2}{1.5}=993.3 \mathrm{mmHg}\)

The additional pressure required = 993.3 – 745 = 248.3 mmHg.

Bonding And Molecular Structure

The atom is considered to be the fundamental particle of an element, but apart from a few cases atoms cannot exist in the free state in nature. They combine with each other to form molecules, which are capable of independent existence. This implies that a molecule must be more stable than the individual atoms. The properties of a molecule are different from those of the constituent atoms.

• What holds these atoms together in a molecule? There must be some force which does so. When atoms are held together in a molecule, we say that there is a chemical bond between them.
• Not all atoms are held together by the same type of force of attraction. In fact, all atoms do not combine. It seems that only certain combinations are allowed in nature. Only certain atoms combine in certain specific ways to form molecules.
• For example, two oxygen atoms combine to form an oxygen molecule (O₂), two hydrogen atoms combine to form a hydrogen molecule (H₂), and though O₃ exists, H₃ does not. Noble gas atoms generally do not combine.

Octet Rule

Scientists were curious to know the cause of the combination of atoms and made several attempts to explain how atoms are held together. The first satisfactory explanation was provided on the basis of the configuration of the noble gases. These gases do not combine with each other and are known to be monatomic, i.e., they exist as single atoms. Their atomic numbers and electronic configurations are as follows.

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Helium: 2 – 2

Neon: 10 – 2,8

Argon: 18 –  2,8,8

Krypton: 36 –  2,8,18,8

Xenon: 54 – 2,8,18,18,8

Radon: 86 – 2, 8,18,32,18,8

• As you can see, other than helium, all the noble gases have 8 electrons in their outermost shell. Since all other elements have less than 8 electrons in their outermost shell, scientists concluded that the chemical inertness of the noble gases is related to the presence of 8 electrons in their outermost shell.
• In other words, they concluded that if an element has 8 electrons in its outermost shell, it is chemically unreactive, or its electronic configuration is stable.
• These observations led an American chemist, G N Lewis, to put forward a generalization known as the octet rule.
• According to this rule, atoms of elements combine with each other so as to attain the stable configuration of 8 electrons in their outermost shell. An atom may attain this configuration by gaining, losing, or sharing electrons with other atoms.

Valency Of Nitrogen

Exceptions to the octet rule: The octet rule is a simple and useful generalisation that can explain the formation of a large number of compounds. However, it cannot be applied in some cases.

1. Hydrogen has only one electron in its valence or outermost shell, and it needs only one more electron to fill this shell and attain a stable configuration. This is possible because by acquiring one more electron, the hydrogen molecule attains a configuration like that of the noble gas helium. It may be noted that there is no octet here as the first shell can accommodate only two electrons.

2. According to the octet rule, elements of groups 1, 2, and 13 should not form covalent bonds because, having less than 4 electrons in their valence shells, they should not be able to attain stable configurations by sharing electrons. But some elements of these groups do form covalent compounds which are called electron-deficient compounds, for example, BeCl₂ (beryllium has only 4 electrons in the outermost shell) and BF₃ (boron has only 3 electrons).

3. The octet rule cannot explain the formation of compounds such as PCI₅ and SF₆ in which the central atom contains more than eight electrons. The central atom is said to have an expanded octet. An expanded octet is mostly found in the elements beyond the third period when d orbitals are available for bonding.

4. The basic assumption on which the octet rule is formulated is that the noble gases are inert. However, compounds of xenon like XeF₂, XeF₄, XeF₆ have been prepared.

5. Some molecules like NO and NO₂ are called odd-electron molecules as the octet rule is not applicable to all the atoms in the molecule.

The Lewis dot representation of these two molecules shows that the octet of the nitrogen atom in both molecules is incomplete (there are 7 electrons).

⇒ \(\ddot{\mathrm{N}}=\ddot{O} \quad \ddot{O}=\dot{\mathrm{N}}^{+}-\ddot{\mathrm{O}}^{-}\)

Valency Of Nitrogen

Lewis Structure The Concept Of Valence Electrons

Another observation that helped scientists understand the ways in which atoms combine was that atoms which f have the same number of electrons in their outermost shells exhibit similar chemical properties. This led them to conclude that only the electrons in the outermost shell participate in the process of chemical combination.

• The electrons in the inner shells do not participate in bonding. The outermost shell electrons are called valence electrons. The chemical properties of an element depend on the number of valence electrons.
• Lewis introduced a simple notation to represent the valence electrons in an atom. In this notation the valence electrons are represented by dots (or crosses) surrounding the chemical symbol of the element.
• These symbols, called Lewis symbols or electron dot symbols, do not show the electrons in the inner shells. The chemical symbol of the element represents the nucleus and the inner electrons. The Lewis symbols for the elements of the second period are given below.

The number of dots in the Lewis symbol of an element indicates the number of electrons in the outermost shell (valence shell) electrons. Consider the symbols for Li, Be, B, and C. The number of dots in these symbols indicates the usual valence of these elements. Thus lithium is monovalent, beryllium is divalent, boron is trivalent and carbon is tetravalent.

• The valencies of nitrogen, oxygen, and neon are 3,2 and 0 respectively. The valence of fluorine is 1. That is to say the valence of these elements can be obtained from the Lewis symbols by subtracting the number of dots from 8.
• Whatever may be the drawbacks of the octet rule, most atoms do tend to combine in order to attain a more stable configuration of 8 electrons in the valence shell. They do so by the redistribution of their valence electrons. This can occur in one of two ways.

Valency Of Nitrogen

1. By the complete transfer of one or more electrons from one atom to another. The chemical bond thus formed is called an ionic bond.
2. By the sharing of electrons between atoms. The bond formed is called a covalent bond.

Electrovalent Or Ionic Bond

Kossel, while studying chemical bonding, observed that the highly electronegative halogens form a negative by gaining an electron whereas the highly electropositive alkali metals form a positive ion by readily losing an electron. The elements of the two groups (halogens and alkali metals) gain or lose electrons respectively and attain a stable outer shell configuration of eight electrons.

The noble gases (Group lb elements) have the stable outer shell configuration of eight electrons except helium, which has just two electrons. Kossel explained that the negative and positive ions (formed by gain or loss of electrons) are hold together by electrostatic attraction among them. This electrostatic attraction was later termed the electrovalent bond.

• Kossel’s observations led to the understanding of ionic bonding in ionic compounds. It also provided the basis of the concept of electron transfer in ion formation and thereafter the formation of ionic compounds. Besides providing a concept of ionic bonding, Kossel realised that not all compounds are formed by electron transfer between the atoms.
• Atoms which contain 1, 2, or 3 electrons in their valence shell have a tendency to lose electrons to acquire a stable configuration. It is easier for such atoms to lose 1,2 or 3 electrons than to gain 7,6 or 5 electrons respectively. Similarly, atoms that have 5, 6, or 7 valence electrons have a tendency to complete their valence shell by gaining 3,2 or 1 electrons respectively.
• Atoms are electrically neutral as they possess an equal number of electrons and protons. However, when an atom loses electrons to attain a stable configuration, it becomes positively charged because it contains a greater number of protons than electrons.

A positively charged atom is called a cation. When, on the other hand, an atom gains electrons to attain a stable configuration, it becomes negatively charged because it has a greater number of electrons than protons. A negatively charged atom is called an anion.

• When one atom acquires the electrons which another atom loses, the oppositely charged ions so formed are drawn together by an electrostatic force of attraction. When two ions are held together in this way, an ionic bond or an electrovalent bond is said to exist between them. In other words, the bond between oppositely charged ions, formed by tire transfer of electrons from one atom to another, is called an ionic bond.
• Such bonds are generally formed between a metal and a nonmetal (as discussed above, alkali metals and halogens respectively). For example, the sodium atom (atomic number 11) has only one electron in its valence shell. In order to acquire the nearest noble gas configuration (2, 8) it has to lose one electron.
• The chlorine atom (atomic number 17) has seven electrons in its valence shell and it is easier for it to gain an electron to acquire the nearest noble gas configuration (2, 8, 8). So, the sodium atom transfers its valence electron to the chlorine atom, resulting in tire formation of a sodium ion (Na⁺) and a chloride ion (Cl^–), which are held together by a force of electrostatic attraction.

The number of electrons which an atom loses or gains while forming an ionic bond is called its electrovalency. Atoms that tend to lose electrons are called electropositive because they form positively charged ions after losing electrons. Atoms which gain electrons are called electronegative because they form negatively charged ions after gaining electrons. One more example of an ionic compound is given here.

Ionic bonds are formed due to electrostatic forces of attraction, which are non-directional. An ion has a uniform electrostatic field of influence around it, i.e., it attracts ions all around it. The attraction between two ions depends on the distance between them, and not on the direction of one from the other.

• Each positively charged ion attracts several negatively charged ions around it and each negative ion attracts several positive ions around it. The number of oppositely charged ions a particular ion will attract depends on its size and charge.
• This aggregation of oppositely charged ions around one ion results in the formation of a three-dimensional aggregate of ions called an ionic crystal. Oppositely charged ions alternate in a regular geometrical pattern in such a crystal. The cations and anions are held together in the crystal by attractive forces.
• An ionic compound consists of an aggregate of ions. These ions held together by an ionic bond cannot be looked upon as a molecule. Thus, an ionic compound cannot be assigned a molecular formula. For example, the formula of sodium chloride can be written as NaCl or Na⁺Cl^–. This is its empirical formula and not its molecular formula.
• The empirical formula of an ionic compound can be derived from the valency of the elements or the ratio of cations to anions. In NaCl, the ratio is 1:1. So, the (empirical) formula of sodium chloride is NaCl.

Formation of an ionic bond: The formation of an ionic bond broadly involves the formation of the gaseous cation, the gaseous anion, and the packing of ions to form the ionic solid.

Formation of cations and anions You have already studied in Chapter 3 that energy is required to remove an electron from an isolated neutral gaseous atom, and is defined as the ionisation enthalpy of the element. Thus, the formation of a positive ion or removal of electron(s) involves the process of ionisation. Let us consider the formation of NaCl.

⇒ \(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} ; \Delta H\) (Ionisation enthalpy)

On the other hand, chlorine atom accepts an electron to form a negative ion (Cl^–). The enthalpy change in this reaction involving addition of electron(s) is termed as electron gain enthalpy.

Limitations Of Octet Rule For Class 11

⇒ \(\mathrm{Cl}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}(\mathrm{g}) ; \Delta H\)(Electron gain enthalpy)

The ionisation enthalpy of an atom is always positive (energy is always required to remove an electron) but the electron gain enthalpy of an atom may be positive or negative, depending on the nature of the element.

• Considering the enthalpy changes during formation of ions, we may conclude that ionic bonding is most likely to occur between atoms with low ionisation enthalpies and high negative electron gain enthalpies.
• Generally, ionic compounds are formed by cations from metallic elements (metals have low ionisation enthalpies) and anions from nonmetals (nonmetals have high negative electron gain enthalpies).

The conditions required to form an ionic bond are best satisfied in alkali halides. The cation NH⁺₄ (ammonium ion) is an exception, it is made up of two nonmetals.

Packing of oppositely charged ions The last step in the formation of an ionic compound is the packing of negative and positive ions to form a lattice. The energy released in this process is known as lattice energy.

More formally, the energy released in the formation of 1 mole of an ionic solid by the packing of oppositely charged ions is called lattice enthalpy. The lattice enthalpy may also be defined as the energy needed to separate one mole of a solid ionic compound into the constituent gaseous ions. The lattice enthalpy of sodium chloride is 788 kJ mol^-1. This means that 788 kJ of energy is needed for the following reaction to occur.

⇒ \(\mathrm{NaCl}(\mathrm{s}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g})\)

The higher the value of the lattice enthalpy of an ionic compound, the more stable it is, and the more readily will such a compound be formed. Ionic compounds are held together by electrostatic attraction and the force of attraction depends directly on the magnitude of the charges and inversely on the square of the distance between the charges.

F = \(\frac{q_1 q_1}{d^2}\)

The greater this force, the more the chance of formation of the ionic compound. In other words, a greater magnitude of charge and a smaller ionic size favours the formation of an ionic solid. A smaller ionic size would mean a smaller internuclear distance.

• For stable ionic bonding, the total energy released during the formation of the crystal lattice should be greater than the energy required for the formation of ions.
• In the case which we have just considered, the first ionisation enthalpy for sodium is 495.8 kJ mol^-1 and the electron gain enthalpy for chlorine is -348.7 kJ mol^-1. Their sum comes to 147.1 kJ mol^-1 which is actually the energy required for the formation of ions. This energy is much less than that released during the formation of the crystal lattice, which is equal to 788 kJ mol^-1. Thus, a greater lattice enthalpy implies greater stability of the compound.

Properties of ionic compounds: By and large, ionic compounds have certain common characteristics.

Limitations Of Octet Rule For Class 11

Physical state They are usually crystalline solids, in which the ions are arranged in a well-defined geometrical pattern. The force of attraction between the oppositely charged ions is very strong, which is why these compounds exist in the solid state at room temperature.

Melting and boiling points Ionic compounds generally have high melting and boiling points. This is because a large amount of energy is required to overcome the strong electrostatic force of attraction which holds the ions together in such compounds.

Electrical conductivity Ionic compounds are not good conductors of electricity in the solid state because the ions are held in fixed positions in the crystal lattice by electrostatic force of attraction between ions. The ions are, thus, not free to move. However, they become free to move in the molten state and in aqueous solutions, and carry charge. So ionic compounds are good conductors of electricity in the molten or dissolved state.

Solubility Ionic compounds are soluble in water and other polar solvents. Such solvents can weaken the force of attraction that holds together the ions in an ionic compound. Such compounds are insoluble in nonpolar solvents like benzene and carbon tetrachloride.

Ionic reactions Chemical reactions between ionic compounds are called ionic reactions. Such reactions take place between the ions of the two compounds produced in an aqueous solution. These reactions involve oppositely charged ions which combine readily, so they occur almost instantaneously.

 

Hydrogen Bond

The hydrogen atom has a unique structure. It has only one proton and one electron. When a hydrogen atom forms a covalent bond with a highly electronegative atom, the bonded electron pair shifts towards the latter. Consequently, the nucleus of the hydrogen atom gets exposed and behaves like a proton.

It exerts a strong electrostatic force of attraction on the electronegative atom of an adjacent molecule. This force of attraction between the (covalently bonded) hydrogen atom of one molecule and the electronegative atom of another molecule is called a hydrogen bond. It is represented by a dotted line.

The following conditions are necessary for hydrogen bonding.

1. The molecule must contain a highly electronegative atom linked to the hydrogen atom. The greater the electronegativity of the atom linked to the hydrogen atom, the stronger is the hydrogen bond.

2. The size of the electronegative atom should be small. The smaller the size, the greater is the attraction between the atom and the hydrogen atom and the stronger is the bond.

Only F, O, and N satisfy both these conditions.

Fluorine, having the highest value of electronegativity, forms the strongest hydrogen bond.

The water molecule contains the highly electronegative oxygen atom linked to the hydrogen atom. The oxygen atom attracts the shared pair of electrons more strongly and this end of the molecule becomes negative, while the hydrogen-end becomes positive.

In ammonia, the more electronegative nitrogen atom attracts the shared electrons towards itself, leaving the hydrogen atom positively charged. This hydrogen atom forms a hydrogen bond with the nitrogen atom of another molecule.

Nature of hydrogen bond: The hydrogen bond is a weak bond. Unlike the covalent bond, which involves the sharing of electrons by two atoms of elements of the same or different electronegativities, the hydrogen bond arises due to the interaction between dipoles.

The strength of the hydrogen bond lies between those of the van der Waals force and the covalent bond. The dissociation energy of the hydrogen bond depends upon the electronegativity of the other atom. The bond energy is highest in H—F because fluorine is the most electronegative element.

Impact of hydrogen bonding: Although the hydrogen bond is weak, it influences the physical properties of the substances in which it acts. Association The hydrogen bond links molecules to form aggregates of molecules.

For example, HF molecules become linked by hydrogen bonding and the formula of hydrogen fluoride can be written as (HF)π. This changes the molecular mass of the compound. For instance, carboxylic acids exist as dimers because of hydrogen bonding and their molecular mass is double the molecular mass predicted by their simple formula. The molecular mass of acetic acid is 120 but its mass according to its simple formula should be 60.

Dissociation In an aqueous solution, hydrogen fluoride dissociates and produces the difluoride ion (HF^–₂) instead of the fluoride ion (F^–). This is due to H-bonding in HF. This explains why KHF₂ exists, while KHCl₂, KHBr_2, and KHI₂ do not. The other halogens cannot form hydrogen bonds because of low electronegativity and bigger size.

High melting and boiling points Compounds in which the molecules are linked by hydrogen bonds have much higher melting and boiling points than compounds formed by other members of their groups. Tins is because a greater amount of energy is required to break the hydrogen bonds.

• Let us consider the hydrogen halides, for example. Hydrogen fluoride exists as a liquid (high boiling point) at room temperature, while HCl is a gas. The boiling points of the hydrogen halides decrease as molecular mass decreases, but there is a sudden increase in the boiling point of HF because of the presence of the hydrogen bond.
• Similarly, H₂O is a liquid, but H₂S is a gas at room temperature. In fact water has an abnormally high boiling point as compared to the other hydrides of its group. NH₃ has an abnormally high boiling point as compared to the hydrides of the other elements of its group.
• This difference is seen among organic compounds too. For example, ethanol has a higher boiling point than diethyl ether because of the presence of hydrogen bonding in the former. Both compounds have oxygen atoms, but in diethyl ether, the oxygen atom is not bonded to hydrogen.

Solubility Covalent compounds are generally insoluble in polar solvents like water. However, compounds which can form hydrogen bonds with water are soluble in water, example, alcohol, acetone, and urea.

Similarly, ammonia (NH₃) dissolves in water but PH₃ (phosphine) does not. This is because of the ability of the nitrogen atom to form a hydrogen bond with water.

Viscosity Viscosity is the resistance offered to the flow of a liquid. The viscosity of a liquid increases if there is hydrogen bonding between its molecules because this is an additional intermolecular force which aggregates molecules. For example, glycerol is more viscous than ethylene glycol and ethyl alcohol. All alcohols have at least one —OH group which can form a hydrogen bond. Glycerol has three —OH groups, ethylene glycol has two, while ethyl alcohol has only one.

Density of ice Ice has a lower density than water, though hydrogen bonding is present in both water and ice. In ice, each oxygen atom is tetrahedrally bonded with four hydrogen atoms. Two hydrogen atoms are bonded by covalent bonds and another two through hydrogen bonds. This gives rise to an open cage-like structure, which occupies a larger volume than does water. In the liquid, the water molecules are more closely packed. When ice melts, this cage-like structure collapses and the water molecules come closer to each other. Thus, for the same mass of water, the volume decreases and density increases.

Types of hydrogen bonding: There are two types of hydrogen bonding—intermolecular and intramolecular.

Intermolecular When hydrogen bonding takes place between different molecules of the same or different compounds, it is called intermolecular hydrogen bonding. For example, HF, H₂O, NH₃, and alcohol.

Intramolecular Hydrogen bonds between atoms of the same molecule are called intramolecular hydrogen bonds. Such bonds are formed in many organic compounds, example, o-nitrophenol and o-hydroxy benzoic acid (salicylic acid).

Intramolecular H-bonding prevents the association of molecules, making the molecule contract. Hence, this kind of H-bonding decreases melting and boiling points and solubility. For example, o-nitrophenol has a lower boiling point than p-nitrophenol.

Resonance

The properties of certain molecules containing a π bond cannot be explained by writing one Lewis structure. For example, the structure of ozone can be written as either (1) or (2) here, but neither can explain its experimentally determined properties.

• According to either of these structures, the central oxygen atom is bonded to one oxygen atom by a double bond and the other oxygen atom by a single bond. Since a double bond (121 pm) is shorter than a single bond (148 pm), the two bond lengths in this molecule should be different.
• However experimental evidence shows that the bonds are equal and that the bond length is 128 pm, between the length of a single bond and that of a double bond. Thus, neither structure (1) nor (2) can explain why O₃ has two equal bonds.
• It is assumed that tire actual structure is between the two Lewis structures. This phenomenon is called resonance and the two (or more) individual structures are called resonating structures or canonical forms. To define this more clearly, when a molecule can be represented by more than one Lewis structure none of which can individually explain all its observed properties, the actual structure of the molecule is the intermediate of the various Lewis structures and is called the resonance hybrid.

This is a theoretical concept. Tire-resonating structures have no physical reality, that is to say, they do not actually exist. Only the hybrid structure exists. There is no equilibrium between the canonical forms. In theory the various resonating structures have the following characteristics.

1. The same arrangement of atoms
2. The same of number of paired and unpaired electrons, but different electronic arrangement
3. They have nearly the same energy but the energy of the resonance hybrid is less than that of each of the canonical forms. Thus, resonance stabilises the molecule
4. The same positions of sigma bonds but different positions of pi bonds

Some examples are given below.

1. Carbon dioxide (CO₂) The C—O bond length in the molecule is experimentally determined as 115 pm. The value is between the length of C=O (121 pm) and C≡O (110 pm). Therefore, the structure of CO₂ can be described as a resonance hybrid of three canonical forms.

∴ \(: \mathrm{O}=\mathrm{C}=\mathrm{O}: \longrightarrow \stackrel{+}{\mathrm{O}} \equiv \mathrm{C}-\overline{\mathrm{O}}: \longrightarrow: \overline{\mathrm{O}}-\mathrm{C} \equiv \stackrel{+}{\mathrm{O}}:\)

2. Carbonate ion (\(\mathrm{CO}_3^{2-}\)) Since the experimentally determined bond length values show that all the carbon- oxygen bonds are equal in a carbonate ion, its structure can be described as the resonance hybrid of the three canonical forms as follows.

3. Phosphate ion \((\left(\mathrm{PO}_4^{3-}\right))\) Experimental findings show that all the phosphate oxygen bonds are equal in the phosphate ion. Therefore, the structure of the ion is best described as the resonance hybrid of the following four canonical forms.

4. Benzene (C₆H₆) It has been found that all carbon-carbon bond lengths are identical and the two important canonical forms are

 

 

Bonding And Molecular Structure Multiple Choice Questions

Question 1. Which among the following contains both covalent and ionic bonds?

1. CCl₄
2. CaCl₂
3. NH₄Cl
4. NaCl

Answer: 3. NH₄Cl

Question 2. Which of the following contains coordinate as well as covalent bonds?

1. NH⁺₄
2. PH₃
3. CHCl₃
4. PCI₅

Answer: 1. NH⁺₄

Question 3. The numbers of sigma (σ) and pi (π) bonds in 2, 3-butadiene (CH₂=CH—CH=CH₂) are

1. 3σ and 2π
2. 9σ and 2π
3. 5σ and 2π
4. 2σ and 5π

Answer: 2. 9a and 2n

Question 4. In which of the following is the central atom sp³ hybridised?

1. PCl₃
2. SO₃
3. BF₃
4. BrF₃

Answer: 1. PCl₃

Question 5. Which among the following has the lowest dipole moment?

1. HF
2. HCl
3. HBr
4. HI

Answer: 4. HI

MCQs on Chemical Bonding and Molecular Structure

Question 6. The C—C bond distance is the longest in

1. C₂H₄
2. C₂H₂
3. C₂H₆
4. C₆H₆

Answer: 3. C₂H₆

Question 7. The dipole moment of BeF₂ is

1. Very low
2. Zero
3. Very high
4. Variable

Answer: 2. Zero

Question 8. The carbon atom in the ethyne molecule has the following hybridisation:

1. sp³
2. sp²
3. sp
4. None of these

Answer: 3. sp

Question 9. The nitrogen molecule has

1. 1σ and 1π bond
2. 2σ and 2π bonds
3. 1σ bond
4. 1σ and 2π bonds

Answer: 4. 1σ and 2π bonds

Question 10. Which of the following does not have any dipole moment?

1. NH₃
2. CCl₄
3. H₂O
4. CHCl₃

Answer: 2. CCl₄

MCQs on Chemical Bonding and Molecular Structure

Question 11. Which among the following lacks hydrogen bonding?

1. H₂S
2. C₂H₅OH
3. HF
4. Ammonia

Answer: 1. H2S

Question 12. Which of the following are isostructural?

1. BF₃ And NH₃
2. CO₂ and H₂O
3. PF₅ and BrF₅
4. CCI₄ and NH₄⁺

Answer: 4. CCI and NH₄⁺

Question 13. Which of the following hybridisations leads to bonds of unequal length?

1. sp²
2. sp³d²
3. sp³d
4. sp³

Answer: 3. sp³d

Question 14. In which of the following is the octet rule violated?

1. H₃O⁺
2. NaCl
3. AICI₃
4. HCl

Answer: 3. AICI₃

Question 15. In which of these is the octet rule followed?

1. XeF₂
2. NO
3. BeH₂
4. MgCl₂

Answer: 4. MgCl₂

Question 16. How many electrons are present in the NH₄^– ion?

1. 10
2. 11
3. 9
4. 12

Answer: 1. 10

Question 17. Which bond among the following is the least ionic?

1. P-F
2. S-F
3. Cl-F
4. F-F

Answer: 4. F-F

Question 18. A bond angle of 120° is associated with

1. sp hybridisation
2. sp² hybridisation
3. sp³ hybridisation
4. sp³d² hybridisation

Answer: 2. sp² hybridisation

Question 19. Which hybridisation leads to an octahedral shape?

1. sp³
2. sp³d
3. sp³d²
4. sp²d

Answer: 3. sp³d²

Question 20. The bond order in the hydrogen molecule is

1. 1/2
2. 3/2
3. zero
4. 0

Answer: 4. 0

Question 21. Which of the following orbitals has the highest energy?

1. \(\sigma^{\circ}(1 \mathrm{~s})
2. [latex]\sigma\left(2 \mathrm{p}_x\right)\)
3. \(\sigma(2 s)\)
4. \(\pi^*\left(2 p_y\right)\)

Answer: 4. \(\pi^*\left(2 p_y\right)\)

Question 22. Which of the following is not paramagnetic?

1. \(\mathrm{N}_2^{+}\)
2. \(\mathrm{Li}_2\)
3. \(\mathrm{O}_2\)
4. \(\mathrm{H}_2^{+}\)

Answer: 2. \(\mathrm{Li}_2\)

Question 23. Among the following pairs, which species have the same bond order?

1. B₂ and C₂
2. H₂ and O₂
3. H₂ and He₂
4. \(\mathrm{N}_2^{-} \text {and }\mathrm{O}_2^{+}\)

Answer: 4. \(\mathrm{N}_2^{-} \text {and }\mathrm{O}_2^{+}\)

Question 24. In which of the following is C*sp³ hybridised?

1. HC*OOH
2. HC*HO
3. (CH₃)₃C*OH
4. CH₃C*HO

Answer: 3. (CH₃)₃C*OH

Question 25. Which has a maximum number of sp hybridised carbon atoms?

1. CO₂
2. CH=C= CH—CN
3. HC≡C-CH₂=C=C=CH₂
4. HC≡C—CN

Answer: 4. HC=C—CN

Question 26. Among LiCl, BeCl₂, BCl₃ and CCl₄, the covalent character varies as

1. LiCl < BeCl₂ < BCl₃ < CCl₄
2. LiCl > BeCl₂ < BCl₃ > CCl₄
3. LiCl > BeCl₂ < BCl₃ < CCl₄
4. LiCl < BeCl₂ > BCl₃ > CCl₄

Answer: 1. LiCl < BeCl₂ < BCl₃ < CCl₄

Question 27. Which among the following has a pyramidal shape?

1. PCl₃
2. BF₃
3. CO^–₂
4. SO₃

Answer: 1. PCl₃

Question 28. The volatility of HF is low because of

1. Low polarisability
2. Low molar mass
3. Strong hydrogen bonding
4. Strong ionic bonding

Answer: 3. Strong hydrogen bonding

Polarity Of Covalent Bonds

The electronegativity of an element usually depends upon its atomic size. The smaller the size of an atom, the more it attracts the bonded electrons (the nucleus of the atom being closer), so the greater is its electronegativity. The value of electronegativity also depends upon the state of hybridisation (sp > sp² > sp³).

In the periodic table, the value of electronegativity decreases down a group because the atomic size Increases. The value of electronegativity increases across a period until Group 17. The electronegativity of group 17 elements (the halogens) is the highest in every period.

• A covalent bond is formed between two atoms when they share a pair of electrons. This shared pair is under the influence of both the atoms and one might expect the two atoms to have an equal hold over the shared pair of electrons. But this is not the case always.
• How much hold each atom has over the shared pair of electrons depends on the value of its electronegativity. And when one atom has n greater hold over the shared pair of electrons and pulls it more towards itself, it acquires a partial negative charge while the other atom acquires a partial positive charge. If, on the other hand, the two atoms share the electron pair equally, no charge develops over either atom.

• Thus in molecules like HF, HBr, or KCl, the halogen atom being highly electronegative attracts the shared electron pair towards itself. On the other hand, in case of a homonuclear diatomic molecule like H₂, Br_2, or N₂, the electron pair is shared equally between the two atoms due to the same electronegativity.
• Thus, electronegativity is an important factor affecting the nature of covalent bonds. Another way of saying this is that a covalent bond may be polar or nonpolar, depending upon the electronegativities of the bonded atoms.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

Nonpolar covalent bond: If two atoms of similar electronegativities form a bond by sharing a pair of electrons, the shared pair of electrons is equally attracted by the two and lies midway between the nuclei of the two atoms. Such a covalent bond is called a nonpolar covalent bond. The bonds formed between similar atoms, as in the case of H₂,O₂, N_2, and are nonpolar covalent bonds.

Polar covalent bond: When two atoms of different electronegativities share a pair of electrons, the shared pair of electrons does not lie midway between the two bonded atoms. It shifts towards the more electronegative atom. In other words, the distribution of electrons gets distorted, or the electron cloud is displaced towards the more electronegative atom.

• The slightly higher electron density around the more electronegative atom makes it acquire a partial negative charge (indicated as δ-). Tine less electronegative atom acquires a partial positive charge (δ+). Thus, positive and negative poles are developed in the molecule. This type of bond is called a polar covalent bond, i.e., a covalent bond with a partial ionic character.
• In the hydrogen chloride molecule, for example, chlorine is more electronegative than hydrogen, so the shared pair of electrons is displaced towards chlorine, i.e., the electron density is higher around the chlorine atom, and it acquires a fractional negative charge. The hydrogen atom, on the other hand, acquires an equal positive charge.
• Molecules containing polar covalent bonds are called polar molecules. A molecule may contain more than one covalent bond, for example, NH₃ and H₂O.

Ionic character of polar covalent bond: One can look upon an ionic bond as an extreme case of a polar covalent bond. When two atoms of different electronegativities are bonded together by a covalent bond, the bond is polar as the shared electron pair shifts slightly towards the more electronegative atom. When the difference in the electronegativities of the two atoms is very high, the shared pair of electrons is almost in complete possession of the more electronegative atom.

• In such a case, it is more valid to say that an electron has been transferred from the less electronegative to the more electronegative atom, and that a positive ion and a negative ion have been created. The two ions are then held together by mutual (electrostatic) attraction or an ionic bond.
• When two atoms are linked together by a covalent bond having different electronegativities, the bond formed is polar, or the bond is said to possess a partial ionic character. The degree of ionic character of a bond is determined by the difference in electronegativities of the combining atoms. The greater the difference in the electronegativities, the greater is the ionic character.

1. If the difference between the electronegativities of the two atoms is 1.9, the bond is said to have 50% ionic and 50% covalent character.
2. If the difference between the electronegativities of the two atoms is more than 1.9, the partial ionic character of the bond is more than 50% and the bond is taken to be ionic.
3. If the difference between the electronegativities of the two atoms is less than 1.9, the bond is predominantly covalent.

Dipole moment: One end of a polar molecule is negative and the other is positive, so the molecule behaves like an electric dipole. Since the negative charge is always equal in magnitude to the positive charge, the molecule as a whole is electrically neutral.

The dipole moment of a molecule is a measure of its polarity. It is the product of the magnitude of the negative or the positive charge (q) and the distance (d) between the charges, i.e., intemuclear distance. It is usually denoted by ji. Mathematically this can be expressed as follows.

μ = q x d.

The charge q is of the order of 10^-10 esu (electrostatic unit) and the intemuclear distance d is of the order of 10^-8 cm (A). Therefore, the dipole moment μ is of the order of 10^-10 x 10^-8 = 10^-18 esu cm. The dipole moment is also expressed in debye (D).

1 D = 1 x 10^-18 esu cm.

In SI units, 1 esu = 3.336 x 10^-10 C

and 1 cm = 10^-2 m.

∴ 1D = 3.336x 10^-10 x 10^-2 x 10^-18 = 3.336 x 10^-30 Cm.

• Dipole moment is a measure of the polarity, i.e., the extent of displacement of the electron cloud. The greater the difference between the electronegativities of the bonded atoms, the higher is the value of the dipole moment.
• For example, the dipole moment of HCI is greater than that of HBr though they have the same structure. This is because the difference between the electronegativities of chlorine and hydrogen is greater than the difference between the electronegativities of bromine and hydrogen.

Dipole moment and molecular structure: Dipole moment is a vector quantity, i.e., it has a magnitude and direction. It isrepresented by an arrow with its tail at the positive end and head pointing towards negative end.

⇒ H-Cl

A molecule may have more than one polar bond. The polarity of a molecule with more than one polar bond is the resultant or the vector sunt of the dipole moments of all the bonded atoms, or the resultant of all the bond dipoles (the dipole moment of a bond is often referred to as the bond dipole).

Diatomic molecule In the case of a diatomic molecule in which the two atoms are bonded to each other by a polar covalent bond, the dipole moment of the molecule is just the dipole moment of that bond. For example, in the case of HC1, the molecular dipole moment is equal to the dipole moment of the H—Cl bond, i.e., 1.03 D.

The greater the difference between the electronegativities of the bonded atoms, the greater is the dipole moment of the molecule. For example, the dipole moment of hydrogen halides are in the order,

H—F (1.91 D) > H—Cl (1.03 D) > H—Br (0.79 D) > H—I (0.38 D).

Polyatomic molecule If a polyatomic molecule has more than one polar bond, its dipole moment is equal to the resultant of the dipole moments of all the individual bonds (or bond dipoles).

• Since dipole moment is a vector quantity, the magnitude and direction of the resultant depends not only on the values of the individual dipole moments but also on their orientation or direction.
• Thus, it may happen that the dipole moment of a molecule is zero even though it contains more than one polar bond. What the resultant of the individual dipole moments of the bonds in a molecule will be depends on the structure (spatial arrangement of atoms) of the molecule.

Dipole moments of CO₂ and SO₂ The carbon dioxide molecule has two C=0 polar bonds but its dipole moment is zero. This is because CO₂ is a linear molecule in which the two C=O bonds are at 180° to each other, so the dipole moments are equal in magnitude and opposite in direction. The SO₂ molecule has two polar bonds but unlike the CO₂ molecule, its dipole moment is not zero. This is because the shape of the molecule is such that the resultant of the dipole moments of the two bonds is not zero.

Dipole moments of BF₃ and NF₃ In BF₃, there are three polar B—F bonds but the dipole moment of the molecule is zero because the resultant of the dipole moments of any two of the bonds is equal and opposite to the third (parallelogram law of forces). The NF₃ molecule also has three polar bonds. But because of the presence of a lone pair of electrons, the arrangement of the electron pairs is distorted tetrahedral and the orientation of the bond dipoles is such that their resultant is 0.8 x 10^-30 Cm.

Generally, in symmetrical molecules, the resultant of all the dipole moments is zero.

• If we compare the dipole moments of NF₃  and NH₃ (both have pyramidal shape with a lone pair of electrons) the dipole moment of NF₃ is expected to be more as fluorine is more electronegative than hydrogen.
• But the dipole moment of NH₃ (4.90 x 10^-30 C m) is more than that of NF₃. This is because in NH₃ the dipole due to the lone pair and the resultant dipole moments of N—H bonds are in the same direction. In NF₃ the direction of the dipoles is opposite.

The usefulness of dipole moment: Knowing the value of the dipole moment of a molecule is useful in many ways.

Polarity of a molecule The value of the dipole moment is an indicator of the polarity of the molecule. The higher the value of the dipole moment the greater is the polarity of the molecule (the greater the polarity of the bond if the molecule contains only one polar bond). The dipole moment of a nonpolar molecule is zero. So the dipole moment can help ascertain whethera molecule is polar or nonpolar.

Ionic character of a molecule All covalent bonds have a partially ionic character. We can predict the ionic character of a molecule by considering the magnitude of its dipole moment.

Shape of molecule The dipole moment of a molecule helps to predict the shape of the molecule. For example, experiments show that the dipole moment of BeF₂ is zero. This can be possible only if the resultant of the dipole moments of the two Be—F bonds is zero. In other words, the two bond dipoles must be oriented in opposite directions, which is possible only if the molecule is linear. The H₂O molecule, on the other hand, has a dipole moment of 1.85 D. Since the resultant of the dipole moments of the two bonds in the H₂O molecule is not zero, the molecule cannot be linear, it must be angular.

Partial covalent character of ionic bonds: Just as partial ionic character is attained in a covalent bond, similarly a partial covalent character is induced in an ionic compound. Consider a system of two ions A⁺ and B^– together at equilibrium distance (i.e., when the system acquires minimum energy).

The cation attracts the electron of the anion (B^–) and repels the nucleus; thus the anion gets distorted or polarised. The anion being larger gets more readily polarised than the cation, and thus the effect of an anion on a cation is less pronounced. If the extent of polarisation is large, the electrons of the anion are drawn towards the cation (this means electrons are shared by the ions), and covalent character results.

The extent of distortion of an anion depends on two factors:

1. The polarising power of the cation
2. The polarisability of the anion (ease of distortion of anion).

Generally, small cations with high positive charge have greater polarising power and large anions with loosely held electrons are easily polarisable. The factors favouring covalency (in ionic compounds) have been summarised by Fajans and are referred to as Fajans’ rules. According to these rules covalency is favored by:

1. More charge on either ion
2. Small cation
3. Large anion

• Many transition metal ions and a few main group elements have high polarising power. Since a noble gas configuration (ns²np⁶, typical of alkali and alkaline earth metal cations) is most effective at shielding nuclear charge the ions with this configuration are not significantly polarising.
• Let us consider NaCl and AlCl₃. In both the cases the anion is Cl^– whereas the cations are Na⁺ and Al³⁺. Al³⁺ (small size; high charge) has greater polarising power than Na⁺ (big size; small charge).
• Therefore, AlCI₃ is more covalent than NaCl. Now, let us consider NaCl, NaBr, and Nal. The cation being the same, covalency depends on the aruons. In all the cases tire charge is the same, but the size follows the order: Cl^– < BP^– < I^–. Thus polarisability also follows this order and the covalent character is maximum for Nal and least for NaCl.

Directional properties of bonds: As you already know, different orbitals have different shapes. The shape of the s orbital is spherical so the orbital is represented as a circle.

• The p orbital consists of two lobes in contact at the origin. As you can see in the figure, p orbitals have a marked directional character, they have axes at right angles to each other. Hence they are known as p_x, p_y, and p_z orbitals respectively.
• The orbitals are equivalent to each other but for the directional property. One lobe of the p orbital is marked with a positive sign and the other -with a negative sign. These signs have no physical significance, they arise from the fact that a wave function can have both positive and negative regions.
• In this context, it is important to remind you that the electron has dual nature and it cannot be both particle and wave at the same time. According to wave mechanics, an electron is represented by a wave function \(\psi\).
• It does not matter at all which lobe of the p orbital has been given a positive or negative sign. Similarly, an s orbital can be + or -. When \(\psi\) is zero, there is no likelihood of finding an electron in the region.

Let us apply these relative signs to understand the overlap of atomic orbitals. Consider two hydrogen atoms, A and B combining with each other. The 1s atomic orbitals of the two atoms overlap to give rise to a molecular orbital. The two electrons, one in each atomic orbital, can be represented by wave functions \(\psi_{\mathrm{A}} \text { and } \psi_{\mathrm{B}}\). Two combinations of the wave functions are then possible:

1. Overlap of the wave functions with same sign
2. Overlap of the wave functions with opposite sign

• Wave functions with same signs can be regarded as waves in the same phase which may add up to give a large resultant wave. On the other hand, wave functions with opposite signs cancel each other. In other words, when the overlap occurs between two regions with the same sign it results in a bonding orbital whereas in case of overlap between regions of the opposite sign, a nonbonding orbital results.
• As it is dear from the Figure, the electron charge is concentrated in the region between two nuclei in case of the bonding orbital, thereby holding the two nuclei together. On the other hand, in the nonbonding orbital, the electron charge is withdrawn from the region between the two nuclei, which results in repulsion between the two nuclei.
• Figure shows different combinations of s and p orbitals with positive, negative, and zero overlap. Overlap is said to be positive or negative if the signs of the overlapping lobes are the same or different respectively. In case of zero overlap, the extent of stabilization obtained by the overlap of orbitals of the same sign is equal to the extent of destabilisation obtained by overlap of orbitals of the opposite sign, so the net overlap is said to be zero.

Molecular Orbital Theory: In valence bond theory, a chemical bond is identified as a shared electron pair. Electrons in atoms are present in atomic orbitals. The atomic orbitals of different atoms overlap to give rise to a chemical bond.

This theory could not explain certain properties like relative bond strengths, and the paramagnetic and the diamagnetic characters of a molecule. Another theory which is widely used to explain bonding in molecules is referred to as molecular orbital theory.

This theory was proposed by Hund and Mulliken (1932). Unlike the valence bond theory, the molecular orbital theory considers the valence electron to be associated with the nuclei of all combining atoms. The main features of this theory are as follows.

1. Atomic orbitals that match in energy and symmetry combine to give molecular orbitals.
2. The total number of molecular orbitals formed is equal to the total number of combining atomic orbitals.
3. Unlike the electron in an atomic orbital the electron in a molecular orbital is influenced by two or more nuclei corresponding to the number of atoms in the molecule. (An atomic orbital is called monocentric, while a molecular orbital is called polycentric.)
4. When two atomic orbitals combine, they form two molecular orbitals—the bonding molecular orbial which has lower energy than the atomic orbitals, and the antibonding molecular orbial which has higher energy than the atomic orbitals.
5. Like atomic orbitals the molecular orbitals represent electron probability distribution.
6. The filling up of molecular orbitals is done in accordance to the aufbau principle, Pauli exclusion principle, and Hund’s rule.
7. The maximum number of electrons present in a molecular orbital is 2.

Linear Combination of Atomic Orbitals (LCAO) Method: You already know that according to wave mechanics, atomic orbitals are represented by a wave function vy which is obtained by solving the Schrodinger wave equation. It is very difficult to solve the Schrodinger wave equation for a molecule (since a molecular orbital is polycentric). Here the difficulty in finding a solution arises from the dependence of the wave function on intemuclear distances or because an electron interacts simultaneously with more than one nuclei.

• Therefore, certain approximations have to be made to solve wave equations for molecules. The most common approximation is adopting the procedure of linear combination of atomic orbitals (LCAO). Why is a molecular orbital considered to be a linear combination of atomic orbitals? The answer to this lies in the fact that the electron with a larger region for movement or more freedom has lower energy.
• Now the larger region for the movement is possible only when the atomic orbitals overlap with each other. The overlap is maximum when the orbitals are in the same plane. Thus the molecular orbital is considered to be a linear combination of atomic orbitals.

Consider two atoms A and B whose atomic orbitals are represented by the wave functions \(\psi_{\mathrm{A}} \text { and } \psi_{\mathrm{B}}\) respectively. During bond formation, these orbitals combine to give molecular orbitals. According to the LCAO method, the atomic orbitals of the individual atoms may combine in two ways—the wave functions may add to or subtract from each other.

The molecular orbital obtained by the addition of the wave functions may be represented as

∴ \(\psi_{(\mathrm{MO})}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}\).

The resultant molecular orbital (\(\psi_{(\mathrm{MO})}\)) is bonding. Here signs of both the wave functions are the same, this implies that two waves are in phase. When such waves combine with each other the resultant wave has larger amplitude (amplitudes of both the waves add—constructive interference).

The molecular orbital obtained by subtracting the wave functions is represented as \(\psi_{(\mathrm{MO})}^*=\psi_{\mathrm{A}}+\left(-\psi_{\mathrm{B}}\right)\).

The resultant molecular orbital obtained by adding two wave functions of opposite signs is antibonding. This corresponds to two waves which are completely out of phase and cancel each other by destructive interference. We have already discussed the bonding and antibonding molecular orbitals taking the H₂ molecule as an example, earlier in the chapter.

• Going back to the H₂ molecule again, the bonding molecular orbital is denoted by \(\sigma_{15}\) while the antibonding molecular orbital is represented by \(\sigma_{1 s}^*\). It is called a because the overlap of atomic orbitals is along the internuclear axis and Is because the combining atomic orbitals are Is orbitals.
• In a bonding molecular orbital there is an increase in electron density in the intemuclear region. The nuclei are shielded from each other so that the repulsion between them is very low. This results in lowering of energy and leads to boning.
• In the case of the antibonding molecular orbital, the electron density in the intemuclear region is severely reduced. In fact, there is a nodal plane in the intemuclear region. Therefore, the repulsion between the nuclei is high. The antibonding molecular orbital is higher in energy than the individual atomic orbitals.

The respective energy levels of the two combining atomic orbitals (1s) and the bonding and antibonding molecular orbitals are given in Figure.

• The energy of the bonding molecular orbitals is lower than that of either of the combining atomic orbitals by an amount ΔE. The energy of antibonding molecular orbital is more than that of the atomic orbitals by the same amount ΔE.
• In H₂ molecule the two electrons are present in the bonding molecular orbital and the system is stabilised by an energy factor corresponding to 2ΔE. As already stated, the bonding molecular orbital is lower in energy than the atomic orbital by a factor ΔE, so putting an electron in a bonding molecular orbital stabilises the system by an energy factor ΔE.
• As the hydrogen molecule has two electrons, the stabilisation energy is 2ΔE. This energy parameter corresponds to the bond energy. The total energy of the two molecular orbitals is equal to the total energy of the two atomic orbitals. This is in accordance with the principle of conservation of energy.

The main differences between molecular orbitals and atomic orbitals as well as between a bonding molecular orbital and an antibonding molecular orbital are as follows.

Conditions for LCAO: The atomic orbitals which combine to form molecular orbitals should satisfy the following three conditions.

1. The combining atomic orbitals should not differ significantly in energy. For example, a 1s orbital cannot overlap with a 2s orbital.
2. The extent of overlap must be large, i.e., the atoms must be sufficiently close.
3. The combining atomic orbitals must have the same symmetry along the intemuclear axis. By convention, the z axis is considered to be the intemuclear axis. Thus an s orbital cannot overlap with a p_x or p_y orbital, and a p_z orbital cannot overlap with a p_x or p_y orbital.

Types of molecular orbitals: Molecular orbitals of diatomic molecules are designated as σ, π, and δ. We will restrict our discussion to σ and π orbitals only. A σ orbital is formed when the overlap of atomic orbitals occurs along the intemuclear line (z-axis).

• An s orbital is spherically symmetrical, so when two s orbitals overlap a a molecular orbital is formed. A σ molecular orbital is also formed when an s orbital overlaps with a p_z orbital or when two p_z orbitals overlap with each other. In both these cases overlap takes place along the intemuclear line. A σ molecular orbital is symmetrical about the bond axis. If the overlaping lobes have opposite signs, it gives rise to antibonding orbitals.
• You already know that lateral overlap of orbitals results in the formation of π bonds. Therefore, when a p_x orbital overlaps with another p_x orbital (or p_y with p_y) a π molecular orbital is formed. In this combination both the lobes of each combining orbitals are perpendicular to the intemuclear line.
• The overlap does not occur along the intemuclear line. In other words, a π molecular orbital is not symmetrical about the intemuclear axis. In case of π bonding molecular orbital, the electron density is large, above and below the intemuclear line. The π* molecular orbital (antibonding) has a node between the two nuclei. All antibonding orbitals possess a nodal plane lying between the nuclei and perpendicular to the intemuclear axis.

Here it should be understood that the overlap of an s orbital with p_x or p_y or a p_z orbital with a p_x or p_y orbital gives a nonbonding combination. The orientation is not proper for a bond to be formed.

Order of energy of molecular orbitals: In the same way as atomic orbitals have definite energies and can be distinguished by quantum numbers, each molecular orbital has a definite energy and is specified by four quantum numbers.

• You are already familiar with the filling of atomic orbitals in accordance to the Pauli exclusion principle and Hund’s rule. The same applies to molecular orbitals. In the case of the molecular orbital method, the whole molecule is considered and the total number of electrons from all the constituent atoms are filled in the molecular orbitals.
• The order of energy of molecular orbitals has been determined experimentally by spectroscopy for the elements of the second period. The increasing order of energies of the molecular orbitals in homonudear diatomic molecules is

⇒ \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\left(\pi 2 p_x=\pi 2 p_y\right)<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)

The energies of \(\pi 2 p_x \text { and } \pi 2 p_y\) molecular orbitals are the same and they form a pair of degenerate orbitals. Similarly the \(\pi^* 2 p_x \text { and } \pi^* 2 p_y\) orbitals are also degenerate. The energies of \(\sigma 2 \mathrm{p}_z \text { and } \pi 2 \mathrm{p}_x / \pi 2 \mathrm{p}_y\) molecular orbitals are very close and the order given above is correct for 02 and F2 molecules. For other lighter elements of the second period of the periodic table, there is a reversal in the order of the energy of \(\pi 2 \mathrm{p}_x / \pi 2 \mathrm{p}_y \text { and } \sigma 2 \mathrm{p}_z\) orbitals. The order now becomes

⇒ \(\sigma 1 \mathrm{~s}<\sigma^* 1 \mathrm{~s}<\sigma 2 \mathrm{~s}<\sigma^* 2 \mathrm{~s}<\left(\pi 2 \mathrm{p}_x=\pi 2 \mathrm{p}_y\right)<\sigma 2 \mathrm{p}_z<\left(\pi^* 2 \mathrm{p}_x=\pi^* 2 \mathrm{p}_y\right)<\sigma^* 2 \mathrm{p}_z\).

The main difference between the two is that for the lighter elements the energy of the \(\sigma 2 \mathrm{p}_z\), molecular orbital is higher than that of the \(\pi 2 \mathrm{p}_x \text { and } \pi 2 \mathrm{p}_y\) molecular orbitals.

• While establishing molecular orbitals we deal with the concept that orbitals of atom A mix with orbitals of atom B if they are matched in energy and symmetry. However mixing occurs in all orbitals of comparable energy having proper symmetry. The energy differences between Is and 2s orbitals is large and we are justified in neglecting this mixing. The energy difference between 2s and 2p orbitals is less and varies with effective nuclear charge.
• It increases drastically from 200 kJ mol^-1 for lithium to 2500 kJ mol^-1 for fluorine. The 2s orbital and 2p_z orbital are matched in symmetry, and thus for the lighter elements mixing occurs. This phenomenon is the equivalent of hybridisation in valence bond theory.

Molecular Orbital Treatment of Homonudear Diatomic Molecules: The distribution of electrons among various molecular orbitals of a molecule is called the electronic configuration of the molecule. The stability and magnetic behaviour of the molecule depend on its electronic configuration.

Stability: A bonding orbital is lower in energy whereas an antibonding orbital is higher in energy than either of the two atomic orbitals which combine. Therefore, an electron in a bonding orbital leads to stabilization, and conversely an electron in an antibonding orbital leads to destabilisation. Thus, the relative number of electrons in bonding and antibonding orbitals of the molecule give an idea about its stability. If the number of electrons in bonding orbitals is given by N_b and that in antibonding orbitals by N_a, then the molecule will be stable if N_b>N_a.

Bond order Tire stability of a molecule is given in terms of bond order. The bond order of a molecide is expressed as half of the difference between the number of electrons present in bonding and antibonding orbitals.

Bond order = \(\frac{1}{2}\left(N_b-N_a\right)\).

A positive bond order indicates a stable molecule whereas a negative or zero bond order indicates an unstable molecule. The value of the bond order decides the nature of the bond. Integral bond order values of one, two, or three correspond to single, double, and triple bonds respectively. The bond order is a useful parameter indicating characteristics of bonds. Bond order is inversely proportional to bond length.

Magnetic behaviour: The electronic configuration of a molecule indicates the magnetic behaviour of a molecule. The macroscopic magnetic properties of a substance arise from the total of the magnetic moments of its component atoms and molecules. If all the electrons in the orbitals are paired then the molecule is diamagnetic. But in case of unpaired electrons present in the orbitals, the molecule is paramagnetic.

In diamagnetic substances, the magnetisation is in the opposite direction to that of the applied field. In paramagnetic substances, the net spin magnetic moments due to impaired electrons in the molecule can be aligned in the direction of the applied field.

Let us now discuss bonding in a few homonuclear diatomic molecules.

1. \(\mathbf{H}_2^{+}\) molecule ion This is the simplest diatomic molecule obtained as a transient species when hydrogen gas [H₂(g)] is subjected to electric discharge at reduced pressure. The single electron is present in the bonding, σ1s molecular orbital. The molecular orbital configuration is σ1s¹. The bond order is 1/2(1- 0) = 1/2. A positive bond order implies a/ certain amount of stability. The molecule is paramagnetic in nature due to the presence of single unpaired electron.

2. H₂ molecule The total number of electrons in the molecule is two (one from each hydrogen atom). The electrons are placed in the bonding molecular orbital and the electronic configuration is σ1s². The molecular orbital diagram is shown in Figure. The bond order of the molecule = 1/2 (2 – 0) = 1. This implies that the two atoms are held together by a single bond. Since there is no unpaired electron, the molecule is diamagnetic.

The bond order of the molecular ion, \(\mathrm{H}_2^{+} \text {is } 1 / 2\) whereas that of hydrogen molecule is 1. This shows that the H₂ molecule is more stable than the \(\mathrm{H}_2^{+} ion\). This is also obvious from their bond lengths which is 106 pm for \(\mathrm{H}_2^{+} ion\) ion and 74 pm for H₂ molecules.

3. He₂ molecule Each He atom has two electrons. Therefore, the He₂ molecule has four electrons arranged as \((\sigma 1 s)^{\frac{2}{2}}\left(\sigma^* 1 s\right)^2\). The bond order is 1/2(2 – 2) = 0, which indicates that the molecule does not exist. Similarly it can be shown that the Be2 molecule does not exist.

4. Li₂ molecule The electronic configuration of Li is 1s²2s¹. Thus in Li_2, there are six electrons and the electronic configuration of the molecule is \((\sigma 1 s)^2(\sigma 1 s)^2(\sigma 2 s)^2(a2s)2\). This is shown in Figure.

The inner shell, i.e., σ1s and \(\sigma^{\circ} 1 \mathrm{~s}\), does not contribute much to the bonding. Therefore, the electronic configuration of the molecule can be written as KK(σ2s)². KK refers to a closed K shell which includes two molecular orbitals \((\sigma 1 s)^2 \text { and }\left(\sigma^{\circ} 1 s\right)^2\). The bond order is 1 [1/2(4 – 2)1 which shows that the Li₂ molecule is stable. In fact, diamagnetic, Li₂ molecules do exist in vapour phase.

5. B₂ molecule Each boron atom has five electrons and the electronic configuration is \(1 s^2 2 s^2 2 p^1\). This is the first instance in the discussion where we have come across a molecular system with an electron in a p orbital. As stated earlier for lighter molecules like B₂ and C₂ the doubly degenerate π2p orbitals are lower in energy than the σ2p_z orbital.

Therefore, the electronic configuration of the B₂ molecule is \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2 \pi^1 2 p_x=\pi^1\) or \(operatorname{KK}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2 \pi^1 2 p_x=\pi^1 2 p_y\). The molecular orbital diagram is given in Figure. The bond order is 1/2(6 – 4) = 1, which shows that the molecule is stable and, having two unpaired electrons, paramagnetic in nature. The bond energy is 290 kJ mol^-1 and the bond length is 159 pm.

6. C₂ molecule Since the carbon atom has six electrons \(\left(1 s^2 2 s^2 2 p^2\right)\), the C₂ molecule will have 12 electrons and the electronic configuration may be written as \(\left(\sigma_{1 s}\right)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\)

or \(\mathrm{KK}\left(\sigma^2\right)^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\)

The bond order is 1/2(8 – 4) = 2, which shows the presence of a double bond in the molecule. But both pairs of electrons constituting the double bond are present in the π molecular orbitals.

Therefore, the double bond comprises two π bonds and not one σ and one π bond as in other molecules. There is no unpaired electron, so that the molecule is diamagnetic and exists in the vapour phase only. The bond energy is 620 kJ mol^-1 and the bond length is 131 pm.

7. N₂ molecule Each nitrogen atom has seven electrons (1s² 2s² 2p³) therefore, the N₂ molecule will have fourteen electrons and the electronic configuration is

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma^* 1 s\right)^2\left(\sigma_2\right)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\sigma^2 \mathrm{p}_z\right)^2\)

or \(\mathrm{KK}(\sigma 2 s)^2\left(\sigma^{\circ} 2 s\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\sigma 2 \mathrm{p}_z\right)^2\).

The bond order of the N₂ molecule is 1/2(10 – 4) = 3.

Therefore, in the diamagnetic nitrogen molecule, the two atoms are held by a triple bond. The bond energy is 945 kJ mol^-1 and the bond length is 110 pm.

Some ionic species formed by the N₂ molecule The \(\mathrm{N}_2^{+}\) ion is formed by removing an electron from the N₂ molecule present in the highest occupied molecular orbital, i.e., \(\sigma 2 \mathrm{p}_{\mathrm{z}}\). The electronic configuration for the ion is

⇒ \(\mathrm{KK}(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2 \sigma^1 2 \mathrm{p}_z\)

The presence of an unpaired electron indicates that the ion will be paramagnetic. The bond order is 2.5, suggesting that it is less stable than N₂. This is expected as an electron has been removed from a bonding molecular orbital.

The \(\mathrm{N}_2^{-}\) ion is formed by adding an electron to the lowest unoccupied molecular orbital of N₂, i.e., the \(\pi^{\circ} 2 p_x\) molecular orbital. The electronic configuration of \(\mathrm{N}_2^{-}\) is

⇒ \(\mathrm{KK}(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\sigma^2 2 \mathrm{p}_z\right)\left(\pi^* 2 \mathrm{p}_x\right)^1\)

The bond order is 2.5 and the molecule is paramagnetic.

Considering the bond order values of the species N₂, N⁺_2, and N^–₂, they can be arranged in order of increasing stability as
\(\mathrm{N}_2, \mathrm{~N}_2^{+} \text {and } \mathrm{N}_2^{-} \text {, }\).

8. O₂ molecule Each oxygen atom has eight electrons \(\left(1 s^2 2 s^2 2 p^4\right)\). Therefore the total number of electrons in the O₂ molecule is 16. The electronic configuration is

⇒ \((\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}\right)^2(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma^2 \mathrm{p}_z\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\pi^* 2 \mathrm{p}_x\right)^1=\left(\pi^* 2 \mathrm{p}_y\right)^1\)

or \(\mathrm{KK}(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma 2 \mathrm{p}_z\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\pi 2 \mathrm{p}_x\right)^1=\left(\pi 2 \mathrm{p}_y\right)^1\).

The energy level diagram of the molecule is shown in Figure. As oxygen is a heavier molecule the energy level sequence of molecular orbitals is different from that of B₂, C₂, etc. The \(\sigma 2 p_z\), molecular orbital is lower in energy than the \(\pi 2 p_x\) molecular orbital.

The \(\pi^{\circ} 2 p_x \text { and } \pi^* 2 p_y\) orbitals are singly occupied (by Hund’s rule). The presence of unpaired electrons gives rise to paramagnetism which is in accordance with experimental findings. Thus, the molecular orbital theory successfully explains the paramagnetic nature of oxygen molecule, which was not explained by the valence bond depiction \((: \ddot{\mathrm{O}}=\ddot{\mathrm{O}}:)\) of the molecule. The bond order is 1/2(8-4) = 2, which corresponds to a double bond. The bond energy is 498 kJ mol^-1 and the bond length is 121 pm.

Some ionic species formed by O₂ molecule The \(\mathrm{O}_2^{+}\) ion contains one electron less than the O₂ molecule whereas the \(\mathrm{O}_2^{-} \text {and } \mathrm{O}_2^{2-}\) ions contain one and two electrons more than the O₂ molecule respectively. The electronic configuration, magnetic behaviour, and bond order of the ionic species are summarised below.

The stability of the O₂ molecule and the ionic species follows the order

∴ \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-} \text {. }\)

Similarly, the §ond length which is inversely proportional to bond order follows the sequence

∴ \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{+}\)

9. F₂ molecule The fluorine atom has nine electrons \(\left(1 s^2 2 s^2 2 p^5\right)\). Therefore, the F₂ molecule contains 18 electrons arranged in the molecular orbitals as \(\mathrm{KK}(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma 2 \mathrm{p}_z\right)^2\left(\pi 2 \mathrm{p}_x\right)^2=\left(\pi 2 \mathrm{p}_y\right)^2\left(\pi^* 2 \mathrm{p}_x\right)^2=\left(\pi^* 2 \mathrm{p}_y\right)^2\).

The bond order is one and the molecule is diamagnetic. The bond energy is 159 kJ mol^-1 and the bond length is 143 pm.

10. Ne₂ molecule The electronic configuration of the Ne₂ molecule with 20 electrons (since each neon atom has 10 electrons) is

⇒ \(\begin{aligned}
\operatorname{KK}(\sigma 2 s)^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\sigma 2 \mathrm{p}_z\right)^2\left(\pi 2 \mathrm{p}_x\right)^2 =\left(\pi 2 \mathrm{p}_y\right)^2 \\
\left(\pi^* 2 \mathrm{p}_x\right)^2 =\left(\pi^* 2 \mathrm{p}_y\right)^2\left(\sigma^* 2 \mathrm{p}_z\right)^2
\end{aligned}\)

The bond order is zero, which implies that the Ne₂ molecule does not exist. In other words, neon is 1 monoatomic.

Example 1. Two p orbitals of one atom and two p orbitals of another atom combine to form molecular orbitals. How mam/ molecular orbitals will result from this combination? Give all the possibilities.
Solution:

We know that there are three degenerate p orbitals and the electronic configuration of each atom may be \(\mathrm{p}_x \mathrm{p}_y \text { or } \mathrm{p}_x \mathrm{p}_z \text { or } \mathrm{p}_y \mathrm{p}_z\).

Thus various probabilities are possible which are shown below.

The overlap of p_x or p_y with p_z or p_x with p_y will give a nonbonding combination of orbitals.

Example 2. Arrange the following in order of increasing stability. \(\mathrm{H}_2, \mathrm{H}_2^{+}, \mathrm{H}_2^{-}, \mathrm{H}_2^{2-}\)
Solution: 

Thus, the order of increasing stability is \(\mathrm{H}_2^{2-}<\mathrm{H}_2^{-}=\mathrm{H}_2^{+}<\mathrm{H}_2\)

Example 3. Use the molecular orbital theory to explain why He₂ does not exist.
Solution:

Each He atom has 2 electrons. Thus He₂ has 4 electrons and the electronic configuration of the molecule is \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2\).

BO = 1/2 (2 – 2) = 0. Thus He₂ does not exist.

Example 4. What is the bond order of the \(\mathbf{H}_2^{+}\) ion and the H₂ molecule? Why is the bond length in H₂ shorter than that in \(\mathbf{H}_2^{+}\)?
Solution:

The bond order of \(\mathbf{H}_2^{+}\) is 0.5 and that of H₂ is 1. The bond order value for the H₂ molecule is more than that for the \(\mathbf{H}_2^{+}\) ion. Since bond length decreases as bond order increases, the bond length of H₂ is shorter.

Hybridisation

According to the valence bond theory, a covalent bond is formed by the overlapping of half-filled atomic orbitals of the valence shell. This does explain the formation of some molecules, like O₂ and N₂, but cannot explain the tetravalency of carbon, the bivalency of beryllium, or the trivalency of boron.

• Carbon, with the electronic configuration of \(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}_x^1, 2 \mathrm{p}_y^1\) has only two unpaired electrons and should have been bivalent. The electronic configuration of boron is \(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}_x^1\) which means that it has only one half-filled orbital but it forms compounds like BF₃. Beryllium, with the electronic configuration of 1s², 2s² has no impaired electron. Its covalency should have been zero, but it forms compounds like BeCl₂.
• To explain this apparent contradiction, the modem theory of bond formation assumes that these and other atoms acquire excited states before participating in bond formation. In the excited state, one of the electrons in the 2s orbital in the examples we are considering gets promoted to a vacant 2p orbital. Thus, for example, in the excited state, carbon has four half-filled orbitals. The energy used for excitation is compensated for by the energy released during bond formation.

The assumption that atoms acquire an excited state before forming bonds can explain some things like the tetravalency of carbon. But it leaves certain other questions unanswered. Take the formation of methane, for example. The carbon atom has four unpaired electrons in the excited state, one electron in the s orbital, and three in the p orbitals.

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• If these orbitals overlap with the Is orbitals of four hydrogen atoms, three p-s sigma bonds and one s-s sigma bond should be formed. In that case the four bonds should not be equivalent, they should have different strengths, bond lengths, and bond angles. However, it has been seen experimentally that the four bonds are equivalent and that the bond length is 109 pm, while the H—C—H bond angle is 109°28′.
• Pauling and Slater, two American scientists, introduced the concept of hybridisation to explain this apparent discrepancy. According to them, before an atom forms bonds with other atoms, its orbitals get mixed and redistributed to form new, equivalent orbitals.
• These equivalent orbitals are called hybrid orbitals and the phenomenon is called hybridisation. Thus, hybridisation is the intermixing of atomic orbitals of comparable energies to form new orbitals of equivalent energy and identical shape. The energies of the orbitals get redistributed in the process.

Going back to methane, the three 2p orbitals of the carbon atom intermix with the 2s orbital to produce four sp³ hybridised orbitals, which form four equivalent bonds with the Is orbitals of four hydrogen atoms.

Rules of hybridisation: Hybridisation is a theoretical concept that is used to explain some structural properties of molecules. It would be useful to remember the following points about hybridisation.

1. The orbitals taking part in hybridisation must have comparable (small difference) energies.
2. It is not necessary for an orbital to be half-filled in order to participate in hybridisation. Vacant and completely filled orbitals can also be involved in hybridisation. This means electrons may or may not be promoted from a lower to a higher subshell for the formation of hybrid orbitals.
3. The electron density of a hybrid orbital is concentrated on one side of the nucleus, which means one lobe of the orbital is relatively larger than the other. Therefore, it Hybridised orbital can overlap to a large extent than an s orbital or a p orbital.
4. Hybridised orbitals have equivalent energies and identical shapes.
5. The number of hybrid orbitals formed is always equal to the number of atomic orbitals taking part in hybridisation.
6. Hybrid orbitals form stronger bonds than pure atomic orbitals because they can overlap to a greater extent. They always form a bonds because their electron cloud is parallel to the internuclear axis.
7. A molecule adopts a particular shape not because of hybridisation but to have the lowest favorable energy.

Types of hybridisation: Hybridisation is a concept which is applicable to all types of atomic orbitals, viz., s, p, d, and f. Let us discuss a few of the simpler types.

sp hybridisation When one s and one p orbital belonging to the same shell of an atom hybridise to form two new orbitals, the type of hybridisation is called sp hybridisation. The new orbitals formed are called sp hybrid orbitals. The two sp hybrid orbitals orient themselves at an angle of 180° to reduce repulsion to the minimum.

Let us consider beryllium chloride (BeCl₂)- The central atom, Be, has two electrons in the 2s subshell. One of these electrons is promoted to the 2p subshell. Then the 2s and 2p orbitals undergo hybridisation to form two sp-hybridised orbitals, which are linear (180° apart) and contain one electron each. The two sp-hybridised orbitals overlap axially with the p orbitals of two chlorine atoms to form two σ bonds.

sp² hybridisation When one s and two p orbitals of the same shell of an atom hybridise to form three hybrid orbitals of equivalent energy and identical shape, the hybrid orbitals are called sp²-hybridised orbitals.

In order to reduce repulsion, the three sp² hybrid orbitals orient themselves in such a way as to form angles of 120° with each other and lie in the same plane. In other words, the orbitals are directed towards the vertices of an equilateral triangle. If the sp²-hybridised central atom of a molecule is linked directly to three other atoms, the molecule is trigonal planar in shape.

Take the case of BF₃, in which the central atom, B, has three electrons in the valence shell (1s², 2s², 2p¹). One electron is unpaired in the ground state. In the excited state there are three unpaired electrons, one in the s and two in the p orbital. Then one s and two p orbitals undergo sp² hybridisation, to form three sp- hybridised orbitals, which form σ bonds with the 2p orbitals of three fluorine atoms. The shape of the BF₃ molecule is trigonal planar.

sp³ hybridisation When one s and three p orbitals belonging to the same shell of an atom hybridise to form four hybrid orbitals of equivalent energy and identical shape, the new orbitals are called sp³-hybrid orbitals.

This type of hybridisation is called sp³ hybridisation. Molecules in which the sp³– hybridised orbitals of the central atom form bonds with four other atoms are tetrahedral in shape. This is because the sp³ -hybridised orbitals are directed towards the vertices of a tetrahedron. The angle between sp³ – hybridised orbitals, and consequently the bond angle of tetrahedral molecules, is 109°28′.

In methane, the carbon atom is sp³ – hybridised. It has four sp³ -hybridised orbitals containing one electron each. The carbon atom shares these electrons with four hydrogen atoms. In other words, all the hybridised orbitals of the central atom form bonds with four atoms. This is why the methane molecule has a regular tetrahedral shape and the bond angle is 109°28′.

However, all molecules in which the central atom is sp³ hybridised do not have a regular tetrahedral shape. In NH₃, for example, the central atom, nitrogen, has three unpaired electrons in its 2p subshell. To make three covalent bonds with three hydrogen atoms, nitrogen needs only three unpaired electrons.

• Therefore, in this case, the question of the promotion of an electron from a lower subshell does not arise. Nitrogen undergoes hybridisation in the ground state. One s and three p orbitals hybridise to form sp³-hybridised orbitals. However, one orbital contains a pair of electrons which the atom does not share. The other three orbitals form sigma bonds with the Is orbitals of three hydrogen Atoms.
• The shape of the molecule Is, thus, not regular tetrahedral because the orbital (of nitrogen) with the lone pair of electrons does not participate in bond formation. On account of unequal repulsions, the bond angle changes from the regular angle of 109°28′ to 107°, and the shape of the molecule is pyramidal.

In water, the central oxygen atom sp³ hybridised, The configuration of the oxygen atom is 1s², 2s², 2p⁴. f wo p orbitals contain unpaired electrons which is all the oxygen atom needs to form bonds with two hydrogen atoms. The oxygen atom undergoes sp³ hybridisation in the ground state.

An s orbital containing two electrons, a p orbital containing two electrons, and two p orbitals containing an electron each undergoes hybridisation. Two of the hybrid orbitals contain a pair of electrons each and do not participate in bond formation. The other two (containing one electron each) form σ bonds with the 1s orbitals of two hydrogen atoms. The unequal repulsion on account of the two lone pairs of electrons reduces the bond angle to 105°.

sp³d hybridisation This type of hybridization entails the intermixing of one s, one d, and three p orbitals. Five equivalent sp³d hybrid orbitals are formed. The central atoms of PCl₅, SF_4, and ICl₃ are sp³d hybridised.

In PCI₅, the central atom has only three unpaired electrons in its 3p subshell in the ground state.

⇒ \(\left({ }_{19} \mathrm{P}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 2 \mathrm{p}_x^1, 3 \mathrm{p}_y^1, 3 \mathrm{p}_z^1\right) \text {. }\).

To make five covalent bonds with five chlorine atoms, phosphorus needs five unpaired electrons which are present in the excited state. Then one 3s, three 3p and one 3d orbital undergo sp³d hybridisation. The five hybrid orbitals form sigma bonds with the orbitals of five chlorine atoms. The shape of the molecule is regular trigonal bipyramidal and bond angles arc 120° and 90°.

In SF₄, the central atom, sulphur, has only two unpaired electrons \(\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}_x^2, 3 \mathrm{p}_y^1, 3 \mathrm{p}_z^1\right)\).

To make four covalent bonds the sulphur atom needs four unpaired electrons. Therefore, one electron from the 3p_x orbital is promoted to the 3d orbital. Then one s, three p, and one d orbitals undergo sp³d hybridisation. Four of the hybrid orbitals form sigma bonds with four fluorine atoms. The fifth hybrid orbital contain a lone pair of electrons.

The central atom is, thus, surrounded by a lone pair and four bonded pairs of electrons. In order to reduce repulsion to the minimum, the lone pair occupies an equatorial position. ‘Hie unequal repulsion on account of the presence of a lone pair gives rise to a distorted trigonal bipyramidal geometry.

sp³d² hybridisation This involves the intermixing of one s, three p, and two d orbitals. The sp³d² hybridised f central atom of a molecule is linked directly to six other atoms, the shape of the molecule is regular octahedral.

⇒ \({ }_{16} s-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}_x^2, 3 \mathrm{p}_y^1, 3 \mathrm{p}_z^1\)

• Let us consider the formation of sulfur hexafluoride (SF₆). The central atom in the ground state has two unpaired electrons in the 3p subshell. In the excited slate, the electrons from the 3s and the 3p subshell are promoted to the 3d subshell.
• Thus one s, three p, and two d orbitals intermix to form six sp³d²– hybridised orbitals. All the hybrid orbitals participate in bond formation with six fluorine atoms. Hence the shape of the SF₆ molecule is regular octahedral.

sp³d³ hybridisation This involves the hybridisation of one s, three p, and three d orbitals to form seven hybrid orbitals. When sp³d³ -hybridised central atom of a molecule is bounded with seven other atoms, the geometry of the molecule is pentagonal bipyramidal.

• For example, in the IF₇ the central atom has seven valence electrons but only one unpaired electron. To make seven covalent bonds, iodine needs seven unpaired electrons. Therefore, three electrons are promoted to d orbitals. Then one s, three p, and three d orbitals hybridise to form seven hybridised orbitals. The IF₇ molecule has a pentagonal bipyramidal shape and the bond angles are 90° and 72°.
• We have already mentioned earlier in the chapter that there are many exceptions to the octet rule. While going through the discussion on hybridization, you would have observed that generally, the molecules in which bonds are formed using s-p-d hybrid orbitals are exceptions to the octet rule. This implies that the octet rule is broken when atoms have an extra energy level, which is close in energy to the p level. This means the octet rule is more likely to be violated in elements having vacant d orbitals.

Molecules containing π bond: Pi bonds are formed by the sideways overlapping of atomic orbitals which have electron clouds perpendicular to the internuclear axis. Obviously, hybrid orbitals cannot form π bonds since their electron densities are parallel to the internuclear axis. Only pure p and d orbitals with electron densities perpendicular to the internuclear axis can overlap laterally to form π bonds.

The π bonds are not responsible for the shape of a molecule. That it is determined by the σ bonds. Pi bonds just shorten the bond lengths in a molecule. Let us consider a few examples of the formation of π bonds.

Structure of ethene In the ethene (C₂H₄) molecule, each carbon atom is sp² hybridised. One of the 2s electrons gets promoted to the 2p subshell. Then one 2s and two 2p orbitals of the excited carbon atom undergo hybridisation to form three sp²  hybridised orbitals. The remaining p orbital does not participate in hybridisation.

⇒ \({ }_6 \mathrm{C}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}_x^1, 2 \mathrm{p}_y^1\)

The three sp²– hybridised orbitals lie In the some pi one and are at an angle of 120° with each other, One sp² -hybridized orbital of one carbon atom overlaps axially with one sp²– hybridised orbital of the other carbon atom to form a σ bond. The remaining two sp²– hybrld orbitals of each carbon atom overlap with the half-filled Is orbital of H atoms along their respective Internuclear axes to form four σ bonds.

The unhybridised 2p orbital of each carbon atom Is perpendicular to the plane of the hybridised orbitals, or the internuclear axis. These two pure p orbitals overlap sideways to form a rc bond between the two carbon atoms. The C—C bond length in the ethanol molecule is 134 pm as compared to 154 pm in the ethane (C₂H₂) molecule. In the ethane molecule, like methane, the two carbon atoms are sp³ hybridised. All the bonds in the ethane molecule—one C—C and six C—H— are σ bonds.

Structure of acetylene (ethyne) In acetylene (C₂H₂), each carbon atom undergoes sp hybridisation. Each carbon atom has two sp- hybridised orbitals oriented at an angle of 180° to each other and two pure p orbitals which arc perpendicular to each other and to the plane of the sp-hybrid orbitals.

• One sp-hybrid orbital of one carbon atom overlaps axially with one sp-hybrid orbital of the other carbon atom to form a C—C σ bond. The second sp-hybrid orbital of each carbon atom overlaps axially with the half-filled Is orbital of a hydrogen atom to form two C—H σ bonds.
• The unhybridised 2p_y orbital of the first carbon atom overlaps sideways with the 2p_y orbital of the second carbon atom, forming a π bond between the two carbon atoms. Similarly, the unhybridised 2p_z orbital of one carbon atom overlaps sideways with the 2p_z orbital of the other carbon atom to form another n bond between the two carbon atoms.
• All the carbon and hydrogen atoms lie linearly in the same plane. The electron clouds of one n bond lie above and below the internuclear axis, while the electron clouds of the other π bond lie in front and at the back of the axis. The C—C bond length in the ethyne molecule is 120 pm. This shows that pi bonds shorten the bond length in a molecule.

Modern Concept Of Covalent Bond

So far we have discussed Lewis structures and geometrical shapes of molecules. Perhaps, the Lewis theory was the first explanation of a covalent bond in terms of electrons. The theory, however, fails to explain the difference in energies of similar bonds. Similarly, the VSEPR theory fails to explain the formation of chemical bonds. To overcome these limitations two modem theories were proposed—valence bond theory and molecular orbital theory—which are based on quantum mechanics.

This concept is based upon the fact that all systems in the universe tend to have the lowest possible potential energy because the lowering of energy leads to stability. According to the modern theory of bonding, atoms form bonds with each other in order to have the minimum possible energy. Another way of putting this is that atoms combine with each other only if such a combination is accompanied by a decrease in energy, or that bond formation is accompanied by the release of energy.

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Valence bond theory: This theory was proposed by Linus Pauling who was awarded the Nobel Prize for chemistry in 1954. The basis of this theory is the Lewis concept of electron-pair bond.

• To understand the theory qualitatively and know why the formation of a bond between two (or more) isolated atoms leads to the lowering of the energy of the ‘system’ of atoms of the molecule, let us consider the formation of a simple molecule—hydrogen.
• The hydrogen atom has only one electron, which is under the field of influence of the nucleus of the atom. Let us consider two such hydrogen atoms, H_A  and H_B, which have electrons eA and eB in their valence shells respectively. When these two atoms are far apart there are no forces of attraction or repulsion between them. Now suppose the two atoms start approaching each other.
• At some point, the electron of one will start being attracted by the nucleus of the other. On the other hand, each nucleus will repel the other and the two electrons will also repel each other.

It has been found that in the beginning the magnitude of the forces of attraction is greater than that of the forces of repulsion. So, the atoms keep drawing closer and the potential energy of the system (i.e., the two atoms) keeps decreasing.

• However, as the atoms approach each other, the magnitude of the repulsion forces keeps increasing and a point is reached where the forces of repulsion just balance the forces of attraction. This is the point at which the system has the lowest potential energy, and consequently, the greatest stability.
• Let us call the distance between the two atoms at this point d0. If the atoms were to come closer than this distance, the magnitude of the repulsive forces would exceed that of the attractive forces and the energy of the system would increase, leading to the instability of the molecule. Figure shows the variation of the energy of the system with intemuclear distance (distance between the two atoms).
• As shown in Figure, the energy of the two-atom system (or the molecule) is the least when the distance between them is d₀. It is far less than the energy of the two isolated atoms. This is why atoms form bonds—to gain stability. The greater the number of bonds formed, the greater is the lowering of energy and the stability of the system.
• In fact, some atoms even form more bonds than required to complete their octet, for example, phosphorus in PCI₅, and sulphur in SF₆. The quantum theory of bonding can, thus, explain why the octet rule is not always obeyed in bond formation.

Bond length: The distance d₀ between the centres of the nuclei of two bonded atoms is called the bond length. At this distance the forces of attraction and repulsion between the bonded atoms just balance each other, also the orbitals of the two atoms overlap partially. The greater the extent of overlapping, the shorter is the bond length, and the stronger the bond. The bond length depends upon

1. The size of the atoms and
2. The nature of the bonds.

• It increases with the size of atoms because the extent of overlapping of the orbitals decreases as the size of the atoms increases. For example, the H—Cl bond length in the HCl molecule is 127 pm, whereas the C—Cl bond length is 177 pm.
• The bond length decreases with the multiplicity of bonds because the greater the number of electrons shared by two atoms, the greater is the force of attraction between them.

Each atom in a bond contributes to the bond length. The contribution of each atom, in case of a covalent bond, is the covalent radius of that atom. Consider a molecule of chlorine, where the two chlorine atoms are bonded together by a single covalent bond. The bond length in the molecule is 198 pm. Here r_Cl is the covalent radius of the chlorine atom, which is 99 pm. Therefore, we may conclude that the bond length in a covalent molecule is equal to the sum of the covalent radii of the atoms which are bonded together.

Bond enthalpy: The energy corresponding to the lowest point (minima) in the potential energy curve is called the bond energy. It is the amount of energy released when one mole of covalent bonds is formed. The amount of energy released during the formation of a bond between two atoms is also the amount of energy required to break the bond between the two atoms or to separate them.

This is called the bond dissociation enthalpy (or bond enthalpy). Thus, the bond dissociation enthalpy is the amount of energy required to break one mole of bonds of the same kind so as to separate the bonded atoms in the gaseous state. For a homonuclear diatomic molecule, the bond enthalpy is equal to the bond dissociation enthalpy.

Consider an example of a homonuclear diatomic molecule like H₂–

⇒ \(\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{H}(\mathrm{g}) ; \Delta_{\mathrm{g}} H^{\ominus}=435.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Similarly, for a heteronuclear diatomic molecule like HCl, we have

⇒ \(\mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 4_{\mathrm{L}} H^{\ominus}=4310 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

• The magnitude of the bond enthalpy depends upon the (1) size of the participating atoms and (2) the multiplicity of bonds. The larger the size of the atoms involved in bond formation, the lesser is the extent of overlapping and the smaller the amount of energy released.
• For example, the bond enthalpy of the C—C bond is 348 kJ mol^-1, whereas that of the H—H bond is 433 kJ mol^-1. Tire magnitude of the bond enthalpy increases with the multiplicity of bonds.
• This is because the greater the number of shared electron pairs, the more is the attraction between two atoms, and the greater is the energy released during bond formation. The bond enthalpies for oxygen molecule (O=O) and nitrogen molecule (N≡N) are 498 kJ mol^-1 and 946 kJ mol^-1 respectively. A larger bond enthalpy implies a stronger bond.

Bond order: The bond order in a covalent molecule or polyatomic ion is the number of bonds present between the atoms of the molecule or ionic species.

• If we consider the examples, O₂ and N₂, the bond orders will be 1, 2, and 3 respectively. It is obvious that the bond orders correspond to the number of shared electron pairs, which is one in hydrogen (H—H), two in oxygen (O=O), and three in nitrogen (N≡N).
• Isoelectronic species have the same bond orders. Consider the example of the peroxide ion, \(\mathrm{O}_2^{2-}\) (Na₂O₂ contains a peroxide ion), and the fluorine molecule, F₂. Both \(\mathrm{O}_2^{2-}\) and F₂ have 18 electrons. The oxygen molecule contains a total of 16 electrons (2 + 6 = 8 in each oxygen atom).
• Therefore, the \(\mathrm{O}_2^{2-}\) ion has 18 electrons. The atomic number of fluorine is 9. Therefore, the F₂ molecule has 18 electrons (2 + 7 = 9 electrons in each atom). Both \(\mathrm{O}_2^{2-}\) ion and F₂ molecule are formed by sharing of one electron pair between the two atoms. Thus, both \(\mathrm{O}_2^{2-}\) and F₂ have bond order 1.
• Generally speaking, with the increase in bond order, bond enthalpy increases and bond length decreases.

Bond angle: Covalent bonds are directional in nature. The atoms constituting a molecule have a three-dimensional arrangement which can be inferred from the spectral data of that compound. The positions of constituent atoms in a molecular structure are fixed.

The position of each atom is determined by the nature of the chemical bond(s) that exist between that atom and the neighboring atoms. The angle that is formed between bonds sharing a common atom is called the bond angle. The bond angle is expressed in degrees/minute/seconds. For example, the C—H bond angle in methane is 109°28′ and the N—H bond angle in ammonia is 107°.

Concept of orbital overlap: When we say that two atoms are covalently bonded we mean that they share electrons. Take the case of the formation of the hydrogen molecule. The two isolated atoms form a molecule when their internuclear distance is d₀ and the potential energy of the two atoms together is the least possible. At this point, the electrons of the atoms are under the influence of both the nuclei. In other words, the two atoms share a pair of electrons.

• Now two atoms cannot share electrons if the electrons occupy two different regions of space. In other words, the two electrons are contained by the molecular orbital formed by the overlapping of the atomic orbitals. Tire filling of the molecular orbital must take place in accordance with the Pauli exclusion principle, i.e., since the two electrons occupy the same region of space, they must have opposite spins. The two atomic orbitals which participate in overlapping thus must be half-filled.
• This simple principle of orbital overlapping can explain not only the formation of the H₂ molecule, but also why H₃ and H₄ are not formed, or why He₂ is not formed. The hydrogen atom has a half-filled Is orbital. When two hydrogen atoms combine to form a molecule, the two Is orbitals (containing electrons with opposite spins) combine to form a molecular orbital.
• This molecular orbital gets completely filled by the two available electrons and no electron is left over. So, the H₂ molecule does not have the capacity to form bonds with more hydrogen atoms. This is why H₃ and H₄ do not exist.
• The next question is why helium exists in the atomic state, or why He₂ does not exist. The Is orbital of the helium atom is completely filled with the two electrons it contains. The Pauli exclusion principle forbids the overlapping of this orbital with the Is orbital of another helium atom. We can now summarise the orbital concept of the formation of covalent bonds.

1. Covalent bonds are formed due to the overlapping of half-filled atomic orbitals.
2. The atomic orbitals which overlap must contain electrons with opposite spins.
3. Overlapping of atomic orbitals results in a decrease of energy.

The greater the extent of overlapping, the greater is the energy released during bond formation and the stronger is the bond.

Types of covalent bonds: Covalent bonds are classified into two types on the basis of the kind of overlapping that takes place between the atomic orbitals. These are

1. The sigma (σ) bond and
2. The pi (π) bond.

Sigma (σ) bond A σ bond is said to be formed between two atoms when their orbitals overlap along the internuclear axis. In other words, the axial or head-on overlapping of the orbitals of two atoms results in the formation of a σ bond between them. The molecular orbital formed as a result of axial overlapping is symmetrical about the internuclear axis. This allows the rotation of the bonded atoms about the σ bond without breaking the bond.

The electrons constituting a σ bond are called sigma electrons. σ bonds can be formed by the overlapping of the following atomic orbitals.

s-s overlapping The s-s σ bond is formed when the half-filled (s) orbitals of two atoms overlap.

s-p overlapping The overlapping of the half-filled s orbital of one atom with the half-filled p orbital of another, leads to the formation of the s-p σ bond.

p-p overlap The p-p σ bond involves the overlapping of the half-filled p orbitals of two atoms. The first bond formed between two p orbitals is a sigma bond because these atomic orbitals can approach each other head on.

Pi (π) bond The lateral or sideways overlapping of atomic orbitals is called a π bond. The overlapping occurs in such a way that the orbitals are parallel to each other, but perpendicular to the intemuclear axis. Two electron clouds are formed in this kind of bond formation—one above and one below the plane of the atoms involved in the formation of the bond.

The electrons participating in the formation of a π bond are called π electrons. The formation of a π bond between two atoms restricts the rotation of the atoms about the π bond.

The first bond formed between two atoms is always a σ bond because the atomic orbitals are free to approach head on. The extent of overlapping is greater in a σ bond than in a π bond, so σ bonds are stronger. Pi bonds are formed in addition to sigma bonds, if two atoms share more than one pair of electrons.

Formation of N₂ and O₂ molecules The electronic configuration of the oxygen atom is \(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}_x^2, 2 \mathrm{p}_y^1, 2 \mathrm{p}_z^1\). The p_y and p_z orbitals, containing one electron each, are half-filled. One of the p orbitals overlaps axially with a half- filled p orbital of another oxygen atom and forms a p-p σ sigma bond. After the formation of the sigma bond, the remaining half-filled orbitals of the two atoms are not free to approach head on. So, these orbitals overlap sideways to form a π bond.

The electronic configuration of the nitrogen atom is \(I1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}_x^1 2 \mathrm{p}_y^1 2 \mathrm{p}_z^1\). All the three p orbitals are half-filled. The p_x orbital of one nitrogen atom overlaps with the p_x orbital of another nitrogen atom along the intemuclear axis to form a σ bond. The p_y and p_z orbitals of tire two atoms overlap sideways to form two π bonds.

Shapes Of Molecules

Molecules of different compounds exist in certain characteristic shapes. They exhibit linear, square planar, octahedral, trigonal planar, tetrahedral, or pentagonal bipyramidal geometric forms. Many physical and chemical properties of a compound depend on the shape of its molecule.

• For example, the double spiral shape of DNA, a biomolecule, is responsible for many of its properties. Covalent bonds involve the overlapping of atomic orbitals. The bond is directional and tire shared electron pairs are localised in a region between the nuclei of the bonded atoms. So, the atoms constituting the molecule occupy definite positions with respect to each other.
• This definite arrangement of atoms in a molecule is known as the geometry of the molecule and the three-dimensional model obtained by joining the points representing the bonded atoms represents the shape of the molecule.
• The study of molecular geometry is a very interesting branch of chemistry. One of the theories that explain the shapes of simple molecules is the valence shell electron pair repulsion (VSEPR) theory.

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VSEPR theory: This theory was proposed by Gillespie and Nyholm in 1957. According to them, the orientation of bonds around the central atom of a molecule depends upon the total number of electron pairs (bonding as well as nonbonding) in its valence shell. These electron pairs repel each other, so they would try to be as far away from each other as possible. In other words, the most favourable geometrical arrangement would be that in which the electron pairs are as far apart as possible in order to minimise repulsion so that the molecule can have the maximum stability. This theory rests on the following four points.

1. The electron pairs surrounding the central atom are as far apart as possible in order that the repulsion between them may be the least and the molecule may have the maximum stability.

2. The shape of the molecule depends upon the number of electron pairs (bonded and nonbonded) surrounding the central atom.

3. The repulsion between two lone pairs of electrons is the maximum and that between two bonded electron pairs is the minimum. The repulsion between electron pairs increases in the following order.

Bond pair-bond pair < bond pair-lone pair < lone pair-lone pair The difference in the force of repulsion exerted by electron pairs arises because a lone pair is under the field of influence of only one nucleus while a bond pair is under the influence of two nuclei.

4. Triple bonds cause maximum repulsion, followed by double and single bonds. Molecules have a regular geometry if the forces of repulsion between the various electron pairs around the central atom are equal. If all the electrons present in the valence shell of the central atom of a molecule participate in bonding then the central atom is surrounded only by bonded electron pairs which repel each other equally, so the molecule has a regular geometry.

If, on the other hand, the central atom has both bonded and nonbonded electrons (lone pairs) in its valence shell, the molecule has an irregular geometry. This is because the different electron pairs do not repel each other equally.

Shapes of BF₃ and NF₃ The electronic configuration of boron is 2, 3. In the BF3 molecule, the central atom (boron) has three electrons in its valence shell. It shares these with three fluorine atoms and forms three covalent bonds with these atoms. Hence, all the electron pairs in the valence shell of the central atom of the BF₃ molecule are bonded electron pairs, which repel each other equally. This is why the BF₃ molecule has a regular trigonal planar shape and its bond angle is 120°.

• The electronic configuration of nitrogen is 2, 5. There are five electrons in the valence shell of the nitrogen atom but in the NF₃ molecule it uses only three of these to form covalent bonds with three fluorine atoms.
• Thus the valence shell of the central atom of the NF₃ molecule contains 4 pairs of electrons—3 bonded pairs and a lone pair. The electron pairs surrounding the central atom repel each other unequally, so the electron pairs are arranged in the shape of a distorted tetrahedron. The bond angle (107°) is less than 109.5°, which is the bond angle of a regular tetrahedral molecule.
• Since one of the tetrahedral positions is occupied by a lone pair, the shape of the molecule is pyramidal.

Shapes of CH₄, NH₃, and H₂O In methane (CH₄), the central carbon atom (Z = 6) has four electrons in its valence shell. It uses all of them to form bonds with four hydrogen atoms. Thus, the central atom in methane is surrounded by four bonded electron pairs, which repel each other equally, resulting in a regular tetrahedral geometry.

• In the ammonia molecule, the central nitrogen atom (Z = 7) has five electrons in its valence shell. It uses only three of these to form covalent bonds with three hydrogen atoms and is left with a lone pair of electrons. Thus, of the four pairs of electrons surrounding the central atom, three are bonded and one is the lone pair.
• Had all four electron pairs been bonded, the shape would have been tetrahedral. But due to unequal repulsion, the different pairs of electrons are arranged in the shape of a distorted tetrahedron and the bond angle reduces to 107°. One of the tetrahedral positions is occupied by a lone pair, so the shape of the molecule is pyramidal.
• In the water molecule, the central oxygen atom has six electrons in its valence shell (the electronic configuration is 2, 6). It uses two of these to form bonds with two hydrogen atoms and is left with two lone electron pairs. Thus, the central atom has four pairs of electrons in its valence shell and the shape of the molecule would have been tetrahedral, had all the pairs been bonded.

But there are two bonded electron pairs and two nonbonded electron pairs which repel each other more strongly than the bonded pair. Tire electron pairs arrange themselves in the shape of a distorted tetrahedron and the angle reduces to 105°. Two of the tetrahedral positions are occupied by lone pairs, so the shape of the molecule is bent.

Shapes of PCl₃ and PCl₅ In PCl₃ the central phosphorus atom has five electrons in its valence shell (electronic configuration 2, 8, 5). It uses only three of these to make three covalent bonds and is left with a lone pair of electrons. Due to the unequal repulsion between the bonded electron pairs and the nonbonded electron pair, the arrangement of electron pairs is distorted tetrahedral with bond angle 107°. The shape of the molecule is pyramidal.

• In PCl₃, phosphorus uses all the five valence electrons to form covalent bonds with five chlorine atoms. Thus, the valence shell of the central atom of PCl₃ contains five bonded electron pairs which repel each other equally. Hence, the shape of the molecule is (regular) trigonal bipyramidal.
• Note that the bond angles in a trigonal bipyramidal arrangement arc not equal. This is one of the few cases (like a pentagonal bipyramid) where the bond angles arround an atom are not the same. Out of the five P—Cl sigma bonds, three lie in one plane and make an angle of 120° with each other.
• These are equatorial bonds. The other two P—Cl sigma bonds are oriented at right angles to the equatorial plane. One lies above and the other below the equatorial plane. In other words, the two bonds occupy axial positions and are known as axial bonds.

Shape of SF₄, SF_6, and ClF₃ In SF4, the central atom (sulfur) has six electrons in the valence shell (electronic configuration, 2, 8, 6). It uses four of these to form covalent bonds with four fluorine atoms and is left with one unshared pair of electrons. The total number of electron pairs surrounding the central atom is five, and the electron pairs arrange themselves in the shape of a trigonal bipyramid.

• In a trigonal bipyramidal structure, each axial bond experiences repulsion from three equatorial bonds at 90° and one axial bond at 180° to it, whereas each equatorial bond experiences repulsion from two equatorial bonds at 120° to it and two axial bonds at 90° to it.
• This means that the axial bond experiences greater repulsion from the other bonds. Therefore, if lone pairs are present in the molecule, they occupy equatorial positions to reduce repulsion. Also, the axial bonds are slightly longer and weaker than the equatorial bonds.

In SF₄, the unshared pair of electrons occupies the equatorial position, and due to unequal repulsion, the arrangement of electron pairs is in the shape of a distorted trigonal bipyramid (see-saw shaped).

In sulphur hexafluoride (SF₆), the central sulphur atom shares all six valence electrons to form covalent bonds with six fluorine atoms. Thus, there are six pairs of bonded electrons in the valence shell of the central atom of the molecule, which has an octahedral (regular) shape. The bond angle is 90°.

ClF₃ or chlorine trifluoride is isoelectronic with SF₄. Tire central atom of the molecule is chlorine with 7 valence electrons. It uses three electrons to form covalent bonds with three fluorine atoms and is left with four electrons or two lone pairs.

• Thus, Cl atom in the ClF₃ molecule has five electron pairs and is therefore expected to have a basic trigonal bipyramidal shape. Since all the bond angles are not the same trigonal bipyramidal is not a regular shape. Lone pairs occupy two corners and fluorine atoms occupy the other three comers. Theoretically, three different arrangements are possible for this molecule.
• As a rule, if more than one lone pair exists in a trigonal bipyramid they will be located in the equatorial position rather than the axial positions to minimise the repulsive forces. On applying this to ClF₃ molecule, we find the structure shown in Figure (T-shaped) to be the most stable, as lone pairs are in the equatorial position. Now let us summarise the shapes of few simple molecules with the central atom having one or more than one lone pair of electrons.

The VSEPR theory can also be extended to molecules with multiple bonds. For example, CO₂ and SO₂. Carbon is sp hybridised in CO₃ and the shape of the molecule is linear. Sulphur is sp² hybridised in SO₂ and the shape of the molecule is bent. The bond angle reduces only a little from 120° to 119.5°, and although bond pair-lone pair repulsion exists, the double bond occupies more space.

Covalent Bond

Ionic bonds are formed when one atom loses electrons to achieve a stable configuration and the other atom accepts those electrons to complete the octet in its outermost shell. But what about atoms that have 4 electrons in their valence shell?

• It would be difficult for such an atom to either lose or gain 4 electrons. And what happens when both the atoms are short of electrons? Neither can gain electrons from the other to complete its octet, for example, when two chlorine atoms combine to form a Cl₂ molecule.
• Lewis suggested that such atoms can complete their octets by sharing electrons. The bond formed between atoms of the same or different elements due to the mutual sharing of electrons is called a covalent bond.
• In the formation of a covalent bond between two atoms, both atoms contribute an equal number of electrons. When two atoms contribute one electron each to form a bond, this shared pair of electrons is common to both atoms.
• These two electrons which are responsible for the formation of the bond are called the bonded pair or shared pair of electrons. You have already studied about the Lewis structures earlier in the chapter. Let us draw Lewis dot structures for a few covalent molecules.

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The chlorine molecule Each chlorine atom (Z = 17) has seven electrons in its valence shell (2, 8, 7) and needs one more electron to complete its octet. In the formation of the chlorine molecule, two chlorine atoms contribute one electron each and share two electrons. In this way both the chlorine atoms achieve the stable configuration of eight electrons in their outermost shell. The sharing of electrons by two chlorine atoms or the formation of a bond between them is shown in Figure.

• The shared pair of electrons completes the outermost shells of both the bonded atoms and is present in a region between the two bonded atoms. These two electrons are attracted by the nuclei of both the atoms. Thus they are held together by a strong attractive force, and form a covalent bond.
• In the formation of a covalent bond, like the one between two chlorine atoms, there is a partial overlapping of atomic orbitals. In other words, a part of the electron cloud of each of the two half-filled atomic orbitals overlaps. Thus, the covalent bond is formed in the direction of overlapping, and unlike ionic bond, it is directional.
• All the electrons in the valence shell may or may not be involved in bonding. In the formation of the chlorine molecule, for example, there is only one shared pair of electrons. Each chlorine atom is left with three pairs of unshared electrons. Such electron pairs are called nonbonding, unshared, or lone pairs of electrons.

The Ammonia Molecule A nitrogen atom has five electrons in the outermost shell. It shares three of these and attains an octet. Hydrogen has only one electron, it needs one more electron to complete its Is orbital. The nitrogen atom and each of the hydrogen atoms contribute one electron each to the bonded pair. The electron pair left unshared on the nitrogen atom is the lone pair.

The Carbon Tetrachloride Molecule A carbon atom has four valence electrons. It shares all the four electrons with four chlorine atoms to attain an octet in the carbon tetrachloride molecule. Each of the chlorine atoms also attains an octet.

Multiple Covalent Bonds: When two atoms share only one pair of electrons, the bond formed between them is called a single covalent bond. The chlorine, ammonia, and carbon tetrachloride molecules contain single covalent bonds. There is one single covalent bond in a chlorine molecule. The ammonia and carbon tetrachloride molecules contain three and four single covalent bonds respectively.

Now, it is quite possible for two atoms to share more than one pair of electrons. When two atoms share more than one pair of electrons, the bond between the atoms is called a multiple covalent bond. When two atoms share two pairs of electrons, the bond between the atoms is called a double covalent bond and when three pairs of electrons are shared by two atoms, the bond formed between them is called a triple covalent bond.

The Oxygen Molecule The oxygen atom (Z = 8) has six electrons in its valence shell (2,6), so it needs two more electrons to complete its outermost cell. In the formation of the 02 molecule, the two oxygen atoms, therefore, contribute two electrons each and share two pairs of electrons. Thus, in the oxygen molecule, the two oxygen atoms are held by a double bond.

The Nitrogen Molecule The nitrogen atom (Z = 7) has five electrons in its valence shell (2,5) and needs three more electrons to complete its outermost shell. In the nitrogen molecule, therefore, each nitrogen atom contributes three electrons and the two atoms share six electrons, or three electron pairs. Thus, the nitrogen atoms in the N₂ molecule are held by a triple covalent bond.

The Ethene Molecule An ethene molecule is formed by bonding between two carbon atoms and four hydrogen atoms. Each of the two carbon atoms combines with two hydrogen atoms forming two single covalent bonds (by sharing two of its electrons). The remaining two electrons of each carbon atom form a double bond between e two carbon atoms.

The valency of an element in a covalent compound is known as covalency or the number of electrons which it contributes to form a covalent bond in that compound. For example, the valency of Cl in Cl₂ is 1, of O in O₂ is 2. Similarly, the covalency of C in CCl₄ is 4 and that of N in N₂ is 3.

Coordinate Covalent Bond: Certain atoms which have a complete octet and contain a lone pair of electrons can donate this pair to another atom which is short of electrons. When two atoms participate in this kind of sharing of electrons they are said to be bound by a coordinate bond. The coordinate bond can be looked upon as a special type of covalent bond.

• The difference is that in the formation of a coordinate bond, the pair of electrons is contributed by only one of the bonded atoms. The atom that donates an electron pair is called the donor, while the atom which only shares the electron pair is called the acceptor. The bond is represented by an arrow (→) pointing from the donor to the acceptor.
• From the point of view of the orbital theory, a coordinate bond involves the overlapping of an orbital (of an atom) containing a lone pair of electrons with a vacant orbital of another atom.

Formation of O₃ A molecule of oxygen contains two oxygen atoms which share two pairs of electrons to complete their octets. In the formation of ozone, one of these oxygen atoms donates a pair of electrons to a third oxygen atom, which contains only six electrons.

A combination that contains of only NH₃ and BF₃ In ammonia, nitrogen has five valence electrons. Three of these are shared with threei hydrogen atoms to form three covalent bonds. Ammonia still has a lone pair of electrons which can be donated to any electron-deficient atom or molecule like BF₃.

Formation of SO₂ Sulphur and oxygen both have six electrons in their valence shell. Each needs two more electrons to complete its octet, so they share two electrons each, thus forming a double bond with each other.

• Now the sulphur atom still has two unshared pairs of electrons. It donates one of these pairs to another oxygen atom, which is short of two electrons.
• Thus, in the sulphur dioxide molecule, there is a covalent bond between one of the oxygen atoms and the sulphur atom and a coordinate covalent bond between the other oxygen atom and the sulphur atom.
• We have used Lewis dot structures to represent bonding in molecules in this chapter. These structures help to understand bonding in molecules in terms of the number of shared pairs of electrons. Thus, we get an idea about the formation of a molecule, so that a few of its properties can be predicted.
• Remember the following basic steps while writing the Lewis dot structures of a molecule or ionic species for which the molecular formula, along with the charge (in case of ionic species) has been given.

1. Write the valence shell configurations of the combining atoms. Add their valence electrons to obtain the total number of electrons required for respresenting the structure.
2. For cations, subtract one electron for each positive charge from the total number of valence electrons. For anions add one electron for each negative charge to the total number of valence electrons.
3. Write the skeletal structure of the molecule. Generally, the least electronegative atom occupies the central position in the molecule. The hydrogen atom usually occupies a terminal position in a molecule.
4. Distribute the total number of electrons as shared pairs (for single bonds) between the atoms. The remaining electron pairs are either involved in multiple bonds or constitute lone pairs.
5. Ensure that each bonded atom gets an octet of electrons. Now, let us write the Lewis dot structures for a few molecules.

The Nitric Acid Molecule

1. To find the total number of valence electrons available for bonding in a nitric add (HNO₃) molecule, write the valence shell configurations of the combining atoms.

⇒ \(\mathrm{H}\left(1 \mathrm{~s}^1\right), \mathrm{N}\left(2 \mathrm{~s}^2 2 \mathrm{p}^3\right), \mathrm{O}\left(2 \mathrm{~s}^2 2 \mathrm{p}^4\right)\)

The total number of valence electrons is [1 + 5 + 3 x (6)] = 24.

2. The skeletal structure of HNO₃ is

You can write the above skeletal structure by considering the atomic structure and elemental properties of the combining atoms, which we have already studied in previous classes. Although hydrogen is the least electronegative atom, nitrogen is placed in the centre of the molecule. The valency of nitrogen is 3, whereas that of hydrogen is 1. For the same reason, N is bonded with three oxygen atoms, and the hydrogen atom is bonded to one of the oxygen atoms.

3. Draw a single bond between each pair of bonding atoms and complete the octets on each oxygen atom.

4. This does not complete the octet on the nitrogen atom. Therefore, we resort to multiple bonding between one of the oxygen atoms and the nitrogen atom so that each bonded atom gets an octet of electrons.

The carbonate ion \(\left(\mathrm{CO}_3^{2-}\right))\)

1. To find the total number of valence electrons available for bonding, write the valence shell configurations of the combining atoms.

∴ \(C\left(2 s^2 2 p^2\right) O\left(2 s^2 2 p^4\right)\)

The total number of valence electrons is [4 + 3 x (6)] = 22. The total number of electrons available is 24 since the carbonate anion carries two negative charges.

2. The skeletal structure of carbonate is

Here carbon is the least electronegative atom and therefore centrally located in the ion.

3. Draw a single bond between each pair of bonding atoms and complete the octet on each of the oxygen atoms.

4. In the structure, the octet of carbon is incomplete. Therefore, we resort to multiple bonding between carbon and one of the oxygen atoms.

The octet of each bonded atom is now complete in the structure of the carbonate ion.

• In the carbonate anion the net negative charge is possessed by the ion as a whole and not by any individual atom (here carbon or oxygen). It is however feasible that each atom carries a charge and the charges on each atom of the polyatomic ion sum up to give the net charge.
• Thus each atom constituting a polyatomic ion or a molecule has a formal charge. The total of formal charges in case of a polyatomic ion should be equal to the net charge on the ion. The total of the formal charges of each atom in a neutral molecule should be equal to zero. The formal charge on an atom is actually the number of electrons of that atom involved in bonding. While working out the formal charge on an atom,

1. Any nonbonding electrons associated with an atom are considered to belong to that atom, and
2. The electrons in a bond are assigned half and half to the two atoms in the bond.

• There is a difference between the number of valence electrons of an atom in the isolated state and the number of electrons assigned to that atom when it combines with other atoms in a molecule. This number is counted “by assuming that the atom in a molecule owns one electron of each shared pair and both the electrons of the lone pair.
• This number varies for the same atom in different molecules. Therefore, the formal charge on an atom will not always be the same, though the number of valence electrons in the free atom is always the same. Also, different atoms of the same element in a molecule may have different formal charges.

Consider the example of the ozone molecule. Its Lewis dot representation is

Let us calculate the formal charge on each oxygen atom in the molecule. All the oxygen atoms have six valence electrons. The central oxygen atom marked 1 has one lone pair and three bond pairs.

∴ formal charge on oxygen atom marked 1 = 6 – 2- 1/2 x 6 = +1

The oxygen atom marked 2 has two lone pairs and two bond pairs,

∴ formal charge on oxygen atom marked 2 = 6 – 4- 1/2 x 4 = 0.

The oxygen atom marked 3 has three lone pairs and one bond pair.

∴ formal charge on oxygen atom marked 3 = 6 – 6 -1/2 x 2 = -1

Since O₃ is a neutral molecule, the sum of the formal charges is zero.

In general, this simple formula can be used to calculate the formal charge (FC) on an atom in a molecule:

FC = (total number of valence electrons in the free atom) – (total number of electrons in the lone pairs) – 1/2 (total number of shared electrons)

Formal charge is calculated for the species involved in covalent bonding. If there are a number of possible Lewis structures for a molecule, then the values of the formal charges on its atoms help to determine the lowest energy structure. Generally, the structure with the smallest formal charges on the atoms is the one with lowest energy.

Properties of covalent compounds Covalent compounds, or compounds formed by covalent bonding, have the following common characteristics.

State Covalent compounds, unlike ionic compounds, exist as individual molecules in which the atoms are held together by the sharing of electrons. The intermolecular force of attraction in such compounds is generally weak. So most of these compounds exist in the liquid or gaseous state at room temperature.

Melting and boiling points Covalent compounds generally have low melting and boiling points because the inter-molecular force of attraction in such compounds is weak and not much energy is required to overcome this force.

Conductivity They are generally poor conductors of electricity because they do not contain free electrons or ions to conduct electricity.

Solubility They are usually insoluble in water because of the lack of interaction between the polar molecules of water and the nonpolar molecules of such compounds. But they dissolve in nonpolar solvents like benzene.

Molecular reactions Covalent compounds do not produce ions when dissolved. So, when such compounds react with other reagents, the reaction does not involve the combination of ions. It involves the cleavage of the covalent bond in the reacting species and the formation of new covalent bonds in the product molecules. Such reactions are naturally much slower than ionic reactions.

Some Basic Concepts of Chemistry

Chemistry is the study of the materials that make up the universe. It deals with the composition, structure, properties, and interaction of substances. The universe is made up of a large variety of materials. Chemists determine how chemical transformations occur among these substances.

Chemistry finds application in diverse areas. For instance, chemical fertilisers have helped increase crop yield to a large extent. Chemical industries also manufacture acids, alkalis, salts, drugs, dyes, soaps, detergents, alloys, polymers, etc.

• Chemistry has provided many life-saving drugs. Cisplatin and Taxol are very effective in cancer therapy, and AZT (Azidothymidine) is used to treat AIDS victims.
• In recent years chemistry has given us many new materials with specific magnetic, electric, and optical properties. These types of materials include conducting polymers, superconducting ceramics, and optical fibers.
• Environmentally hazardous refrigerants like CFCs (chlorofluorocarbons), which are responsible for the depletion of the ozone layer, have been replaced by alternatives.
• Like the study of all other sciences, the study of chemistry is based on the careful observation of various phenomena (like a reaction between substances) under controlled conditions.
• In other words, experiments form the foundation of chemistry. The observations made during an experiment may be qualitative or quantitative in nature. Quantitative observations involve the measurement of one or more quantities and are discussed later in the chapter.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

Classification Of Matter

We are often struck by the diversity of the things that surround us. The diversity of plant and animal life, the diversity of climate, and the diversity of geographical features. And yet, what is perhaps more wonderful is that all that surrounds us, in fact, everything in the entire universe, of which our surroundings comprise an infinitesimal part, is made of two things—matter and energy. Energy is something that we cannot hold in our hands or feel directly.

• We can experience its effects like when we feel the warmth of sunlight on our bodies.
• Matter, on the other hand, is something we can feel and see directly. Expressed in a more formal way, matter is anything that occupies space, has mass, offers resistance, and can be perceived of directly by our senses.
• Matter can be classified according to its physical state or according to its chemical composition. Physically, matter can exist in three states, viz., solid, liquid, and gaseous.

Chemical classification of matter: On the basis of chemical composition, matter can be broadly divided into pure substances and mixtures. Pure substances can be of two types, viz., elements and compounds, while mixtures can be homogeneous or heterogeneous. An element can be a metal, a nonmetal, or a metalloid, while a compound can be either inorganic or organic in nature. The following chart should make you more familiar with this classification of matter.

Elements, Compounds And Mixtures

The substances that we see around us are either elements (like gold), compounds (like sugar), or mixtures (like air). Molecules are the units of matter. They may be made up of two or more atoms of the same element or of different elements combined in a definite ratio (compounds).

A mixture is a combination of elements, compounds, or element(s) and compound(s) in any ratio. Elements and compounds are pure substances that have a particular set of properties. Early chemists found it very difficult to distinguish an element from a compound. It was Lavoisier who made a breakthrough and showed how elements can be distinguished from compounds.

1. Elements: An element is the simplest form of a pure substance, which can neither be decomposed into nor built from simpler substances by ordinary physical or chemical methods. This definition of an element arose out of Lavoisier’s work, about which you will leam later in this chapter.

• Of course, there were difficulties related to defining an element this way. Until a substance could be disintegrated into simpler substances, it was considered to be an element, and when someone showed that it could be broken up, it was no longer considered to be an element. For example, water was considered to be an element until Sir Humphrey Davy showed that it could be decomposed by passing an electric current through it.
• A more exact definition of an element came up after the discovery of radioactivity, which showed that elements can be broken up and also that they can be synthesised from simpler substances. The modem definition, thus, is that an element is a pure substance that contains only one kind of atoms. Mono-, di- or polyatomic molecules with the same kinds of atoms constitute elements, for example, He, O₂, N₂, and S₈.
• We know of 111 elements so far. Only 90 of these occur in nature and just 20 elements make up 99% of the Earth’s crust. Some of the most abundant elements on the Earth’s crust are oxygen, silicon, aluminum, iron, carbon, calcium, sodium, and potassium. An element can be a metal, a nonmetal, or a metalloid.

Metals With the exception of mercury, metals are generally solid at room temperature. They are malleable (can be beaten into thin sheets) and ductile (can be drawn into wires), and are good conductors of heat and electricity. Most metals have a high tensile strength and are lustrous. About 75-80 percent of the known elements are metals, gold, silver, copper, iron, and aluminum being some of the common ones.

Nonmetals These elements are poor conductors of electricity and heat. They are generally brittle (if solid), and without luster except iodine which shows a metallic luster. Carbon, sulfur, hydrogen, oxygen, and nitrogen are some common nonmetals

Metalloids These elements have some properties of metals and some of nonmetals. Some common examples of metalloids are silicon, arsenic, antimony, and selenium.

2. Compounds: A compound is a pure substance which is formed by the union of two or more elements in a definite proportion by mass and which can be decomposed into its constituent elements by suitable chemical methods. The composition of a compound is always the same regardless of the source from which it is obtained or the method by which it is produced.

• For example, water obtained from different sources (example seas, lakes, rivers, or wells) is always made of hydrogen and oxygen chemically combined in the ratio of 1: 8 by mass.
• A compound is homogeneous (uniform throughout) and has fixed properties. For example, sodium chloride is a compound of sodium and chlorine and has a fixed melting point, density, and solubility in water.
• The properties of a compound are completely different from those of its constituent elements. For example, hydrogen and oxygen are gases at room temperature, also hydrogen is combustible and oxygen is a supporter of combustion.
• These two gases combine to form water, which is liquid at room temperature and is used to extinguish fire.

The basic differences between elements and compounds are listed in Table, while some compounds and the ratios of their constituents are mentioned in Table.

Differences between elements and compounds

Some compounds and their constituents

Compounds can be of two types, inorganic and organic.

Inorganic compounds These compounds are derived from nonliving sources like rocks and minerals. Common salt, washing soda, and lime are some inorganic compounds we use in everyday life.

Organic compounds These compounds are derived from living sources like plants and animals, or their remains buried under the earth (for example, petroleum). All organic compounds, like carbohydrates, fats, oils, waxes, and proteins contain carbon.

3. Mixtures: Mixtures, like compounds, contain more than one element, but the constituents of a mixture are not present in any particular ratio. Also, the components of a mixture do not lose their identity, so the properties of a mixture are not completely different from those of its constituents.

For example, when we dissolve sugar in water, it still tastes sweet. The components of a mixture do not combine chemically. They can be separated by some physical means. Mixtures can be homogeneous or heterogeneous.

Homogeneous mixtures A mixture that has the same composition throughout is called a homogeneous mixture. Air is a homogeneous mixture of CO₂, O₂, N₂, etc. A solution of sugar in water, natural gas, and alloys, like brass, bronze, and steel, are all homogeneous mixtures. Obviously, homogeneous mixtures can be solids, liquids, or gaseous. Such mixtures are also called solutions. The components of a homogeneous mixture cannot be distinguished even under a microscope.

Heterogeneous mixtures Such mixtures do not have the same composition throughout. In other words, the components of a heterogeneous mixture are not distributed evenly. These components (called phases) can be distinguished with the naked eye, for example, a mixture of sand and sugar, a mixture of salt and pepper, or clay and water. Some mixtures look homogeneous to the naked eye, but are actually heterogeneous, for example, milk. Seen under a microscope, milk looks like a clear liquid in which droplets of fat arc suspended.

Differences between mixtures and compounds

Some of the characteristics of mixtures mentioned in Table do not apply to homogeneous mixtures like bronze. Bronze is an alloy (a solution in the solid state) of copper and zinc in a definite proportion (3:2). The properties of bronze are different from those of copper or zinc. Besides, bronze cannot be separated into copper and zinc by ordinary physical methods. And yet, bronze is a mixture and not a compound because the constituents do not combine chemically.

Properties of Matter: The characteristics of matter are called properties of matter. Such characteristics may include solubility, melting point, odour and chemical behaviour. Properties of matter can be classified as physical or chemical depending on whether the property involves a change in the chemical make-up of a substance.

• The physical properties are those characteristics that do not involve a chemical reaction. Examples are colour, odour, boiling point, solubility and electrical conductivity. The chemical properties, in contrast, describe the chemical reactions a substance will undergo. For example, sodium reacts violently with water to form sodium hydroxide solution and hydrogen gas. This is a chemical property of sodium.
• Properties can also be classified as either intensive or extensive depending on whether their values change with the size/amount of the sample. Properties like melting point and boiling point are intensive—they do not change with the change in amount of a substance. As you know, the boiling point of 1 g or 1 kg water is the same—373 K. Length and volume are extensive properties. The length and volume of an ice cube are certainly much smaller than those of an iceberg.

Measurements

When any quantity has to be measured, it is compared with a fixed standard, known as the unit of measurement. As an illustration, if we say that the height of a person is 150 centimeters (usually written as cm), it means that this height is 150 times the unit of measurement, which in this case is 1 cm. The height is measured using a scale with centimetre and millimetre markings. Thus the result of any measurement has two parts—a number (150) and a unit (cm). One does not make sense without the other.

• Different types of units have been used for the measurement of physical quantities around the world. Even in the first half of the twentieth century people used different units to measure the same physical quantity. For example, weight was measured in pounds and ounces, and seers; distance was measured in feet and yards, and furlongs and miles.
• The units used varied according to the profession, too. For example, jewellers used tolas and raties to measure weight. The use of different units to measure the same physical quantity was confusing and caused complications. Besides, these systems of units were rather cumbersome. For example, 1 stone = 14 pounds j and 1 pound —16 ounces.
• To get around these difficulties, the French Academy of Science devised the metric system in 1791. In this simple system, the different units of a physical quantity are related to each other by powers of 10, and these powers are indicated by prefixes used with the unit for the particular physical quantity. The unit of length, for example, is k the metre. The kilometer, a larger unit, is 103 meters, while the centimeter, a smaller unit, is 10-2 metre. Soon, scientists across the world adopted the metric system, and today, the whole world uses it.

The Measurement International System of Units (SI): Though almost the whole world was using the metric system by the middle of the twentieth century, scientists noticed that different metric units were being used for the same physical quantity.

• In 1960, the General Conference of Weights and Measures adopted a set of units to be used by scientists all over the world to measure six basic physical quantities.
• Mole, as a unit of amount of substance, was added in the year 1971. This system, known as the International System of Units, is popularly referred to as the SI units (after the French Sysfeme International d’ Unites).

The seven basic units from which all other units are derived are given in Table Some derived units are listed in Table, while the standard prefixes used to reduce or enlarge the basic units are mentioned in Table.

Measuring quantities: Some of the quantities which are often used in experimentation in chemistry at this level have been discussed below.
Mass The mass of a substance is the quantity of matter possessed by it.

• Though the SI unit of mass is the kilogram, it is too large to be used for many purposes in a chemical laboratory. The milligram (1 mg = 0.001 g = 10^-6 kg) and the microgram (1 μg = 0.001 mg = 10^-6 g = 10^-9 kg), are more commonly used in the laboratory.
• Mass is measured during chemical transformations. To measure with high accuracy, one needs an analytical balance. The standard kilogram is defined as the mass of a cylindrical bar of a platinum-sodium alloy stored in a vault in a suburb of Paris. Here it is worthwhile to mention that mnss is a physical property measuring the amount of matter in a substance whereas weight is the measure of the pull of gravity on that object by the Earth (or any other celestial body).
• Volume It is a derived quantity, defined as the amount of space occupied by an object. The volume is measured in ST units in cubic metres (m³). But for convenience, measurements in chemistry are usually made in cubic decimetres (1 dm³ = 0.001 m³) and cubic centimetres (1 cm³ = 0.001dm³ = 10^-6 m³). The litre is a common unit of volume in everyday life (1L = 1000 mL = 1000 cm³ = 1 dm³). In the laboratory, mostly a graduated cylinder, a volumetric flask, a burette, and pipette are used for measuring volumes of liquids.
• Density It is an intensive physical property of a substance and relates the mass of an object to its volume. The density of a substance is its mass per unit volume. The SI-derived unit of density is kg m^-3, which is too large to measure densities such as that of water (1.00 g cm^-3). Chemists often use smaller units—g cm^-3 for solids and g/mL for liquids.
  • The densities of substances change with change in temperature (since volume of a substance varies with temperature). Therefore, it is better to specify the temperature while stating the density of a substance. Temperature The three common scales to measure temperature are °C (degree Celsius), °F (degree Fahrenheit) and K (kelvin).

The temperature on the Celsius and the Fahrenheit scales are related to each other as:

∴ C =5(F-32)/9.

The magnitude of the kelvin is equal to that of the degree Celsius but the two are related as:

∴ K=°C + 273.15.

Practically speaking, the lowest temperature permitted in nature is -273.15 °C (0 K). This temperature is known as absolute zero.

Chemists measure time as it helps to know how long it takes for a chemical process or reaction to occur. Some reactions like the formation of fossil fuels take millions of years. On the other hand, when hydrochloric add is added to a solution of silver nitrate, silver chloride is precipitated in a fraction of a second.

Measurement Dimensional analysis: All derived quantities, like area, volume, density, and speed, can be derived from the seven basic quantities. The units of the derived quantities can be obtained from the seven basic units, as shown in Table. For example, if the side of a square is expressed in metres (m), the area is expressed in square metres (m²). Similarly, if the distance covered is expressed in metres and the time taken in seconds, the velocity is expressed in m s^-1.

• Quite often it becomes necessary to convert one set of units into another. This can be done quite simply by dimensional analysis. The dimensions of a derived quantity are the powers to which the basic quantities have to be raised in a product defining the quantity.
• Dimensional analysis involves calculations based on the fact that if two quantities have to be equated, they must have the same dimensions or the same units. Some calculations involving the dimensions of physical quantities follow.

Measurement Conversion of one unit into another

1. Suppose you want to convert 10 minutes into seconds.

First find the unit conversion factor.

1 min = 60 s.

∴ 1 = \(\frac{60 \mathrm{~s}}{1 \mathrm{~min}}\)

Thus, 60 s/1 min is the conversion factor. Note that the quantities dimensionless, since time is divided by time on the right-hand side.

Let us convert 10 minutes into seconds.

10 min = 10 min x \(\frac{60 \mathrm{~s}}{1 \mathrm{~min}}\) = 600 s. ….(1)

Dimensionally both sides of the Equation (1) are the same. Also, multiplying 10 minutes by (60 s/1 min) does not change its value, since (60 s/1 min) is equal to unity. A great advantage of using this method is that any mistake made in writing the conversion factor can be spotted easily.

In Equation (1) the unit minutes in the numerator and denominator of the right-hand side cancel, leaving behind the unit seconds. This would not have been so if the wrong conversion factor had been used and the mistake would have been noticed immediately. For example, if 1 min/60 s had been used as the conversion factor

10 min = 10 min \(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}=\frac{1 \mathrm{~min}^2}{6 \mathrm{~s}}\)

2. Similarly, to convert 170 pounds (lb) into kilograms (kg), first find the unit conversion factor.

1 kg = 2205 lb.

∴ 1 = \(\frac{2.205 \mathrm{lb}}{1 \mathrm{~kg}}=\frac{1 \mathrm{~kg}}{2.205 \mathrm{lb}}\)

Mass(kg) = 170 lb x \(\frac{1 \mathrm{~kg}}{2.205 \mathrm{Ib}}\) = 77.09 kg ……..(2)

If the wrong conversion factor had been used, the units on the two sides of Equation would not have been the same.

Measurement Validity of an equation: Dimensional analysis is also used to check the validity of an equation. If the dimensions on both sides of the equation do not match, the equation cannot be correct. Let us take an example.

∴ E = mc² …….(3)

Using SI units, E is in joules, m in kg and c in m s^-1. Let us see if the units on both sides of Equation (3) match.

J = kg (ms^-1)² or

or, Nm = kg m² s^-2

or kg m² s^-2 = kg m² s^-2.

Uncertainty in measurement: You can be very certain about the measurement of a quantity when it involves the counting of objects. For example, when you count the number of chairs in your class you can be certain about the exactness of your answer. You can also be sure that if someone else were to count the number of chairs, one would get the same result.

However, can you be as certain about the result you get when you measure your friend’s height with a measuring tape? You may read it as 150.1 cm and another friend may measure it as 150.3 cm. At the most you and your friend can reach the conclusion that his height is greater than 150 cm and less than 151 cm.

• There is always a certain amount of uncertainty in such measurements. This uncertainty arises not because the height of your friend (or any other measured quantity) is uncertain, but because of the way the measuring instrument is calibrated (accuracy of the scale) and because two people may record the same reading differently (human error).
• The accuracy of a measurement thus, depends on the accuracy of the measuring instrument and the skill of the person making the measurement. The difference between the two kinds of measurements that we have just considered (counting chairs and measuring a person’s height) is that in one case, we used a discrete variable and in the other we used a continuous variable.
• This means that while the number of chairs in your class can be 30 or 31 but not between 30 and 31 (discrete value), the height of your friend can be anything between 150 cm and 151 cm (continuous variable).

Measurement Precision and accuracy: If you repeat a particular measurement, you may not obtain precisely the same result (experimental error). Precision refers to the closeness of a set of values obtained from two or more measurements of the same quantity.

Accuracy, on the other hand, refers to the closeness of a single measurement to its true value. For example, three students A, B, and C separately determine the weight of the same person which is actually 50.45 kg. The results obtained are given below. (The measurements are made in kilograms.)

The values obtained by student A differ widely from one another and the average value is incorrect. Thus the data is neither precise nor accurate. The values determined by student 13 are more precise as they deviate little from one another. The average weight is still not accurate. The data obtained by student C arc both precise and accurate.

Measurement Significant figures: How does one express the uncertainty about the accuracy of a measurement while writing the result of an observation? The convention is to include all the digits which are certain and a last digit that is uncertain while writing the result of a measurement.

• The total number of digits is called the number of significant figures. It represents the accuracy and precision with which a quantity has been measured. Remember that the number of significant figures is the number of digits in the result, including the last digit which is uncertain.
• Suppose the length of a piece of wood is reported by three students as 123 cm, 123.0 cm, and 123.00 cm. Thu results may seem equivalent to you but the scientific significance of the three results is very different. The number of significant figures in the three results is 3, 4, and 5 respectively.
• In other words, in the first case (123 cm), the digits 1 and 2 are certain but not 3—it is only the best estimate. This generally means that the actual value lies between 122 and 124 or that the length is 123 ± 1 cm. When the result of a measurement is expressed in this fashion, it implies that the measurement has been made with a crude scale.
• The second result has four significant numbers, which means that the length lies between 122.9 cm and 123.1 cm. In other words, in this case, the third digit is also certain, so obviously the scale used for this measurement (123.0 cm) is more precise than the one used for the first measurement.
• The last result shows that the length lies between 122.99 cm and 123.01 cm (5 significant figures) and implies that the scale used is the most precise of the three.
• While reporting the result of a measurement, one must be very careful about indicating the precision with which the measurement has been made. To write more significant figures than the scale of measurement allows would be wrong, while writing fewer significant figures than the situation allows for would be holding back information that could be useful. It would be useful to remember the following rules in this context.

1. All digits are significant except zeros in the beginning of a number. For example, 132 cm, 0.132 cm, and 0. 0132 cm have three significant figures each. Zeros to the left of the first nonzero digit are not significant, even if such zeros follow the decimal point.
2. Zeros to the right of the decimal point are significant. Thus, 151 g, 151.0 g, and 151.00 g have 3, 4, and 5 significant figures respectively.

The following examples should help you understand these rules better.

• 678 has 3 significant figures
• 0. 42 has 2 significant figures
• 40.5 has 3 significant figures
• 2004 has 4 significant figures
• 0. 02 has one significant figure
• 0. 0022 has two significant figures
• 2.20 has three significant figures
• 0. 0200 has three significant figures

In numbers that do not contain a decimal point, trailing zeros may or may not be significant. In case of a very large number of significant figures, for instance the value of n (n = 3.1415926) one may choose tlie number of digits according to the calculation. In other words, all the numeral values involved in a calculation should be so expressed that they have almost the same number of significant figures.

Measurement Calculations involving significant figures: Several calculations may have to be made in the course of an experiment and these calculations may involve the addition, subtraction, multiplication or division of different measured quantities. The measurement of all the quantities (used to arrive at the final result of the experiment) may not be made with the same precision.

The precision of the final result depends on the precision of the least accurate of the measurements. In other words, the final result cannot be more precise than the least precise of the quantities involved in the calculations. Remembering a couple of rules will be helpful while making calculations involving quantities of different precisions.

Rule 1: In additions or subtractions the final result should be reported to the same number of decimal places as that of the term with the least number of decimal places. The number of significant figures of the different numbers have no role to play.

Example 1. The three numbers to be added have 3,1 and 2 decimal places respectively, while the significant figures in the three numbers are 4,2 and 3 respectively.
Solution:

⇒ \(\begin{aligned}
2.512 \\
2.2 \\
5.23 \\
\hline 9.942
\end{aligned}\)

∴ Actual sum 9.942

∴ Reported sum 9.9

The reported sum is 9.9 because the term with the least number of decimal places is 2.2, which has only one decimal place.

Example 2. Each number has three decimal places, so the answer is to be reported up to three decimal places. The significant figures in each of the terms is 4 but the result has 5 significant figures.
Solution:

⇒ \(\begin{array}{r}
6.612 \\
5.234 \\
2.020 \\
\hline 13.866
\end{array}\)

∴ Actual sum 13.866

∴ Reported sum 13.866

Example 3. The first number has 2 decimal places and the second has 4 decimal places, so the answer is reported up to 2 decimal places.
Solution: 

⇒ \(\begin{array}{r}
19.26 \\
(-)11.02534 \\
\hline 8.23466
\end{array}\)

∴ Actual difference 8.23466

∴ Reported difference 8.23

Rule 2: In multiplications and divisions, the result is reported to the same number of significant figures as the least precise term or the term with the least number of significant figures.

Example 1: The first term has 5 significant figures, while the second term has 3 significant figures, so the result can have only 3 significant figures.
Solution:

⇒ \(\begin{aligned}
41.012 \\
\times 1.21 \\
\hline 49.62452
\end{aligned}\)

∴ Actual product 49.62452

∴ Reported product 49.6

Example 2: In this division, 0.41 has 2 significant figures, while 15-724 has 5 significant figures. The result can have only two significant figures.
Solution:

⇒ \(\begin{aligned}
& 15.724 \\
& +0.41 \\
& \hline 0.0260747
\end{aligned}\)

∴ Actual result 0.0260747

∴ Reported result 0.026

Rule 3: If one of the term is in an expression is an exact number, the result should have the same number of significant figures as the least precise term other than the exact number. The presence of exact numberfs) in an expression atfect the number of significant figures that should be included in the answer because an exact number is supposed to have an infinite number of significant figures.

Example 1: All the terms other than the two exact numbers have 3 significant figures. So the answer should also have 3 significant figures.
Solution:

⇒ \(\frac{4.23 \times 0.141 \times 3}{0.0214 \times 2}\)

∴ Actual result 41.8058

∴ Reported reult 41.8

• Rounding off As seen in the above examples, the final result often has more figures than die number of significant figures in the least precise quantity involved in the calculations. lVhile reporting the result, only the significant figures are retained. The others are dropped. This procedure is called rounding off. The convention followed while rounding off a number can be summarised as follow’s.
• If the digit which follows the last digit to be reported is less than 5, the last digit is left unchanged and all the digits to its right are dropped. If, however, the digit following the last digit is equal to or more than 5, the last significant figure is increased by one. Thus 1.234,1277 and 12252 become 122,13 and 13 respectively, if the fire result is to be reported up to the first decimal place.
• While solving a problem remember to take up all significant digits used in calculations and round off only the final result to the desirable number of significant digits.

Scientific (exponential notation): Consider the quantity 10600 g. Written as such, it implies that there are 5 significant figures. Suppose the measurement was made in such a way that the number is precise to only 3 significant figures? Writing the result as 10600 g would then not be correct because it would not indicate the precision of the measurement To remove such ambiguity, results are expressed in the scientific notation.

Thus, 10600 g may be expressed in one of the following ways, depending on the precision of the measurement

1.06 x 10⁴ g (3 significant figures)

1.060 x 10⁴ g (4 significant figures)

1.0600 x 10⁴ g (5 significant figures)

• You are familiar with the Avogadro number, which is 6.022 x 10²³. If you write it in the ordinary way and not in terms of 10 raised to a power, it will be 602,213,700 000 000 000 000 000. It is difficult to write such numbers in an ordinary way and errors may creep in while doing so. The scientific or exponential notation is a better way to represent such numbers.
• In scientific notation, all numbers, large or small, are expressed as a number between 1.000 and 9.999 multiplied or divided by 10 an appropriate number of times. For example, the number 2484.32 may be expressed as 2.48432 x 10 x 10 x 10 or 24832 x 10³. Here 3 is the power or exponent, to which 10 is raised. In general in scientific notation a number is represented as N x 10ⁿ,

• where N is a number between 1.000 and 9.999 and n is a number (not necessarily a single digit) called exponent. To express a number smaller than 1.000 in scientific notation, the decimal point is moved to the right until there is only one nonzero digit before the decimal point.
• The number is then divided by 10 an appropriate number of times. For example, to express 0.00032481 in scientific notation, we move the decimal point to the right. The number is divided 4 times by 10, as the decimal has shifted four places. In this example, the exponent n = -4.
• Take another example of a number larger than 9.999. To express it in scientific notation the decimal point is moved to the left until there is only one nonzero digit before the decimal point. Suppose the number is 14872.5. When this number is expressed in scientific notation, the decimal point is moved to the left and the number thus obtained is multiplied by 10.

⇒ 00032481 = \(\frac{3.2481}{10 \times 10 \times 10 \times 10}\) =3.2481 x 10^-4

⇒ 14872.5 = \(1.48725 \times 10 \times 10 \times 10 \times 10=1.48725 \times 10^4\)

∴ The exponent n = 4.

To subtract or add numbers in scientific notation the exponent n or power of 10 should be the same in all the numbers. If the exponent is not the same, it has to be made the same by shifting the decimal place in any of the numbers before adding or subtracting. For example, suppose we have to add 6.426 x 10³ and 2.045 x 10⁴. We can transform 6.426 x 10³ to 0.6426 x 10⁴ and then add,

∴ 2.045 x 10⁴ + 0.6426 x 10⁴ = (2.045 + 0.6426) x 10⁴ = 2.6876 x 10⁴.

To multiply two numbers in scientific notation, we make use of the relation, (10)^x x(10)^y =10^(x+y).

In this case, the exponent need not be the same. For example, (2.0456 x 10⁴) (4.132 x 10^-7) = 8.452 x 10^[4+(-7)] = 8.452 x 10^-3.

To divide two numbers in scientific notation, we make use of the relation

∴ \(\frac{10^x}{10^y}=10^{x-y}\)

The exponent need not be the same here too. For example,

∴ \(\frac{2.45 \times 10^{14}}{9.24 \times 10^{24}}=\frac{2.45}{9.24} \times 10^{14-24}=0.265 \times 10^{-10}\)

In scientific notation, the number is expressed as 2.65 x 10^-11. In similar example,

∴ \(\frac{4.65 \times 10^{-4}}{2.92 \times 10^{-10}}=\frac{4.65}{2.92} \times 10^{[-4-(-10)]}=1.59 \times 10^{(-4+10)}=1.59 \times 10^6\)

Historical Approach To The Particulate Nature Of Matter

The concept that matter is made up of tiny bits of material—particles—originated in Greek natural philosophy. The Greek philosopher Democritus in the fifth century BC proposed the “particulate nature of matter” He believed that matter is composed of indivisible particles called atoms (from the Greek atoms, meaning indivisible).

However, now we know that atom is divisible into smaller subatomic particles like protons, neutrons, and electrons. Nevertheless, the word “atom” as indivisible still makes sense as once split, an atom loses its identity. Also, it is the smallest particle of an element that takes part in a chemical combination.

• The atomic theory of Democritus was not appreciated for many years. Aristotle’s theory that matter is composed of four elements—air, wind, fire, and water—was widely accepted. However, after a series of experiments in more recent times, the particulate nature of matter was established.
• The French chemist Antoine Laurent Lavoisier, in the late 1700s, discovered that during a chemical reaction, the components involved change in terms of appearance and form only, and the total mass remained the same. In other words, the total mass of the reactants is equal to that of the products (law of conservation of mass).
• This prompted him to conclude that some basic part of the components which was not visible, did not change. Later, in 1800, scientists working on chemical reactions found that something ruled the behaviour of matter that even their microscopes could not perceive.
• The law of conservation of mass helped establish the law of constant composition or the law of definite proportions. These laws were a result of Joseph Proust’s extension of Lavoisier’s work.
• In 1803, John Dalton proposed the atomic theory of matter on the basis of the laws of chemical combination and other related chemical observations. The theory stated that all matter is made of indivisible and indestructible ultimate particles called atoms. Thus the start of the nineteenth century brought back Democritus’ view to the forefront of science.

Percentage Composition And Molecular Formula

To study a chemical compound, mainly its chemical properties, it is essential to know its chemical formula. The chemical formula of a compound can be determined by analyzing the compound for the amount of elements (moles) in a given mass of the compound. The result may also be expressed in percentage composition.

The mass percentage composition of a compound is defined as the number of grams of different elements present in 100 g of the compound. The mass percentage of an element may be calculated by dividing the mass of the element in one mole of the compound by the molar mass of the compound and multiplying the result by 100.
Let us consider CO₂.1 mol of carbon dioxide will always contain 1 mol of carbon and 2 mol of oxygen atoms.

The molar mass of CO₂ is 12 + (2 x 16) = 44.0 g.

Now 44.0 g of CO₂ contains 12.0 g of C and 32.0 g of O.

Mass percentage of C in CO₂ = \(\frac{\text { mass of } \mathrm{C} \text { in } 1 \mathrm{~mole} \mathrm{CO}_2}{\text { molar mass of } \mathrm{CO}_2} \times 100\)

= \(\frac{12.0 \mathrm{~g}}{44.0 \mathrm{~g}} \times 100=27.27\)

Similarly, we may calculate the mass percentage of O in 1 mole of CO₂.

Mass percentage of O in CO₂ = \(\frac{32.0 \mathrm{~g}}{44.0 \mathrm{~g}}\) x 100 = 72.73.

The molecular composition of a compound can be expressed in any of the following ways.

1. A chemical formula giving the number of atoms of each type per molecule, i.e., CO₂.
2. The number of moles of each element per mole of a compound.
3. The mass of each element per 100 g of a compound.

We may conclude that if the formula of a compound is known we can determine its percentage composition and vice versa.

Empirical Formula And Molecular Formula

Tire empirical formula of a compound represents the simplest whole-number ratio of the atoms of the various elements present in a molecule. For example, the empirical formula of benzene is CH. This indicates that carbon and hydrogen are present in benzene in a the ratio 1 :1, The empirical formula of ethanoic acid is CH₂O which means that that a molecule of ethanoic acid contains carbon, hydrogen, and oxygen atoms in the ratio 1:2:1

The molecular formula of a compound, on the other hand, shows the actual number of atoms of the various elements present in one molecule of the compound. For example, the molecular formula of benzene is C₆H₆, which means that one molecule of benzene contains 6 atoms of carbon and 6 atoms of hydrogen. Similarly, the molecular formula of ethanoic acid is C₂H₄O₂, which indicates that 1 molecule of ethanoic acid contains 2 atoms of carbon, 4 atoms of hydrogen, and 2 atoms of oxygen.

The molecular formula of a compound is a simple whole-number multiple of its empirical formula. This can be expressed mathematically as follows.

Molecular formula = n x empirical formula

∴ where n is an integer. The value of n can be determined from the relation

∴n = \(\frac{\text { molecular mass }}{\text { empirical formula mass }}\)

For example, the molecular mass of glucose (C₆H₁₂O₆) is 180 and its empirical formula (CH₂O) mass is 30.

∴ n = \(\frac{180}{30}\)

When the value of n = 1, the empirical formula and molecular formula are the same.

Determination of empirical formula: The empirical formula of a compound can be determined from the percentage of the different elements present in it and their atomic masses. Such calculations involve the following steps.

1. Divide the percentage of each element by its atomic mass. This will tell you the relative number of moles of the elements present in a molecule of the compound.
2. Divide the quotients obtained by the smallest in value to get a simple ratio of moles of various elements present.
3. If the results obtained in step 2 are not whole numbers, raise the values to the nearest whole numbers or multiply all of them by a suitable integer to obtain whole numbers.
4. Write the symbols of the elements side by side and insert the corresponding numerical values you have obtained in step 3 at the lower right-hand comer of each symbol. This is the empirical formula of the compound.

Determination of molecular formula: The determination of the molecular formula of a compound involves the following steps.

1. Determine the empirical formula as described above.
2. Find the empirical formula mass by adding the atomic masses of the atoms in the empirical formula.
3. Divide the molecular mass of the compound (determined by a suitable method) by the empirical formula mass to obtain n.
4. Multiply the empirical formula by n to obtain the molecular formula.

Example 1. The percentages of carbon and hydrogen in an organic compound are 92.5 and 7.5. Determine the molecular formula of the compound if its molecular mass is 78.
Solution:

Thus the empirical formula of the compound is CH.

Therefore, its empirical formula mass = 12×1 + 1×1 = 13

Given that its molecular mass is 78.

∴ n = \(\frac{\text { molecular mass }}{\text { empirical formula mass }}=\frac{78}{13}=6\)

Therefore, the molecular formula of the compound is 6 x (CH) = C₂H₆.

Example 2. Calculate the molecular formula of a compound on the basis of the following data.

1. 4.24 mg of an organic compound yielded 8.45 mg of CO₂ and 3.46 mg of H₂O on combustion.
2. The moleadar mass of the compound is 88 g.

Solution :

1. Calculate the percentage of carbon in the compound as follows.

44 mg of CO₂ contains 12 mg of C.

Therefore, 8.45 mg of CO₂ contains 12/44 x 8.45 mg of carbon.

∴ percentage of carbon =\(\frac{\text { weight of carbon }}{\text { weight of compound }}\) x 100

= \(\frac{12}{44}\) x \(\frac{8.45}{4.24}\) x 100 = 543.

2. Calculate the percentage of hydrogen in the compound as follows.

18 mg of H₂O contains 2 mg of hydrogen.

Therefore, 3.46 mg of H₂O contains 2/18 x 3.46 mg of hydrogen.

∴ percentage of hydrogen = \(=\frac{\text { weight of hydrogen }}{\text { weight of compound }}\) x 100

= \(\frac{2}{18}\) x \(\frac{3.46}{4.24}\) x 100 = 9.0

3. Percentage of oxygen = 100 – (54.3 + 9.0) = 36.7.

4. Determine the empirical formula as usual.

Thus the empirical formula of the compound is C₂H₄O.

∴ empirical formula mass = 2 x 12 + 4 x 1 +16 = 44 g.

Given that the molecular mass of the compound is 88 g.

∴ n = \(\frac{88}{44}\)= 2.

∴ molecular formula = 2(C₂H₄O) = C₄H₈O₂.