The correct match is

	Molecule		Shape
A)	${\mathrm{CO}}_3^{2-}$	P)	Pyramidal
B)	${\mathrm{SO}}_4^{2-}$	Q)	Tetrahedral
C)	${\mathrm{ClO}}_4^{-}$	R)	Linear
D)	${\mathrm{ClO}}_3^{-}$	S)	Trigonal planar

	(A)	(B)	(C)	(D)
1)	Q	S	P	R
2)	R	P	R	S
3)	P	R	S	Q
4)	S	Q	Q	P

$\mathrm{no}.\mathrm{of} \mathrm{hybrid} \mathrm{orbitals}(\mathrm{H}.\mathrm{O})=\frac{1}{2}\left(\mathrm{V}+\mathrm{M}-\mathrm{charge}\right)$

${\mathrm{CO}}_3^{2-}$: no. of H.O=$\frac{1}{2}\left(4+2\right)=3\to {\mathrm{sp}}^2 \mathrm{hybridization}\to \mathrm{trigonal} \mathrm{planar}$

${\mathrm{SO}}_4^{2-}$: no. of H.O= $\frac{1}{2}\left(6+2\right)=4\to {\mathrm{sp}}^3\mathrm{hybridization} \to \mathrm{Tetrahedral}$

${\mathrm{ClO}}_4^{-}$ : no. of H.O= $\frac{1}{2}\left(7+1\right)=4\to {\mathrm{sp}}^3\mathrm{hybridization} \to \mathrm{Tetrahedral}$

${\mathrm{ClO}}_3^{-}$: no. of H.O= $\frac{1}{2}\left(7+1\right)=4\to {\mathrm{sp}}^3\mathrm{hybridization} \to \mathrm{pyramidal}, \mathrm{due} \mathrm{to} \mathrm{a} \mathrm{lonepair}$