Q17 / 54 · JEE Advanced 2022 (Paper 1) · Chemistry

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

  LIST-I   List-II
I CrCN64- P t2g orbitals contain 4 electrons
II RuCl62- Q μspin-only=4.9BM
III CrH2O62+ R low spin complex ion
IV FeH2O62+ S metal ion in 4+ oxidation state
    T d4 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

  • Correct answer
    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

Solution

I  CrCN64-

Cr2=Ar3d44s0

It is d2sp3 hybridised low spin complex as CN- is a strong field ligand.

II  RuCl62-

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=Kr4d45s0

t2g set contains 4 electrons.

III  CrH2O62+

Cr+2=Ar3d44s0

Its configuration is t2g3eg1

It has 4 unpaired e- as H2O is weak field ligand.

So, its Magnetic moment, μ=4.9 B.M

IV  FeH2O62+

Fe2+=Ar3d64 s0

=t2g4eg2

It has 4 unpaired e-, its Magnetic moment, μ=4.9 B.M.

Hence, the correct answer is option: 'A'.

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