Question

According to Molecular Orbital Theory:
THIS QUESTION HAS MULTIPLE CORRECT OPTIONS
A. $C_2^{2 - }$ is expected to be diamagnetic
B. $O_2^{2 + }$ is expected to have a longer bond length than ${O_2}$.
C. ${N^{2 + }}$ and ${N^{2 - }}$ have the same bond order
D. $He_2^{4 - }$ has the same energy as two isolated $He$ atoms

Answer 1

According to this, in a molecule formation, all atomic orbitals of participating atoms overlap or mix up to form an equivalent number of new orbitals called molecular orbitals. In doing so the atomic orbitals lose their individual identity and all the electrons get distributed in a molecular orbital in accordance with Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule.

Complete step by step solution:
(a) For finding whether $C_2^{2 - }$ is diamagnetic, we just have to write their molecular orbital configuration.
In $C_2^{2 - }$, we have 14 electrons
So their molecular orbital configuration is
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2$
Since all the electrons are paired in the molecule, it is said to be diamagnetic.
Therefore, option (a) is correct

(b) We know that,
$Bond\,Order\,(B.O)\, = \dfrac{1}{2}({N_b} - {N_a})$
Where ${N_b}$ is the number of electrons present in Bonding molecular orbital
And ${N_a}$ is number of electrons present in Antibonding molecular orbital
Also, we know that bond order is inversely proportional to bond length
$Bond\,Order\,(B.O)\, \propto \,\dfrac{1}{{Bond\,length}}$
Greater the bond order, shorter the bond length and vice-versa.
So the molecular orbital configuration of $O_2^{2 + }$
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2$
From this, we get
${N_b}$ = 10
${N_a}$=4
So, the Bond Order will be,
$B.O = \dfrac{1}{2}(10 - 4) = 3$
The molecular orbital configuration of ${O_2}$ will be,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^1\pi^* 2p_y^1$
So the ${N_b}$ = 10
And ${N_a}$ = 6
So the bond order will be
$B.O = \dfrac{1}{2}(10 - 6) = 2$
Since the bond order is inversely proportional to bond length,${O_2}$ will have longer bond length.
Therefore, option(b) is incorrect.

(c) Again to find the bond order we can write their molecular orbital configuration
${N^{2 + }}$ Will have a configuration of,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^1$
From this we get
${N_b}$ = 9 and ${N_a}$ =4
Therefore, $Bond\,Order\,(B.O)\, = \dfrac{1}{2}({N_b} - {N_a})$
$B.O = \dfrac{1}{2}(9 - 4) = 2.5$
i.e., $Bond\,order = 2.5$
${N^{2 - }}$ will have configuration of
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2\pi^*2p_x^1$
From this we get
${N_b}$=10 and ${N_b}$ = 5
Therefore, $Bond\,Order\,(B.O)\, = \dfrac{1}{2}({N_b} - {N_a})$
$B.O = \dfrac{1}{2}(10 - 5) = 2.5$
i.e.,$Bond\,order = 2.5$
So, the bond order of ${N^{2 - }}$ and ${N^{2 + }}$ is the same.
Option (c) is correct.

(d) This statement is false. If we have $H{e_2}$, whatever bonding is obtained is cancelled by its antibonding. Now for $He_2^{4 - }$ the bond order will be $0.5$ therefore, there is some kind of attraction. So, we can conclude that they don’t have the same energy level.
Option (d) is incorrect.

Note:
Shortcut for finding the bond order that has 10 to 18 electrons.
We have to remember that the bond order of ${N_2}$, 14 electrons have a bond order of 3.
Now for every electron added or subtracted to this, reduces the bond order by 0.5
For example, if the number of electrons is 13, then bond order is $3 - 0.5 = 2.5$.