Question

Four diatomic species are listed below. Identify the correct order of increasing bond order:

• A. $NO<O_2^{-}<C_2^{2-}<H{e}_2^{2+}$
• B. $O_2^{-}<NO<C_2^{2-}<H{e}_2^{2+}$
• C. $C_2^{2-}<H{e}_2^{2+}<O_2^{-}<NO$
• D. $H{e}_2^{2+}<O_2^{-}<NO<C_2^{2-}$

Answer 1

The correct option is D $H{e}_2^{2+}<O_2^{-}<NO<C_2^{2-}$

For diatomic species, the correct order in which the bond order is increasing is $H{e}_2^{2+}\left(1\right)<O_2^{-}\left(1.5\right)<NO\left(2.5\right)<C_2^{2-}\left(3\right).$

$C_2^{2-}\Rightarrow$ Total number of electrons = 14
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\pi \left(2p_x{)}^2\pi \right(2p_y{)}^2\sigma (2p_z{)}^2$
$\text{Bond order}=\frac{Nb-Na}{2}=\frac{10-4}{2}=3$

$NO\to$ Total number of electrons = 7 + 8 = 15
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^1$
$\text{Bond order}=\frac{10-5}{2}=2.5$

$O_2^{-}\to$ Total number of electrons = 8 + 9 = 17
Configuration:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^2{\pi }^{\ast }(2p_y{)}^1$
$\text{Bond Order}=\frac{10-7}{2}=1.5$

$H{e}_2^{2+}\to$ Total number of electrons = 1 + 1 = 2
Configuration:
$H{e}_2^{2+}\Rightarrow \sigma (1s{)}^2$
$B.O=\frac{2-0}{2}=1$