Question

In which of the following pair(s), is/are the stronger bond found in the first species?

• A. $O_2^{-},O_2$
• B. $N_2,O_2^{+}$
• C. $N{O}^{+},NO$
• D. $O_2^{-},O_2^{2+}$

Answer 1

Answer: (B), (C)

The correct options are
B $N_2,O_2^{+}$
C $N{O}^{+},NO$

$Bondorder=\frac{noofbondingelectrons-noofanti-bondingelectrons}{2}$

The electronic configuration of the $O_2^{+}$ containing 15 electrons can be written as:

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

Bond Order = $\frac{8-3}{2}$ = 2.5

For $O_2$

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

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

The electronic configuration of the $O_2^{-}$ ion containing 17 electrons can be written as:

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

Bond Order = $\frac{6-3}{2}$ = 1.5

For $NO$,

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^1\ast 2p{y}^0$

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

For $N{O}^{+}$,

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

Bond Order = $\frac{10-4}{2}$ = 3

The electronic configuration of the $O_2^{+2}$ containing 14 electrons can be written as:

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

Bond Order = $\frac{10-4}{2}$ = 3

(A) $O_2^{-}=(10-7)/2=1.5,O_2=(10-6)/2=2$
(B) $N_2=(10-4)/2=3,N_2^{+}=(9-4)/2=2.5$
(C)$N{O}^{+}=(10-4)/2=3,N{O}^{-}=(10-6)/2=2$
(D) $O_2^{2+}=(10-4)/2=3$
Higher the bond order, stronger is the bond.