NCERT Solution Class 11th Chemistry Chapter – 4 Chemical Bonding and Molecular Structure Question & Answer

NCERT Solution Class 11th Chemistry Chapter – 4 Chemical Bonding and Molecular Structure

NCERT Solution Class 11th Chemistry Chapter – 4 Chemical Bonding and Molecular Structure

?Chapter – 3?

✍Chemical Bonding and Molecular Structure✍

?Question & Answer?

Q-1) Explain the formation of a chemical bond

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

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

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

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

a) Mg

b) Na

c) B

d) O

e) N

f) Br

Answer:

a) Mg

Magnesium atom contains only 2 valence electrons. Thus, the lewis dot symbols for Mg is

b) Na

Sodium atom contains only 1 valence electron. Thus, the lewis dot symbols for Na is Na\cdot

c) B

Boron atom contains only 3 valence electrons. Thus, the lewis dot symbols for B is

d) O

Oxygen atom contains only 6 valence electrons. Thus, the lewis dot symbols for O is

e) N

Nitrogen atom contains only 5 valence electrons. Thus, the lewis dot symbols for N is

f) Br

Bromine atom contains only 7 valence electrons. Thus, the lewis dot symbols for Br is

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

S and S^2-; Al and Al³⁺; H and H^–

Answer: For S and S^2-

A sulphur atom contains only 6 valence electrons. Thus, the lewis dot symbols for S is

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

Thus lewis dot symbol is

For Al and Al³⁺

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

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

Thus lewis dot symbol is

[Al]^{3+}[Al]3+

For H and H^–

A hydrogen atom contains only 1 valence electrons. Thus, the lewis dot symbols for H is

$H\cdot$

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

Thus lewis dot symbol is

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

H₂S, SiCl₄, BeF₂, CO_{3}^{2-}CO32−​, HCOOH

Answer:

H₂S

SiCl₄

BeF₂

CO_{3}^{2-}CO32−​

HCOOH

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

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

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

Limitations:

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

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

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

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

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

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

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

The factors that are favourable for ionic bond formation:

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

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

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

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

BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃

Answer:

BeCl₂

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

BCl₃

Central atom is not having any lone pair but have 3 bond pairs. Thus, its shape is AB₃. i.e. Trigonal planar.

SiCl₄

Central atom is not having any lone pair but have 4 bond pairs. Thus, its shape is AB₄. i.e. Tetrahedral.

AsF₅

Central atom is not having any lone pair but have 5 bond pairs. Thus, its shape is AB₅. i.e. Trigonalbipyramidal.

H₂S

Central atom is having 1 lone pair and is having 2 bond pairs. Thus, its shape is AB₂E. i.e. Bent shape.

PH₃

Central atom is having 1 lone pair and is having 3 bond pairs. Thus, its shape is AB₃E. i.e. Trigonalbipyramidal.

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

Answer: The geometry of H₂O and NH₃:

Central atom(N) in ammonia is having 1 lone pair and is having 3 bond pairs.

Central atom(O) in water is having 2 lone pair and is having 2 bond pairs.

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

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

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

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

Q-10) Define Bond length.

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

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

Answer:  Experimental results shows that, all the C-O bond in CO_{3}^{2-}CO32−​ are equivalent.

Thus, it is inefficient to represent CO_{3}^{2-}CO32−​ ion by single lewis structure which is having 1 double bond and 2 single bonds.

Thus, the resonance structures of CO_{3}^{2-}CO32−​ is :

Q-12) H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same. 

Answer:

In the given structures the position of atoms are changed, so we cannot take the 2 given structures as a canonical form of resonance hybrid which is representing H₃PO_3.

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

Answer: SO₃

NO₂

NO_{3}^{-}NO3−

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

(i) K and S 

(ii) Ca and O

(iii) Al and N.

Answer:

(i) K and S

Electronic configurations of S and K are:

S: 2, 8, 6

K: 2, 8, 8, 1

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

(ii) Ca and  O

Electronic configurations of O and Ca are:

O: 2, 6

Ca: 2, 8, 8, 2

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

(iii) Al and N

Electronic configurations of N and Al are:

N: 2, 5

Al: 2, 8, 3

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

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

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

$\therefore Resultant,\mu =0$

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

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

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

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

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

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

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

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

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

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

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

Answer: Ionic characteristic of a molecule depends on the difference in electronegativity between constituents atoms. So, higher the difference, the ionic character of a molecule will be higher.

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

N₂< SO₂< ClF₃< K₂O <LiF.

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

Answer:

Correct Lewis structure of CH₃COOH is given below:

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

Answer: Electronic configuration of C- atom:

6C :1s^{2}\;2s^{2}\;2p^{2}1s22s22p2

Orbital picture of C- atom in excited state is:

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

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

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

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

Answer: Lewis structure of BeH₂ is:

H : Be: H

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

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

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

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

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

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

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

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

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

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

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

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

(a) sp hybrid orbital

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

(b) sp² hybrid orbital

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

(c) sp³ hybrid orbital

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

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

AlCl₃ + Cl^{-}Cl− —> AlCl_{4}^{-}AlCl4−​

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

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

Thus, Al -atom in AlCl₃ undergoes sp² hybridisation and forms trigonal planar geometry. For the formation of AlCl_{4}^{-}AlCl4−​ the vacant 3p_z orbital will also get involved. Thus, sp² hybridisation is converted into sp³ hybridisation and forms a tetrahedral structure.

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

BF₃ + NH₃ —> F₃B.NH₃

Answer:  N- atom in NH₃ is having sp3 hybridization. Orbital picture of N- atom is shown below:

B- atom in NF₃ is having sp2 hybridisation . Orbital picture of B- atom is shown below:

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

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

Answer:  C₂H₄

Electronic configuration of carbon atom in excited state is given below:

₆C: 1s^{2}2s^{1}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}1s22s12px1​2py1​2pz1​

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

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

C₂H₂

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

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

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

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

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

(a) C₂H₂

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

(b) C₂H₄

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

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

(a) 1s and 1s   (b) 1s and 2p_x   (c) 2p_y and 2p_y   (d) 1s and 2s

Answer:  (c) 2p_y and 2p_y

2p_y and 2p_y orbitals won’t form a $\sigma$ as it will undergo lateral over lapping and will form a $\pi$ bond.

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

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

Answer: (a) CH₃-CH₃

Here, C₁ and C₂ are having sp³ hybridization.

(b) CH₃-CH=CH₂

Here, C₃ and C₂ are having sp² hybridization and C₁ is having sp³ hybridization.

(c) CH₃-CH₂-OH

Here, C₁ and C₂ are having sp³ hybridization.

(d) CH₃-CHO

Here, C₁ is having sp³ hybridization and C₂ is having sp² hybridization.

(e) CH₃COOH

Here, C₁ is having sp³ hybridization and C₂ is having sp² hybridization.

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

Answer: A covalent bond is formed when 2 atoms combine with each other by sharing their valence electrons.

“The shared pairs of electrons present between the bonded atoms are called bond pairs”. Each and every electron cannot participate in bonding. “The pairs of electrons which do not participate in bonding are called lone pairs”.

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

b) Water is having 2 bond pairs and 2 lone pairs on O- atom

.

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

Answer:

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

Answer: Assuming 2 H- atoms X and Y with nuclei N_X and N_Y and electrons e_X and e_Y, respectively.

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

The repulsive forces are:

(i) Between electrons of both the atoms i.e. e_X and e_Y.

(ii) Between nuclei of both the atoms i.e. N_X and N_Y.

The attractive forces are:

(i) Between the electron and nucleus of the same atom i.e. N_X – e_X and N_Y -e_Y.

(ii) Between the electron of one atom and nucleus of other atom i.e. N_X–e_Y and N_Y–e_X.

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

Repulsive forces:

Attractive forces:

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

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

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

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

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

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

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

Ans.) Electronic configuration of Be:

1s² 2s²

Molecular orbital electronic configuration of Be₂ is:

\sigma _{1s}^{2}\; \sigma _{1s}^{\cdot 2}\;\sigma _{2s}^{2}\;\sigma _{2s}^{\cdot2}σ1s2​σ1s⋅2​σ2s2​σ2s⋅2​

Thus, bond order of Be₂: 0.5(N_b – N_a).

N_b: No. of electrons in the bonding orbitals

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

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

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

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

O₂, O_{2}^{+}O2+​, O_{2}^{-}O2−​ (Superoxide), O_{2}^{2-}O22−​ (Peroxide)

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

• Electronic configuration of O₂ is:

[\sigma – (1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (1p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}[\pi^{*} (2p_{y})]^{1}[σ–(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]1[π∗(2py​)]1

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

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of O_{2}^{+}O2+​ is:

KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}KK[σ(2s)]2[σ∗(2s)]2[σ(2pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]1

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

As, bonddissociation energy $\propto$ bond order

Hence, higher the bond order, higher stability will be there.

Arrangement according to decreasing order of stability is given as:

O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}O2+​>O2​>O2−​>O22−​

• Electronic configuration of O_{2}^{-}O2−​ (Superoxide) is:

KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{1}KK[σ(2s)]2[σ∗(2s)]2[σ(2pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]2[π∗(2py​)]1

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of O_{2}^{2-}O22−​ (peroxide) is:

KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{2}KK[σ(2s)]2[σ∗(2s)]2[σ(2pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]2[π∗(2py​)]2

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 6

Now,

Bond order = 0.5(8 – 6) = 1

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

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

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

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

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

Answer: The electronic configuration of outer orbital of phosphorus in excited state and in ground state is given below:

Ground State:

Excited State:

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

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

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

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

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

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

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

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

4^{\delta } – X^{\delta }\;………..H^{\delta +} – X^{\delta -}\;………….H^{\delta +} – X^{\delta -}4δ–Xδ………..Hδ+–Xδ−………….Hδ+–Xδ−

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

Two types of hydrogen bonds are there:

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

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

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

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

Answer:  Bond Order: It is defined as 0.5 times the difference between the “No. of electrons present in bonding orbitals and No. of electrons present in anti-bonding orbitals” of a molecule.

Bond Order = 0.5(N_b – N_a);

Na: No. of anti-bonding electrons

Nb: No. of bonding electrons

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

• Electronic configuration of O₂ is:

[\sigma – (1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (1p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}[\pi^{*} (2p_{y})]^{1}[σ–(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]1[π∗(2py​)]1As 1s- orbital of each O- atom does not involve in the bonding,

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of O_{2}^{-}O2−​ (superoxide) is:

KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{1}KK[σ(2s)]2[σ∗(2s)]2[σ(2pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]2[π∗(2py​)]1

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of O_{2}^{+}O2+​ is:

KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}KK[σ(2s)]2[σ∗(2s)]2[σ(2pz​)]2[π(2px​)]2[π(2py​)]2[π∗(2px​)]1

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

• Electronic configuration of N₂ is:

[\sigma(1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi (2p_{z})]^{2}[σ(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[π(2px​)]2[π(2py​)]2[π(2pz​)]2

No. of bonding electrons = N_b = 10

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(10 – 4) = 3