Class 11 - Chemistry - Chemical Bonding Structure

Question 4.1

Explain the formation of a chemical bond.

Answer.

Chemical bond is an attractive force that bounds the constituents of a chemical species together.

Theories like valence shell electron pair repulsion theory, electronic theory,

molecular orbital theory and valence bond theory have explained the formation of chemical bond.

The formation of a chemical bond is the tendency of the system to achieve stability.

An atom is said to be stable when it attains noble gas configuration that is the outermost orbitals are completely filled.

Atoms combine with one another and fill their separate octets.

 

 

Question 4.2

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

Answer.

Mg:

Na.

.B:

:ö:

:N̈.

 

Question 4.3

Write Lewis symbols for the following atoms and ions:

S and S2-, Al and Al3+, H and H-

Answer.

Electronic configuration of S = 2, 8, 6

Electronic configuration of Al = 2, 8, 3

Electronic configuration of H = 1

   /u-img/00/01/12/00011219.jpg      

 

 

Question 4.4

Draw the Lewis structures for the following molecules and ions:

H2S, SiCl4, BeF2, CO32-, HCOOH

Answer.

 Class_11_Chemistry_Chemical_BondingClass_11_Chemistry_Chemical_Bonding_Structure

 

Question 4.5

Define octet rule. Write the significance and limitations.

Answer.

Atoms of elements combine with each other in order to complete their respective octets so as to acquire stable gas configuration,

this is octet rule.

This is also called as electronic theory. It was given by Kossel and Lewis.

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

Different atoms combine with each other to form ionic or covalent compounds.

Limitations of Octet rule:

  • Octet rule fails to predict the relative stability and shape of the molecules.
  • Octet rule is based on the inert nature of noble gases. But some inert gases like Kr and Xe forms compounds.
  • This does not account for shape of molecules.
  • If compound is having less than 8 electrons surrounding central atom, then octet rule cannot be applied to such compound. For example: BeH2

 

 

Question 4.6

Write the favourable factors for the formation of ionic bond.

Answer.

Factors affecting the formation of ionic bond:

  • High electron affinity of atoms of non-metal.
  • High lattice energy of compound which is formed.
  • Low ionization enthalpy of metal atom.

 

 

Question 4.7

Discuss the shape of the following molecules using the VSEPR model.

BeCl2, BCl3, SiCl4, AsF5, H2S, PH3

Answer.

BeCl2

Cl: Be: Cl

 

  Shape of  BCl3  is triangular planar.

Shape of  BeCl2  is linear.

Shape of SiCl4  is tetrahedral.

Shape of  AsF5 is trigonal bipyramidal.

Shape of H2S is bent/V-shaped.

Shape of PH3 is Trigonal.

 Class_11_Chemistry_Chemical_Bonding_Structure_1

 

Question 4.8

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

Answer.

Electron dot structure of these molecules are given below:

Since, there are two lone pairs of electrons on O-atom, repulsion on bond pairs is greater in water as compared to ammonia.

Thus, bond angle is less in water molecules.

 Class_11_Chemistry_Chemical_Bonding_Structure_3

 

Question 4.9

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

Answer.

Bond strength is directly proportional to bond order. Greater the bond order more is the bond strength.

In order words, as bond strength increases, the bond becomes stronger and bond order 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.

Bond lengths are measured by spectroscopic methods.

 

 

Question 4.11

Explain the important aspects of resonance with reference to the CO32- ions.

Answer.

Resonance in CO32-. I, II and III represents three canonical forms.

Class_11_Chemistry_Chemical_Bonding_Structure_4

  • In all these structures, position of nuclei is same.
  • All of them have almost equal energy.
  • They have same number of paired and impaired electrons, they differ only in their position.

 

 

Question 4.12

H3PO4 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 H3PO4 ?

If not, give reasons for the same.

Class_11_Chemistry_Chemical_Bonding_Structure_5

Answer.

These are structures cannot be taken as canonical forms because the positions of atoms have been changed.

 

 

Question 4.13

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

Answer.

SO3

NO2

NO3-

 Class_11_Chemistry_Chemical_Bonding_Structure_6

 

Question 4.14

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

Answer.

 

 Class_11_Chemistry_Chemical_Bonding_Structure_7

Question 4.15

Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear.

Explain this on the basis of dipole moment.

Answer.

In CO2 molecule, there are two C=O bonds. Each C-O bond is polar bond. The net dipole of CO2 molecule is zero.

This is possible if the shape of molecule is linear as dipole moments of bond between C-O is equal and opposite, so they cancel out each other.

In water molecule, net dipole moment is 1.84 D. Water molecule has a bent structure because O-H bonds are oriented at an angle of 104.5o

and do not cancel each other.

 

 

Question 4.16

Write the significance/application of dipole moment.

Answer.

  • It helps in predicting the nature of molecules.
  • It helps in determining the shapes of molecules.
  • It helps in calculating the percentage ionic character.

 

 

Question 4.17

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

Answer.

Electronegativity is the tendency of an atom to attract shared pair of electrons. It is not constant for any element.

It is not a measurable quantity.

Electron gain enthalpy is the tendency of an atom to attract outside electron.It is constant for any element.

It is a measurable quantity.

 

 

Question 4.18

Explain with the help of suitable example polar covalent bond.

Answer.

When two atoms with different electronegativity are linked to each other by covalent bond, bond pair of electrons is not shared equally.

For example:

HCl is having polar covalent bond. In HCL, Cl-atom is having more electronegativity than H-atom.

Bond pair shift towards Cl-atom due to which it acquires positive charge.

 Class_11_Chemistry_Chemical_Bonding_Structure_8

 

Question 4.19

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

Answer.

Ionic character of a molecule depends on different in electronegativity between constituent atoms.

N2 < SO2 < ClF3 < K2O < LiF

 

 

Question 4.20

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.

Class_11_Chemistry_Chemical_Bonding_Structure_9

Answer.

Class_11_Chemistry_Chemical_Bonding_Structure_9 

 

Question 4.21

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 atoms at its centre.

Explain why CH4 is not square planar?

Answer.

According to VSEPR theory, if methane (CH4) is square planar, the bond angle would be 90o.

For a tetrahedral structure, bond angle is 109o.28.

In square planar structure, repulsion between bond pairs would be more and thus stability is less.

 

 

Question 4.22

Explain why BeH2 molecule has a zero dipole moment although the be-H bonds are polar.

Answer.

BeH2 is a linear molecule, bond angle is 180o. Be-H bonds are polar due to difference in their electronegativity.

But bond polarities cancel out each other. Dipole moment of BeH2 is zero.

 

 

Question 4.23

Which out of NH3 and NF3 has higher dipole moment and why?

Answer.

In NH3 and NF3, difference in electronegativity is nearly same but there is difference in dipole moment.

In NH3, dipole moment of the three N-H bonds are in the same directions as the lone pair of electron.

In NF3, dipole moment of three N-F bonds are in the direction opposite to that of the lone pair.

The resultant dipole moment in NH3 is more than in NF3.

 

 

Question 4.24

What is meant by hybridization of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.

Answer.

Hybridization is defined as the intermixing of a set of atomic orbitals of slightly

different energies to give rise to new hybridized orbitals having equivalent energy and identical shapes.

Shape of sp hybridization:

1s –orbital hybridizes with 1p-orbitals to form 2 sp hybrid orbitals. Sp hybrid orbital is having linear shape.

Shape of sp2 hybridization:

1s-orbital hybridizes with 2p-orbitals to form 3 sp2 hybrid orbitals. Shape of sp2 orbital is trigonal planar.

Shape of sp3 hybrid orbital

1s-orbital hybridizes with 3 p orbitals to form 4 sp3 hybrid orbitals. Shape of sp3 orbital is tetrahedron.

 

 

Question 4.25

Describe the change in hybridization (if any) of the Al atom in the following reaction.

AlCl3 + Cl- --> AlCl4-

Answer.

Electronic configuration of Al is 1s2 2s2 3p6 3s1 3px1 3py1

Hybridisation will be sp2.

In AlCl3, there is empty 3pz obital is also involved. So the hybridization in sp3 and the shape is tetrahedral.

 

 

Question 4.26

Is there any change in the hybridization of B and N atoms as a result og the following reaction?

BF3 + NH3 --> F3B. NH3

Answer.

In BF3, B atom is sp2 hybridization. In NH3, N is sp3 hybridized. After combining with F, hybridization of B changes from sp2 to sp3.

 

 

Question 4.27

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

Answer.

In case of ethene (C2H4)

 Class_11_Chemistry_Chemical_Bonding_Structure_of_Ethene

In case of ethyne (C2H2)

 Class_11_Chemistry_Chemical_Bonding_Structure_of_Ethyne

 

Question 4.28

What is the total number of sigma and pi bonds in the following molecules?

  • C2H2 (b) C2H4

Answer.

  • H – C = C – H

Sigma bond = 3, π bonds = 2

(b) Sigma bond = 5, π bonds = 1

 Class_11_Chemistry_Chemical_Bonding_Number_Of_Bonds_in_Ethene

 

Question 4.29

Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1 s and 1s

  • 1s and 2px
  • 2py and 2py
  • 1s and 2s

Answer.

Correct option is (c)

It will not form a s-bond because taking x-axis as the internuclear axis, lateral overlap between two 2py orbitals will occur forming a pi bond.

 

 

Question 4.30

Which hybrid orbitals are used by carbons atoms in the following molecules?

  • CH3-CH3
  • CH3-CH=CH2
  • CH3-CH2-OH
  • CH3-CHO
  • CH3COOH

Answer.

Class_11_Chemistry_Chemical_Bonding_Hybrid_Orbitals

  • C1 has sp3 hybrid orbitals and C2 has sp2 hybrid orbitals.

Class_11_Chemistry_Chemical_Bonding_Hybrid_Orbitals_1

  • C1 has sp3 hybrid orbitals and C2 has sp2 hybrid orbitals.

 Class_11_Chemistry_Chemical_Bonding_Hybrid_Orbitals_2

 

Question 4.31

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

Answer.

The electron pair involved in sharing of electrons between two atoms during covalent bonding is called bond pair.

Class_11_Chemistry_Chemical_Bonding_Bond_Angles

The electron pair which is not involved in sharing is called lone pair of electrons.

 Class_11_Chemistry_Chemical_Bonding_Bond_Angles_1

 

Question 4.32

Distinguish between a sigma and a pi bond.

Answer.

Pi BOND

SIGMA BOND

It is comparatively weak bond.

It is comparatively strong bond.

It is formed by lateral overlapping of orbitals.

It is formed by end to end overlapping of orbitals.

There is only one overlapping orbital is p-p.

The overlapping orbitals are s-s. s-p, p-p.

 

Rotation around pi bond is restricted.

Rotation is possible around sigma bond.

Electron cloud is not symmetrical.

Electron cloud is symmetrical.

 

 

 

Question 4.33

Explain the formation of H2 molecule on the basis of valence bond theory.

Answer.

Let us consider the combination between atoms of hydrogen HX and HY and eX and eY their respective electrons.

As they come closer, two difference forces operate between nucleus and electron of the other and vice versa.

The nuclei of the atoms as well as their electrons repel each other. Energy is required to overcome the force of repulsion.

Here, the number of new attractive and repulsive forces are same.

When two H-atoms approach each other, overall potential energy of system decreases, finally stable molecule of hydrogen is formed.

 

 

Question 4.34

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

Answer.

  1. Joining of orbital orbitals must have approximately same energy.
  2. Joining of atomic orbitals must have legitimate orientations to ensure the maximum overlap.
  3. Overlapping must be in a large extent.

 

 

Question 4.35

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

Answer.

Electronic configuration of Be = 1s2 2s2

Molecular Orbital Electronic Configuration = σ21s σ*21s σ22s σ*22s

Bond order = ½ (4-4) = 0

Be2 does not exist.

 

 

Question 4.36

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

O2, O2+, O2- (superoxide), O22- (peroxide)

Answer.

O2 – Bond order = 2 (Paramagnetic)

O+2 – Bond order = 2.5 (Paramagnetic)

O-2 – Bond order = 1.5 (Paramagnetic)

O2-2 – Bond order = 1 (Diamagnetic)

 

Order of relative stability will be

O2+ > O2 > O2- > O22-

 

 

Question 4.37

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

Answer.

Molecular orbitals are represented by the wave function.

Positive sign represents a molecular orbital indicates positive wave function.

Negative sign represents a molecular orbital indicates negative wave function.

 

 

Question 4.38

Describe the hybridization in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?

Answer.

Electronic configuration of phosphorous in ground and excited state is represented as

One s- three p and one d-orbital hybridize to yield five sets of sp3d hybrid orbitals which are directed

towards the five corners of a trigonal bipyramidal.

This is because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.

Axial bond are slightly longer and hence, weaker than equatorial bonds.

 Class_11_Chemistry_Chemical_Bonding_Hybridization_of_PCl5

 

Question 4.39

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

Answer.     

Hydrogen bond is defined as an attractive force acting between hydrogen attached to an electronegative atom of one

molecule and an electronegative atom of a different molecule.

Hydrogen bond is stronger than the Van der Waals forces.

 

 

Question 4.40

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

Answer.

Bond order 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 is the number of anti-bonding electrons.

Nb is the number of bonding electrons.

  • Molecular Orbital configuration of N2 =

Class_11_Chemistry_Chemical_Bonding_Molecular_Orbital_Configuration1

Bond Order (B.O.) = 0.5 [8 – 2] = 3

  • Molecular Orbital configuration of O2 =

 Class_11_Chemistry_Chemical_Bonding_Molecular_Orbital_Configuration2

B.O. = 0.5 [8 – 4] = 2

(iii)  Molecular Orbital configuration of O2+ =

Class_11_Chemistry_Chemical_Bonding_Molecular_Orbital_Configuration3

B.O. = 0.5 [8 – 3] = 2.5

(iv) Molecular orbital configuration of O2- =

Class_11_Chemistry_Chemical_Bonding_Molecular_Orbital_Configuration4

B.O. = 0.5 [8 – 5] = 1.5

Share this with your friends  

Download PDF


You can check our 5-step learning process


.