Lattice energy of an ionic compound depends upon :

Consider the following statements

${\mathrm{S}}_1:{\mathrm{N}}_2{\mathrm{H}}_4$ is pyramidal about each $\mathrm{N}$ atom.

${\mathrm{S}}_2:{\mathrm{NH}}_2\mathrm{OH}$ is pyramidal about the $\mathrm{N}$ atom and bent about the $\mathrm{O}$ atom.

${\mathrm{S}}_3:{\mathrm{CH}}_3\mathrm{COCl}$ is trigonal about the carbon atom (attached to $\mathrm{O}$ and $\mathrm{Cl}$ ) and out of these select the correct one.

Which of the following statements are correct?

($\mathrm{I}$)

 structure is not allowed because octet around 'O' can not be expanded.

($\mathrm{II}$) ${\mathrm{H}}_2{\mathrm{O}}_2$ is ionic compound.

($\mathrm{III}$) In ${\mathrm{B}}_2$ motecule, the highest occupied molecular orbital is $\sigma$ molecular orbital.

($\mathrm{IV}$) The $\mathrm{Ip}-\mathrm{bp}$ repulsion is stronger than $\mathrm{bp}-\mathrm{bp}$ repulsion.

$\mathrm{S}-1:$ simple ionic compounds show isomerism and isomorphism due to the directional nature of the electrovalent bond.

$\mathrm{S}-2:$ Covalent bond formed by hybrid orbitals are more stronger than those of formed by pure atomic orbitals.

$\mathrm{S}-3: {\mathrm{CO}}_3^{2-}$ anion has

$\left(\mathrm{i}\right)$ bonds of unequal length 

$\left(\mathrm{ii}\right)$ ${\mathrm{sp}}^2$shybridisation of carbon atom with same $\mathrm{OCO}$ bond angles

$\mathrm{S}-4:$ The number of lone pair$\left(\mathrm{s}\right)$ of electrons present on $\mathrm{Xe}$ in ${\mathrm{XeO}}_2{\mathrm{F}}_2,{\mathrm{XeOF}}_4,{\mathrm{XeO}}_3$ and ${\mathrm{XeO}}_4$ are $1 1,1$ and $0$, respectively.

Which of the following is not an ionic compound?