Question

Which charges for ${\mathit{N}}_{\mathbf{2}}$molecule would give a bond order of 2.5

Short answer

The charge would be -1

The bond filling order is${\sigma }_{1s}{{\sigma }^{*}}_{1s}{\sigma }_{2s}{{\sigma }^{*}}_{2s}{\pi }_{2px}{\pi }_{2py}{\sigma }_{2pz}{{\pi }^{*}}_{2px}{{\pi }^{*}}_{2px}$

The bond order is$\frac{1}{2}(n_b-n_a)$

Where

$$\begin{gathered} n_b=bondingelectrons \\ {n}_a=noofantibondingelectron \end{gathered}$$

Step-by-step solution

Step 1

In a nitrogen molecule, there are fourteen electrons.

So For this, the filling will be as follow

${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2({\sigma }_{2s}{)}^2({{\sigma }^{*}}_{2s}{)}^2{{\pi }^2}_{2px}{{\pi }^2}_{2py}{{\sigma }^2}_{2pz}$

So bond order will be

role="math" localid="1663761495908" $$\begin{gathered} B.O=\frac{1}{2}(n_b-n_a) \\ B.O=\frac{1}{2}(10-4) \\ B.O=3 \end{gathered}$$

Step 2

So to make bond order 2.5

The number of antibonding electrons should be five

To add one more electron - charge should be there

Final answer:

The charge would be -1