Question

The correct order regarding bond length:
(i) $N_2{H}_2<N_2{H}_4$ (N - N Bond length)
(ii) $NO>N{O}^{+}$ (N - O Bond length)
(iii) $C_2{H}_4>K\left[PtC{I}_3\right(C_2{H}_4\left)\right]$ (C - C Bond length)
(iv) $PO{F}_3>POC{I}_3$ (P - O Bond length)

• A. (i) & (ii)
• B. (iii) & (iv)
• C. (i), (ii) & (iv)
• D. (ii) & (iii)

Answer 1

The correct option is A (i) & (ii)

From the respective bond diagrams, we can say that

The Bond length of (II)$>$(I) because of $N-N$ bond in II and double bond in (I).

The bond order of (II) $>$ (I). So, Bond length of $NO$ is greater than $N{O}^{+}$.