CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

${\mathrm{N}}_3^{-}\left(\mathrm{Azide}\right): \overline{\mathrm{N}}=\stackrel{+}{\mathrm{N}}=\overline{\mathrm{N}}\leftrightarrow \overline{\mathrm{N}}-\stackrel{+}{\mathrm{N}}\equiv \mathrm{N}$

It has $\mathrm{sp}$ hybridised central atom. Hence it is linear in shape.

${\mathrm{I}}_3^{-}$(triiodide):

                          

It has linear geometry, ${\mathrm{sp}}^3\mathrm{d}$ hybridisation with three lone pairs at central atom.

${\mathrm{NO}}_2^{-}$ (nitrite) : 

 

It is (nonlinear) bent shaped as it has ${\mathrm{sp}}^2$ hybridisation with one lone pair at central atom.

${\mathrm{O}}_3$ (ozone) : 

 

It Is bent in shape with ${\mathrm{sp}}^2$ hybridisation and one lone pair at central atom.

${\mathrm{SO}}_2$(sulphur dioxide) : 

                    

It is bent in shape with ${\mathrm{sp}}^2$ hybridisation and one lone pair at central atom.

So $3$ among the given molecules is bent in shape.