$$N{H_3}$$  does not show any resonating structure due to the absence of double bond.

Polarizing power of $$Z{n^{2 + }}$$  is more than others, hence $$ZnC{l_2}$$  is least ionic in nature.

$$Xe{F_4}$$  has square planar structure, while $$NH_4^ + ,BF_4^ - $$   and $$CC{l_4}$$  have tetrahedral structure.

According to Fajan’s rule smaller, highly charged cation has greatest covalent character while large cation with smaller charge has greatest ionic character.

$$\because $$ after forming the bonds, $$C$$ has only $$6\,{e^ - }$$ in its valence shell.

Structure of $$Xe{F_4}$$  is as follows

It involves $$s{p^3}{d^2}$$  hybridisation in $$Xe - {\text{atom}}{\text{.}}$$   The molecules has square planar structure. $$Xe$$  and four $$F - {\text{atoms}}$$   are coplanar. The lone pairs are present on axial positions, minimise electron pair repulsion.

A bond length is the average distance between the centres of nuclei of two bonded atoms.
A multiple bond (double or triple bonds) is always shorter than the corresponding single bond.
The $$C - {\text{atom}}$$   in $$CO_3^{2 - }$$  is $$s{p^2}$$  hybridised as shown:

The $$C - {\text{atom}}$$   in $$C{O_2}$$  is $$sp$$  hybridised with bond distance carbon-oxygen is $$122\,pm.$$

The $$C - {\text{atom}}$$   in $$CO$$  is $$sp$$  hybridised with $$C - O$$  bond distance is $$110\,pm.$$

So, the correct order is
$$CO < C{O_2} < CO_3^{2 - }$$

Equilateral or triangular planar geometry is formed by $$s{p^2}$$  hybridisation.

Statement (iii) is not true since $$N{H_3}$$  has higher boiling point than $$P{H_3}$$  due to hydrogen bonding.

$$N{F_3}$$  ( pyramidal ) and $$B{F_3}$$  ( trigonal planar )
$$BF_4^ - $$  ( tetrahedral ) and $$NH_4^ + $$  ( tetrahedral )
$$BC{l_3}$$  ( trigonal planar ) and $$BrC{l_3}$$  ( trigonal bipyramidal )
$$N{H_3}$$  ( pyramidal ) and $$NO_3^ - $$  ( trigonal planar )
Thus, $$BF_4^ - $$  and $$NH_4^ + $$  are isostructural species.