The pH of aqueous solution of $N{H}_{4}CN$ ($K_a$of HCN is $9.2\times {10}^{–10}$ & $K_b$ of $N{H}_{4}OH$ is 1.8 $\times {10}^{–5})$

 HCN is a weak acid & $N{H}_{4}OH$ is a weak base. While $N{H}_{4}CN$ is a salt of weak acid & weak base.

Therefore, pH = 7 +$\frac{1}{2}(p{K}_a- p{K}_b)$

= 7 + $\frac{1}{2}$ [($-log{K}_a$) - (-logKb)]

= 7 + $\frac{1}{2}$[9.04 - 4.74]

= 7 + 2.15 = 9.15

Hence, the correct option is option (A) > 7.