Chemistry - CHEMICAL BONDING Question with Solution | TestHub

A] The molecular orbital energy configuration of${\mathrm{C}}_2^{2-}$is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },\mathrm{ }{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^2$
In the MO of${\mathrm{C}}_2^{2-}$there is no unpaired electron hence it is diamagnetic
B] Bond order of${\mathrm{O}}_2^{2+}$is 3 and${\mathrm{O}}_2$is 2 therefore bond length of${\mathrm{O}}_2$is greater than${\mathrm{O}}_2^{2+}$
C] The molecular orbital energy configuration of${\mathrm{N}}_2^{+}$is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },\mathrm{ }{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^1$
Bond order of${\mathrm{N}}_2^{+}=\frac{1}{2}\left(9-4\right)=2.5$
The molecular orbital energy configuration of${\mathrm{N}}_2^{-}$is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^2\mathrm{ },{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{\mathrm{*}1}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{\mathrm{*}}$
Bond order of${\mathrm{N}}_2^{-}=\frac{1}{2}\left(10-5\right)=2.5$
D]$\mathrm{H}{\mathrm{e}}_2^{+}$has less energy as compare to two isolated He atoms