Question

For diatomic species are listed below, identify the correct order in which the bond order is increasing in them.

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

Answer 1

The correct option is C ${He}_2^{+}<O_2^{-}<NO<C_2^{2-}$

For diatomic species the correct order in which the bond order is increasing is ${He}_2^{+}<O_2^{-}<NO<C_2^{2-}$.

The bond order is increasing in the order $0.5<1.5<2.5<3$

$C_2^{2-}\Rightarrow$ total electron $=14$
Configuration:

$KK\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p{x}^2\pi 2p{y}^2\sigma 2p{z}^2$

Bondorder $=\frac{Nb-Na}{2}=\frac{8-2}{2}=3$

$NO\Rightarrow$ total electron $=5+6=11$

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

Bond order $=\frac{(8-3)}{2}=2.5$

$O_2^{-}\Rightarrow KK\sigma {\left(2s\right)}^2{\sigma }^{\ast }{\left(2s\right)}^2\sigma {\left(2pz\right)}^2\pi {\left(2px\right)}^2\pi {\left(2py\right)}^2{\pi }^{\ast }{\left(2px\right)}^2{\pi }^{\ast }{\left(2py\right)}^1$

B.O$=\frac{8-5}{2}=1.5$

${He}^{2+}$ $\Rightarrow$ $\sigma 1s^2{\sigma }^{\ast }1s^1$

B.O$=\frac{2-1}{2}=0.5$