Verified Solution • 2025–26 Session moderate

QJEEM21Ch55ec: JEE Main 2021 Chemistry - Class 12 JEEM21Ch55ec Solution: The calculated magnetic moments (spin only value) for species , …

The calculated magnetic moments (spin only value) for species [FeCl4]2−{[FeC{l_4}]^{2 - }} , [Co(C2O4)3]3−{[Co{({C_2}{O_4})_3}]^{3 - }} and MnO42−MnO_4^{2 - } respectively are :

Multiple Choice Options

  • A 5.92, 4.90 and 0 BM
  • B 4.90, 0 and 1.73 BM
  • C 5.82, 0 and 0 BM
  • D 4.90, 0 and 2.83 BM

Step-by-Step Verified Solution

(i) [FeCl4]2−⟶Fe2+⟶[Ar]3d6

∴\therefore Cl is weak field ligand so does not pairing occur

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Coordination Compounds Question 183 English Explanation 1
So, magnetic moment (μ)=n(n+2)BM(\mu ) = \sqrt {n(n + 2)} BM

=4(4+2)BM= \sqrt {4(4 + 2)} BM (n = Number of total unpaired e −- = 4)

=24BM=4.90 BM= \sqrt {24} BM = 4.90\,BM

(ii) [Co(C2O4)3]3−⟶Co3+⟶[Ar]3d6{[Co{({C_2}{O_4})_3}]^{3 - }}\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} C{o^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} [Ar]3{d^6}

C2O4 is strong field ligand so pairing occur.

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Coordination Compounds Question 183 English Explanation 2
All electrons are paired, n = 0

hence, μ\mu = 0

(iii) MnO42−⟶Mn+6⟶[Ar]3d1MnO_4^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} M{n^{ + 6}}\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} [Ar]3{d^1}

n=1n = 1

=n(n+2)BM= \sqrt {n(n + 2)} BM

=1(1+2)BM= \sqrt {1(1 + 2)} BM

=3BM= \sqrt 3 BM

=1.73 BM= 1.73\,BM

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