Question

Which of the following orders is correct for the bond dissociation energy of $O_2^{+},O_2,O_2^{-}$ and $O_2^{2-}$?

• A. $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$
• B. $O_2^{+}>O_2<O_2<O_2^{2-}$
• C. $O_2^{+}<O_2<O_2<O_2^{2-}$
• D. $O_2^{+}>O_2>O_2^{2-}>O_2^{-}$

Answer 1

The correct option is A $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$

$\text{Bond Order}=\frac{\text{Bonding electrons}-\text{Antibonding electrons}}{2}$

$O_2^{2-}\to$ Total electron = 16 + 2 = 18
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^2{\pi }^{\ast }(2p_y{)}^2$
$\text{Bond Order}=\frac{10-8}{2}=1$

$O_2^{-}\to$ Total electron = 16 +1 = 17
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^2{\pi }^{\ast }(2p_y{)}^1$
$\text{Bond Order}=\frac{10-7}{2}=1.5$

$O_2\to$ Total electron = 8 + 8 = 16
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^1{\pi }^{\ast }(2p_y{)}^1$
$\text{Bond Order}=\frac{10-6}{2}=2$

$O_2^{+}\to$ Total electron = 16 -1 = 15
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^1$
$\text{Bond Order}=\frac{10-5}{2}=2.5$

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

$\text{Bond order}\propto \text{Bond dissociation enthalpy}$

Order: $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$