Chemical Bonding and Molecular Structure Test

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Chemical Bonding and Molecular Structure Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following statements is not true regarding molecular orbital theory?

A
The atomic orbitals of comparable energies combine to form molecular orbitals

B
An atomic orbital is monocentric while a molecular orbital is polycentric

C
Bonding molecular orbital has higher energy than antibonding molecular orbital

D
Molecular orbitals like atomic orbitals obey Aufbau principle for filling of electrons

Solution

Bonding molecular orbital has lower energy than the antibonding molecularorbital.
Question 2
4 / -1

2s and 2p-atomic orbitals combine to give how many molecular orbitals?

A
2

B
4

C
8

D
6

Solution

2s-> (σ2s) (0*2s) (2)
2p -> (σ2p_z) (σ*2p_z)(π2p_x)(π*2p_x) (π2p_y) (π*2p_y) (6)
Question 3
4 / -1

The conditions for the combination of atomic orbitals to form molecular orbitals are stated below. Mark the incorrect condition mentioned here.

A
The combining atomic orbitals must have nearly same energy

B
The combining atomic orbitals must overlap to maximum extent

C
Combining atomic orbitals must have same symmetry about the molecular axis

D
Pi (π) molecular orbitals are symmetrical around the bond axis

Solution

Sigma molecular orbitals are symmetrical around the bond axis while pi molecular orbitals are not symmetrical.
Question 4
4 / -1

The electronic configuration of carbon is Is² 2s² 2p². There are 12 electrons in C₂. The correct electronic configuration of C₂ molecule is

A
(σls²) (σ*1s²) (σ2s²) (σ*2s²) (σ2p²_z) (π2p²_x)

B
(σls²)(σ*1s²) (σ2s²) (σ*2s²) (π2p²_x=π2p²_y)

C
(σls²) (σ*1s²) (σ2s²) (σ*2s²)(σ2p²_z)(π2p¹_x=π2p¹_y)

D
(σls²) (σ*1s²) (σ2s²)(σ*2s²) (π2p¹_x=π2p¹_y)

Solution

For atoms like nitrogen and below N in atomic number have higher energy for p_z -orbital, hence the configuration for C₂ molecule is
(σls²)(σ*1s²) (σ2s²) (σ*2s²) (π2p²_x=π2p²_y)
Question 5
4 / -1

The increasing order of energies of various molecular orbitals of N₂ is given below:
σls <σ*ls <σ2s <σ*2s <π2p_x = π2p_y <σ2p_z < π*2p_x = π*2p_y<σ*2p_z
The above sequence is not true for the molecule

A
C₂

B
B₂

C
O₂

D
Be₂

Solution

For allel ements which have atomic number more than 7(beyond nitrogen) the energy of σ2p_z is lower than
π2p_x and π2p_y -orbitals.
Question 6
4 / -1

Which of the following species has unpaired electrons?

A
N₂

B
F₂

C
O^-₂

D
O^2-₂

Solution

N₂, F₂ and O^2-₂have paired electrons
Question 7
4 / -1

What is the order of stability of N₂ and its ions?

A
N₂> N⁺₂= N⁻₂> N²⁻₂

B
N⁺₂> N⁻₂> N₂> N²⁻₂

C
N⁻₂> N⁺₂> N₂ > N²⁻₂

D
N²⁻₂> N⁻₂= N⁺₂> N₂

Solution

Bond order of N₂ = 3, N⁺₂ = 2.5 N^-₂ = 2.5 and N₂^2- is 2. Higher the bond order, more is the stability.
Question 8
4 / -1

What will be the bond order of the species with electronic configuration 1s²2s² 2p⁵?

A
One

B
Two

C
Three

D
Zero

Solution

Electronic configuration of atom: 1s²2s² 2p⁵
M.O. configuration:
Question 9
4 / -1

Which of the following bond orders is indication of existence of a molecule?

A
Zero bond order

B
Negative bond order

C
Positive bond order

D
All of these

Solution

A molecule exists only if the bond order is positive. If bond order is zero or negative, the molecule does not exist.
Question 10
4 / -1

Which of the following pairs will have same bond order?

A
F₂ and O^2-₂

B
N₂ and CO₂

C
O₂ and O^-₂

D
N₂ and N⁺₂

Solution

F₂ and O^2-₂are isoelectronic
Question 11
4 / -1

Bond order of N⁺₂,N^-₂ and N₂ will be

A
2.5, 2.5 and 3 respectively

B
2,2.5 and 3 respectively

C
3, 2.5 and 3 respectively

D
2.5, 2.5 and 2.5 respectively

Solution
Question 12
4 / -1

According to molecular orbital theory, which of the following will not exist?

A
H⁺₂

B
Be₂

C
B₂

D
C₂

Solution

Be₂: (σ1s²) (σ*1s²) (σ2s²) (σ*2s²)
Since B.O. is zero, the molecule does not exist.
Question 13
4 / -1

Which of the following facts regarding bond order is not valid?

A
Bond order is given by the number of bonds between the two atoms in a molecule

B
With increase in bond order, bond enthalpy of the molecule decreases

C
Isoelectronic molecules and ions have identical bond orders

D
With increase in bond order, bond length decreases

Solution

Higher the bond order, higher is the bond enthalpy.
Question 14
4 / -1

Which of the following formulae does not show the correct relationship?

A

B

C

D
N_b> N_a,B.O.= +ve

Solution

Bond order ∝ Bond dissociation energy.
Question 15
4 / -1

Fill in the blanks with the appropriate choice.
Bond order of N⁺₂ is ___P___ while that ofN₂ is ___Q___ . Bond order of O⁺₂ is ___R___ while that of O₂is ___S___ . N−N bond distance ___T___ , when N₂ changes to N⁺₂and when O₂changes to O⁺₂, the O-O bond distance ___U___ .

A
a

B
b

C
c

D
d

Solution

Since N⁺₂ has lower bond order than N₂, bond length of N - N in N⁺₂ increases .In O⁺₂ , bond order increases form 2 to 2.5 hence , bond length decreases.