Question

Incorrect order of bond length is:

• A. $CO<C{O}^{+}$
• B. $NO>N{O}^{+}$
• C. $N_2^{+}<N_2^{-}$
• D. $H_2^{+}<H_2^{-}$
• E. None of the above

Answer 1

Answer: (C), (D)

$N_2^{+}<N_2^{-}$
$H_2^{+}<H_2^{-}$

(A) The bond orders of $CO$ and $C{O}^{+}$ are $3$ and $2.5$ respectively.

Electronic configuration of $CO$ is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^{2}2p{z}^2$ Bond Order = $\frac{6-0}{2}=3$

Electronic configuration of $C{O}^{+}$ is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^{2}2p{z}^1$ Bond Order = $\frac{5-0}{2}=2.5$

(B) The bond orders of $NO$ and $N{O}^{+}$ are $2.5$ and $3$ respectively.

Electronic configuration of $NO$ is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^1$ Bond Order = $\frac{6-1}{2}$ = 2.5

Electronic configuration of $N{O}^{+}$ is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^1$

Bond Order = $\frac{6-0}{2}=3$

Since, the bond order of $N{O}^{+}$ is higher, its bond length will be lower. Hence, the order of bond length is $NO>N{O}^{+}$

(C) For $N_2^{-}$

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^{2}2p{z}^2\ast 2p{x}^1$

Bond Order = $\frac{6-1}{2}=2.5$

For $N_2^{+}$

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^{2}2p{z}^1$

Bond Order = $\frac{5-0}{2}=2.5$

The bond orders of $N_2^{+}$ and $N_2^{-}$ are $2.5$ each. Hence, they are expected to have equal bond lengths.
(D) For $H_2^{+}$,

$1s^1$

Bond Order = $\frac{1-0}{2}$ = 0.5

For $H_2^{-}$,

$1s^2\ast 1s^1$

Bond Order = $\frac{2-1}{2}=0.5$

The bond orders of $H_2^{+}$ and $H_2^{-}$ are $0.5$ each. Hence, they are expected to have equal bond lengths.