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 which 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

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 stable configuration of the nearest noble gases. As it was seen that the

noble gases are very stable and were inert to react to 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 4.2(a) Write Lewis dot symbols for atoms of the following elements : Aak

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Answer :

The Lewis dot symbol of Mg atom is;

As there are two valence electrons in Mg atom.

Hence, the Lewis dot symbol for Mg is: .

Question 4.2(b) Write Lewis dot symbols for atoms of the following elements :

Answer:

The Lewis dot symbol of Na atom is;

As there is only one valence electron in Na atom of Na.

Hence, the Lewis dot structure is .

Question 4.2(c) Write Lewis dot symbols for atoms of the following elements :

Answer :

The Lewis dot symbol of atom is;

As there are three valence electrons in atom.

Hence, the Lewis dot structure is

Question 4.2(d) Write Lewis dot symbols for atoms of the following elements :

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,

Answer :

The Lewis dot symbol of atom is;

As there are six valence electrons in an atom of .

Hence, the Lewis dot structure is

Question 4.2(e) Write Lewis dot symbols for atoms of the following elements :

Answer :

The Lewis dot symbol of atom is;

As there are five valence electrons in an atom of .

Hence, the Lewis dot structure is

Question 4.2(f) Write Lewis dot symbols for atoms of the following elements :

Answer :

The Lewis dot symbol of atom is;

As there are seven valence electrons in an atom of .

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Hence, the Lewis dot structure is

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

Answer :

As the number of valence electrons in sulphur is six.

Therefore its Lewis dot symbol of sulphur(S) is

And of 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:

Answer :

As the number of valence electrons in aluminium is three.

Therefore its Lewis dot symbol of aluminium(Al) is

And of is, if it has donated three electrons because of its tripositive charge.

Hence. the Lewis symbol is

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

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Answer :

As the number of valence electrons in hydrogen is one.

Therefore its Lewis dot symbol of hydrogen (H) is

And of is, if it has one more electron because of its a negative charge develops.

Hence. the Lewis symbol is

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

Answer :

The Lewis structure of is:

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

Answer :

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The Lewis structure of is:

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

Answer :

The Lewis structure of is:

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

Answer :

The Lewis structure of is:

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

Answer :

The Lewis structure of 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.

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• Odd-electron molecules - the octet rule is not satisfied for all the atoms

in .

• 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 :

Formation of 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:

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,

Answer :

Using the VSEPR model we have,

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

is of the type

hence, it has a linear shape .

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

(b)

Answer :

Using the VSEPR model we have,

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

is of the type

hence, it has trigonal planar shape. a

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

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(c)

Answer :

Using the VSEPR model we have,

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

is of the type

hence, it has tetrahedral shape.

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

(d)

Answer :

Using the VSEPR model we have,

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

is of the type

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Hence, it has trigonal bipyramidal shape.

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

Answer :

Using the VSEPR model we have,

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

is of the type

Hence, it has a bent shape.

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

(f)

Answer :

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Using the VSEPR model we have,

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

is of the type

Hence, it has trigonal bipyramidal shape.

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

Answer :

Bond Strength gives us that amount of energy needed to break a bond between atoms

forming 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. Aak

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The bond length in a covalent molecule AB.

where (R is the bond length and and are covalent radii of atoms A

and B respectively.

Question 4.11 Explain the important aspects of resonance with reference to

the 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. Aakas

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Question 4.12 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 .? If not, give reasons for the same.

Answer :

As per the rule, it is not having the same position as the atoms and is changed.

Hence the given structures cannot be taken as the canonical forms of the resonance

hybrid.

Question 4.13 Write the resonance structures for and

Answer :

The resonance structures Aakas

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The resonance structures

The resonance structures

Question 4.14(a) Use Lewis symbols to show electron transfer between the following

atoms to form cations and anions :

Answer :

K and S:

We have the electronic configurations of both:

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having 1 electron in the valence shell, and it can donate 1 electron to get

to the nearest noble gas configuration.

having 6 electrons in the valence shell, and it wants to 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 :

Answer :

:

We have the electronic configurations of both:

having 2 electrons in the valence shell, and it can donate 2 electrons to

get to the nearest noble gas configuration.

having 6 electrons in the valence shell, and it wants to complete its octet by

accepting 2 more electrons.

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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 :

Answer :

:

We have the electronic configurations of both:

having 3 electron in the valence shell, and it can donate 3 electron to get to

the nearest noble gas configuration.

having 3 electrons in the valence shell, and it wants to complete its octet by

accepting 2 more electrons.

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

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Question 4.15 Although both and are triatomic molecules, the shape

of molecule is bent while that of 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. Net dipole moment of 6.17 × 10–30 C m is the resultant of the dipole

moments of two O–H bonds.

While 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 has opposite

dipole moments cancelling each other.

Question 4.16 Write the significance/applications of dipole moment.

Answer :

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

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 bent shape or some angular shapes.

For determining the polarity of the molecules. Greater the dipole moment value, more

will be the polarity and vice-versa.

We can say that if a molecule has 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, , 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 4.18 Explain with the help of suitable example polar covalent bond.

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A polar covalent bond, when two different atoms are linked to each other by covalent

bond, then the shared electron pair will not be in the centre just because the bonding

atoms differ in electronegativities.

For examples, in ,

Here slightly positive charges are developed in hydrogen atoms and slightly negative

charge developed in oxygen atom as oxygen is more electronegative than the

hydrogen. Thus, opposite poles are developed in the molecule.

Hence the bond pair lies towards oxygen atom.

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

molecules:

Answer :

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The ionic character in a molecule depends on the electronegativity difference between

the constituting atoms. More the difference more will be the ionic character of the

molecule.

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

molecules.

.

Question 4.20 The skeletal structure of 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.

Therefore, the correct skeletal structure of as shown below:

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Question 4.21 Apart from tetrahedral geometry, another possible geometry for is

square planar with the four atoms at the corners of the square and the atom at its

centre. Explain why is not square planar ?

Answer :

The electronic configuration of carbon atom is .

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

Hence the carbon atom undergoes hybridization in methane molecule and takes a

tetrahedral shape.

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

But here the absence of d-orbital, as a result, it does not undergo hybridization, the

structure of methane cannot be square planar.

Also the reason that bond angle in square planar which makes the molecule more

unstable because of repulsion between the bond pairs.

Hence according to VSEPR theory molecule take a tetrahedral structure.

Question 4.22 Explain why molecule has a zero dipole moment although

the bonds are polar.

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Answer :

molecule has a zero dipole moment because the two equal bond dipoles point in

the opposite directions and cancel the effect of each other.

Question 4.23 Which out of and has higher dipole moment and why ?

Answer :

Here both have central atom Nitrogen 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 to the hydrogen. Hence it is

expected that the net dipole moment of is greater than .

However has the net dipole moment of 1.46D and has the net dipole

moment of 0.24D. which is greater than .

This is because of the direction of the dipole moments of each individual bond

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The moments of the lone pair in partly cancel out. But in the resultant

moment add up to the bond moment of the lone pair.

Question 4.24 What is meant by hybridisation of atomic orbitals ? Describe the shapes

of , , hybrid orbitals.

Answer :

Hybridisation which can be defined as the process of intermixing of 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 , , hybrid orbitals are shown:

hybrid orbital: It is linear in shape and formed by intermixing of s and p

orbitals.

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hybrid orbital: It is the trigonal planar shape and is formed by the intermixing

of one s-orbital and two 2p-orbitals.

hybrid orbital: It is tetrahedron 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 atom in the

following reaction.

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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 was hybridisation and hence having a trigonal planar

shape.

Then it reacts with chloride ion to form . Where it has the empty orbital which

gets involved and the hybridisation changes from .

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

Question 4.26 Is there any change in the hybridisation of and atoms as a result of

the following reaction?

Answer :

Initially boron atom was in hybridised. The valence orbital of boron in the

excited state can be shown as:

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And nitrogen atom in is hybridised. The valence orbital of nitrogen in the

excited state can be shown as:

Then after the reaction has occured the product is formed by the

hybridisation of 'B' changes to . 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 and molecules.

Answer :

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

Formation of an ethane molecule by overlapping of a hybridized orbital of

another carbon atom, thereby forming a sigma bond.

The remaining two orbitals of each carbon atom from a 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.

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Formation of 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 sp hybrid orbital of the

other carbon atom to form C–C sigma bond, while the other hybridised orbital of each

carbon atom overlaps axially with the half-filled s orbital of hydrogen atoms forming σ

bonds

Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise 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

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

molecules?

Answer :

Given molecule :

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So, there is three sigmas (2C-H bonds + 1 C-C bond) and two pi-bonds (2 C-C bonds)

in .

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

molecules?

Answer :

Given molecule :

So, there are five sigma (4C-H bonds + 1 C-C bond) and one pi-bonds (C-C bonds)

in .

Question 4.29(a) Considering x-axis as the internuclear axis which out of the following

will not form a sigma bond and why?

Answer :

Orbitals will form a sigma bond as both orbitals are spherical and can

combine along x-axis as the internuclear axis.

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Question 4.29(b) Considering x-axis as the internuclear axis which out of the following

will not form a sigma bond and why?

Answer :

Orbitals will form a sigma bond as 1s orbital and 2p x orbital are align such

that they can combine along 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?

Answer :

Orbitals will not form a sigma bond as both 2p y orbital are align in y -

direction but the internuclear axis is x-axis. Aakas

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Formation of 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?

Answer :

Orbitals 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?

Answer :

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There are 4 sigma bonds (single bond) each with the help of one s hybrid orbital and 3 p

hybrid orbital, Hence C 1 and C 2 are hybridized.

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

molecules?

Answer :

is making 4 sigma bonds (single bond) therefore it is hybridised.

While are making a double bond.

Therefore they both are hybridized.

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

molecules?

Answer :

is making 4 sigma bonds (single bond) therefore it is hybridised.

and is also making a 4 sigma bonds. therefore it is also hybridised.

Therefore they both are hybridized.

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

molecules?

Answer :

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is making 4 sigma bonds (single bond) therefore it is hybridised.

and is making a 3 sigma bonds with hydrogen, carbon and oxygen. and one pi bond

with oxygen therefore it is hybridised.

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

molecules?

Answer :

is making 4 sigma bonds (single bond) therefore it is hybridised.

and is making a 2 sigma bonds with carbon and 1 sigma bond with oxygen and one

pi bond with oxygen therefore it is hybridised.

Question 4.31 What do you understand by bond pairs and lone pairs of electrons?

Illustrate by giving one exmaple of each type.

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Answer :

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

And all valence electrons may not participate in bonding that electron pairs that do not

participate in bonding are called lone pairs of electrons.

For examples,

In ethane, there are seven bond pairs but no lone pair is present.

In , 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 :

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

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Sigma Bond Pi Bond

(a) Formed by end to end overlapping of orbitals. Formed by the lateral overlapping of orbitals

(b) Sigma bonds are stronger than the pi bond. Weak bond.

(c) The orbitals involved in the overlapping are s-s,

s-p, p-p.

Bonds are formed only with overlapping of p-

p orbitals.

(d) The electron cloud is symmetrical about an

internuclear axis.

The electron cloud is not symmetrical.

(e) Free rotation is possible in case of a sigma bond. Rotation is restricted in case of pi-bonds.

Question 4.33 Explain the formation of molecule on the basis of valence bond

theory.

Answer :

Formation of molecule:

Assume that two hydrogen atoms with nuclei and

electrons 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.

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Attractive force arises between:

(a) The nucleus of one atom and its own electron i.e., and .

(b) The nucleus of one atom and electron of another atom i.e., and

Repulsive force arises between:

(a) Electrons of two atoms i.e., .

(b) Nuclei of two atoms i.e., .

The force of attraction brings the two atoms together, whereas the force of repulsion

tends to push them apart.

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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 molecule does not

exist.

Answer :

The electronic configuration of Be is .

From the molecular orbital electronic configuration, we have for molecule,

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We can calculate the bond order for is where,

is the number of electrons in bonding orbitals and is the number of electrons in

anti-bonding orbitals.

So, therefore we have,

Bond order of

that means that the molecule is unstable.

Hence, molecule does not exist.

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

magnetic properties;

Answer :

The electronic configuration of molecule can be written as:

Here the number of bonding electrons is and the number of antibonding

electrons is .

Therefore,

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The electronic configuration of molecule can be written as:

Here the number of bonding electrons is and the number of antibonding

electrons is .

Therefore,

The electronic configuration of molecule can be written as:

Here the number of bonding electrons is and the number of antibonding

electrons is .

Therefore,

The electronic configuration of molecule can be written as:

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Here the number of bonding electrons is and the number of antibonding

electrons is .

Therefore,

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:

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 while negative wave function respectively.

Question 4.38 Describe the hybridisation in case of 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:

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So, the phosphorus atom is hybridized in the excited state.The donated electron

pairs by five Cl atoms are filled and make .

The resultant shape is trigonal bipyramidal and the five hybrid orbitals are directed

towards the five corners.

The five P-Cl sigma bonds, three lies in one plane and make with each other

are equatorial bonds and the two P-Cl bonds lie above and below the equatorial plane

makes an angle of with the plane are axial bonds.

So, just because of more repulsion from the equatorial bond pairs, the axial

bonds are slightly longer than equatorial bonds.

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

forces?

Answer :

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 gets drifted towards a more electronegative atom. As a

result, the hydrogen atom becomes slightly positively charged.

Hydrogen bonds are stronger than the 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 :

,

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.

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Where are the number of electrons occupying bonding orbitals and the

number occupying the antibonding orbitals respectively.

So, bond order for different molecules are:

: The electronic configuration

is

Where, the number of bonding electrons and number of antibonding

electrons,

So, Bond order of nitrogen molecule

: The electronic configuration is

Where, the number of bonding electrons and number of antibonding

electrons,

So, Bond order of nitrogen molecule

: The electronic configuration

is

Where, the number of bonding electrons and number of antibonding

electrons,

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So, Bond order of molecule

: The electronic configuration is

Where, the number of bonding electrons and number of antibonding

electrons,

So, Bond order of molecule

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