Question

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

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

Answer 1

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

A/c to Molecular orbital energy diagram, bond order is calculated from the formula,

$\frac{N_b-N_a}{2}$

$N_b=$ no. of electrons is bonding $M.O.$.

$N_a=$ no. of electrons in antibonding $MO$.

For ${He}_2^{+}$ :- ${\left(\sigma 1S\right)}^2{(\sigma \ast 1S)}^1$

$\Rightarrow \frac{2-1}{2}=0.5$

For $NO$ :- ${\left(\sigma 2S\right)}^2{(\sigma \ast 2S)}^2{\left(\pi 2px\right)}^2{\left(\pi 2py\right)}^2{\left(\sigma 2pz\right)}^2{\left({\pi }^{\ast }2px\right)}^1$

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

For $O_2^{-}$ :- ${\left(\sigma 2S\right)}^2{\left({\sigma }^{\ast }2S\right)}^2{\left(\sigma 2pz\right)}^2{\left(\pi 2px\right)}^2{\left(\pi 2py\right)}^2{\left({\pi }^{\ast }2px\right)}^2{\left({\pi }^{\ast }2py\right)}^1$

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

For $C_2^{2-}$ :- $\frac{6-0}{2}=3$

So, the correct option is 'D'.