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Coordination Compounds question

2021 · 24 Feb · Shift 2 · Q8
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  5. /2021 · 24 Feb · Shift 2 · Q8

Coordination Compounds question

2021 · 24 Feb · Shift 2 · Q8

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
The calculated magnetic moments (spin only value) for species [FeCl4]2−{[FeC{l_4}]^{2 - }}[FeCl4​]2−, [Co(C2O4)3]3−{[Co{({C_2}{O_4})_3}]^{3 - }}[Co(C2​O4​)3​]3− and MnO42−MnO_4^{2 - }MnO42−​ respectively are :
  1. A
    5.92, 4.90 and 0 BM
  2. B
    4.90, 0 and 1.73 BM
  3. C
    5.82, 0 and 0 BM
  4. D
    4.90, 0 and 2.83 BM
View written solutionFree

Correct answer: B

  1. Use the spin-only magnetic moment formula

For a species with nnn unpaired electrons, the spin-only magnetic moment is

μ=n(n+2)  BM\mu = \sqrt{n(n+2)}\; \text{BM}μ=n(n+2)​BM

We find nnn for each species.

  1. For [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−
  • Let oxidation state of Fe be xxx.
  • Since each Cl−Cl^-Cl− is −1-1−1,
x+4(−1)=−2x + 4(-1) = -2x+4(−1)=−2 x−4=−2⇒x=+2x - 4 = -2 \Rightarrow x = +2x−4=−2⇒x=+2

So, Fe2+Fe^{2+}Fe2+ is:

Fe:[Ar]3d64s2⇒Fe2+:[Ar]3d6Fe: [Ar]3d^64s^2 \Rightarrow Fe^{2+} : [Ar]3d^6Fe:[Ar]3d64s2⇒Fe2+:[Ar]3d6

Now, Cl−Cl^-Cl− is a weak field ligand and [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2− is tetrahedral. Tetrahedral complexes are generally high spin.

So d6d^6d6 tetrahedral has 4 unpaired electrons.

Thus,

μ=4(4+2)=24≈4.90  BM\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\; \text{BM}μ=4(4+2)​=24​≈4.90BM
  1. For [Co(C2O4)3]3−[Co(C_2O_4)_3]^{3-}[Co(C2​O4​)3​]3−

Let oxidation state of Co be xxx.

Each oxalate ligand C2O42−C_2O_4^{2-}C2​O42−​ has charge −2-2−2.

x+3(−2)=−3x + 3(-2) = -3x+3(−2)=−3 x−6=−3⇒x=+3x - 6 = -3 \Rightarrow x = +3x−6=−3⇒x=+3

So, Co3+Co^{3+}Co3+ is:

Co:[Ar]3d74s2⇒Co3+:[Ar]3d6Co: [Ar]3d^74s^2 \Rightarrow Co^{3+}: [Ar]3d^6Co:[Ar]3d74s2⇒Co3+:[Ar]3d6

This is an octahedral complex with three bidentate oxalate ligands. For Co3+Co^{3+}Co3+, the crystal field splitting is large enough to give a low-spin d6d^6d6 configuration:

t2g6eg0t_{2g}^6e_g^0t2g6​eg0​

Hence, number of unpaired electrons n=0n=0n=0.

Therefore,

μ=0(0+2)=0  BM\mu = \sqrt{0(0+2)} = 0\; \text{BM}μ=0(0+2)​=0BM
  1. For MnO42−MnO_4^{2-}MnO42−​

Let oxidation state of Mn be xxx.

x+4(−2)=−2x + 4(-2) = -2x+4(−2)=−2 x−8=−2⇒x=+6x - 8 = -2 \Rightarrow x = +6x−8=−2⇒x=+6

So, Mn6+Mn^{6+}Mn6+ is:

Mn:[Ar]3d54s2⇒Mn6+:[Ar]3d1Mn: [Ar]3d^54s^2 \Rightarrow Mn^{6+}: [Ar]3d^1Mn:[Ar]3d54s2⇒Mn6+:[Ar]3d1

Thus there is 1 unpaired electron.

Hence,

μ=1(1+2)=3≈1.73  BM\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\; \text{BM}μ=1(1+2)​=3​≈1.73BM
  1. Match with the options

The magnetic moments are:

[FeCl4]2−:4.90  BM,[Co(C2O4)3]3−:0  BM,MnO42−:1.73  BM[FeCl_4]^{2-} : 4.90\; \text{BM}, \quad [Co(C_2O_4)_3]^{3-} : 0\; \text{BM}, \quad MnO_4^{2-} : 1.73\; \text{BM}[FeCl4​]2−:4.90BM,[Co(C2​O4​)3​]3−:0BM,MnO42−​:1.73BM

This matches Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

So they agree.

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