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- In solutions of NCERT class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure chapter, you will deal with a concept like what is an ionic bond and a covalent bond, the polar character of covalent bonds, valence electrons, Lewis structure, VSEPR theory, the concept of hybridization that involves s, d and p orbitals, the molecular orbital theory (MOT) and more. In CBSE NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure, there are 40 questions in the exercise. The solutions of NCERT for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure are designed and solved by chemistry experts. These NCERT solutions of class 11 can help you in your preparation for class 11 final examination as well as in the various competitive exams like NEET, JEE Mains, BITSAT etc. If you are looking for an answer from any other chapter even from any other class then go with NCERT Solutions, there you will get all the answers of NCERT easily.

The NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure is an important chapter for both class 11 final examination and for various competitive exams like JEE and NEET. After completing NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure, you will be able to understand Kossel-Lewis approach to chemical bonding; explain the octet rule, draw Lewis structures of simple molecules; explain the formation of different types of bonds, able to describe VSEPR theory; explain the valence bond approach for the formation of covalent bonds; explain the concept of different types of hybridization that involves s, d and p orbitals; molecular orbital theory(MOT) and hydrogen bonding.

 

What is Chemical Bond?

The matter is made up of one or more than one type of elements but under normal conditions, except noble gases, no other element exists as an independent atom in nature. As independent entities, molecules are the smallest particles of matter. These molecules are cluster or group of atoms of same or different elements which are found exist together as a single unit having characteristic properties. So, a chemical bond is the force of attraction which holds various atoms ( constituents, ions etc.) together in different chemical species.

Topics of NCERT Grade 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

4.1 Kössel-Lewis Approach to Chemical Bonding

4.2 Ionic or Electrovalent Bond

4.3 Bond Parameters

4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory

4.5 Valence Bond Theory

4.6 Hybridisation

4.7 Molecular Orbital Theory

4.8 Bonding in Some Homonuclear Diatomic Molecules

4.9 Hydrogen Bonding

NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure - Exercise Questions

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 :

 Mg

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: \ddot{Mg}.

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

Na

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 \dot{Na}.

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

B

Answer:

The Lewis dot symbol of B atom is;

As there are three valence electrons in B atom.

Hence, the Lewis dot structure is Boron Octet

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

O,

Answer:

The Lewis dot symbol of O atom is;

As there are six valence electrons in an atom of O.

Hence, the Lewis dot structure is Oxygen Octet

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

 N,

Answer:

The Lewis dot symbol of N, atom is;

As there are five valence electrons in an atom of N,.

Hence, the Lewis dot structure is Nitrogen octet

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

 Br.

Answer:

The Lewis dot symbol of Br atom is;

As there are seven valence electrons in an atom of Br.

Hence, the Lewis dot structure is 

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

S \; and\; S^{2-} 

Answer:

As the number of valence electrons in sulphur is six.

Therefore its Lewis dot symbol of sulphur(S) is Sulphur octet

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

Sulphur dinegative

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

 Al\; and\; Al^{3+}

Answer:

As the number of valence electrons in aluminium is three.

Therefore its Lewis dot symbol of aluminium(Al) is sluminum

And of Al^{3+} 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:

 H \; and \; H^{-}

Answer:

As the number of valence electrons in hydrogen is one.

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

And of H^{-} is, if it has one more electron because of its a negative charge develops.

Hence. the Lewis symbol is hydrogen negative

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

 H_{2}S

Answer:

The Lewis structure of H_{2}S is:

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

  SiCl_{4}

Answer:

The Lewis structure of SiCl_{4} is:

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

  BeF_{2}

Answer:

The Lewis structure of BeF_{2} is:

Bef2 new

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

  CO_{3}^{2-}

Answer:

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 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

LimitationsThere 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\ LiCl, BeH_{2}\ and\ BCl_3.

  • Odd-electron molecules - the octet rule is not satisfied for all the atoms in  NO\ and\ NO_{2}.nono2

  • 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.pf5

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:

 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.  becl2

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 trigonal planar shape.    a

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}  sicl4

hence, it has tetrahedral shape.  sicl4  

 

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 trigonal bipyramidal shape.    asf5

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 trigonal bipyramidal shape.    ph3

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.

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.

carbonante resonance

Question 4.12     H_{3}PO_{3}  can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H_{3}PO_{3}.? 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 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, 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 :

 Ca \; and \; O

Answer:

Ca \; and \; O:

We have the electronic configurations of both:

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

O = 2,6   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(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 having 3 electron in the valence shell, and it can donate 3 electron to get to the nearest noble gas configuration.

N = 2,5   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:

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. 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:

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 bent shape or some angular shapes.

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

3. 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, \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 4.18     Explain with the help of suitable example polar covalent bond.

Answer:

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 H_{2}O,

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:

                 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. 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.

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.

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 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 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 dsp^2 hybridization, the structure of methane cannot be square planar.

Also the reason that 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 take 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 the opposite directions and cancel the effect of each other.

Question 4.23     Which out of NH_{3}  and NF_{3}  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 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 individual 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 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 SP, SP^{2}, SP^{3}  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 SP, SP^{2}, SP^{3}  hybrid orbitals are shown:

SP hybrid orbital: It is linear in shape and formed by 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 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 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:

excited state

So, initially, aluminium (AlC_{3}) was sp^2 hybridisation and hence having a trigonal planar shape.

Then it reacts with chloride ion to form AlC_{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 occured 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 C-atom in the excited state is:

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

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

The remaining two sp^2 orbitals of each carbon atom from 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.

c2h4 ethene

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 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

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 is three sigmas (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-bonds (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 x-axis as the internuclear axis.

ss overlapping

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 2px orbital are align such that they can combine along x-axis as the internuclear axis.

sp overlapping

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 2porbital are align in y -direction but the internuclear axis is x-axis. 

pp overlapping 2

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? 

 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}

c2h6

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

c3h6

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 bond) therefore it is sp^3 hybridised.

and C_{2} is also making a 4 sigma bonds. 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

c2h4o

C_{1} is making 4 sigma bonds (single bond) 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 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 that electron pairs that do not participate in bonding are called lone pairs of electrons.

For examples,

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:

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

Sigma (\sigma) Bond

Pi (\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  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.

Attractive force arises between:

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

(b) The nucleus of one atom and 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.

H2 molecule

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 while 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.

pcl54

The five P-Cl sigma bonds, three lies 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 makes an angle of 90^{\circ} 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.

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.

hbond2

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 :

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, bond order 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

 

NCERT solutions for class 11 chemistry

Chapter 1

NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry

Chapter-2

CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom

Chapter-3

Solutions of NCERT class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties

Chapter-4

NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure

Chapter-5

CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter

Chapter-6

Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics

Chapter-7

NCERT solutions for class 11 chemistry chapter 7 Equilibrium

Chapter-8

CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction

Chapter-9

Solutions of NCERT class 11 chemistry chapter 9 Hydrogen

Chapter-10

NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements

Chapter-11

CBSE NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements

Chapter-12

Solutions of NCERT class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques

Chapter-13

NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons

Chapter-14

CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

NCERT solutions for class 11 subject wise

NCERT solutions for class 11 biology

Solutions of NCERT class 11 maths

CBSE NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

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