Match List - I with List - II :

List-I		List-II
(A)	$$[PtCl_4]^{2-}$$	(I)	$$sp^3d$$
(B)	$$BrF_5$$	(II)	$$d^2sp^3$$
(C)	$$PCl_5$$	(III)	$$dsp^2$$
(D)	$$[Co(NH_3)_6]^{3+}$$	(IV)	$$sp^3d^2$$

Choose the most appropriate answer from the options given below

We need to match each species with its correct hybridization.

(A) $$[PtCl_4]^{2-}$$:

Pt is in +2 oxidation state ($$Pt^{2+}$$) with electronic configuration $$[Xe]5d^8$$.

$$Pt^{2+}$$ is a $$d^8$$ system. With 4 $$Cl^-$$ ligands (weak field for most metals, but Pt²⁺ being a 5d metal with large crystal field splitting always forms square planar complexes).

Square planar geometry corresponds to $$dsp^2$$ hybridization → matches (III).

(B) $$BrF_5$$:

Br has 7 valence electrons. With 5 bond pairs and 1 lone pair (total 6 electron pairs around Br).

6 electron pairs require $$sp^3d^2$$ hybridization → square pyramidal geometry → matches (IV).

(C) $$PCl_5$$:

P has 5 valence electrons. With 5 bond pairs and 0 lone pairs (total 5 electron pairs around P).

5 electron pairs require $$sp^3d$$ hybridization → trigonal bipyramidal geometry → matches (I).

(D) $$[Co(NH_3)_6]^{3+}$$:

Co is in +3 oxidation state ($$Co^{3+}$$) with electronic configuration $$[Ar]3d^6$$.

$$NH_3$$ is a strong field ligand, so all 6 electrons in $$3d$$ pair up, leaving two empty $$3d$$ orbitals available.

The hybridization is $$d^2sp^3$$ (inner orbital complex) → octahedral geometry → matches (II).

Final matching:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

The correct answer is Option B.