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

## NCERT Solutions Class 11 Chemistry Chemical Bonding and Molecular Structure – Free PDF Download

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure are provided in this article for CBSE students. Comprehensive answers to every question listed in the NCERT Class 11 Chemistry textbook can be found here. NCERT Solutions are drafted by the highly experienced faculty at BYJU’S to help students who are preparing for the board exam.

Chemistry is the basis of things we see in our environment and is known as the “central science”. As Chemistry is a mandatory subject for students, it requires more focus from the exam perspective. The solutions provided here are equipped with all the basic details for questions that might be asked in the board exam. Furthermore, the NCERT Solutions for Class 11 Chemistry can also be downloaded in a PDF format (for free) by clicking the download button provided above.

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

Chemical Bonding and Molecular Structure” is the fourth chapter of the term – I CBSE Class 11 Chemistry Syllabus for session 2022-23. This chapter touches on several fundamental concepts in the field of Chemistry (such as hybridization and the modern theories on chemical bonding). The concepts covered in this chapter are listed below.

• VSEPR theory
• Lewis structures
• Valence bond theory
• The polar character of covalent bonds
• The concept of hybridization
• The molecular orbital theory of homonuclear diatomic molecules
• Hydrogen bonding

The NCERT Solutions for Class 11 Chemistry (Chapter 4) provided on this page feature the following types of questions:

• Drawing Lewis dot symbols for atoms, molecules and polyatomic ions
• Questions on bond parameters
• Expressing resonance with the help of Lewis structures
• Questions related to dipole moment, bond polarity and polar covalent bonds
• Questions on hybridization
• Questions on VBT, MOT and the VSEPR theory

The concept-focused approach followed during the creation of the Class 11 Chemistry NCERT Solutions makes them extremely useful during chapter revisions. Crafted by highly experienced subject experts, the Chemistry NCERT Solutions for Class 11 Chapter 4 provided by BYJU’S enable students to solve similar questions with ease. Moreover, solutions to the relatively complex questions feature step-by-step algorithms that can be followed while solving them.

### Subtopics of Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure

1. Octet Rule
• Covalent Bond
• Lewis Representation of Simple Molecules (The Lewis Structures)
• Formal Charge
• Limitations of the Octet Rule
1. Ionic or Electrovalent Bond
• Lattice Enthalpy
2. Bond Parameters
• Bond Length
• Bond Angle
• Bond Enthalpy
• Bond Order
• Resonance Structures
• Polarity of Bonds
3. The Valence Shell Electron Pair Repulsion (VSEPR) Theory
4. Valence Bond Theory
• Orbital Overlap Concept
• Directional Properties Of Bonds
• Overlapping of Atomic Orbitals
• Types of Overlapping and Nature of Covalent Bonds
• The Strength of Sigma and Pi Bonds
5. Hybridisation
• Types of Hybridisation
• Other Examples of Sp3, Sp2 and Sp Hybridisation
• The Hybridisation of Elements Involving D Orbitals
6. Molecular Orbital Theory
• Formation of Molecular Orbitals: Linear Combination of Atomic Orbitals (LCAO)
• Conditions for the Combination of Atomic Orbitals
• Types of Molecular Orbitals
• Energy Level Diagram for Molecular Orbitals
• Electronic Configuration and Molecular Behaviour
7. Bonding in Some Homonuclear Diatomic Molecules
8. Hydrogen Bonding
• Cause of Formation of Hydrogen Bond
• Types of H-bonds.

## Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 4

Q-1) Explain the formation of a chemical bond.

Ans.)

“A chemical bond is an attractive force that bonds the constituents of a chemical species together.”

So many theories are suggested for chemical bond formation, such as valence shell electron pair repulsion theory, electronic theory, molecular orbital theory and valence bond theory.

The formation of a chemical bond is credited to the tendency of a system to achieve stability. It was noticed that the inertness of noble gasses is the direct result of their completely filled outermost orbitals. Consequently, it was proposed that the elements having a deficiency of electrons in outermost shells are unstable. Thus, atoms combine with one another and finish their separate octets or duplets to achieve the stable configuration of the closest inert gasses. So, this combination may occur either by sharing of electrons. The chemical bond formed as a result of sharing of electrons among atoms is known as a covalent bond.  Also, an ionic bond is formed as a result of sharing of electrons among atoms.

Q-2) Write Lewis dot symbols for atoms of the following elements :

a) Mg

b) Na

c) B

d) O

e) N

f) Br

Ans.)

a) Mg

The magnesium atom contains only 2 valence electrons. Thus, the Lewis dot symbols for Mg is

b) Na

The sodium atom contains only 1 valence electron. Thus, the Lewis dot symbol for Na is $Na\cdot$

c) B

The boron atom contains only 3 valence electrons. Thus, the Lewis dot symbols for B is

d) O

The oxygen atom contains only 6 valence electrons. Thus, the Lewis dot symbol for O is

e) N

The nitrogen atom contains only 5 valence electrons. Thus, the Lewis dot symbol for N is

f) Br

The bromine atom contains only 7 valence electrons. Thus, the Lewis dot symbols for Br is

Q-3) Write Lewis symbols for the following atoms and ions:

S and S2-; Al and Al3+; H and H

Ans.)

For S and S2-

The sulphur atom contains only 6 valence electrons. Thus, the Lewis dot symbol for S is

The bi-negative charge on sulphur indicates that it has gained 2 electrons. So, there are six valance electrons plus two gained electrons.

Thus the Lewis dot symbol is

For Al and Al3+

An aluminium atom contains only 3 valence electrons. Thus, the Lewis dot symbols for Al is

The tri-positive charge on aluminium indicates that it has donated 3 electrons.

Thus, the Lewis dot symbol is

$[Al]^{3+}$

For H and H

The hydrogen atom contains only 1 valence electron. Thus, the Lewis dot symbol for H is

$H\cdot$

The single negative charge on hydrogen indicates that it has gained 1 electron. So, one valance electron plus one gained electron.

Thus lewis dot symbol is

Q-4) Draw the Lewis structures for the following molecules and ions :

H2S, SiCl4, BeF2$CO_{3}^{2-}$, HCOOH

Ans.)

H2S

SiCl4

BeF2

$CO_{3}^{2-}$

HCOOH

Q-5) Define the octet rule. Write its significance and limitations.

Ans.)

Octet rule says, “Atoms can combine either by transfer of valence electrons from one atom to another or by sharing their valence electrons in order to achieve the nearest inert gas configuration by having an octet in their valence shell.”

Octet rule explains chemical bond formation depending upon the nature of the element.

Limitations:

(a) Octet rule fails to predict the relative stability and shape of the molecules.

(b) It is based on the inert nature of noble gases. But, some inert gases, say krypton(Kr) and xenon(Xe), form compounds like KrF2, XeF2 etc.

(c) For elements beyond the 3rd period, the octet rule cannot be applied. Elements present beyond the 3rd period have more than 8 valence electrons surrounding the central atom. E.g. SF6, PF6 etc.

(d) For atoms in a molecule having an odd number of electrons, the octet rule is not applied. E.g. For No2 and NO octet rule is not applicable.

(e) If a compound has less than 8 electrons surrounding the central atom, then the octet rule cannot be applied to that compound. E.g. BeH2, AlCl3, LiCl etc., does not obey the octet rule.

Q-6) Write the favourable factors for the formation of an ionic bond.

Ans.)

The formation of an ionic bond takes place through the transfer of 1 or more electrons from one atom to another. Thus, ionic bond formation depends on the flexibility of neutral atoms to lose or gain electrons. The formation of an ionic bond also depends on the lattice energy of the compound which is formed.

The factors that are favourable for ionic bond formation:

(a) High electron affinity of atoms of non-metals.

(b) The high lattice energy of the compound which is formed.

(c) Low ionization enthalpy of an atom of metal.

Q-7) Discuss the shape of the following molecules using the VSEPR model:

BeCl2, BCl3, SiCl4, AsF5, H2S, PH3

Ans.)

BeCl2

The central atom does not have any lone pair but has 2 bond pairs. Thus, its shape is AB2, i.e., linear shape.

BCl3

The central atom does not have any lone pair but has 3 bond pairs. Thus, its shape is AB3, i.e., trigonal planar.

SiCl4

The central atom does not have any lone pair but has 4 bond pairs. Thus, its shape is AB4,  i.e., tetrahedral.

AsF5

The central atom does not have any lone pair but has 5 bond pairs. Thus, its shape is AB5, i.e., trigonal bipyramidal.

H2S

The central atom has 1 lone pair and has 2 bond pairs. Thus, its shape is AB2E, i.e., bent shape.

PH3

The central atom has 1 lone pair and has 3 bond pairs. Thus, its shape is AB3E, i.e.,trigonalbipyramidal.

Q-8) Although geometries of NH3 and H2O molecules are distorted tetrahedral, the bond angle in water is less than that of  ammonia. Discuss.

Ans.)

The geometry of H2O and NH3:

The central atom(N) in ammonia has 1 lone pair and has 3 bond pairs.

The central atom(O) in water has 2 lone pairs and has 2 bond pairs.

Thus, these 2 lone pairs on O- atom in the water molecule repels the 2 bond pairs. And this repulsion between the lone pair and the bond pair on O- atom of H2O is stronger than the repulsion between the lone pair and bond pair on the N-atom of NH3.

Thus, the bond angle in H2O is less than NH3, even though they have a distorted tetrahedral structure.

Q-9) How do you express the bond strength in terms of bond order?

Ans.)

The extent of bonding which occurs between two atoms while forming a molecule is represented by bond strength. As the bond strength increases, the bond becomes stronger, and the bond order increases.

Q-10) Define Bond length.

Ans.)

“Bond length is defined as the equilibrium distance between the nuclei of 2 bonded atoms in a molecule.”

Q-11) Explain the important aspects of resonance with reference to the CO32− ion.

Ans.)

Experimental results shows that, all the C-O bond in $CO_{3}^{2-}$ are equivalent.

Thus, it is inefficient to represent $CO_{3}^{2-}$ ion by a single Lewis structure, which has 1 double bond and 2 single bonds.

Thus, the resonance structures of $CO_{3}^{2-}$ is :

Q-12) H3PO3 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 H3PO3? If not, give reasons for the same.

Ans.)

In the given structures, the position of atoms is changed, so we cannot take the 2 given structures as a canonical form of resonance hybrid which represents H3PO3.

Q-13) Write the resonance structures for SO3, NO2, and NO3.

Ans.)

SO3

NO2

$NO_{3}^{-}$

Q-14) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :

(i) K and S

(ii) Ca and O

(iii) Al and N.

Ans.)

(i) K and S

Electronic configurations of S and K are:

S: 2, 8, 6

K: 2, 8, 8, 1

Here, it is clear that K has 1 more electron than the nearest inert gas, i.e., Ne, whereas S needs 2 electrons to complete its octet. Thus, the transfer of electrons takes place in the following way,

(ii) Ca and  O

Electronic configurations of O and Ca are:

O: 2, 6

Ca: 2, 8, 8, 2

Here, it is clear that Ca has 2  more electrons than the nearest inert gas Ar, whereas O needs 2 electrons to complete its octet. Thus, the transfer of electrons takes place in the following way,

(iii) Al and N

Electronic configurations of N and Al are:

N: 2, 5

Al: 2, 8, 3

Here, it is clear that Al has 3 more electrons than the nearest inert gas Ne, whereas N needs 3 electrons to complete its octet. Thus, the transfer of electrons takes place in the following way,

Q-15) Although both CO2 and H2O are triatomic molecules, the shape of the H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.

Ans.)

Experimental results show that the dipole moment of CO2 is 0. And it is possible only if the shape of the molecule is linear as dipole moments of a bond between C-O are equal and opposite. So, it nullifies each other.

$∴ Resultant,\; \mu = 0$

H2O has 1.84 D dipole moment. The value of dipole moments indicates that the structure of water molecule is bent as dipole moments of the bond between O-H are unequal.

Q-16) Write the significance/applications of dipole moment.

Ans.)

The following are some of the key significance of the dipole moment:

• The molecule’s shape can be determined. Symmetrical molecules, such as linear, have zero dipole moments, whereas non-symmetrical molecules take on varied shapes, such as bent or angular.
• In order to determine the polarity of molecules. The polarity will be greater if the dipole moment is greater, and vice versa.
• We can say that if a molecule has zero dipole moment, it is non-polar, and if it has some polar character, it is non-polar.

Q-17) Define electronegativity. How does it differ from electron gain enthalpy?

Ans.)

“Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself”.

 Sr. No Electronegativity Electron affinity 1 A tendency to attract the shared pairs of electrons for an atom which is in a chemical compound is its electronegativity. A tendency to gain electrons for an isolated gaseous atom is its electron gain enthalpy. 2 It varies according to the element with which it is bounded. It does not vary according to the element with which it is bounded. 3 It is not constant for any element. It is constant for an element. 4 It is not a measurable quantity. It is a measurable quantity.

Q-18) Explain with the help of suitable example polar covalent bond.

Ans.)

When two unique atoms having distinct electronegativities join to form a covalent bond, the bond pair of electrons are not shared equally. The nucleus of an atom having greater electro-negativity attracts the bond pair. So, the electron distribution gets distorted and an electronegativity atom attracts the electron cloud.

Thus, the electronegative element gets slightly negatively charged and on the other hand, the other atom gets slightly positively charged. As a result of this, two opposite poles are developed in a molecule, and this type of bond formed is termed a ‘polar covalent bond’.

E.g. HCl is having a polar covalent bond. In HCl, Cl- atom has more electronegativity than H- atom. Thus, bond pair shifts towards Cl- atom and because of that, it acquires a positive charge.

Q-19) Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2, and ClF3.

Ans.)

The ionic character of a molecule depends on the difference in electronegativity between constituents atoms. So, the higher the difference is, the higher the ionic character of a molecule will be.

So, the required order of ionic character of the given molecules is

N2< SO2< ClF3< K2O <LiF.

Q-20) The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Ans.)

Correct Lewis structure of CH3COOH is given below:

Q-21) Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar.

Ans.)

Electronic configuration of C- atom:

6C :$1s^{2}\;2s^{2}\;2p^{2}$

The orbital picture of C- atom in excited state is:

Thus, C- atom undergoes sp3 hybridization in methane molecule and forms a tetrahedral structure.

For square planer geometry, C-atom should have dsp2 hybridization. But as C- atom does not have a d- orbital, it cannot undergo dsp2 hybridization. Thus, Methane’s geometry cannot be square planer.

Also in square planar geometry the bond angle is $90^{\circ}$ so the stability is not there because of the repulsion between bond pairs. So, as per VSEPR theory, methane’s tetrahedral structure is perfect.

Q-22) Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.

Ans.)

Lewis structure of BeH2 is:

Central atom is not having any lone pair but has 2 bond pairs. Thus, its shape is AB2, i.e., linear shape.

Thus, the dipole moment of Be- H bond is equal and opposite in direction, nullifying one another. Thus, the dipole moment of BeH2 is 0.

Q-23) Which out of NH3 and NF3 has higher dipole moment and why?

Ans.)

N- atom is the central atom of NF3 and NH3.

Central atom has 1 lone pair and is having 3 bond pairs. Thus,  for both, the shape is AB3E, i.e., pyramidal.

As, the F-atom has more electronegativity than the H- atom, NF3 should have a higher dipole moment than NH3. But the dipole moment of NH3 is 1.46D which is higher than the dipole moment of NF3, which is 0.24D.

It gets clear from the directions of dipole moments of individual bonds in NF3 and NH3.

As both the N-H bonds are in the same direction, it adds to the bond moment of the lone pair, while N-F bonds are in the opposite direction. So they partly cut the bond moment of lone pair.

Thus, dipole moment of NH3 is higher than that of NF3.

Q-24) What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.

Ans.)

“Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes”.

E.g. 1 s- orbital hybridises with 3 p- orbitals to form 4 sp3 hybrid orbitals.

(a) sp hybrid orbital

1 s- orbital hybridises with 1 p- orbitals to form 2 sp hybrid orbitals. sp hybrid orbital has a linear shape. The formation of sp orbital is:

(b) sp2 hybrid orbital

1 s- orbital hybridises with 2 p- orbitals to form 3 sp2 hybrid orbitals. The shape of sp2 orbital is trigonal planar.

(c) sp3 hybrid orbital

1 s- orbital hybridises with 3 p- orbitals to form 4 sp3 hybrid orbitals. The shape of sp3 orbital is a tetrahedron.

Q-25) Describe the change in hybridisation (if any) of the Al atom in the following reaction.

AlCl3 + $Cl^{-}$ —> $AlCl_{4}^{-}$

Ans.)

The ground state of the valence orbital of Al –atom is:

In excited state the orbital picture of Al- atom is:

Thus, Al -atom in AlCl3 undergoes sp2 hybridisation and forms trigonal planar geometry. For the formation of $AlCl_{4}^{-}$ the vacant 3pz orbital will also get involved. Thus, sp2 hybridisation is converted into sp3 hybridisation and forms a tetrahedral structure.

Q-26) Is there any change in the hybridisation of B and N atoms as a result of the following reaction?

BF3 + NH—> F3B.NH3

Ans.)

N- atom in NH3 has sp3 hybridization. The orbital picture of N- atom is shown below:

B- atom in NF3 has sp2 hybridisation . The orbital picture of B- atom is shown below:

On the reaction of NH3 and BF3, F3B.NH3 is obtained as a product, as hybridization of B-atom is changed to sp3. Although, hybridization of N- atoms remain unchanged.

Q-27) Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2
molecules.

Ans.)

C2H4

The electronic configuration of a carbon atom in excited state is given below:

6C: $1s^{2}2s^{1}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}$

In the formation of C2H4 (ethane) molecule 1 sp2 orbital of the C- atom overlaps sp2 orbital of the other C- atom, thus forming a C-C sigma bond.

The 2 remaining sp2 orbitals of every C- atom forms sp2-s $\sigma$ bond with 2 H- atoms. One c- atom having unhybridized orbital overlaps with the unhybridized orbital of the other C- atom and forms a pie bond.

C2H2

In the formation of the ethyne(C2H2) molecule, C- atom has sp hybridization with 2 2p- orbitals in an unhybridized state.

1 sp orbital of each C- atom overlaps the inter-nuclear axis and forms C-C sigma bond. The 2ndsp orbital of each C- atom overlaps half-filled 1s orbital so as to form a sigma bond.

The triple bond between the 2 C- atoms has 1 sigma bond and 2 Pie bonds. This is because 2 unhybridized 2p- orbitals overlap with the 2p- orbital of other C- atom, thus forming 2 pie bonds.

Q-28) What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4

Ans.)

A single bond is formed as the axis of bonding orbital overlaps. Thus, it forms a $\sigma$ bond. By sidewise overlapping of orbital double and triple bonds, i.e., multiple bonds are formed.$\pi$ bond is always present in the multiple bonds. Triple bond consist of 2 $\pi$ and 1$\sigma$ bond.

(a) C2H2

Thus, there are 2 $\pi$ bonds and 3 $\sigma$ in C2H2.

(b) C2H4

Thus, there are 1 $\pi$ bonds and 5 $\sigma$ in C2H4.

Q-29) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?

(a) 1s and 1s   (b) 1s and 2px   (c) 2py and 2py   (d) 1s and 2s

Ans.)

(c) 2py and 2py

2py and 2py orbitals won’t form a $\sigma$ as it will undergo lateral overlapping and will form a $\pi$ bond.

Q-30) Which hybrid orbitals are used by carbon atoms in the following molecules?

(a) CH3-CH3; (b) CH3-CH=CH2; (c) CH3CH2-OH; (d) CH3-CHO; (e) CH3COOH.

Ans.)

(a) CH3-CH3

Here, C1 and C2 have sp3 hybridization.

(b) CH3-CH=CH2

Here, C3 and C2 have sp2 hybridization and C1 has sp3 hybridization.

(c) CH3-CH2-OH

Here, C1 and C2 have sp3 hybridization.

(d) CH3-CHO

Here, C1 has sp3 hybridization and C2 has sp2 hybridization.

(e) CH3COOH

Here, C1 has sp3 hybridization and C2 has sp2 hybridization.

Q-31) What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Ans.)

A covalent bond is formed when 2 atoms combine with each other by sharing their valence electrons. The shared pairs of electrons present between the bonded atoms are called bond pairs. Each and every electron cannot participate in bonding. The pairs of electrons which do not participate in bonding are called lone pairs.

E.g. a) Ethane has 7 bond pairs but zero lone pair.

b) Water has 2 bond pairs and 2 lone pairs on O- atom.

Q-32) Distinguish between a sigma and a pi bond.

Ans.)

 Sr. No. Pi bond Sigma bond 1 Pi bond is formed by lateral overlapping of orbitals. Sigma bond is formed by end-to-end overlapping of orbitals. 2 It is a comparatively weak bond. It is a comparatively strong bond. 3 There is only one overlapping orbital is p-p. The overlapping orbitals are s-s, s-p, p-p. 4 Rotation around pi- bond is restricted. Rotation is possible around sigma bond. 5 Electron cloud is not symmetrical about the line joining 2 nuclei. Electron cloud is symmetrical about the line joining 2 nuclei. 6 It has 2 electron clouds, one above the plane of atomic nuclei and one below the plane of atomic nuclei. It has 1 electron cloud and that is symmetrical about the inter-nuclear axis.

Q-33) Explain the formation of H2 molecule on the basis of valence bond theory.

Ans.)

Assuming 2 H- atoms X and Y with nuclei NX and NY and electrons eX and eY, respectively.

When X and Y are far from each other, there is no interaction between them. As soon as they come closer, the attractive force and repulsive force become active.

The repulsive forces are:

(i) Between electrons of both the atoms, i.e., eX and eY.

(ii) Between nuclei of both the atoms, i.e., NX and NY.

The attractive forces are:

(i) Between the electron and nucleus of the same atom i.e. NX – eX and NY -eY.

(ii) Between the electron of one atom and the nucleus of the other atom, i.e., NX–eY and NY–eX.

The repulsive force pushes the 2 atoms apart, whereas the attractive force tends to bring them together.

Repulsive forces:

Attractive forces:

The values of repulsive forces are less than that of attractive forces. Thus, 2 atoms approach each other. There is a decrease in potential energy. In the end, a stage is reached when the repulsive forces balance the attractive forces, and the system achieves the minimum energy, which leads to the formation of H2 molecule.

Q-34) Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Ans.)

The condition that is required for the linear combination of atomic orbitals to form molecular orbitals are as follows:

(i) The joining of atomic orbitals must have approximately the same energy. This implies in a homo-nuclear molecule, the 1s-orbital of one atom can join with the 1s- orbital of another atom but cannot join with the 2s-orbital.

(ii) The joining atomic orbitals must have legitimate orientations to ensure the maximum overlap.

(iii) The overlapping must be to a large extent.

Q-35) Use molecular orbital theory to explain why the Be2 molecule does not exist.

Ans.)

Electronic configuration of Be:

1s2 2s2

The molecular orbital electronic configuration of Be2 is:

$\sigma _{1s}^{2}\; \sigma _{1s}^{\cdot 2}\;\sigma _{2s}^{2}\;\sigma _{2s}^{\cdot2}$

Thus, bond order of Be2: 0.5(Nb – Na).

Nb: No. of electrons in the bonding orbitals

Na: No. of electrons in the anti-bonding orbitals

Therefore, bond order of Be2 = 0.5(4 – 4) = 0

Zero value of bond order indicates that the given molecule is unstable. Thus, Be2 doesn’t exist.

Q-36) Compare the relative stability of the following species and indicate their magnetic properties:

O2$O_{2}^{+}$$O_{2}^{-}$ (Superoxide), $O_{2}^{2-}$ (Peroxide)

Ans.)

O2 contain 16 electrons i.e. 8 electrons from each O- atom.

• The electronic configuration of O2 is:
$[\sigma – (1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (1p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}[\pi^{*} (2p_{y})]^{1}$

As 1s- orbital of each O- atom does not involve in the bonding,

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of $O_{2}^{+}$ is:
$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

As, bond dissociation energy $\propto$ bond order

Hence, the higher the bond order, the higher stability.

The arrangement according to decreasing order of stability is given as:

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

• Electronic configuration of $O_{2}^{-}$ (Superoxide) is:
$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{1}$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of $O_{2}^{2-}$ (peroxide) is:
$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{2}$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 6

Now,

Bond order = 0.5(8 – 6) = 1

Q-37) Write the significance of a plus and a minus sign shown in representing the orbitals

Ans.)

Generally, molecular orbital is represented by the ‘wave function’.

Positive (+) sign in representing a molecular orbital indicates positive wave function.

Negative (-) sign in representing a molecular orbital indicates negative wave function.

Q-38) Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?

Ans.)

The electronic configurations of the outer orbital of phosphorus in excited state and in ground state are given below:

Ground State:

Excited State:

Phosphorus atom has sp3d hybridization. These orbitals are filled due to the donation of electron pairs by 5 Cl- atoms: PCl5

The 5 sp3d hybrid orbitals present here are directed towards 5 corners of trigonal bipyramidal. Thus, the geometry of PCl5 is given below:

PCl5 contains 5 P- Cl sigma bonds. Out of which, 3 P-Cl bonds lie in only 1 plane, and they are making $120^{\circ}$ with each other. And as these bonds lie in 1 plane, they are known as equatorial bonds.

Out of 2 remaining P-Cl bonds, one bond lie above the equatorial plane, and one bond lie below the equatorial bond. And they are making $90^{\circ}$ with each other. These bonds are called an axial bond.

Equatorial bond pairs repel axial bond pairs to a large extent. So, equatorial bonds are slightly shorter than axial bonds.

Q-39) Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Ans.)

“H-bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule”.

As there is a difference in the electronegativities between the atoms, the bond pair electronegative atom and hydrogen atom is drifted away from H- atom. Therefore, the hydrogen atom gets electropositive w.r.t. the other atom and procures a positive charge.

$4^{\delta } – X^{\delta }\;………..H^{\delta +} – X^{\delta -}\;………….H^{\delta +} – X^{\delta -}$

The value of H- bond is minimum in gaseous state and maximum in the solid state.

Two types of hydrogen bonds are there:

(a) Intramolecular hydrogen bond, e.g., o- nitrophenol

(b) Intermolecular hydrogen bond, e.g., HF, H2O etc.

H- bonds are stronger than Van der Waals forces as H- bonds are regarded as an extreme form of the dipole-dipole interaction.

Q-40) What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+ and O2.

Ans.)

Bond Order: It is defined as 0.5 times the difference between the “number of electrons present in bonding orbitals and number of electrons present in anti-bonding orbitals” of a molecule.

Bond Order = 0.5(Nb – Na);

Na: No. of anti-bonding electrons

Nb: No. of bonding electrons

O2 contain 16 electrons, i.e., 8 electrons from each O- atom.

• Electronic configuration of O2 is:
$[\sigma – (1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (1p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}[\pi^{*} (2p_{y})]^{1}$As 1s- orbital of each O- atom does not involve in the bonding,

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of $O_{2}^{-}$ (superoxide) is:
$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{1}$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of $O_{2}^{+}$ is:
$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

• Electronic configuration of N2 is:
$[\sigma(1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi (2p_{z})]^{2}$

No. of bonding electrons = Nb = 10

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(10 – 4) = 3

 Also Access NCERT Exemplar for Class 11 Chemistry Chapter 4 CBSE Notes for Class 11 Chemistry Chapter 4

We at BYJU’S believe in creating seamless opportunities to help students excel at the highest level. Therefore, we have created our NCERT Solutions with the utmost care, aiming to help students clear their board examinations with flying colours. The NCERT Solutions for Class 11 Chemistry Chapter 4 provided here include relevant exercises that have been solved by our highly qualified subject experts. Detailed explanations have been provided in these solutions in order to give students a crystal-clear understanding of the concepts behind the questions.

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## Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 4

### Give me a summary of NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure.

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure has
1. VSEPR theory
2. Lewis structures
3. Valence bond theory
4. The polar character of covalent bonds
5. The concept of hybridization
6. The molecular orbital theory of homonuclear diatomic molecules
7. Hydrogen bonding

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Yes, if you want to score good in your final exams, you have to practise all the questions and formulae related to them. The solutions present on BYJU’S website are very accurate and clear. Students can start practising NCERT Solutions for Class 11 Chemistry Chapter 4 to score higher marks. These solutions can be helpful not only for the board exam preparation but also in solving homework and assignments.

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No, not at all. If you practise regularly, NCERT Solutions for Class 11 Chemistry Chapter 4 is not much difficult to understand. The main aim of these solutions is to provide a fundamental knowledge of Chemistry, which helps the students understand every concept clearly.
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