JEE Advanced2022ChemistryCoordination CompoundsActual
LIST- I contains metal species and LIST- II contains their properties. LIST- I List- II I Cr CN 6 4 - P t 2 g orbitals contain 4 electrons II RuCl 6 2 - Q μ spin - only = 4 . 9 BM III Cr H 2 O 6 2 + R low spin complex ion IV Fe H 2 O 6 2 + S metal ion in 4 + oxidation state T d 4 species [Given: Atomic number of Cr = 24 , Ru = 44 , Fe = 26 ] Match each metal species in LIST- I with their properties in LIST- II ,
Options
- AI → R , T ; II → P , S ; III → Q , T ;   IV → P , Q
- BI → R , S ; II → P , T ; III → P , Q ; IV → Q , T
- CI → P , R ; II → R , S ; III → R , T ; IV → P , T
- DI → Q , T ; II → S , T ; III → P , T ; IV → Q , R
Correct answer
A. I → R , T ; II → P , S ; III → Q , T ;   IV → P , Q
Step-by-step solution
I     Cr CN 6 4 - Cr 2 = Ar 3 d 4 4 s 0 It is d 2 sp 3 hybridised low spin complex as CN - is a strong field ligand. II     RuCl 6 2 - Low spin complex as Ru belongs to 4d series, as elements belonging to 4d and 5d series forms low spin complexes even with weak field ligand. Ru + 4 = Kr 4 d 4 5 s 0 t 2 g set contains 4 electrons. III     Cr H 2 O 6 2 + Cr + 2 = Ar 3 d 4 4 s 0 Its configuration is t 2 g 3 e g 1 It has 4 unpaired e - as H 2 O is weak field ligand. So, its Magnetic moment