Question

Which of the following options represents the correct bond order?

• A. $O_2{}^{-}>O_2<O_2^{+}$
• B. $O_2{}^{-}<O_2>O_2^{+}$
• C. $O_2{}^{-}>O_2>O_2^{+}$
• D. $O_2{}^{-}<O_2<O_2^{+}$

Answer 1

The correct option is D $O_2{}^{-}<O_2<O_2^{+}$

$BondOrder=\frac{BondingOrbitalelectrons-AntibondingOrbitalelectrons}{2}$

The electronic configuration of the $O_2$ containing 16 electrons can be written as:

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

Bond Order = $\frac{8-4}{2}$ = 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

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 = $\frac{8-5}{2}$ = 1.5

$O_2^{-}$contains 3 electrons in the anti-bonding orbital. It requires less energy to remove an electron from anti-bonding orbital.

Hence, the correct bond order in the following species is $O_2^{2+}>O_2>O_2^{-}$.

Therefore option D is correct.