Verified Solution • 2025–26 Session moderate

QJEEAdv22Chde64: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Chde64 Solution: LIST-I contains metal species and LIST-II contains their…

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

List-I List-II
(I) [Cr(CN)6]4−\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}
(P) t2gt_{2 \mathrm{g}} orbitals contain 4 electrons
(II) [RuCl6]2−\left[\mathrm{RuCl}_{6}\right]^{2-} (Q) μ\mu (spin-only )=4.9BM)=4.9 \mathrm{BM}
(III) [Cr(H2O)6]2+\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}
(R) low spin complex ion
(IV) [Fe(H2O)6]2+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}
(S) metal ion in 4+4+ oxidation state
(T) d4d^{4} species

[Given: Atomic number of Cr=24,Ru=44,Fe=26\mathrm{Cr}=24, \mathrm{Ru}=44, \mathrm{Fe}=26 ]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

Multiple Choice Options

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

Step-by-Step Verified Solution

(I) [Cr(CN)6]4−{[Cr{(CN)_6}]^{4 - }}

Cr+2=[Ar] 3d4 4s0C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}

It is d2sp3 hybridised as CN −- is a strong field ligand.

(II) [RuCl6]2−{[RuC{l_6}]^{2 - }}

Ru+4=[Kr] 4d4 5s0R{u^{ + 4}} = [Kr]\,4{d^4}\,5{s^0}

t2g set contains 4 electron.

(III) [Cr(H2O)6]2+{[Cr{({H_2}O)_6}]^{2 + }}

Cr+2=[Ar] 3d4 4s0C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}

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

So, its μ\mu = 4.9 B.M.

(IV) [Fe(H2O)6]2+{[Fe{({H_2}O)_6}]^{2 + }}

Fe+2=[Ar] 3d6 4s0F{e^{ + 2}} = [Ar]\,3{d^6}\,4{s^0}

=t2g4 eg2= t_{2g}^4\,e_g^2

It has 4 unpaired e −- , its μ\mu = 4.9 B.M.

Related Learning Resources & Exam Preparation