Statement-I: In C₂ vapours, the bond order is two with one sigma bond and one pi bond.

Statement-II: Formal charge on terminal nitrogen atom of $:\stackrel{..}{\mathrm{N}}=\mathrm{N}=\stackrel{..}{\mathrm{O}}:$  is “+1”.

 $\begin{array}{l}\mathrm{I}) {\mathrm{C}}_2...{\mathrm{KK}}^{\text{'}}{\left({\mathrm{\sigma }}_{2\mathrm{s}}^{ }\right)}^2{\left({\mathrm{\sigma }}_{2\mathrm{s}}^{*}\right)}^2{\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^2{\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^2\\ \mathrm{Bond}\mathrm{order}=\frac{8-4}{2}=2;\end{array}$

Both the bonds are pi bonds because there are four electrons in two pi bonding molecular orbitals.

II) $:\stackrel{..}{\mathrm{N}}=\mathrm{N}=\stackrel{..}{\mathrm{O}}:$

Formal charge = Number of electrons in free state - number of electrons of lone pairs - number of bonds

Formal charge on terminal nitrogen = 5 - 4 - 2 = -1