Question

Match List - I with List - II :

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

Choose the most appropriate answer from the options given below :

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

Hint:

Decide the geometry of each species: square planar for \(d^8\) Pt(II), steric number 6 for \(\mathrm{BrF_5}\), steric number 5 for \(\mathrm{PCl_5}\), and inner-orbital octahedral for the strong-field Co(III) complex.

Answer 1

The Correct Option is B

(A) In \([\mathrm{PtCl_4}]^{2-}\), Pt is +2 (\(5d^8\)). The complex is square planar, so the hybridisation is \(dsp^2\) : (III).
(B) \(\mathrm{BrF_5}\) has 5 bond pairs + 1 lone pair = 6 electron pairs around Br, so \(sp^3d^2\) : (IV).
(C) \(\mathrm{PCl_5}\) has 5 bond pairs and no lone pair, so \(sp^3d\) (trigonal bipyramidal) : (I).
(D) In \([\mathrm{Co(NH_3)_6}]^{3+}\), Co is +3 (\(3d^6\)). \(\mathrm{NH_3}\) is a strong-field ligand, so the electrons pair up and two \(3d\) orbitals are used: \(d^2sp^3\) (inner orbital complex) : (II).
So (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Answer: (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)