JEE Main 2022 — Chemical Bonding and Molecular Structure Question with Solution

The species which have ${\mathrm{sp}}^2$ hybridisation with no lone pairs, ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridisation with two lone pairs are planar molecules.

The number of hybrid orbitals can be calculated as follows,

No.of hybrid orbitals = 1/2(V+M-C+A)

V = number of valence electrons

M = number of monovalent atoms attached to central atom.

C = Number of positive charges on the molecule

A = Number of negative charges on the molecule

 

${\mathrm{SO}}_3$	${\mathrm{sp}}^2$	Planar
${\mathrm{BF}}_3$	${\mathrm{sp}}^2$	Planar
${\mathrm{NO}}_3^{-}$	${\mathrm{sp}}^2$	Planar
${\mathrm{SF}}_4$	${\mathrm{sp}}^3\mathrm{d}$	Non-Planar
${\mathrm{H}}_2{\mathrm{O}}_2$	${\mathrm{sp}}^3$	Non-Planar
${\mathrm{PCl}}_3$	${\mathrm{sp}}^3$	Non-Planar
${\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right]}^{-}$	${\mathrm{sp}}^3$	Non-Planar
${\mathrm{XeF}}_4$	${\mathrm{sp}}^3{\mathrm{d}}^2$	Planar
${\mathrm{XeO}}_3$	${\mathrm{sp}}^3$	Non-Planar
${\mathrm{PH}}_4^{+}$	${\mathrm{sp}}^3$	Non-Planar