Chemistry - CHEMICAL BONDING Question with Solution | TestHub

1. Electronic configuration of ${\mathrm{C}}_2$ molecule is $\left(\mathrm{\sigma } 1{\mathrm{s}}^2\right)\left({\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\right)\left(\mathrm{\sigma }2{\mathrm{S}}^2\right)\left({\mathrm{\sigma }}^{*}2{\mathrm{S}}^2\right)\left(\mathrm{\pi }2{\mathrm{P}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{py}}^2\right)$

Hence it has $4$ electrons in $\mathrm{\pi }$ molecular orbitals.

2. ${{\mathrm{H}}_2}^{+}$ has one electron as one electron is removed from ${\mathrm{H}}_2$ molecule.

3. Electronic configuration of ${{\mathrm{O}}_2}^{+}$ ion is

$\mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_2^2 \left({\mathrm{\pi }}_{2{\mathrm{P}}_{\mathrm{x}}^2}={\mathrm{\pi }}_{2{\mathrm{Py}}^2}\right){\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1{\mathrm{\pi }}^{*}2{\mathrm{P}}_{\mathrm{y}}^0$

It has one unpaired electron so it is para-magnetic.

4. Bond orders of ${\mathrm{O}}_2^{+}, {\mathrm{O}}_2, {\mathrm{O}}_2^{-}$ and ${\mathrm{O}}_2^{2-}$ are $2.5, 2, 1.5$ and $1$ respectively.