Question

LIST-I contains metal species and LIST-II contains their properties.
 

LIST-I		LIST-II
I	\(\text{[Cr(CN)}_6\text{]}^{4-}\)	P	t2g orbitals contain 4 electrons
II	\(\text{[RuCl}_6\text{]}^{2-}\)	Q	\(\mu\)(spin-only) = 4.9 BM
III	\(\text{[Cr(H}_2\text{O)}_6\text{]}^{2+}\)	R	low spin complex ion
IV	\(\text{[Fe(H}_2\text{O)}_6\text{]}^{2+}\)	S	metal ion in 4+ oxidation state
		T	d4 species

[Given : Atomic number of Cr = 24, Ru = 44, Fe = 26] 
Metal each metal species in LIST-I with their properties in LIST-II, and choose the correct option.

• A. I → R, T; II → P, S; III → Q, T; IV → P, Q
• B. I → R, S; II → P, T; III → P, Q; IV → Q, T
• C. I → P, R; II → R, S; III → R, T; IV → P, T
• D. I → Q, T; II → S, T; III → P, T; IV → Q, R

Answer 1

The Correct Option is A

(I) \(\text{[Cr(CN)}_6\text{]}^{4-}\)

The Cr²⁺ ion, with the electron configuration [Ar] 3d⁴ 4s⁰, undergoes d²sp³ hybridization due to the strong field ligand CN^–. 

(II) \(\text{[RuCl}_6\text{]}^{2-}\):

The Ru4+ ion, with the electron configuration [Kr] 4d⁴ 5s⁰, has a t_2g set that contains four electrons.

(III) \(\text{[Cr(H}_2\text{O)}_6\text{]}^{2+}\):

The Cr2+ ion, with the electron configuration [Ar] 3d⁴ 4s⁰, exhibits four unpaired electrons due to the weak field ligand H₂O, resulting in a magnetic moment of 4.9 B.M.

(IV) \(\text{[Fe(H}_2\text{O)}_6\text{]}^{2+}\)

The Fe²⁺ ion, with the electron configuration [Ar] 3d⁶ 4s⁰, also possesses four unpaired electrons, resulting in a magnetic moment of 4.9 B.M.
Hence option A is Correct