Chapter 4 – Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure

Question 1.
Explain the formation of a chemical bond.

Solution.
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond. The formation of chemical compounds takes place as a result of combination of atoms in different ways. The atoms interact with each other on account of following reasons :
(i) Lowering of energy of combining atoms : The process of bonding between the atoms decreases the energy of the combining atoms and gives rise to the formation of a system which has lower energy and hence has greater stability.
(ii) Lewis octet rule : Octet theory of valency or electronic theory of valency states that in the formation of a chemical bond, atoms interact with each other by losing, gaining or sharing of electrons so as to acquire a stable outer shell of eight electrons. The tendency of atoms to achieve eight electrons in their outermost shell is known as Lewis octet rule.

Question 2.
Write Lewis dot symbols for atoms of the following elements:
Mg, Na, B, O, N, Br.

Solution.
Lewis dot symbols:

Question 3.
Write Lewis symbols for the following atoms and ions:
S and S^2-; Al and Al³⁺; H and H^–

Solution.
Lewis symbols:

Question 4.
Draw the Lewis structures for the following molecules and ions:
H²S, SiCl⁴, BeF², CO2−3, HCOOH.

Solution.
Lewis structures :

Question 5.
Define octet rule. Write its significance and limitations.

Solution.
Octet rule : The tendency of atoms in molecules to have eight electrons in their valence shells (two for hydrogen atom) is known as the octet rule.
Significance :

1. It explains the formation of most of the compounds.
2. It is quite useful for understanding the structures of most of the organic compounds.

Limitations:

1. In some compounds, the number of electrons surrounding the central atom is less than eight e.g., LiCl, BeH₂ and BCl₃.
2. In molecules with an odd number of electron like nitric oxide (NO) and nitrogen dioxide (NO₂), the octet rule is not satisfied for all the atoms.
   
3. In a number of compounds 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 6.
Write the favourable factors for the formation of ionic bond.
Solution.
The following factors facilitate the formation of an ionic bond between a metal and a non-metal:

1. Ionization energy : Lesser the ionization energy, greater is the ease of formation of a cation.
2. Electron affinity : High electron affinity favours formation of an anion.
3. Lattice Energy : It is defined as the amount of energy released when cations and anions are brought close to each other from infinity to their respective equilibrium sites in the crystal lattice to form one mole of the ionic compound. The higher the magnitude of the lattice energy, the greater is the tendency of the formation of an ionic bond.

Question 7.
Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.

Solution.
According to VSEPR theory, the shape of molecule depends upon the number of valence shell electron pairs (bonded or non bonded) around the central atom.
(i) BeCl₂: Presence of two bond pairs and lack of lone pairs gives it linear shape. Cl: Be : Cl
(ii) BCl₃ : Presence of three bond pairs and lack of lone pairs gives it trigonal planar geometry.

(iii) SiCl₄ : Presence of four bond pairs and lack of lone pairs gives it tetrahedral geometry.

(iv) AsF₅ : Presence of five bond pairs and lack of lone pairs gives it trigonal bipyramidal geometry.

(v) H₂S : Presence of two bond pairs and two lone pairs gives it angular or bent shape.

(vi) PH₃ : Presence of three bond pairs and one lone pair gives it trigonal pyramidal shape.

Question 8.
Although geometries of NH3and H20 molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Solution.
Bond angle in NH3 is 107.5° and in H₂O, angle is 104.5°. In H₂O, oxygen has two lone pairs and in NH₃, nitrogen has only one lone pair. In H₂O, two lone pairs on oxygen repel each other and the bonded pairs more strongly and cause them to come closer thereby reducing the angle.

Question 9.
How do you express the bond strength in terms of bond order?

Solution.
Greater the bond order, shorter is the bond length and so, higher is the bond strength.

Question 10.
Define the bond length.

Solution.
The equilibrium distance between the centers of the nuclei of two atoms bonded together is termed as bond length or bond distance.
Bond length in ionic compounds = r_c⁺+r_a^–
Bond length in covalent compounds (AB) = r_A+ r_B

Question 11.
Explain the important aspects of resonance with reference to the CO₃²⁻ ion.

Solution.
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons, known as the canonical forms of the hybrid are taken, which describe the molecule accurately. This phenomenon is known as resonance.
Resonance hybrid structures of carbonate ion are shown as follows :

1. The relative position of all the atoms in each of the canonical forms is same.
2. The number of unpaired and paired electrons in each of the canonical forms is same.
3. Negative charge is residing on more electronegative atom.
4. As a result of resonance, bond length and bond order is same.

Question 12.
H₃PO₃ can be represented by structures (i) and(ii) 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.

Solution.
These two cannot be called as canonical forms as relative position of hydrogen is changed. In resonance, canonical forms should differ only in the position of electrons and not in the position of atoms.

Question 13.
Write the resonance structures for SO₃, NO₂ and NO₃^–.

Solution.
SO₃

NO₂

Question 14.
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
(a) K and S
(b) Ca and O
(c) Al and N.

Solution.

Question 15.
Although both CO₂ and H₂O are triatomic molecules, the shape of H₂0 molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.

Solution.
Even though both CO₂ and H₂O are triatomic, but CO₂, has zero dipole moment. It implies that CO, has a symmetrical structure which is possible only when it is linear. However, H₂0 has a net dipole moment suggesting that the molecule is non-linear.

Question 16.
Write the significance/applications of dipole moment.

Solution.
Applications of dipole moment are :

1. To decide polarity of the molecule : Molecules having zero dipole moment are said to be non-polar molecules and those having µ_R ≠ 0 are polar in nature.
2. To determine percentage of ionic character :
   Percentage of ionic character
   =Experimental value of dipole moment/Theoretical value of dipole moment × 100
3. To determine geometry of molecules: The values of dipole moments provide valuable information about the structures of molecules.
   • CO₂, CS₂ are linear as values of their dipole moments are zero.
   • H₂O is non-linear as it has a net dipole moment.

Question 17.
Define electronegativity. How does it differ from electron gain enthalpy?

Solution.
The relative tendency of a bonded atom to attract the shared electron pair towards itself is called electronegativity while electron gain enthalpy is the energy change that occurs for the process of adding an electron to a gaseous isolated atom to convert it into a negative ion i.e., to form, a monovalent anion. Electron gain enthalpy and electronegativity both measure the power of attracting electrons, but electron gain enthalpy is concerned with an isolated gaseous atom while electronegativity is concerned with the atom in combination.

Question 18.
Explain with the help of suitable example polar covalent bond.

Solution.
In a hetero-atomic molecule the shared pair of electrons between the two atoms gets displaced more towards the electronegative atom. The resultant covalent bond is polar covalent bond e.g., in H – F the electron pair is displaced towards fluorine due to highly electronegative F atom. Hence H develops a partial +ve charge while F develops a partial

Question 19.
Arrange the bonds in order of increasing ionic character in the molecules : LiF, K₂O, N₂, SO₂ and CIF3.

Solution.
As per Fajan’s rule, N₂ < SO₂ < CIF₂ < K₂O < LiF

Question 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

Solution.
In the given figure, hydrogen has valency of two, which is not possible and also octet of oxygen is not complete.
So, the correct structure is

Question 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?

Solution.
In CH₄ only bond pairs are there and no lone pairs are present so, to attain most stabilized structure it arranges itself in tetrahedral geometry with greater bond angle of 109° rather than in square planar geometry with lesser bond angle of 90°.

Question 22.
Explain why BeH₂molecule has a zero dipole moment although the Be—H bonds are polar?

Solution.
The dipole moment in case of BeH₂ is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

Question 23.
Which out of NH₃ and NF₃ has higher dipole moment and why?

Solution.
Both the molecules NH₃ and NF₃ have pyramidal shape with a lone pair of electron on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH₃ (4.90 x 10^-30 C m) is greater than that of NF₃ (0.8 x 10<sup.-30C m). This is because, in case of NH₃ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N—H bonds whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N—F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N—F bond moment which results in low dipole moment of NF₃ as represented below:

Question 24.
What is meant by hybridization of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

Solution.
Hybridization 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 new set of orbitals of equivalent energies and shape.
Salient features of hybridization: The main features of hybridization are as under:

1. The number of hybrid orbitals is equal to the number of the atomic orbitals that undergo hybridisation.
2. The hybrid orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred directions to have minimum repulsion between electron pairs and thus have a stable arrangement. Therefore, the type of hybridisation indicates the geometry of the molecules.

Important conditions for hybridisation :

1. Only those orbitals that are present in the valence shell of the atom undergo hybridisation.
2. The orbitals undergoing hybridisation should have almost equal energy.
3. Promotion of electron is not essential condition prior to hybridisation.
4. It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Shapes of hybrid orbitals :

(i) sp-hybridization : One ‘s’ and one ‘p’ orbital of an atom intermix giving two sp-hybrid orbitals making an angle of 180° with each other. Each hybrid orbital has 50% s and 50% p-orbital character.
The molecules having such type of hybridization in their central atom have linear geometry with bond angles 180°. Examples : BeCl₂, C₂H₂, CO₂, HgCl₂, CS₂, N₂O, BeH₂, etc.

(ii) sp²– hybridization : One ‘s’ and two ‘p’ orbitals of an atom intermix giving three sp2-hybrid orbitals, which are directed towards the corners of an equilateral triangle making an angle of 120° with each other. Each hybrid orbital has 33.3% s and 66.7% p-orbital character. The molecules having such type of hybridization in their central atom are expected to have triangular planar geometry with bond angles 120°.
Examples : BCl₃, BF₃, SO₃, C₂H₄, Graphite, SO₂, etc.

(iii) sp²-hybridization : One ‘s’ and three ‘p’ orbitals of an atom intermix giving four sp²-hybrid orbitals, which are directed towards the corners of a regular tetrahedron making an angle of 109°28′. Each hybrid orbital has 25% s- and 75% p-orbital character. The molecules having such type of hybridization in their central atom are expected to have tetrahedral geometry with bond angles 109°28′.
Examples : CH₄, C₂H₆, CCl₄, H₂O, NH₃, diamond, etc.

Question 25.
Describe the change in hybridization (if any) of the Al atom in the following reaction.
Alcl₃ + cl^– → AlCl₄^–

Solution.
There is change in hybridization from sp² → sp³. In AlCl₃, Al is sp² hybridized. AlCl₃ thus has three sp² hybrid orbitals and a vacant Un hybrid p-orbital. When AlCl₄^– is formed, Al undergoes sp³ hybridization and thus Cl^– overlaps with the fourth orbital.

Question 26.
Is there any change in the hybridization of B and N atoms as a result of the following reaction? BF₃ + NH₃ → F₃B.NH₃

Solution.
In BF₃, B is sp² hybridized and has one vacant p-orbital which gets filled by accepting a lone pair of electron present on the N-atom of NH₃. Nitrogen in this adduct acts as donor atom and BF₃ acts as an acceptor. So, hybridization of B in BF₃ changes from sp² to sp³ whereas there is no change in hybridization of N in NH₃ and in the adduct.

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

Solution.
C₂H₄ molecule

C₂H₂ molecule

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

Solution.

Question 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 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s

Solution.
A bond formed between two atoms by the overlap of singly occupied orbitals along their axes (end to end overlap) is called c bond. Hence, 1s and 1s, 1s and 2p_x and 1s and 2s will form σ bonds. π bonds are formed by the sidewise or lateral overlapping of p-orbitals. Hence, p_yand p_y orbitals will form π bond.

Question 30.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃CH₃
(b) CH₃CH = CH₂
(c) CH₃CH₂OH
(d) CH₃CHO
(e) CH₃COOH

Solution.
(a) sp³, sp³
(b) sp³, sp², sp²
(c) sp³, sp³
(d) sp³, sp²
(e) sp³, sp²

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

Bond pairs are the pairs of electrons taking part in bonding e.g.,

A pair of electrons in a molecule which is not shared by any of the two constituent atoms i.e., does not take part in the direct bonding is called a lone pair e.g.,

Question 32.
Distinguish between a sigma and a pi bond.

Solution.
Comparison between σ and π bonds

S. No.	Sigma bond	Pi bond
1.	The bond is formed by the overlap of orbitals along their axes (end to end overlap). It includes s-s and p-p overlapping.	The bond is formed by sidewise overlapping of orbitals (lateral overlapping).It includes p-p overlapping.
2.	It is a strong bond.	It is a weak bond.
3.	There can be free rotation of atoms around this bond.	Free rotation is not possible around this bond.
4.	The shape of the molecule is determined by these bonds.	These bonds do not affect the shape of the molecule.
5.	o-electrons are referred to as localized electrons.	71-electrons are referred to as mobile electrons.

Question 33.
Explain the formation of H2 molecule on the basis of Valence Bond Theory.

Solution.
We can consider two hydrogen atoms A and B approaching each other having nuclei N_A and N_A and electrons present in them are represented by e_A and e_A. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms begin to approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between :

1. nucleus of one atom and its own electron that is N_A – e_A and N_B – e_B.
2. nucleus of one atom and electron of other atom i.e., N_A – e_B, N_B – e_A.

Similarly repulsive forces arise between :

1. electrons of two atoms like e_A – e_B.
2. nuclei of two atoms N_A – N_B.
   Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
   
3. Experimentally it has been found that the magnitude of new attractive forces is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm.
   Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than isolated hydrogen atoms. The energy so released is caljed bond enthalpy, which corresponds to the minimum in the curve depicted in fig. Conversely, 435.8 kj of energy is required to dissociate one mole of H2 molecule.
   H_2(g) + 435.8 kj mol^-1 → H_(g) + H_(g)

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

Solution.
Necessary conditions for linear combination of atomic orbitals:

1. The symmetry and relative energies of combining atomic orbitals _A and _B must be close to each other.
2. The extent of overlap between the charge clouds of _A and _B should be large,
3. Atomic orbitals of the same sign should overlap.

Question 35.
Use molecular orbital theory to explain why Be₂ molecule does not exist.

Solution.
Beryllium atom has electronic configuration 1s² 2s². So, in Be₂ there are a total of eight electrons which must be filled in four molecular orbitals.
M.O. configuration of Be₂ → σ1s²σ*1s² σ2s² σ*2s²
B.O. = 1/2(4 – 4) = 0. Since the bond order is zero it shows that molecular beryllium will not exist.

Question 36.

Solution.

Question 37.
Write the significance of a plus and a minus sign shown in representing the orbitals.

Solution.
The plus acne minus sign indicate the sign of the wave function in the various lobes and not that of the nuclear or electronic charges. The ‘+’ sign (or upward crest) represents one phase and the sign represents second phase of the wave function. Bonding molecular orbital is formed by combination of ‘+’ with ‘+’ and with part of the electron waves whereas antibonding molecular orbital is formed by combination of ‘+’ with part of the electron waves.
As orbitals are represented by wave functions, a plus sign in an orbital represents a +ve wave function and a minus sign represents a -ve wave function.

Question 38.
Describe the hybridization in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?

Solution.
Formation of PCl₅ (sp³d hybridization):
The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented here.

Now the five orbitals (i.e., one s, three p and one d orbital) are available for hybridization to yield a set of five sp³d hybrid orbitals which are directed towards the five corners of a trigonal bipyramid as depicted in the figure.

All the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl₅ the five sp³d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P-Cl sigma bonds. Three P-Cl bonds lie in one plane and make angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P-Cl bonds, one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds.

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

Solution.
Hydrogen bond can be defined as an attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Following conditions should be fulfilled by a molecule for the formation of hydrogen bond.
(a) Hydrogen atom should be linked to a highly electronegative atom, such as F, O or N through a covalent bond.
(b) The size of the electronegative atom bonded covalently to the hydrogen atom should be small.
Types of H-bonding:
Intermolecular hydrogen bond : When hydrogen bond is formed between the hydrogen atom of one molecule and the electronegative atom of another molecule of the same compound (such as HF) or different compounds (such as water and alcohol), it is known as intermolecular hydrogen bond.
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Intermolecular Hydrogen bond : When hydrogen bond is formed between the hydrogen atom and an electronegative atom, present in the same molecule, it is known as intramolecular hydrogen
bond.

H-bonding is stronger than van der Waals forces.

Question 40.
What is meant by the term bond order?
Calculate the bond order of: N₂,O₂,O₂⁺ and O₂^–

Solution.
Bond order of a molecule or an ion is a measure of the strength or stability of the bond. Numerically, it is half of the difference between number of electrons in bonding (N_b) and in anti-bonding (N_a) molecular orbitals.
Bond order = 1/2(Nb−Na)
If B.O. is zero or negative, the bond is highly unstable, i.e. the molecule does not exist. Higher bond order means high bond dissociation energy and greater stability. Greater the bond order, lesser is the bond length.
M.O. configuration of N₂ :