Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II.

	LIST -I		LIST -II
A)	$ds{p}^2$	P)	${\left[Fe{F}_6\right]}^{4-}$
B)	$s{p}^3$	Q)	$\left[Ti{\left(H_{2}O\right)}_{3}C{l}_3\right]$
C)	$s{p}^3{d}^2$	R)	${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$
D)	$d^{2}s{p}^3$	S)	${\left[FeC{l}_4\right]}^{2-}$
		T)	$Ni{\left(CO\right)}_4$
		U)	${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

• Correct answer
  A

  a-u;b-s,t;c-p;d-q,r;
• B

  a-s,t;b-s,t,u;c-r,s;d-q,r;
• C

  a-u;b-t,u;c-u;d-s,t,u;
• D

  a-s,t;b-q,r;c-r;d-r,s,t,u;

Solution

$\left(P\right)$ ${\left[Fe{F}_6\right]}^{4-}$

$\left(Q\right)$ $\left[Ti{\left(H_{2}O\right)}_{3}C{l}_3\right]$

$\left(R\right)$ ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$

$\left(S\right)$ ${\left[FeC{l}_4\right]}^{2-}$

$\left(T\right)$ $\left[Ni{\left(CO\right)}_4\right]$
$Ni :3d^{8}e{S}^2$

Back pairing of elements due to pressure of strong field ligand

$\left(U\right)$ ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$
$N{I}^{2+} :3d^8$

Hence, the correct answer is option: 'A'.