Question

Assertion
${{F}_{2}}$ and $O_{2}^{2-}$ have bond order 1 while ${{N}_{2}}$, CO 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

When the atomic orbitals overlap with each other in the region where density of electrons is high, the molecular orbitals are formed. Overlap of the atomic orbitals determines the efficiency of the interaction between the atomic orbitals. Energy of bonding molecular orbitals is less than the nonbonding molecular orbitals.

Complete step by step solution:
The bond order is half of the difference between the total numbers of the bonding electrons
and antibonding electrons in the given molecule.
Formula to calculate Bond order: $Bond\,Order=\dfrac{1}{2}\left( \begin{aligned} & Electron\,in\,bonding\,molecular\,orbital- \\\ & Electron\,in\,anti\,bonding\,molecular\,orbital \\\ \end{aligned} \right)$
Bond order of ${{F}_{2}}$:$\begin{aligned} & {{F}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\pi *{{(2{{p}_{x}})}^{2}}\pi *{{(2{{p}_{y}})}^{2}} \\\ & Bond\,Order=\dfrac{10-8}{2}=1 \\\ \end{aligned}$
Similarly, we can calculate the bond order of $O_{2}^{2-}$ by writing its molecular orbital configuration which is 1.
Bond order of ${{N}_{2}}$:

& {{N}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{x}})}^{2}}\sigma {{(2{{p}_{y}})}^{2}}\sigma {{(2{{p}_{z}})}^{2}} \\\ & Bond\,Order=\dfrac{10-4}{2}=3 \\\ \end{aligned}$$ Similarly, we can calculate for CO and $$N{{O}^{+}}$$, by again by writing its molecular orbital configuration which is 3. So, the assertion is correct. The bond order of a molecule is directly proportional to the stability of that molecule. With increasing bond order, the bond length is decreased. Consequently, the amount of energy to dissociate the shorter bond is higher than that of the larger bond. Hence, the bond order of a molecule is more stable. So, the reason is correct. Since, nothing is stated in the assertion about stability, the reason becomes unrelated to it. Therefore, the correct option is (b). **Note:** If the bond order is low, there will be less attraction between electrons and this causes the atoms to be held together more loosely. As a result, they dissociate easily, meaning they are less stable.