NCERT Solutions for Class 11 Chemistry Chapter 4 - Chemical Bonding and Molecular Structure

Question 4.1: Explain the formation of a chemical bond.

Answer :

The attractive force that holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond. Different theories and concepts have been put forward from time to time to analyze the formation of the bond. These are the Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory, and Molecular Orbital (MO) Theory.

And every system tends to be more stable, and bonding is nature’s way of lowering the energy of the system to attain stability.

Atoms, therefore, combine with each other and complete their respective octets or duplets to attain a stable configuration of the nearest noble gases. It was seen that the noble gases are very stable and inert and do not react with others.

So, there is a sharing of electrons or transferring one or more electrons from one atom to another; as a result a chemical bond is formed known as a covalent bond or ionic bond.

Question 4.3(a) Write Lewis symbols for the following atoms and ions:

$S \; and\; S^{2-}$

Answer :

Sulphur belongs to the oxygen family. So the number of valence electrons in sulphur is six. Therefore, its Lewis dot symbol of sulphur(S) is

And of $S^{2-}$ is, if it has two electrons more because of its dinegative charge.

Question 4.3(b) Write Lewis symbols for the following atoms and ions:

$Al\; and\; Al^{3+}$

Answer :

Al belongs to the boron family, so the number of valence electrons in aluminium is three.

Therefore, the Lewis dot symbol of aluminium(Al) is

And of $Al^{3+}$, it means it has donated three electrons and acquires a tripositive charge. Hence, the Lewis symbol is

Question 4.3(c) Write Lewis symbols for the following atoms and ions:

$H \; and \; H^{-}$

Answer :

As the number of valence electrons in hydrogen is one.

Therefore, its Lewis dot symbol of hydrogen (H) is

And of $H^{-}$, it has one more electron because of its negative charge. Hence, the Lewis symbol is

Question 4.4(c) Draw the Lewis structures for the following molecules and ions :

$BeF_{2}$

Answer :

Be has two valence electrons, with which it forms two single bonds with each F to complete its octet. F has seven electrons in its valence shell. So after bonding, there will be three lone pairs on each F. Beryllium forms two single bonds with two fluorine atoms and has no lone pairs, resulting in a linear shape.

The Lewis structure of $BeF_{2}$ is:

Question 4.4(d) Draw the Lewis structures for the following molecules and ions :

$CO_{3}^{2-}$

Answer :

C has 4 valence electrons, and O has 6. So to complete its octet, C needs four more electrons. Carbon forms one double bond and two single bonds with oxygen atoms with a -2 charge delocalized over the three oxygens.

The Lewis structure of $CO_{3}^{2-}$ is:

Question 4.4(e) Draw the Lewis structures for the following molecules and ions :

$HCOOH$

Answer :

The carbon atom forms a single bond with an oxygen (bonded with a hydrogen), a double bond with one oxygen atom (=O), and a single bond with a hydrogen atom. So, both the oxygens have 2 lone pairs on them.

The Lewis structure of $HCOOH$ is:

Question 4.5 Define octet rule. Write its significance and limitations.

Answer:

Atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as the octet rule.

Significance: It is quite useful for understanding the structures of most of the organic compounds and it applies mainly to the second-period elements of the periodic table

Limitations: There are three types of exceptions to the octet rule.

The incomplete octet of the central atom - the number of electrons surrounding the central atom is less than eight. Examples are $\mathrm{LiCl}, \mathrm{BeH}_2, \mathrm{BCl}_3$.

Odd-electron molecules - the octet rule is not satisfied for all the atoms in $NO\ and\ NO_{2}$.

The expanded octet - there are more than eight valence electrons around the central atom. This is termed as the expanded octet. Some of the examples of such compounds are PF₅, SF₆, H₂SO₄ and a number of coordination compounds.

Question 4.6 Write the favourable factors for the formation of ionic bond.

Answer :

The formation of an ionic bond takes place by the transfer of one or more electrons from one atom to another. So, ionic bond formation mainly depends upon the ease with which neutral atoms can lose or gain electrons.

The bond formation also depends upon the lattice energy of the compound formed.

Ionic bonds will be formed more easily between elements with comparatively low ionization enthalpies and elements with a comparatively high negative value of electron gain enthalpy.

Question 4.7(a) Discuss the shape of the following molecules using the VSEPR model:

$BeCl_{2}$ ,

Answer :

Using the VSEPR model, we have $BeCl_{2}$

The central atom has no lone pair, and there are two bond pairs.

$BeCl_{2}$ is of the type $AB_{2}$

Hence, it has a linear shape.

Question 4.7(b) Discuss the shape of the following molecules using the VSEPR model:

(b) $BCl_{3}$

Answer :

Using the VSEPR model, we have $BCl_{3}$

The central atom has no lone pair and there are three bond pairs.

$BCl_{3}$ is of the type $AB_{3}$

Hence, it has a trigonal planar shape.

Question 4.7(c) Discuss the shape of the following molecules using the VSEPR model:

(c) $SiCl_{4}$

Answer :

Using the VSEPR model, we have $SiCl_{4}$

The central atom has no lone pair, and there are four bond pairs.

$SiCl_{4}$ is of the type $AB_{4}$.

Hence, it has a tetrahedral shape.

Question 4.7(d) Discuss the shape of the following molecules using the VSEPR model:

(d) $AsF_{5}$

Answer:

Using the VSEPR model, we have $AsF_{5}$

The central atom has no lone pair and there are five bond pairs.

$AsF_{5}$ is of the type $AB_{5}$

Hence, it has a trigonal bipyramidal shape.

Question 4.7 (e) Discuss the shape of the following molecules using the VSEPR model:

$H_{2}S$

Answer:

Using the VSEPR model we have, $H_{2}S$

The central atom has no lone pair and there are two bond pairs.

$H_{2}S$ is of the type $AB_{2}E$

Hence, it has a bent shape.

Question 4.7(f) Discuss the shape of the following molecules using the VSEPR model:

(f) $PH_{3}$

Answer :

Using the VSEPR model, we have $PH_{3}$

The central atom has no lone pair and there are three bond pairs.

$PH_{3}$ is of the type $AB_{3}E$

Hence, it has a trigonal bipyramidal shape.

Question 4.9 How do you express the bond strength in terms of bond order?

Answer :

Bond strength refers to the amount of energy required to break the bond between two atoms in a molecule.

So, with an increase in bond order, bond enthalpy increases as a result, bond strength increases.

Question 4.10 Define the bond length.

Answer :

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

The bond length in a covalent molecule AB.

$R = r_{A}+r_{B}$ where (R is the bond length and $r_{A}$ and $r_{B}$ are covalent radii of atoms A and B respectively.

Question 4.11 Explain the important aspects of resonance with reference to the $CO_{3}^{2-}$ ion

Answer :

The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately, as it represents unequal bonds. According to the experimental findings, all carbon-to-oxygen bonds in

CO₃^2- are equivalent. Therefore, the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.

Question 4.13 Write the resonance structures for $SO_{3},$ $NO_{2},$ and $NO_{3}^{-}.$

Answer :

The resonance structures $SO_{3}$

The resonance structures $NO_{2}$

The resonance structures $NO_{3}^{-}$

Question 4.14(a) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :

$K \; and\; S$

Answer :

K and S:

We have the electronic configurations of both:

$K = 2,8,8,1$ having 1 electron in the valence shell, and it can donate 1 electron to get to the nearest noble gas configuration.

$S = 2,8,6$ having 6 electrons in the valence shell, so it can complete its octet by accepting 2 more electrons.

So, there will be an electron transfer between them as follows:

Question 4.14(b) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :

$Ca \; and \; O$

Answer :

$Ca \; and \; O$ :

We have the electronic configurations of both:

$Ca = 2,8,8,2$ has 2 electrons in the valence shell, and it can donate 2 electrons to get to the nearest noble gas configuration.

$O = 2,6$ has 6 electrons in the valence shell, so it can complete its octet by accepting 2 more electrons.

So, there will be an electron transfer between them as follows:

Question 4.14(c) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :

$Al\; and\; N.$

Answer :

$Al\; and\; N.$ :

We have the electronic configurations of both:

$Al = 2,8,3$ has 3 electrons in the valence shell, and it can donate 3 electrons to get to the nearest noble gas configuration.

$N = 2,5$ has 3 electrons in the valence shell, so it can complete its octet by accepting 2 more electrons.

So, there will be electron transfer between them as follows:

Question 4.15 Although both $CO_{2}$ and $H_{2}O$ are triatomic molecules, the shape of $H_{2}O$ molecule is bent while that of $CO_{2}$ is linear. Explain this on the basis of dipole moment.

Answer :

H₂O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.50. The net dipole moment of 6.17 × 10^-30 C m is the resultant of the dipole moments of two O–H bonds.

On the other hand, the dipole moment of carbon dioxide is zero. This may be because of linear shape of the molecule as it has two C-O bonds, which have opposite dipole moments cancelling each other.

Question 4.16 Write the significance/applications of dipole moment.

Answer :

Some of the important significance of the dipole moment is as follows:

1. We can determine the shape of the molecule. Symmetrical molecules like linear, etc., do have zero dipole moment, whereas if not symmetrical, then they take different shapes, such as a bent shape or some angular shapes.

2. For determining the polarity of the molecules, the greater the dipole moment value, the greater will be the polarity and vice versa.

3. We can say that if a molecule has a zero dipole moment, then it must be non-polar, and if it is non-zero, then it must have some polar character.

Question 4.17 Define electronegativity. How does it differ from electron gain enthalpy?

Answer :

Electronegativity is the ability of an atom in a compound to attract a bond pair of electrons towards itself. It cannot be measured and it is a relative number.

The electron gain enthalpy, $\triangle_{eg}H$, is the enthalpy change when a gas-phase atom in its ground state gains an electron. The electron gain process may be exothermic or endothermic.

An element has a constant value of the electron gain enthalpy that can be measured experimentally.

Question 4.19 Arrange the bonds in order of increasing ionic character in the molecules:

$LiF$

$K_{2}O$

$N_{2}$

$SO_{2}$

$ClF_{3}$

Answer :

The ionic character in a molecule depends on the electronegativity difference between the constituting atoms. The greater the difference more ionic the character of the molecule.

So, on this basis, we have the order of increasing ionic character in the given molecules.

$N_{2}<SO_{2}<ClF_{3}<K_{2}O<LiF$ .

Question 4.20 The skeletal structure of $CH_{3}COOH$ as shown below is correct, but some of the bonds are shown incorrectly

Answer :

Here hydrogen atom is bonded to carbon with a double bond, which is not possible because hydrogen has only one electron to share with carbon.

Also, the second carbon does not have its valency satisfied means it has formed five bonds instead of four.

Therefore, the correct skeletal structure of $CH_{3}COOH$ as shown below:

Question 4.21 Apart from tetrahedral geometry, another possible geometry for $CH_{4}$ is square planar with the four $H$ atoms at the corners of the square and the $C$ atom at its centre. Explain why $CH_{4}$ is not square planar?

Answer :

The electronic configuration of a carbon atom is $C: 1s^22s^22p^2$.

Where it has s-orbital, p-orbital only, and there is no d-orbital present.

Hence, the carbon atom undergoes $sp^3$ hybridization in the methane molecule and takes a tetrahedral shape.

And for a molecule to have a square planar structure, it must have a d orbital present.

But here the absence of d-orbital, as a result, it does not undergo $dsp^2$ hybridization, the structure of methane cannot be square planar.

Also, the reason that the bond angle in square planar $90^{\circ}$ which makes the molecule more unstable because of repulsion between the bond pairs.

Hence, according to VSEPR theory $CH_{4}$ molecule takes a tetrahedral structure.

Question 4.22 Explain why $BeH_{2}$ molecule has a zero dipole moment, although the $Be-H$ bonds are polar.

Answer :

$BeH_{2}$ molecule has a zero dipole moment because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

Question 4.23 Which out of $NH_{3}$ and $NF_{3}$ has a higher dipole moment and why?

Answer :

Here, both have nitrogen as a central atom, and it has a lone pair of electrons with three bond pairs.

Hence, both molecules have a pyramidal shape.

The electronegativity of fluorine is more as compared than that of hydrogen. Hence, it is expected that the net dipole moment of $NF_{3}$ is greater than $NH_{3}$.

However, $NH_{3}$ has the net dipole moment of 1.46D and $NF_{3}$ has the net dipole moment of 0.24D. which is greater than $NF_{3}$ .

This is because of the direction of the dipole moments of each bond in $NH_{3}$ and $NF_{3}$.

The moments of the lone pair in $NF_{3}$ partly cancel out. But in $NH_{3}$, the resultant moment adds up to the bond moment of the lone pair.

Question 4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of $sp$ , $sp^{2}$ , $sp^{3}$ hybrid orbitals.

Answer :

Hybridisation can be defined as the process of intermixing the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

The shapes of $sp$ , $sp^{2}$ , $sp^{3}$ hybrid orbitals are shown:

$sp$ hybrid orbital: It is linear in shape and formed by the intermixing of s and p orbitals.

$sp^{2}$ hybrid orbital: It is the trigonal planar shape and is formed by the intermixing of one s-orbital and two 2p-orbitals.

$sp^{3}$ hybrid orbital: It is tetrahedral in shape and is formed by the intermixing of one s-orbital and three p-orbitals.

Question 4.25 Describe the change in hybridisation (if any) of the $Al$ atom in the following reaction.

$AlCl_{3}+Cl^{-}\rightarrow AlCl_{4}^{-}$

Answer :

Initially, the aluminium is in the ground state, and the valence orbital can be shown as:

Then the electron gets excited so the valence orbital can be shown as:

So, initially, aluminium $(AlCl_{3})$ had $sp^2$ hybridisation and hence had a trigonal planar shape.

Then it reacts with chloride ion to form $AlCl_{4}^{-}$. Where it has the empty $3p_{z}$ orbital which gets involved and the hybridisation changes from $sp^2 \rightarrow sp^3$.

Hence, there is a shape change from trigonal planar to tetrahedral.

Question 4.26 Is there any change in the hybridisation of $B$ and $N$ atoms as a result of the following reaction?

$BF_{3}+NH_{3}\rightarrow F_{3}B.NH_{3}$

Answer :

Initially boron atom $BF_{3}$ was in $sp^2$ hybridised. The valence orbital of boron in the excited state can be shown as:

And nitrogen atom in $NH_{3}$ is $sp^3$ hybridised. The valence orbital of nitrogen in the excited state can be shown as:

Then, after the reaction has occurred, the product $F_{3}B.NH_{3}$ is formed by the hybridisation of 'B' changes to $sp^3$. However, the hybridisation of 'N' remains unchanged.

Question 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in $C_{2}H_{4}$ and $C_{2}H_{2}$ molecules.

Answer :

We have the electronic configuration of the C-atom in the excited state:

$C= 1s^22s^12p_{x}^12p_{y}^12p_{z}^1$

Formation of an ethane molecule $(C_{2}H_{4})$ by overlapping of a $sp^2$ hybridized orbital of another carbon atom, thereby forming a $C-C$ sigma bond.

The remaining two $sp^2$ orbitals of each carbon atom form a $sp^2-s$ sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak n-bond.

Formation of $C_{2}H_{2}$ molecule- each C-atom is sp-hybridized with two 2p-orbitals in an unhybridized state.

One sp hybrid orbital of one carbon atom overlaps axially with the sp hybrid orbital of the other carbon atom to form a C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half-filled s orbital of the hydrogen atoms forming σ bonds.

Each of the two unhybridised p orbitals of both the carbon atoms overlaps sideways to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds as shown in Fig

Question 4.28(a) What is the total number of sigma and pi bonds in the following molecules?

$C_{2}H_{2}$

Answer :

Given molecule $C_{2}H_{2}$ :

So, there are three sigma (2C-H bonds + 1 C-C bond) and two pi-bonds (2 C-C bonds) in $C_{2}H_{2}$.

Question 4.28(b) What is the total number of sigma and pi bonds in the following molecules?

$C_{2}H_{4}$

Answer :

Given molecule $C_{2}H_{4}$ :

So, there are five sigma (4C-H bonds + 1 C-C bond) and one pi-bond (C-C bonds) in $C_{2}H_{4}$.

Question 4.29(a) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?

$1s \; and\; 1s$

Answer :

Orbitals $1s \; and\; 1s$ will form a sigma bond as both orbitals are spherical and can combine along the x-axis as the internuclear axis.

Question 4.29(b) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?

$1s \; and\; 2p_{x}$

Answer :

Orbitals $1s \; and\; 2p_{x}$ will form a sigma bond as 1s orbital and 2p x orbital are aligned such that they can combine along the x-axis as the internuclear axis.

Question 4.29(c) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?

$2p_{y}\; and\; 2p_{y}$

Answer :

Orbitals $2p_{y}\; and\; 2p_{y}$ will not form a sigma bond as both 2py orbitals are aligned in y-direction but the internuclear axis is the x-axis.

The formation of a pi bond takes place.

Question 4.29(d) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?

$1s \; and \; 2s.$

Answer :

Orbitals $1s \; and \; 2s.$ will form a sigma bond as both 1s and 2s orbitals are spherical and can combine along the x-axis as the internuclear axis.

Question 4.30(a) Which hybrid orbitals are used by carbon atoms in the following molecules?

$CH_{3}-CH_{3}$

Answer :

$CH_{3}-CH_{3}$

There are four sigma bonds (single bond), each with the help of one s hybrid orbital and three p hybrid orbitals. Hence, C₁ and C₂ are $sp^3$ hybridized.

Question 4.30(b) Which hybrid orbitals are used by carbon atoms in the following molecules?

$CH_{3}-CH=CH_{2};$

Answer :

$CH_{3}-CH=CH_{2};$

$C_{1}$ is making 4 sigma bonds (single bond) therefore it is $sp^3$ hybridised.

While $C_{2}\ and\ C_{3}$ are making a double bond. $(1 sigma\ bond + 1\ pi\ bond)$

Therefore, they both are $sp^2$ hybridized.

Question 4.30(c) Which hybrid orbitals are used by carbon atoms in the following molecules?

$CH_{3}-CH_{2}-OH$

Answer :

$CH_{3}-CH_{2}-OH$

$C_{1}$ is making 4 sigma bonds (single bonds), therefore it is $sp^3$ hybridised.

and $C_{2}$ is also making a 4 sigma bond. Therefore, it is also $sp^3$ hybridised.

Therefore, they both are $sp^3$ hybridized.

Question 4.30(d) Which hybrid orbitals are used by carbon atoms in the following molecules?

$CH_{3}-CHO$

Answer :

$CH_{3}-CHO$

$C_{1}$ is making 4 sigma bonds (single bonds), therefore it is $sp^3$ hybridised.

and $C_{2}$ is making a 3 sigma bonds with hydrogen, carbon and oxygen. and one pi bond with oxygen,n therefore it is $sp^2$ hybridised.

Question 4.30(e) Which hybrid orbitals are used by carbon atoms in the following molecules?

$CH_{3}COOH$

Answer :

$CH_{3}COOH$

$C_{1}$ is making 4 sigma bonds (single bond) therefore it is $sp^3$ hybridised.

and $C_{2}$ is making a 2 sigma bonds with carbon and 1 sigma bond with oxygen and one pi bond with oxygen therefore, it is $sp^2$ hybridised.

Question 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.

Answer :

The shared pairs of electrons present between the bonded atoms are called bond pairs.

And all valence electrons may not participate in bonding; those electron pairs that do not participate in bonding are called lone pairs of electrons.

For example,

In $C_{2}H_{6}$ ethane, there are seven bond pairs but no lone pair is present.

In $H_{2}O$ , there are two bond pairs and two lone pairs on the central atom (oxygen).

Question 4.32 Distinguish between a sigma and a pi bond.

Answer :

The difference between the sigma bond and the pi bond is shown in the table below:

Question 4.33 Explain the formation of $H_{2}$ molecule on the basis of valence bond theory.

Answer :

Formation of $H_{2}$ molecule:

Assume that two hydrogen atoms $(A\ and\ B)$ with nuclei $(N_{A}\ and\ N_{B})$ and electrons $(e_A\ and\ e_B)$ are taken to undergo a reaction to form a hydrogen molecule.

When the two atoms are at a large distance, there is no interaction between them. As they approach each other, the attractive and repulsive forces start operating.

An attractive force arises between:

(a) The nucleus of one atom and its electron, i.e., $N_A- e_A$ and $N_B- e_B$ .

(b) The nucleus of one atom and an electron of another atom i.e., $N_A-e_B$ and $N_B-e_A$

Repulsive force arises between:

(a) Electrons of two atoms i.e., $e_A - e_B$ .

(b) Nuclei of two atoms i.e., $N_{A} - N_{B}$ .

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

The attractive force overcomes the repulsive force. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is achieved when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

Question 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Answer :

The important conditions required for the linear combination of atomic orbitals to form molecular orbitals are as follows:

1. The combining atomic orbitals must have the same or nearly the same energy.

2..The combining atomic orbitals must have the same symmetry about the molecular axis.

3. The combining atomic orbitals must overlap to the maximum extent.

Question 4.35 Use molecular orbital theory to explain why the $Be_{2}$ molecule does not exist.

Answer :

The electronic configuration of Be is $1s^22s^2$ .

From the molecular orbital electronic configuration, we have for $Be_{2}$ molecule,

$\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}$

We can calculate the bond order for $Be_{2}$ is $= \frac{1}{2}(N_{b}-N_{a})$ where,

$N_{b}$ is the number of electrons in bonding orbitals and $N_{a}$ is the number of electrons in anti-bonding orbitals.

So, therefore we have,

Bond order of $Be_{2} = \frac{1}{2}(4-4) = 0$

that means that the molecule is unstable.

Hence, $Be_{2}$ molecule does not exist.

Question 4.36 Compare the relative stability of the following species and indicate their magnetic properties;

$O_{2},O^{+}_{2},O^{-}_{2}-(superoxide),\ O_{2}^{2-} (peroxide)$

Answer :

The electronic configuration of $O_{2}$ molecule can be written as:

$(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^1 \equiv\pi^*2p_{y}^1)$

Here, the number of bonding electrons is $N_{b} = 10$ and the number of antibonding electrons is $N_{a} = 6$.

Therefore,

$Bond\ order = \frac{1}{2}(N_{b}-N_{a})$

$= \frac{1}{2}(10-6) = 2$

The electronic configuration of $O_{2}^+$ molecule can be written as:

$(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^1 )$

Here, the number of bonding electrons is $N_{b} = 10$ and the number of antibonding electrons is $N_{a} = 5$.

Therefore,

$Bond\ order = \frac{1}{2}(N_{b}-N_{a})$

$= \frac{1}{2}(10-5) = 2.5$

The electronic configuration of $O_{2}^-$ molecule can be written as:

$(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv\pi^*2p_{y}^1)$

Here, the number of bonding electrons is $N_{b} = 10$ and the number of antibonding electrons is $N_{a} = 7$.

Therefore,

$Bond\ order = \frac{1}{2}(N_{b}-N_{a})$

$= \frac{1}{2}(10-7) = 1.5$

The electronic configuration of $O_{2}^{2-}$ molecule can be written as:

$(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv\pi^*2p_{y}^2)$

Here, the number of bonding electrons is $N_{b} = 10$ and the number of antibonding electrons is $N_{a} = 8$.

Therefore,

$Bond\ order = \frac{1}{2}(N_{b}-N_{a})$

$= \frac{1}{2}(10-8) = 1$

Therefore, the bond dissociation energy is directly proportional to the bond order.

Thus, the higher the bond order, the greater will be the stability.

We get this order of stability:

$O_{2}^+>O_{2}>O_{2}^{-}>O_{2}^{2-}$

Question 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals.

Answer :

Wave functions can be used to represent molecular orbitals. The plus and minus represent the positive wave function and negative wave function, respectively.

Question 4.38 Describe the hybridisation in case of $PCl_{5}.$ Why are the axial bonds longer as compared to equatorial bonds?

Answer :

The initial ground state and final excited state electronic configuration of phosphorus (P) are:

So, the phosphorus atom is $sp^3d$ hybridized in the excited state. The donated electron pairs by five Cl atoms are filled and make $PCl_{5}.$.

The resultant shape is trigonal bipyramidal and the five $sp^3d$ hybrid orbitals are directed towards the five corners.

In the five P-Cl sigma bonds, three lie in one plane and make $120^{\circ}$ with each other, are equatorial bonds and the two P-Cl bonds lie above and below the equatorial plane making an angle of $90^{\circ}$ with the plane, are axial bonds.

So, because of more repulsion from the equatorial bond pairs, the axial bonds are slightly longer than the equatorial bonds.

Question 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Answer :

A hydrogen bond can be defined as the attractive force acting between the hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule.

Because of the difference between electro-negativities, the bond pair between hydrogen and the electronegative atom drifts towards a more electronegative atom. As a result, the hydrogen atom becomes slightly positively charged.

Hydrogen bonds are stronger than van der Waals forces because H-bonds are considered as an extreme form of dipole-dipole interaction.

Question 4.40 What is meant by the term bond order? Calculate the bond order of :

$N_{2}$

$O_{2}$ ,

$O_{2}^{+}$

$O_{2}^{-}$

Answer :

Bond order (B.O.) is defined as one-half the difference between the number of electrons present in the bonding and the antibonding orbitals of a molecule.

$Bond\ Order = \frac{1}{2}(N_{b}-N_{a})$

Where $N_{b}\ and N_{a}$ are the number of electrons occupying bonding orbitals and the number occupying the antibonding orbitals, respectively.

So, the bond orders for different molecules are:

$N_{2}$ : The electronic configuration is $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2$

$(\pi2p_{x})^2(\pi2p_{y})^2(\sigma2p_{z})^2$

Where, the number of bonding electrons $N_{b} =10$ and number of antibonding electrons, $N_{a} =4$

So, Bond order of nitrogen molecule $= \frac{1}{2}(10-4) = 3$

$O_{2}$ : The electronic configuration is $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^1\equiv \pi^*2p_{y}^1)$

Where, the number of bonding electrons $N_{b} =10$ and number of antibonding electrons, $N_{a} =6$

So, Bond order of nitrogen molecule $= \frac{1}{2}(10-6) = 2$

$O_{2}^{+}$ : The electronic configuration is $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^1)$

Where, the number of bonding electrons $N_{b} =8$ and number of antibonding electrons, $N_{a} =3$

So, Bond order of $O_{2}^{+}$ molecule $= \frac{1}{2}(8-3) = 2.5$

$O_{2}^{-}$ : The electronic configuration is $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2$

$(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv \pi^*2p_{y}^1 )$

Where, the number of bonding electrons $N_{b} =8$ and number of antibonding electrons, $N_{a} =5$

So, Bond order of $O_{2}^{-}$ molecule $= \frac{1}{2}(8-5) = 1.5$