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JEE Advanced Chemistry Coordination Compounds 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Chemistry Question (2022) — Solution

Question

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 , and choose the correct option

Options

  1. A. I → R , T ; II → P , S ; III → Q , T ;   IV → P , Q
  2. B. I → R , S ; II → P , T ; III → P , Q ; IV → Q , T
  3. C. I → P , R ; II → R , S ; III → R , T ; IV → P , T
  4. D. I → Q , T ; II → S , T ; III → P , T ; IV → Q , R

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,  μ = 4 . 9   B . M IV     Fe H 2 O 6 2 + Fe 2 + = Ar 3 d 6 4   s 0 = t 2 g 4 e g 2 It has  4  unpaired  e - , its Magnetic moment,  μ = 4 . 9   B . M .

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Related: Chemistry — Coordination Compounds · All PYQ Banks