Question

Match List-I (Species) with List-II (Hybrid orbitals used by the central atom in their formation) and select the correct answer:

List - I	List - II
(A) $Ni(CN{)}_5^{3-}$	(1) $s{p}^3$
(B) $CuCl{)}_5^{3-}$	(2) $ds{p}^2$
(C) $AuC{l}_4^{-}$	(3) $s{p}^3{d}_{z^2}$
(D) $Cl{O}_4^{-}$	(4) $d_{x^2-y^2}s{p}^3$

• A. $\left(A\right)-1,\left(B\right)-3,\left(C\right)-2,\left(D\right)-4$
• B. $\left(A\right)-3,\left(B\right)-4,\left(C\right)-2,\left(D\right)-1$
• C. $\left(A\right)-4,\left(B\right)-2,\left(C\right)-1,\left(D\right)-3$
• D. $\left(A\right)-4,\left(B\right)-3,\left(C\right)-2,\left(D\right)-1$

Answer 1

The correct option is D $\left(A\right)-4,\left(B\right)-3,\left(C\right)-2,\left(D\right)-1$

$Ni(CN{)}_5^{3-}$ involves $d_{x^2-y^2}s{p}^3$ hybridization as shown in the figure which results in trigonal bipyramidal geometry.

Similarly,

$CuC{l}_5^{3-}$ involves $s{p}^3{d}_{z^2}$ hybridization.

$AuC{l}_4^{-}$ involves $ds{p}^2$ hybridization with square planar geometry.

$Cl{O}_4^{-}$ involves $s{p}^3$ hybridization with tetrahedral geometry.
The correct match is $\left(A\right)-4,\left(B\right)-3,\left(C\right)-2,\left(D\right)-1$
Option D is correct.