Question

Assertion: ${{F}_{2}}$ and $O_{2}^{2-}$ have bond order 1 while ${{N}_{2}}$, $CO\,\text{and}\,N{{O}^{+}}$ have bond order 3.
Reason: Higher the bond order, higher is the stability of the molecule.
(A)- Both Assertion and Reason are
correct, and Reason is the correct explanation for Assertion.
(B)- Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
(C)- Assertion is correct but Reason is incorrect.
(D)- Both Assertion and Reason are incorrect.

Answer 1

Hint: The bond order of the given molecules is calculated from the molecular orbital theory, from which the number of electrons present in the bonding and the antibonding orbitals is determined. Also, the bond order determines the number of bonds present in the molecule.

Complete answer:
Firstly, we will see the bond order of the given compounds, in accordance with the molecular orbital theory.
The configuration of the ${{F}_{2}}$ is
\[\sigma {{(1s)}^{2}}{{\sigma }^{*}}{{(1s)}^{2}}\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}{{\pi }^{*}}{{(2{{p}_{y}})}^{2}}\]
where * is for the anti-bonding orbital.
Then the bond order =$\dfrac{1}{2}\text{electrons}\,\text{in}\left( \text{bonding}\text{orbital - antibonding}\text{orbital} \right)$
The bond order for ${{F}_{2}}$ will be $=\dfrac{1}{2}\left( 10-8 \right)=1$
Similarly, for $O_{2}^{2-}$ the configuration is
\[\sigma {{(1s)}^{2}}{{\sigma }^{*}}{{(1s)}^{2}}\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}{{\pi }^{*}}{{(2{{p}_{x}})}^{1}}{{\pi }^{*}}{{(2{{p}_{y}})}^{1}}{{\sigma }^{*}}{{(2{{p}_{z}})}^{2}}\]
The bond order is $=\dfrac{10-8}{2}=1$
So, with the bond order being one, there is a single sigma bond formed between the two fluorine atoms and between the two oxygen atoms, having an extra electron on each oxygen atom.
In ${{N}_{2}}$,with configuration $\sigma {{(1s)}^{2}}{{\sigma }^{*}}{{(1s)}^{2}}\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\sigma {{(2{{p}_{z}})}^{2}}$
The bond order is $=\dfrac{10-4}{2}=3$
Similarly, in $CO\,$and$N{{O}^{+}}$with the same configuration, the bond order is 3. So, with bond order being 3, the molecules have one sigma and two pi- bonds in it.

Hence, it can be seen that with the increase in the bond order of the molecule, due to presence of the pi-bond, the bond length decreases, and the bond strength increases. With the increase in the bond strength, the energy required to dissociate the molecule will be more. Thus, it is a stable molecule.
But since in the assertion, the stability of the molecule in relation to the bond order has not been stated.

So, option (B)- Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion is correct.

Note:
When the bond order is low, the bond can be dissociated easily as the electrons present in orbitals have lower s-character, and it is thus, loosely bound and is less stable.