Chemical Bonding & Molecular Structure Class 11 · Chapter 4 · NCERT Exercise Solutions

Q1. Explain the formation of a chemical bond.

Solution

A chemical bond is the attractive force that holds atoms together in a molecule or an ionic compound. Atoms form chemical bonds in order to attain a more stable electronic configuration, usually similar to that of the nearest noble gas. This stability is achieved by completing the outermost shell with electrons, which lowers the overall energy of the system.

According to modern atomic theory, most atoms are unstable in their isolated state because their valence shells are incomplete. To overcome this instability, atoms interact with one another by either transferring electrons or sharing electrons. This interaction leads to bond formation.

The process can be understood step by step. Consider two atoms approaching each other. As they come closer, attractive forces between nuclei and electrons increase, while repulsive forces between like charges also act. A bond is formed when the net attractive force becomes greater than the repulsive force and the total energy of the system reaches a minimum.

Mathematically, this stabilization may be represented as:

\[ \begin{aligned} \text{Initial state:} \quad & \text{Separate atoms (higher energy)} \\ \text{Interaction:} \quad & \text{Attraction} > \text{Repulsion} \\ \text{Final state:} \quad & \text{Bonded atoms (lower energy)} \end{aligned} \]

There are two common ways in which atoms achieve stability. In ionic bonding, one atom loses electron(s) while another gains them, forming oppositely charged ions that attract each other electrostatically. In covalent bonding, atoms share one or more pairs of electrons so that each atom effectively completes its valence shell.

Thus, the formation of a chemical bond is essentially an energy-lowering process driven by the tendency of atoms to achieve a stable electronic arrangement. The bonded state is always more stable than the separated atoms, which is why chemical bonds form naturally.

Q2. Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.

Solution

Lewis dot symbols represent the valence electrons of an atom by dots placed around the chemical symbol of the element. Since chemical bonding mainly involves valence electrons, these symbols help us understand how atoms combine to form molecules. The number of dots is equal to the number of electrons present in the outermost shell of the atom.

To write the Lewis symbols of the given elements, we first determine the number of valence electrons from their electronic configurations.

Q3. Write Lewis symbols for the following atoms and ions: \(\mathrm{S}\) and \(\mathrm{S^{2–}}\); Al and \(\mathrm{Al^{3+}}\); H and \(\mathrm{H^-}\)

Solution

Lewis symbols show the number of valence electrons present in an atom or ion by placing dots around its chemical symbol. For ions, the dots correspond to the electrons present after loss or gain, and the entire symbol is written with the appropriate charge.

First, we determine the valence electrons for each species using their electronic configurations.

Q4. Draw the Lewis structures for the following molecules and ions : \(\mathrm{H_2S,\ SiCl_4,\ BeF_2,\ CO_3^{2−},\ HCOOH}\)

Solution

Lewis structures are drawn by counting total valence electrons, arranging atoms with the least electronegative atom (except H) at the centre, completing octets (or duplets for hydrogen), and finally checking formal charges. Let us construct each structure step by step.

First, we calculate total valence electrons for every molecule or ion.

Q5. Define octet rule. Write its significance and limitations.

Solution

The octet rule states that atoms tend to combine in such a way that each atom attains eight electrons in its outermost shell, thereby achieving a stable electronic configuration similar to that of the nearest noble gas. This rule is mainly applicable to elements of the second period and explains why atoms form chemical bonds.

The basic idea of the octet rule can be represented schematically as:

\[ \begin{aligned} \text{Isolated atom} & \rightarrow \text{Incomplete valence shell} \\ \text{During bonding} & \rightarrow \text{Loss, gain, or sharing of electrons} \\ \text{After bonding} & \rightarrow \text{Eight electrons in valence shell} \end{aligned} \]

The significance of the octet rule lies in its ability to explain the formation of ionic and covalent bonds. It helps in predicting valency and provides a simple framework for writing Lewis structures of molecules and ions. Using this rule, one can understand why sodium forms Na⁺, chlorine forms Cl⁻, and why atoms like carbon generally form four covalent bonds. Thus, the octet rule serves as a fundamental guideline for explaining chemical combination and molecular stability.

However, the octet rule is not universal and has several important limitations. Many molecules exist in which the central atom contains fewer than eight electrons, such as in BeCl₂ or BF₃, where incomplete octets are observed. There are also compounds like PF₅ and SF₆ in which the central atom accommodates more than eight electrons, showing expanded octets. In addition, molecules containing odd numbers of electrons, such as NO and NO₂, cannot satisfy the octet rule completely. Furthermore, the rule fails to explain the shapes of molecules, bond energies, and magnetic properties.

Therefore, while the octet rule provides a useful first approximation for understanding chemical bonding, it cannot explain all bonding situations and must be supplemented by more advanced concepts such as hybridisation and molecular orbital theory.

Q6. Write the favourable factors for the formation of ionic bond.

Solution

An ionic bond is formed when one atom transfers one or more electrons to another atom, resulting in the formation of oppositely charged ions that are held together by electrostatic attraction. The formation of an ionic bond depends on certain favourable conditions that make this electron transfer energetically possible.

These conditions can be understood by considering the energy changes involved in the process.

\[ \begin{aligned} \text{Metal atom} & \rightarrow \text{Metal}^{+} + e^- \quad (\text{ionisation energy}) \\ \text{Non-metal} + e^- & \rightarrow \text{Non-metal}^{-} \quad (\text{electron gain enthalpy}) \\ \text{M}^{+} + \text{X}^{-} & \rightarrow \text{MX} \quad (\text{lattice energy released}) \end{aligned} \]

The first favourable factor is a low ionisation energy of the metal atom. When the ionisation energy is small, the metal can easily lose electrons to form positive ions. This is why alkali and alkaline earth metals readily participate in ionic bonding.

The second important factor is a high (negative) electron gain enthalpy of the non-metal atom. A non-metal with strong attraction for electrons can easily accept electrons and form negative ions. Halogens, for example, readily gain electrons because of their high electron affinity.

The third and most decisive factor is high lattice energy. After the ions are formed, they arrange themselves in a crystal lattice. If a large amount of energy is released during this process, known as lattice energy, it compensates for the energy spent in forming ions and makes the overall process energetically favourable.

Thus, an ionic bond is most easily formed when the metal has low ionisation energy, the non-metal has high electron affinity, and the resulting ionic compound possesses high lattice energy. When these conditions are satisfied, the total energy of the system decreases and a stable ionic compound is produced.

Q7. Discuss the shape of the following molecules using the VSEPR model: \(\mathrm{BeCl_2,\ BCl_3,\ SiCl_4,\ AsF_5,\ H_2S,\ PH_3}\)

Solution

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, the shape of a molecule depends on the number of electron pairs (bonding as well as lone pairs) present around the central atom. These electron pairs arrange themselves in space in such a way that repulsion among them is minimum, leading to a definite molecular geometry.

Let us determine the shapes of the given molecules by counting electron pairs around the central atom.

\[ \begin{aligned} &\text{BeCl}_2 : &\text{Be has 2 bonding pairs and 0 lone pairs} \\ &\text{BCl}_3 : &\text{B has 3 bonding pairs and 0 lone pairs} \\ &\text{SiCl}_4 : &\text{Si has 4 bonding pairs and 0 lone pairs} \\ &\text{AsF}_5 : &\text{As has 5 bonding pairs and 0 lone pairs} \\ &\text{H}_2\text{S} : &\text{S has 2 bonding pairs and 2 lone pairs} \\ &\text{PH}_3 : &\text{P has 3 bonding pairs and 1 lone pair} \end{aligned} \]

In BeCl₂, the central beryllium atom is surrounded by two bonding pairs and no lone pairs. These two electron clouds repel each other and arrange themselves at an angle of 180°, giving the molecule a linear shape.

In BCl₃, boron is surrounded by three bonding pairs and no lone pairs. The electron pairs lie in one plane and orient themselves at 120° to each other, resulting in a trigonal planar geometry.

In SiCl₄, silicon has four bonding pairs and no lone pairs. The four electron pairs arrange themselves symmetrically in three-dimensional space to minimize repulsion, producing a tetrahedral shape with bond angles close to 109.5°.

In AsF₅, arsenic is surrounded by five bonding pairs and no lone pairs. These five electron clouds adopt a trigonal bipyramidal arrangement, with three fluorine atoms in the equatorial plane and two in axial positions.

In H₂S, sulfur has two bonding pairs and two lone pairs. Although the electron pair geometry is tetrahedral, the presence of two lone pairs increases repulsion and compresses the bond angle, giving the molecule a bent or V-shaped structure.

In PH₃, phosphorus has three bonding pairs and one lone pair. The overall electron pair geometry is tetrahedral, but due to the lone pair, the molecular shape becomes trigonal pyramidal.

Thus, BeCl₂ is linear, BCl₃ is trigonal planar, SiCl₄ is tetrahedral, AsF₅ is trigonal bipyramidal, H₂S is bent, and PH₃ is trigonal pyramidal. These shapes arise from the tendency of electron pairs to stay as far apart as possible in accordance with the VSEPR model.

Q8. Although geometries of \(\mathrm{NH_3}\) and \(\mathrm{H_2O}\) molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Solution

Both NH₃ and H₂O have distorted tetrahedral geometries because the central atom in each molecule is surrounded by four electron pairs. According to VSEPR theory, these electron pairs arrange themselves to minimise mutual repulsion, giving a basic tetrahedral electron pair geometry.

Let us first count the bonding and lone pairs on the central atoms.

\[ \begin{aligned} \text{NH}_3 &: 3 \text{ bonding pairs} + 1 \text{ lone pair} \\ \text{H}_2\text{O} &: 2 \text{ bonding pairs} + 2 \text{ lone pairs} \end{aligned} \]

In ammonia, nitrogen has three bonding pairs and one lone pair. The lone pair exerts a stronger repulsive force than bonding pairs, pushing the N–H bonds slightly closer together. As a result, the H–N–H bond angle decreases from the ideal tetrahedral value of 109.5° to about 107°.

In water, oxygen has two bonding pairs and two lone pairs. Here, the repulsion is even stronger because two lone pairs are present. Lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion, which in turn is greater than bond pair–bond pair repulsion:

\[ \text{LP–LP} > \text{LP–BP} > \text{BP–BP} \]

Due to the presence of two lone pairs on oxygen, the bonding pairs are pushed much closer together than in ammonia. This increased repulsion compresses the H–O–H bond angle further, reducing it to about 104.5°.

Therefore, although both NH₃ and H₂O possess distorted tetrahedral shapes, the bond angle in water is smaller than in ammonia because water contains two lone pairs on the central atom, while ammonia has only one. The greater lone pair repulsion in H₂O leads to a larger angular compression.

Q9. How do you express the bond strength in terms of bond order ?

Solution

Bond strength refers to the stability of a chemical bond and is commonly measured in terms of bond energy, which is the energy required to break a bond. One of the most useful qualitative measures of bond strength is bond order. Bond order represents the number of bonds between two atoms and indicates how strongly the atoms are held together.

In molecular orbital theory, bond order is defined mathematically as:

\[ \boxed{\bbox[indigo, 5pt]{\text{Bond order}=\frac{N_b-N_a}{2}}} \]

where \(N_b\) is the number of electrons present in bonding molecular orbitals and \(N_a\) is the number of electrons present in antibonding molecular orbitals.

The relationship between bond order and bond strength can be expressed stepwise as:

\[ \begin{aligned} \color{red}\text{Higher bond order} & \;\color{red}\Rightarrow\; \color{cyan}\begin{cases} \text{greater electron density between nuclei} \\ \text{stronger attraction between atoms} \\ \text{higher bond energy} \\ \text{greater bond strength} \end{cases} \end{aligned} \]

Thus, as the bond order increases, the bond becomes shorter and stronger. For example, a triple bond is stronger than a double bond, and a double bond is stronger than a single bond. Conversely, a decrease in bond order weakens the bond.

Therefore, bond strength is directly proportional to bond order. A larger bond order signifies a stronger and more stable bond, while a smaller bond order indicates a weaker bond that is easier to break.

Q10. Define the bond length.

Solution

Bond length is defined as the average distance between the nuclei of two bonded atoms in a molecule. It represents the equilibrium position at which the attractive and repulsive forces between the atoms are balanced, resulting in maximum stability of the bond.

When two atoms approach each other, the forces acting between them change continuously. This behaviour can be understood stepwise.

\[ \begin{aligned} \text{Atoms far apart} & \Rightarrow \text{Negligible interaction} \\ \text{Atoms approach} & \Rightarrow \text{Attractive forces dominate} \\ \text{Minimum energy point} & \Rightarrow \text{Stable bond formed} \\ \text{Internuclear distance at this point} & \Rightarrow \text{Bond length} \end{aligned} \]

Thus, bond length corresponds to the distance between the nuclei at which the potential energy of the bonded atoms is minimum. Any decrease or increase from this distance results in increased repulsion or reduced attraction, making the bond less stable.

Bond length depends on factors such as the size of the bonded atoms, the bond order, and the nature of the bond. Generally, a higher bond order leads to a shorter bond length because of greater electron density between the nuclei.

Therefore, bond length is an important structural parameter that provides valuable information about the strength and nature of a chemical bond in a molecule.

Q11. Explain the important aspects of resonance with reference to the \(\mathrm{CO_3^{2−}}\) ion.

Solution

Resonance is a concept used to describe molecules or ions that cannot be represented accurately by a single Lewis structure. Instead, two or more equivalent structures, called canonical forms, are written, and the actual structure is considered to be a hybrid of these forms. This hybrid has lower energy and greater stability than any individual contributing structure.

The carbonate ion, CO₃²⁻, provides a clear illustration of resonance. First, we count the total number of valence electrons.

\[ \begin{aligned} \text{C} &: 4 \\ 3 \times \text{O} &: 3 \times 6 = 18 \\ \text{Extra electrons due to }2- &: 2 \\ \text{Total} &: 24 \end{aligned} \]

When a Lewis structure is drawn using these 24 electrons, one C=O double bond and two C–O single bonds are obtained. However, any one of the three oxygen atoms can form the double bond with carbon, leading to three equivalent canonical structures.

\[ \begin{aligned} &\begin{array}{c} \ce{O^- - C = O} \\ \big| \\ \ce{O^-} \end{array} \quad \leftrightarrow \quad \begin{array}{c} \ce{O = C - O^-} \\ \big| \\ \ce{O^-} \end{array} \quad \leftrightarrow \quad \begin{array}{c} \ce{O^- - C - O^-} \\ \big\Vert \\ \ce{O} \end{array} \end{aligned} \]

None of these individual structures represents the real carbonate ion. The actual CO₃²⁻ ion is a resonance hybrid of all three, in which all C–O bonds are identical and intermediate between a single and a double bond. Experimentally, each C–O bond has the same length and strength, confirming this delocalisation of electrons.

The important aspects of resonance highlighted by CO₃²⁻ are that the ion does not oscillate between structures, the canonical forms differ only in electron arrangement and not in atomic positions, and the true structure is more stable than any single form. The negative charge is also evenly distributed over the three oxygen atoms rather than being localised on any one atom.

Thus, resonance explains the equal bond lengths, enhanced stability, and charge distribution in the carbonate ion by showing that its real structure is a hybrid formed due to delocalisation of π electrons over the entire ion.

Q12. \(\mathrm{H_3PO_3}\) can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing \(\mathrm{H_3PO_3}\) ? If not, give reasons for the same.

Solution

No, the two given structures of H₃PO₃ cannot be treated as canonical forms of a resonance hybrid.

To understand this, recall that resonance structures differ only in the arrangement of electrons, while the positions of atoms remain exactly the same. In other words, during resonance, only π electrons or lone pairs shift; σ bonds and atomic skeletons do not change.

In H₃PO₃, the experimentally established structure contains two O–H bonds and one P–H bond. When the two proposed structures are examined carefully, it is observed that one structure shows hydrogen attached to phosphorus, whereas the other shows hydrogen attached to oxygen.

This difference may be represented schematically as:

\[ \begin{aligned} \text{Structure 1:} & \quad \text{P–H bond present} \\ \text{Structure 2:} & \quad \text{O–H bond present} \end{aligned} \]

Here, a hydrogen atom shifts from phosphorus to oxygen. This involves breaking and forming σ bonds and changing atomic positions, which violates the basic requirement of resonance.

Therefore, these two structures are not resonance forms but represent entirely different molecular arrangements. Resonance is not possible because the nuclei of atoms occupy different positions in the two structures.

Hence, H₃PO₃ cannot be described as a resonance hybrid of the given structures. The correct structure contains one P–H bond and two O–H bonds, and this arrangement is fixed.

Q13. Write the resonance structures for \(\mathrm{SO_3}\) , \(\mathrm{NO_2}\) and \(\mathrm{NO+3^−}\) .

Solution

Resonance structures are written when a single Lewis structure cannot adequately represent the bonding in a molecule or ion. In such cases, two or more canonical forms differing only in the arrangement of electrons are drawn, while the positions of the atoms remain unchanged.

For sulphur trioxide, SO₃, the total number of valence electrons is calculated as follows:

\[ \begin{aligned} \text{S} &: 6 \\ 3 \times \text{O} &: 3 \times 6 = 18 \\ \text{Total} &: 24 \end{aligned} \]

When the Lewis structure is drawn, one S=O double bond and two S–O single bonds can be written. However, the position of the double bond can shift among the three oxygen atoms, giving three equivalent resonance forms.

\[ \begin{aligned} &\begin{array}{c} \ce{O = S - O^-} \\ \big| \\ \ce{O^-} \end{array} \qquad \begin{array}{c} \ce{O^- - S = O} \\ \big| \\ \ce{O^-} \end{array} \qquad \begin{array}{c} \ce{O^- - S - O^-} \\ \big\Vert \\ \ce{O} \end{array} \end{aligned} \]

Thus, the actual structure of SO₃ is a resonance hybrid in which all S–O bonds are equivalent.

For nitrogen dioxide, NO₂, the total number of valence electrons is:

\[ \begin{aligned} \text{N} &: 5 \\ 2 \times \text{O} &: 12 \\ \text{Total} &: 17 \end{aligned} \]

Since the molecule has an odd number of electrons, it is a free radical. Two resonance structures are possible in which the position of the N=O double bond alternates between the two oxygen atoms, while one unpaired electron remains on nitrogen.

\[ \ce{O = N - O^{\cdot}} \qquad \ce{O^{\cdot} - N = O} \]

The real structure is a hybrid of these forms with partial double bond character in both N–O bonds.

For the nitrate ion, NO₃^-, the total valence electrons are:

\[ \begin{aligned} \text{N} &: 5 \\ 3 \times \text{O} &: 18 \\ \text{Extra electron due to } -1 &: 1 \\ \text{Total} &: 24 \end{aligned} \]

The Lewis structure shows one N=O double bond and two N–O single bonds carrying negative charges. The double bond can be placed with any of the three oxygen atoms, giving three equivalent resonance forms.

\[ \begin{aligned} &\begin{array}{c} \ce{O = N - O^-} \\ \big| \\ \ce{O^-} \end{array} \qquad \begin{array}{c} \ce{O^- - N = O} \\ \big| \\ \ce{O^-} \end{array} \qquad \begin{array}{c} \ce{O^- - N - O^-} \\ \big\Vert \\ \ce{O} \end{array} \end{aligned} \]

Q14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.

Solution

Lewis symbols are used to represent valence electrons as dots around atomic symbols. During ionic bond formation, electrons are transferred from metal atoms to non-metal atoms, producing positively charged cations and negatively charged anions. Let us show this electron transfer for each pair.

First, we note the valence electrons of the atoms involved.

\[ \begin{aligned} \text{K} &: 1 \\ \text{S} &: 6 \\ \text{Ca} &: 2 \\ \text{O} &: 6 \\ \text{Al} &: 3 \\ \text{N} &: 5 \end{aligned} \]

In potassium and sulphur, each potassium atom donates one electron, while sulphur requires two electrons to complete its octet. Hence, two potassium atoms transfer electrons to one sulphur atom.

Thus, two K⁺ ions and one S²⁻ ion are formed.

In calcium and oxygen, calcium loses two electrons and oxygen gains two electrons, leading to the formation of Ca²⁺ and O²⁻.

Here, calcium attains a noble gas configuration by losing two electrons, while oxygen completes its octet by gaining them.

In aluminium and nitrogen, aluminium has three valence electrons and nitrogen needs three electrons to complete its octet. Therefore, one aluminium atom transfers three electrons to one nitrogen atom.

As a result, Al³⁺ and N³⁻ ions are produced.

In each case, electron transfer leads to the formation of oppositely charged ions, which attract each other electrostatically to form stable ionic compounds.

Q15. Although both \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\) are triatomic molecules, the shape of \(\mathrm{H_2O}\) molecule is bent while that of \(\mathrm{CO_2}\) is linear. Explain this on the basis of dipole moment.

Solution

Although both CO₂ and H₂O are triatomic molecules, their molecular shapes are different because of the way bond dipoles combine in each case. This difference can be explained clearly using the concept of dipole moment.

Dipole moment arises due to separation of positive and negative charges in a molecule. It is a vector quantity and depends not only on the magnitude of individual bond dipoles but also on their directions in space.

Let us examine both molecules stepwise.

\[ \begin{aligned} \text{CO}_2 &: \text{O=C=O (linear arrangement)} \\ \text{H}_2\text{O} &: \text{H–O–H (bent arrangement)} \end{aligned} \]

In CO₂, carbon forms two double bonds with oxygen atoms. Each C=O bond is polar, but the molecule is linear with a bond angle of 180°. The two equal bond dipoles act in exactly opposite directions and cancel each other.

\[ \begin{aligned} \mu_{\text{net}} &= \mu_1 - \mu_2 = 0 \end{aligned} \]

As a result, the net dipole moment of CO₂ is zero. This complete cancellation is possible only when the molecule is linear and symmetrical, confirming the linear shape of CO₂.

In H₂O , oxygen forms two single bonds with hydrogen and also possesses two lone pairs. Due to lone pair–bond pair repulsions, the molecule adopts a bent geometry with a bond angle of about 104.5°. Each O–H bond is polar, and because the molecule is bent, the bond dipoles do not cancel.

\[ \begin{aligned} \mu_{\text{net}} &= \mu_1 + \mu_2 \neq 0 \end{aligned} \]

Hence, water has a significant net dipole moment. The presence of this permanent dipole moment proves that H₂O cannot be linear and must be bent.

Therefore, even though both molecules are triatomic, CO₂ is linear because its bond dipoles cancel completely, while H₂O is bent because its bond dipoles add up to give a non-zero dipole moment. This difference in dipole moment directly explains their contrasting molecular shapes.

Q16. Write the significance/applications of dipole moment.

Solution

Dipole moment is a quantitative measure of the polarity of a molecule and arises due to the separation of positive and negative charges. It is defined as the product of the magnitude of charge and the distance between the centres of positive and negative charges. Since it is a vector quantity, both magnitude and direction are important.

Its role in understanding molecular behaviour can be explained stepwise as:

\[ \begin{aligned} \text{Presence of unequal charge distribution} &\rightarrow \text{Formation of bond dipoles} \\ \text{Vector addition of bond dipoles} &\rightarrow \text{Net dipole moment} \\ \text{Magnitude of dipole moment} &\rightarrow \text{Measure of molecular polarity} \end{aligned} \]

One important application of dipole moment is in predicting molecular shape. A zero dipole moment generally indicates a symmetrical structure, as in CO₂, whereas a non-zero value suggests an unsymmetrical or bent geometry, as in H₂O.

Dipole moment is also used to distinguish between polar and non-polar molecules. Molecules with appreciable dipole moments are polar, while those with zero dipole moment are non-polar. This helps in understanding solubility, intermolecular attractions, and physical properties such as boiling points.

It is further useful in estimating the percentage ionic character of a bond. A larger dipole moment usually indicates greater ionic character, providing insight into the nature of bonding between atoms.

Dipole moment helps in differentiating between isomeric molecules as well. For example, cis and trans isomers often have different dipole moments due to different spatial arrangements of bonds, even though their molecular formulas are identical.

Therefore, dipole moment serves as an important tool for predicting molecular geometry, identifying polarity, comparing bond character, and understanding the physical and chemical behaviour of substances.

Q17. Define electronegativity. How does it differ from electron gain enthalpy ?

Solution

Electronegativity is defined as the tendency of an atom in a chemical bond to attract the shared pair of electrons towards itself. It is not an absolute property of an isolated atom; rather, it is a relative measure that depends on the bonding environment. An atom with higher electronegativity pulls bonding electrons more strongly, making the bond polar.

Electron gain enthalpy, on the other hand, is the enthalpy change that occurs when an isolated gaseous atom accepts an electron to form a negative ion. It is a measurable thermodynamic quantity and reflects how much energy is released or absorbed during this process.

The distinction between the two can be understood stepwise as:

\[ \begin{aligned} \text{Electronegativity} &\rightarrow \text{Attraction of shared electrons in a molecule} \\ &\rightarrow \text{Relative, unitless quantity} \\ \\ \text{Electron gain enthalpy} &\rightarrow \text{Addition of electron to an isolated atom} \\ &\rightarrow \text{Energy change expressed in kJ mol}^{-1} \end{aligned} \]

Thus, electronegativity describes the behaviour of an atom while it is bonded to another atom, whereas electron gain enthalpy refers to an isolated atom in the gaseous state. Electronegativity has no units and is obtained from scales such as Pauling’s scale, while electron gain enthalpy is an experimentally measurable energy change with definite units.

Moreover, electronegativity helps in predicting bond polarity and molecular structure, whereas electron gain enthalpy mainly indicates the tendency of an atom to form an anion. Although both concepts involve attraction for electrons, they differ fundamentally in definition, measurement, and application.

Q18. Explain with the help of suitable example polar covalent bond

Solution

A polar covalent bond is a type of covalent bond in which the shared pair of electrons is unequally distributed between the bonded atoms. This unequal sharing arises due to a difference in electronegativity between the two atoms, causing one atom to attract the bonding electrons more strongly than the other.

As a result of this unequal attraction, partial charges develop on the atoms. The atom with higher electronegativity acquires a partial negative charge (δ−), while the other atom develops a partial positive charge (δ+). This separation of charges produces bond polarity.

This process can be represented stepwise using the example of hydrogen chloride, HCl.

\[ \begin{aligned} \text{H} + \text{Cl} &\rightarrow \text{H–Cl} \\ \text{Shared electrons} &\rightarrow \text{pulled towards Cl} \\ \text{H} &\rightarrow \delta^{+}, \quad \text{Cl} \rightarrow \delta^{-} \end{aligned} \]

In HCl, chlorine is more electronegative than hydrogen, so it attracts the shared electron pair closer to itself. Consequently, hydrogen becomes slightly positive and chlorine becomes slightly negative. However, the electrons are not completely transferred; they are still shared between the atoms. Hence, the bond remains covalent but becomes polar.

The presence of partial charges and a measurable dipole moment confirms the polar nature of the H–Cl bond. This distinguishes polar covalent bonds from non-polar covalent bonds, where electrons are shared equally, and from ionic bonds, where electrons are completely transferred.

Therefore, a polar covalent bond may be defined as a covalent bond formed between atoms of different electronegativities, resulting in unequal sharing of electrons and the development of partial charges, as illustrated by the HCl molecule.

Q19. Arrange the bonds in order of increasing ionic character in the molecules: \(\mathrm{LiF,\ K_2O,\ N_2,\ SO_2\ and\ ClF_3}\).

Solution

The ionic character of a bond depends mainly on the electronegativity difference between the bonded atoms. A larger electronegativity difference results in greater electron transfer and hence higher ionic character, whereas identical or nearly identical electronegativities lead to purely covalent bonding.

Let us analyse each molecule stepwise on this basis.

\[ \begin{aligned} \text{N}_2 &: \text{Both atoms are identical} \Rightarrow \Delta EN = 0 \\ \text{SO}_2 &: \text{Moderate difference between S and O} \\ \text{ClF}_3 &: \text{Slightly higher difference between Cl and F} \\ \text{K}_2\text{O} &: \text{Large difference between K and O} \\ \text{LiF} &: \text{Very large difference between Li and F} \end{aligned} \]

In N₂, both nitrogen atoms have the same electronegativity, so electrons are shared equally and the bond is purely covalent, giving zero ionic character.

In SO₂, oxygen is more electronegative than sulphur, so the S–O bonds are polar covalent with some ionic character.

In ClF₃, fluorine is more electronegative than chlorine, producing greater bond polarity than in SO₂, and hence higher ionic character.

In K₂O, potassium being a metal readily loses electrons to oxygen, a non-metal. This results in strong ionic bonding with substantial ionic character.

In LiF, lithium transfers its electron almost completely to fluorine because fluorine has extremely high electronegativity. This makes LiF the most ionic among the given compounds.

Therefore, the bonds arranged in order of increasing ionic character are:

\[ \text{N}_2 \;<\; \text{SO}_2 \;<\; \text{ClF}_3 \;<\; \text{K}_2\text{O} \;<\; \text{LiF} \]

This order reflects the progressive increase in electronegativity difference and hence ionic nature of the bonds.

Q20. The skeletal structure of \(\mathrm{CH_3COOH}\) as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. \[\begin{array}{c} &&\mathrm{H}&&\mathrm{:O:}\\ &&|&&|\\ \mathrm{H}&=&\mathrm{C}&-&\mathrm{\underset{\cdot\cdot}{C}}&-&\underset{\cdot\cdot}{O}&-&\mathrm{H}\\ &&|\\ &&\mathrm{H} \end{array} \]

Solution

The given skeletal structure of acetic acid, CH₃COOH, shows the correct arrangement of atoms but some bonds and electron pairs are not represented properly. To obtain the correct Lewis structure, we must first count the total number of valence electrons and then distribute them so that each atom (except hydrogen) satisfies the octet rule.

\[ \begin{aligned} 2\,\text{C} &: 2 \times 4 = 8 \\ 4\,\text{H} &: 4 \times 1 = 4 \\ 2\,\text{O} &: 2 \times 6 = 12 \\ \text{Total valence electrons} &= 24 \end{aligned} \]

Carbon atoms form the backbone of the molecule. One carbon is bonded to three hydrogen atoms forming a CH₃ group, while the second carbon is attached to this group and also bonded to two oxygen atoms. To satisfy the octet of the second carbon, one oxygen must be double-bonded to it, while the other oxygen is single-bonded and further attached to hydrogen, forming the –OH group.

This arrangement can be represented stepwise as:

\[ \begin{array}{c} \ce{&&\mathrm{{:O:}}}\\ \ce{&&||} \\ \ce{H_3C&-& \mathrm{C}& - &\mathrm{\overset{\cdot\cdot}{\underset{\cdot\cdot}{O}}}& - &H} \\ \end{array} \]

Each oxygen atom carries two lone pairs, the carbonyl carbon forms one double bond and two single bonds, and the hydroxyl oxygen forms two single bonds. This completes the octet of both carbon and oxygen atoms, while hydrogen satisfies its duplet.

Hence, the correct Lewis structure of acetic acid is CH₃–C(=O)–OH, with two lone pairs on each oxygen atom. The original skeletal structure becomes accurate only after replacing one C–O single bond with a C=O double bond and ensuring proper lone pair placement on oxygen atoms.

Q21. Apart from tetrahedral geometry, another possible geometry for \(\mathrm{CH_4}\) is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why \(\mathrm{CH_4}\) is not square planar?

Solution

To examine whether methane could exist in a square-planar form, we compare this hypothetical arrangement with the actual tetrahedral structure predicted by valence bond theory and confirmed experimentally.

In square planar geometry, the four hydrogen atoms would lie at the corners of a square, with the carbon atom at the centre. For such a structure, carbon would need to use one s orbital and two p orbitals (similar to dsp² type mixing), leaving one unhybridised p orbital unused. This leads to two major problems.

First, the four C–H bonds would not be equivalent. Bonds formed by pure p orbitals differ in strength and direction from those formed by hybrid orbitals. However, experimental evidence shows that all four C–H bonds in methane are identical in length and energy. A square planar arrangement therefore contradicts this observation.

Second, the square planar structure places the H atoms at 90° to each other. Such small bond angles cause strong repulsion between bonding electron pairs, making the molecule energetically unstable. According to VSEPR principles, electron pairs arrange themselves to minimise repulsion.

Carbon instead undergoes sp³ hybridisation, producing four equivalent hybrid orbitals directed toward the corners of a tetrahedron. This geometry maximises separation between bonding pairs and gives identical C–H bonds.

This can be summarised mathematically as:

Because the tetrahedral arrangement provides maximum bond separation, equal bond energies, and minimum electron-pair repulsion, it is far more stable than a square planar form.

Hence, CH₄ is not square planar because such a geometry would give unequal bonds and greater repulsion, whereas the tetrahedral structure arising from sp³ hybridisation is energetically preferred.

Q22. Explain why \(\mathrm{BeH_2}\) molecule has a zero dipole moment although the \(\mathrm{Be–H}\) bonds are polar.

Solution

The dipole moment of a molecule depends not only on the polarity of individual bonds but also on the spatial arrangement of those bonds. In the case of \(\mathrm{BeH_2}\), each \(\mathrm{Be–H}\) bond is polar because hydrogen is slightly more electronegative than beryllium. Therefore, each bond possesses a bond dipole directed from beryllium towards hydrogen.

To understand why the overall dipole moment is zero, we must examine the geometry of the molecule. Beryllium has the electronic configuration \(1s^2\,2s^2\). In the excited state, one electron is promoted, allowing hybridisation to occur.

The two \(sp\) hybrid orbitals are oriented in exactly opposite directions, forming a linear molecule. As a result, the two \(\mathrm{Be–H}\) bond dipoles are equal in magnitude but act in opposite directions along the same straight line.

Because dipole moment is a vector quantity, these equal and opposite bond dipoles cancel each other completely. Hence, their vector sum becomes zero.

Mathematically, if the magnitude of each bond dipole is \(\mu\), then: \[ \begin{aligned} \mu_{\text{net}} &= \mu - \mu \\ &= 0 \end{aligned} \]

Therefore, although each \(\mathrm{Be–H}\) bond is polar, the linear and symmetrical arrangement of bonds in \(\mathrm{BeH_2}\) results in complete cancellation of dipole moments, making the molecule non-polar with zero net dipole moment.

Q23. Which out of \(\mathrm{NH_3}\) and \(\mathrm{NF_3}\) has higher dipole moment and why ?

Solution

Both \(\mathrm{NH_3}\) and \(\mathrm{NF_3}\) possess trigonal pyramidal geometry due to the presence of one lone pair on nitrogen. Although their shapes are similar, their dipole moments differ significantly because of the direction and mutual reinforcement or cancellation of individual bond dipoles.

In \(\mathrm{NH_3}\), nitrogen is more electronegative than hydrogen, so each \(\mathrm{N–H}\) bond dipole is directed towards the nitrogen atom. The lone pair on nitrogen also contributes to the overall dipole moment in the same general direction. As a result, all three bond dipoles and the lone-pair effect reinforce one another, producing a relatively large net dipole moment.

Therefore, \(\mathrm{NH_3}\) has a higher dipole moment than \(\mathrm{NF_3}\) because, in ammonia, the bond dipoles and lone-pair dipole act in the same direction, whereas in nitrogen trifluoride they oppose each other and reduce the overall dipole moment.

Q24. What is meant by hybridisation of atomic orbitals? Describe the shapes of \(\mathrm{sp,\ sp^2,\ sp^3}\) hybrid orbitals.

Solution

Hybridisation of atomic orbitals is the theoretical concept in which atomic orbitals of nearly equal energy within the same atom intermix to form a new set of equivalent orbitals called hybrid orbitals. These hybrid orbitals possess identical energy, identical shape, and definite orientation in space, which helps explain the geometry of molecules and the equivalence of bonds formed.

Mathematically, hybridisation may be represented as a linear combination of atomic orbitals:

\[ \begin{aligned} \text{Hybrid orbital} &= a(\text{s}) + b(\text{p}) + c(\text{p}) + \dots \\ \text{Number of hybrids formed} &= \text{Number of atomic orbitals mixed} \end{aligned} \]

In sp hybridisation, one s orbital mixes with one p orbital. Thus, two equivalent sp hybrid orbitals are formed.

\[ \begin{aligned} 1s + 1p &\longrightarrow 2(sp) \\ \text{Geometry} &:\ \text{Linear} \\ \angle &= 180^\circ \end{aligned} \]

The two sp hybrid orbitals are oriented in opposite directions along a straight line. Each orbital has a large lobe and a small lobe, with the larger lobe participating in bond formation.

In sp² hybridisation, one s orbital mixes with two p orbitals to form three equivalent sp² hybrid orbitals.

\[ \begin{aligned} 1s + 2p &\longrightarrow 3(sp^2) \\ \text{Geometry} &:\ \text{Trigonal planar} \\ \angle &= 120^\circ \end{aligned} \]

The three sp² hybrid orbitals lie in the same plane and are directed toward the corners of an equilateral triangle, making equal angles of 120° with one another.

In sp³ hybridisation, one s orbital mixes with three p orbitals, producing four equivalent sp³ hybrid orbitals.

\[ \begin{aligned} 1s + 3p &\longrightarrow 4(sp^3) \\ \text{Geometry} &:\ \text{Tetrahedral} \\ \angle &= 109.5^\circ \end{aligned} \]

These four sp³ hybrid orbitals are directed toward the four corners of a tetrahedron. The orientation ensures maximum separation between electron pairs, resulting in bond angles of approximately 109.5°.

Thus, hybridisation explains the formation of equivalent bonds and the specific geometrical arrangements of atoms in molecules through directional hybrid orbitals such as sp (linear), sp² (trigonal planar), and sp³ (tetrahedral).

Q25. Describe the change in hybridisation (if any) of the Al atom in the following reaction. \(\mathrm{AlCl + Cl^- \rightarrow AlCl_4^-}\)

Solution

To understand the change in hybridisation of aluminium in this reaction, we first examine the bonding situation in \(\mathrm{AlCl_3}\) and then in the product \(\mathrm{AlCl_4^-}\).

In gaseous \(\mathrm{AlCl_3}\), the aluminium atom is bonded to three chlorine atoms. Aluminium uses three equivalent hybrid orbitals to form three \(\sigma\)-bonds with chlorine. This corresponds to sp² hybridisation, giving a trigonal planar geometry with bond angles close to \(120^\circ\). At this stage, aluminium has an incomplete octet and possesses a vacant orbital.

When a chloride ion approaches, it donates a lone pair of electrons to aluminium, forming a coordinate bond. This increases the coordination number of aluminium from three to four.

The electronic rearrangement during this process can be represented as:

\[ \begin{aligned} \mathrm{AlCl_3} &: \text{Al is } sp^2 \text{ hybridised (three }\sigma\text{-bonds)} \\ \mathrm{Cl^-} &: \text{donates a lone pair} \\ \mathrm{AlCl_4^-} &: \text{Al becomes } sp^3 \text{ hybridised (four }\sigma\text{-bonds)} \end{aligned} \]

After accepting the electron pair, aluminium now forms four equivalent \(\mathrm{Al–Cl}\) bonds. To accommodate these four bonds symmetrically, aluminium undergoes rehybridisation from sp² to sp³, resulting in a tetrahedral arrangement of chloride ions around the central aluminium atom.

Hence, during the reaction \(\mathrm{AlCl_3 + Cl^- \rightarrow AlCl_4^-}\), the hybridisation of aluminium changes from sp² in \(\mathrm{AlCl_3}\) to sp³ in \(\mathrm{AlCl_4^-}\) due to the formation of an additional coordinate bond and completion of the octet.

Q26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
\(\mathrm{BF_3 + NH_3 \rightarrow F_3B.NH_3}\)

Solution

To determine whether hybridisation changes in this reaction, we examine the electronic state of boron in \(\mathrm{BF_3}\) and nitrogen in \(\mathrm{NH_3}\), and then compare them with their states in the adduct \(\mathrm{F_3B.NH_3}\).

In \(\mathrm{BF_3}\), boron is bonded to three fluorine atoms and forms three \(\sigma\)-bonds using three equivalent hybrid orbitals. This corresponds to sp² hybridisation, giving a trigonal planar structure. Boron has only six electrons in its valence shell and therefore possesses a vacant orbital.

In \(\mathrm{NH_3}\), nitrogen forms three \(\mathrm{N–H}\) bonds and retains one lone pair. Nitrogen uses sp³ hybrid orbitals, leading to a trigonal pyramidal geometry.

When ammonia approaches boron trifluoride, nitrogen donates its lone pair to the electron-deficient boron atom, forming a coordinate bond and producing the adduct \(\mathrm{F_3B.NH_3}\).

The change in hybridisation during this process can be represented as:

\[ \begin{aligned} \text{In } \mathrm{BF_3} &: \text{B is } sp^2 \\ \text{In } \mathrm{NH_3} &: \text{N is } sp^3 \\ \text{After adduct formation} &: \text{B becomes } sp^3,\ \text{N remains } sp^3 \end{aligned} \]

After accepting the lone pair, boron now forms four equivalent \(\sigma\)-bonds and attains a complete octet. To accommodate four bonds symmetrically, boron rehybridises from sp² to sp³, giving a tetrahedral environment around boron. Nitrogen, however, was already sp³ hybridised in ammonia, and it continues to use an sp³ orbital to donate its lone pair, so its hybridisation does not change.

Thus, in the reaction \(\mathrm{BF_3 + NH_3 \rightarrow F_3B.NH_3}\), boron undergoes a change in hybridisation from sp² to sp³, while nitrogen shows no change and remains sp³.

Q27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in \(\mathrm{C_2H_4}\) and \(\mathrm{C_2H_2}\) molecules.

Solution

Thus, in \(\mathrm{C_2H_4}\) the double bond arises from one \(\sigma\) and one \(\pi\) overlap due to sp² hybridisation, while in \(\mathrm{C_2H_2}\) the triple bond is formed by one \(\sigma\) and two mutually perpendicular \(\pi\) overlaps resulting from sp hybridisation of carbon.

Q28. What is the total number of sigma and pi bonds in the following molecules?
(a) \(\mathrm{C_2H_2}\)
(b) \(\mathrm{C_2H_4}\)

Solution

To determine the total number of sigma (\(\sigma\)) and pi (\(\pi\)) bonds, we recall that every single bond consists of one \(\sigma\)-bond, a double bond consists of one \(\sigma\)-bond and one \(\pi\)-bond, and a triple bond consists of one \(\sigma\)-bond and two \(\pi\)-bonds.

For \(\mathrm{C_2H_2}\) (ethyne), the structure is \(\mathrm{H–C \equiv C–H}\). The carbon–carbon triple bond contains one \(\sigma\)-bond and two \(\pi\)-bonds. In addition, there are two \(\mathrm{C–H}\) single bonds, each contributing one \(\sigma\)-bond.

\[ \begin{aligned} \text{C}\equiv\text{C} &: 1\,\sigma + 2\,\pi \\ 2(\text{C–H}) &: 2\,\sigma \\ \text{Total } \sigma &: 1 + 2 = 3 \\ \text{Total } \pi &: 2 \end{aligned} \]

Hence, \(\mathrm{C_2H_2}\) contains three \(\sigma\)-bonds and two \(\pi\)-bonds.

For \(\mathrm{C_2H_4}\) (ethene), the structure is \(\mathrm{H_2C=CH_2}\). The carbon–carbon double bond consists of one \(\sigma\)-bond and one \(\pi\)-bond. There are four \(\mathrm{C–H}\) single bonds, each contributing one \(\sigma\)-bond.

\[ \begin{aligned} \text{C}=\text{C} &: 1\,\sigma + 1\,\pi \\ 4(\text{C–H}) &: 4\,\sigma \\ \text{Total } \sigma &: 1 + 4 = 5 \\ \text{Total } \pi &: 1 \end{aligned} \]

Therefore, \(\mathrm{C_2H_2}\) has 3 \(\sigma\)-bonds and 2 \(\pi\)-bonds, while \(\mathrm{C_2H_4}\) has 5 \(\sigma\)-bonds and 1 \(\pi\)-bond.

Q29. Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
(a) \(\mathrm{1s\ and\ 1s}\)
(b) \(\mathrm{1s\ and\ 2p_x}\);
(c) \(\mathrm{2p_y\ and\ 2p_y}\)
(d) \(\mathrm{1s\ and\ 2s}\).

Solution

A \(\sigma\)-bond is formed by head-on (axial) overlap of atomic orbitals along the internuclear axis. Since the x-axis is taken as the internuclear axis here, only those orbitals that can overlap directly along this axis will produce a \(\sigma\)-bond.

Let us examine each case on this basis.

The overlap of \(\mathrm{1s}\) with \(\mathrm{1s}\) occurs directly along the x-axis, giving effective head-on overlap and hence a \(\sigma\)-bond. Similarly, overlap between \(\mathrm{1s}\) and \(\mathrm{2p_x}\) is also along the internuclear axis because the \(\mathrm{2p_x}\) orbital is oriented along x, so this combination forms a \(\sigma\)-bond. The overlap of \(\mathrm{1s}\) with \(\mathrm{2s}\) is again axial and symmetric about the internuclear axis, therefore it also results in a \(\sigma\)-bond.

However, in the case of \(\mathrm{2p_y}\) and \(\mathrm{2p_y}\), both orbitals are oriented perpendicular to the x-axis. Their overlap occurs sideways rather than head-on, which leads to lateral overlap.

This situation may be summarised as:

\[ \begin{aligned} 1s + 1s &\rightarrow \sigma \\ 1s + 2p_x &\rightarrow \sigma \\ 1s + 2s &\rightarrow \sigma \\ 2p_y + 2p_y &\rightarrow \pi \ (\text{sidewise overlap}) \end{aligned} \]

Hence, option (c) \(\mathrm{2p_y\ and\ 2p_y}\) will not form a \(\sigma\)-bond because these orbitals are perpendicular to the internuclear (x) axis and can only overlap sideways, producing a \(\pi\)-bond instead of a \(\sigma\)-bond.

Q30. Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) \(\mathrm{CH_3–CH_3}\) ;
(b) \(\mathrm{CH_3–CH=CH_2}\) ;
(c) \(\mathrm{CH_3-CH_2-OH}\);
(d) \(\mathrm{CH_3-CHO}\);
(e) \(\mathrm{CH_3COOH}\)

Solution

The hybridisation of a carbon atom depends on the number of sigma bonds and lone pairs around it. A carbon forming four sigma bonds is sp³ hybridised, one forming three sigma bonds and one pi bond is sp² hybridised, and one forming two sigma bonds and two pi bonds is sp hybridised.

In \(\mathrm{CH_3–CH_3}\) (ethane), each carbon forms four single bonds, three with hydrogen and one with the other carbon. Thus, each carbon has four sigma bonds.

\[ \begin{aligned} \text{Number of }\sigma\text{ bonds around each C} &= 4 \\ \Rightarrow \text{Hybridisation} &= sp^3 \end{aligned} \]

In \(\mathrm{CH_3–CH=CH_2}\) (propene), the first carbon (\(\mathrm{CH_3}\)) forms four sigma bonds and is therefore sp³ hybridised. The second and third carbons are involved in a double bond; each forms three sigma bonds and one pi bond.

\[ \begin{aligned} \text{C}_1 &: 4\,\sigma \Rightarrow sp^3 \\ \text{C}_2 &: 3\,\sigma + 1\,\pi \Rightarrow sp^2 \\ \text{C}_3 &: 3\,\sigma + 1\,\pi \Rightarrow sp^2 \end{aligned} \]

In \(\mathrm{CH_3–CH_2–OH}\) (ethanol), both carbon atoms are connected only by single bonds. Each carbon forms four sigma bonds.

\[ \begin{aligned} \text{Each C}: 4\,\sigma \Rightarrow sp^3 \end{aligned} \]

In \(\mathrm{CH_3–CHO}\) (ethanal), the methyl carbon (\(\mathrm{CH_3}\)) forms four sigma bonds and is sp³ hybridised. The carbonyl carbon forms three sigma bonds and one pi bond due to the C=O double bond.

\[ \begin{aligned} \text{CH}_3\text{ carbon} &: 4\,\sigma \Rightarrow sp^3 \\ \text{Carbonyl carbon} &: 3\,\sigma + 1\,\pi \Rightarrow sp^2 \end{aligned} \]

In \(\mathrm{CH_3COOH}\) (acetic acid), the methyl carbon forms four sigma bonds and is sp³ hybridised. The carboxyl carbon is involved in one double bond with oxygen and two single bonds (one with oxygen and one with carbon), making a total of three sigma bonds and one pi bond.

\[ \begin{aligned} \text{Methyl carbon} &: 4\,\sigma \Rightarrow sp^3 \\ \text{Carboxyl carbon} &: 3\,\sigma + 1\,\pi \Rightarrow sp^2 \end{aligned} \]

Thus, carbons with only single bonds are sp³ hybridised, while carbons participating in a double bond are sp² hybridised in the given molecules.

Q31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.

Solution

Solution:

When atoms combine to form molecules, their valence electrons play the central role in bond formation. These electrons may either be shared between atoms or remain localized on a single atom. On this basis, electron pairs are classified into two types: bond pairs and lone pairs.

A bond pair of electrons refers to a pair of valence electrons that is shared between two atoms, resulting in the formation of a covalent bond. This shared pair holds the atoms together and directly participates in bonding. For example, in a hydrogen molecule, each hydrogen atom contributes one electron, and these two electrons are shared to form a single bond:

\[ \begin{aligned} \text{H} \cdot + \cdot \text{H} &\longrightarrow \text{H}:\text{H} \\ (\text{one shared pair }&=\text{ one bond pair}) \end{aligned} \]

Here, the pair of electrons between the two hydrogen atoms is the bond pair.

A lone pair of electrons, on the other hand, is a pair of valence electrons that remains on the same atom and does not take part in bond formation. Although lone pairs do not form bonds, they strongly influence the shape and properties of molecules. A clear example is provided by the ammonia molecule, NH₃. Nitrogen has five valence electrons; three are used to form bonds with hydrogen atoms, while the remaining two stay on nitrogen as a lone pair:

\[ \begin{aligned} \text{Valence electrons of N} &= 5 \\ \text{Electrons used in bonding} &= 3 \\ \text{Remaining electrons} &= 2 \; (\text{one lone pair}) \end{aligned} \]

Thus, in NH₃, the three N–H bonds involve three bond pairs, whereas the extra pair of electrons on nitrogen represents a lone pair.

In summary, bond pairs are shared electron pairs responsible for chemical bonding, while lone pairs are unshared electron pairs residing on a single atom and influencing molecular geometry without forming bonds.

Q32. Distinguish between a sigma and a pi bond.

Solution

In covalent bonding, atoms combine by overlapping their atomic orbitals. Depending on the manner of overlap, two different types of covalent bonds are formed, namely sigma (σ) bonds and pi (π) bonds. Although both arise due to sharing of electrons, they differ clearly in their mode of formation, strength, and spatial distribution of electron density.

A sigma (σ) bond is formed by head-on or axial overlap of atomic orbitals along the internuclear axis. This overlap may occur between two s orbitals, an s and a p orbital, or two p orbitals. Since the electron cloud is concentrated symmetrically around the line joining the nuclei, a σ bond is cylindrically symmetrical. This type of overlap is maximum, making the σ bond stronger and more stable. Every single covalent bond is always a sigma bond.

Mathematically, the axial overlap involved in σ-bond formation may be represented as:

\[ \begin{aligned} \text{s–s overlap} &\rightarrow \sigma \\ \text{s–p overlap} &\rightarrow \sigma \\ \text{p–p (endwise) overlap} &\rightarrow \sigma \end{aligned} \]

A pi (π) bond is formed by sideways overlap of parallel p orbitals. In this case, the electron density lies above and below the internuclear axis rather than directly between the nuclei. Because sideways overlap is less effective than head-on overlap, a π bond is weaker than a σ bond. A π bond is always formed in addition to a σ bond and never exists independently.

The sideways overlap leading to π-bond formation can be expressed as:

\[ \begin{aligned} \text{p–p (sidewise) overlap} &\rightarrow \pi \end{aligned} \]

Thus, a sigma bond results from direct orbital overlap and provides the primary framework of a molecule, whereas a pi bond arises from lateral overlap and is responsible for multiple bonding. Sigma bonds allow free rotation of atoms about the bond axis, while pi bonds restrict this rotation due to their specific orientation in space.

In essence, σ bonds are stronger, formed first, and have electron density along the internuclear axis, whereas π bonds are comparatively weaker, formed after σ bonds, and possess electron density above and below the axis joining the bonded atoms.

Q33. Explain the formation of \(\mathrm{H_2}\) molecule on the basis of valence bond theory.

Solution

According to valence bond theory, a covalent bond is formed when two atoms approach each other in such a way that their half-filled atomic orbitals overlap and the electrons with opposite spins pair up. The formation of the hydrogen molecule, H₂, can be clearly explained on this basis.

Each hydrogen atom has one electron in its 1s orbital. Thus, both atoms possess a half-filled orbital. When two hydrogen atoms come close, their 1s orbitals overlap head-on along the internuclear axis. This overlap allows the two electrons to pair with opposite spins and become shared by both nuclei.

The electronic situation during bond formation may be represented as:

\[ \begin{aligned} \text{H}(1s^1) + \text{H}(1s^1) &\longrightarrow \text{H}_2 \\ \text{Half-filled }1s + \text{Half-filled }1s &\longrightarrow \sigma(1s-1s) \end{aligned} \]

As the atoms approach each other, attractive forces between each nucleus and the shared electron pair increase, while repulsion between the nuclei and between electrons also develops. At a particular internuclear distance, called the equilibrium bond distance, the attractive forces dominate and the potential energy of the system becomes minimum. At this stage, a stable H–H bond is formed.

The overlapping 1s orbitals produce a sigma (σ) bond, since the electron density becomes concentrated symmetrically around the line joining the two nuclei. This shared electron pair holds the two hydrogen atoms together, giving rise to the H₂ molecule. Each hydrogen atom effectively attains a stable duplet configuration similar to that of helium.

Thus, on the basis of valence bond theory, the H₂ molecule is formed by head-on overlap of two half-filled 1s orbitals accompanied by pairing of electrons with opposite spins, resulting in a stable σ bond and a lower energy molecular system.

Q34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Solution

According to the Linear Combination of Atomic Orbitals (LCAO) approach of molecular orbital theory, atomic orbitals combine mathematically to form molecular orbitals. However, not all atomic orbitals can combine effectively. Certain essential conditions must be satisfied for their linear combination to result in the formation of stable molecular orbitals.

The first important condition is that the combining atomic orbitals must have comparable energies. If the energy difference between the atomic orbitals is very large, their interaction becomes negligible and effective overlap does not occur. This requirement may be expressed conceptually as:

\[ \begin{aligned} \Delta E &= |E_1 - E_2| \\ \text{For effective combination:} \quad \Delta E &\approx 0 \end{aligned} \]

The second condition is that the atomic orbitals must have proper symmetry with respect to the internuclear axis. Orbitals should possess the same symmetry so that their wave functions can combine constructively. For example, an s orbital can combine with a p orbital only if the p orbital is oriented along the internuclear axis; otherwise, overlap becomes zero due to mismatch in symmetry.

The third essential condition is that the extent of overlap between the atomic orbitals must be significant. Greater overlap leads to stronger interaction and more stable bonding molecular orbitals. If the overlap is negligible, the resulting molecular orbital formation is ineffective.

The linear combination itself may be represented as:

\[ \begin{aligned} \psi_{\text{bonding}} &= \psi_A + \psi_B \\ \psi_{\text{antibonding}} &= \psi_A - \psi_B \end{aligned} \]

Here, constructive combination results in a bonding molecular orbital, while destructive combination leads to an antibonding molecular orbital. Both arise only when the above conditions of comparable energy, proper symmetry, and appreciable overlap are satisfied.

Thus, for atomic orbitals to combine effectively under the LCAO method, they must be nearly equal in energy, possess compatible symmetry, and overlap to a considerable extent along the internuclear axis.

Q35. Use molecular orbital theory to explain why the Be2 molecule does not exist.

Solution

According to molecular orbital theory, the existence and stability of a diatomic molecule depend on its bond order, which is calculated from the number of electrons present in bonding and antibonding molecular orbitals. A molecule is stable only if its bond order is positive.

Beryllium has the electronic configuration 1s²2s². When two Be atoms combine, a total of four valence electrons (from the two 2s orbitals) take part in molecular orbital formation. These electrons are distributed between the bonding σ(2s) and antibonding σ*(2s) molecular orbitals.

The filling of molecular orbitals for Be₂ may be represented as:

\[ \begin{aligned} \text{Total valence electrons} &= 4 \\ \sigma(2s) &= 2 \text{ electrons} \\ \sigma^{*}(2s) &= 2 \text{ electrons} \end{aligned} \]

Using the bond order expression,

\[ \begin{aligned} \text{Bond order} &= \frac{N_b - N_a}{2} \\\\ &= \frac{2 - 2}{2} \\\\ &= 0 \end{aligned} \]

where \(N_b\) represents the number of electrons in bonding molecular orbitals and \(N_a\) represents the number of electrons in antibonding molecular orbitals.

A bond order of zero indicates that the stabilizing effect of bonding electrons is exactly cancelled by the destabilizing effect of antibonding electrons. As a result, no net bond is formed between the two beryllium atoms.

Hence, according to molecular orbital theory, the Be₂ molecule does not exist because its bond order is zero, implying complete absence of chemical bonding and molecular stability.

Q36. Compare the relative stability of the following species and indicate their magnetic properties;
\(\mathrm{O_2,\ O_2^+,\ O_2^-(superoxide),\ O_2^{2-}(peroxide)}\)

Solution

The relative stability and magnetic behaviour of O₂, O₂⁺, O₂⁻ (superoxide) and O₂²⁻ (peroxide) can be understood clearly using molecular orbital theory. All these species differ only in the number of electrons occupying the molecular orbitals derived mainly from oxygen 2p orbitals. Their stability is judged from bond order, while magnetic nature depends on the presence or absence of unpaired electrons.

For oxygen and its ions, electrons are filled in the sequence …σ2p_z, π2p_x=π2p_y, π*2p_x=π*2p_y. The bond order is calculated using:

\[ \begin{aligned} \text{Bond order}=\frac{N_b-N_a}{2} \end{aligned} \]

For O₂, there are 10 bonding electrons and 6 antibonding electrons:

\[ \begin{aligned} \text{Bond order of O}_2 &= \frac{10-6}{2}\\&=2 \end{aligned} \]

Two electrons occupy the degenerate π* orbitals singly, so O₂ is paramagnetic.

In O₂⁺, one electron is removed from a π* antibonding orbital:

\[ \begin{aligned} \text{Bond order of O}_2^{+} &= \frac{10-5}{2}\\&=2.5 \end{aligned} \]

This increase in bond order makes O₂⁺ more stable than O₂. One unpaired electron remains, hence it is paramagnetic.

In superoxide, O₂⁻, one extra electron enters a π* antibonding orbital:

\[ \begin{aligned} \text{Bond order of O}_2^{-} &= \frac{10-7}{2}\\&=1.5 \end{aligned} \]

The bond order decreases, so O₂⁻ is less stable than O₂. It still contains one unpaired electron and is therefore paramagnetic.

In peroxide, O₂²⁻, two additional electrons occupy both π* orbitals:

\[ \begin{aligned} \text{Bond order of O}_2^{2-} &= \frac{10-8}{2}\\&=1 \end{aligned} \]

All electrons are now paired, making O₂²⁻ diamagnetic. The low bond order indicates the weakest O–O bond among the four species.

Thus, on the basis of bond order, the relative stability is:

\[ \begin{aligned} \text{O}_2^{+} > \text{O}_2 > \text{O}_2^{-} > \text{O}_2^{2-} \end{aligned} \]

Regarding magnetic properties, O₂, O₂⁺, and O₂⁻ are paramagnetic due to unpaired electrons, whereas O₂²⁻ is diamagnetic because all its electrons are paired.

Q37. Write the significance of a plus and a minus sign shown in representing the orbitals.

Solution

While representing atomic orbitals, the plus (+) and minus (−) signs do not indicate electrical charge. Instead, they denote the phase (or algebraic sign) of the orbital wave function. These signs are extremely important in molecular orbital theory because they decide whether orbitals combine constructively or destructively during overlap.

The wave function of an orbital may be written symbolically as ψ. When two orbitals approach each other, their combination depends on these signs:

\[ \begin{aligned} \psi_{\text{bonding}} &= \psi_A + \psi_B \\ \psi_{\text{antibonding}} &= \psi_A - \psi_B \end{aligned} \]

If overlapping lobes carry the same sign (both + or both −), the wave functions add up. This is called constructive overlap, leading to increased electron density between the nuclei and formation of a bonding molecular orbital.

On the other hand, if overlapping lobes have opposite signs (+ and −), the wave functions cancel each other. This destructive overlap produces a node between the nuclei and results in an antibonding molecular orbital.

Thus, the plus and minus signs simply represent opposite phases of the orbital wave function. Their significance lies in predicting whether orbital overlap will stabilize the molecule (same signs, bonding) or destabilize it (opposite signs, antibonding). Without these phase signs, it would not be possible to explain the formation of bonding and antibonding molecular orbitals.

Q38. Describe the hybridisation in case of \(\mathrm{PCl_5}\) . Why are the axial bonds longer as compared to equatorial bonds?

Solution

In phosphorus pentachloride, PCl₅, the central phosphorus atom has the valence shell configuration 3s²3p³. In order to form five covalent bonds with five chlorine atoms, one of the paired 3s electrons is promoted to a vacant 3d orbital. As a result, phosphorus attains five unpaired electrons in its valence shell.

The electronic rearrangement may be represented as:

\[ \begin{aligned} \text{Ground state:} \quad &3s^2\,3p^3 \\ \text{Excited state:} \quad &3s^1\,3p^3\,3d^1 \end{aligned} \]

These five orbitals (one 3s, three 3p and one 3d) then undergo hybridisation to form five equivalent sp³d hybrid orbitals. The hybridisation process can be expressed as:

\[ \begin{aligned} 1s + 3p + 1d &\longrightarrow 5\,sp^3d \end{aligned} \]

The five sp³d hybrid orbitals are arranged in a trigonal bipyramidal geometry. Three orbitals lie in the same plane at 120° to each other, called equatorial positions, while the remaining two occupy positions perpendicular to this plane, known as axial positions. Each hybrid orbital overlaps with a half-filled p orbital of chlorine to form five σ (P–Cl) bonds.

The axial bonds are longer than the equatorial bonds because of greater repulsion experienced in the axial positions. Each axial bond is at 90° to the three equatorial bonds, leading to stronger bond pair–bond pair repulsions. In contrast, equatorial bonds experience only two such 90° interactions and two 120° interactions, which involve relatively less repulsion. Due to this increased repulsion, axial P–Cl bonds are slightly elongated and therefore weaker than equatorial bonds.

Thus, PCl₅ exhibits sp³d hybridisation with trigonal bipyramidal geometry, and the difference in bond lengths arises from unequal bond pair repulsions in axial and equatorial positions.

Q39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Solution

A hydrogen bond is a special type of intermolecular (or sometimes intramolecular) attraction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine, and this hydrogen atom experiences an additional attraction toward another electronegative atom in the same or a neighbouring molecule.

This interaction arises because the electronegative atom pulls electron density away from hydrogen, leaving it with a partial positive charge. The positively polarised hydrogen is then attracted to a lone pair of electrons present on another electronegative atom. The situation may be represented symbolically as:

\[ \begin{aligned} \text{X–H} \cdots \text{Y} \end{aligned} \]

where X and Y are electronegative atoms (such as O, N, or F), X–H represents a covalent bond, and the dotted line indicates the hydrogen bond.

Hydrogen bonding is responsible for many important physical properties, such as the unusually high boiling point of water and the structural stability of biomolecules.

When compared with van der Waals forces, hydrogen bonds are significantly stronger. Van der Waals forces arise from temporary or induced dipoles and are very weak in nature, whereas hydrogen bonding involves strong dipole–dipole attraction due to highly polar X–H bonds. However, hydrogen bonds are still much weaker than ordinary covalent bonds.

Thus, a hydrogen bond is an attraction between a hydrogen atom attached to a highly electronegative atom and another electronegative atom, and it is stronger than van der Waals forces but weaker than covalent bonds.

Q40. What is meant by the term bond order? Calculate the bond order of : \(\mathrm{N_2,\ O_2,\ O_2^{+}\ and\ O^{2–}}\).

Solution

The term bond order refers to the number of effective chemical bonds between two atoms in a molecule. According to molecular orbital theory, bond order is defined as half the difference between the number of electrons present in bonding molecular orbitals and antibonding molecular orbitals. It gives a direct measure of bond strength and molecular stability: the higher the bond order, the stronger and shorter the bond.

The mathematical expression for bond order is:

\[ \begin{aligned} \text{Bond order}=\frac{N_b-N_a}{2} \end{aligned} \]

where \(N_b\) is the number of electrons in bonding orbitals and \(N_a\) is the number of electrons in antibonding orbitals.

For N₂, the molecular orbital configuration shows 10 electrons in bonding orbitals and 4 electrons in antibonding orbitals:

\[ \begin{aligned} \text{Bond order of N}_2 &= \frac{10-4}{2} \\ &= 3 \end{aligned} \]

This high bond order explains the exceptional stability and strength of the N≡N bond.

For O₂, there are 10 bonding electrons and 6 antibonding electrons:

\[ \begin{aligned} \text{Bond order of O}_2 &= \frac{10-6}{2} \\ &= 2 \end{aligned} \]

Thus, O₂ contains a double bond.

For O₂⁺, one electron is removed from an antibonding π* orbital, leaving 10 bonding and 5 antibonding electrons:

\[ \begin{aligned} \text{Bond order of O}_2^{+} &= \frac{10-5}{2} \\ &= 2.5 \end{aligned} \]

The increase in bond order indicates that O₂⁺ is more stable than O₂.

For O₂²⁻, two extra electrons enter antibonding orbitals, giving 10 bonding and 8 antibonding electrons:

\[ \begin{aligned} \text{Bond order of O}_2^{2-} &= \frac{10-8}{2} \\ &= 1 \end{aligned} \]

This low bond order shows that the peroxide ion has the weakest O–O bond among the given species.

Hence, the bond orders are N₂ = 3, O₂ = 2, O₂⁺ = 2.5, and O₂²⁻ = 1. Bond order not only predicts bond multiplicity but also provides insight into relative molecular stability.