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Coordination Compounds question

2020 · 9 Jan · Shift 2 · Q20
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Coordination Compounds question

2020 · 9 Jan · Shift 2 · Q20

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
The correct order of the spin-only magnetic moments of the following complexes is : (I) [Cr(H2O)6]Br2[Cr(H_2O)_6]Br_2[Cr(H2​O)6​]Br2​ (II) Na4[Fe(CN)6]Na_4[Fe(CN)_6]Na4​[Fe(CN)6​] (III) Na3[Fe(C2O4)3]Na_3[Fe(C_2O_4)_3]Na3​[Fe(C2​O4​)3​] (Δ\DeltaΔ 0 >\gt> P) (IV) (Et4N)2[CoCl4](Et_4N)_2[CoCl_4](Et4​N)2​[CoCl4​]
  1. A
    (III) > (I) > (II) > (IV)
  2. B
    (II) ≈\approx≈ (I) > (IV) > (III)
  3. C
    (III) > (I) > (IV) > (II)
  4. D
    (I) > (IV) > (III) > (II)
View written solutionFree

Correct answer: D

  1. Find oxidation state and d-electron count for each complex

The spin-only magnetic moment is μ=n(n+2) BM\mu = \sqrt{n(n+2)}\ \text{BM}μ=n(n+2)​ BM where nnn is the number of unpaired electrons. So we first determine nnn for each complex.


  1. Complex (I): [Cr(H2O)6]Br2[Cr(H_2O)_6]Br_2[Cr(H2​O)6​]Br2​

Since Br−Br^-Br− are counter ions, the complex cation is [Cr(H2O)6]2+[Cr(H_2O)_6]^{2+}[Cr(H2​O)6​]2+.

  • Oxidation state of Cr=+2Cr = +2Cr=+2
  • Cr:[Ar]3d54s1Cr: [Ar]3d^5 4s^1Cr:[Ar]3d54s1
  • Cr2+⇒3d4Cr^{2+} \Rightarrow 3d^4Cr2+⇒3d4

H2OH_2OH2​O is a weak field ligand, so octahedral high-spin configuration: d4⇒t2g3eg1d^4 \Rightarrow t_{2g}^3 e_g^1d4⇒t2g3​eg1​

Number of unpaired electrons: n=4n=4n=4

Thus, μI=4(4+2)=24\mu_I = \sqrt{4(4+2)} = \sqrt{24}μI​=4(4+2)​=24​


  1. Complex (II): Na4[Fe(CN)6]Na_4[Fe(CN)_6]Na4​[Fe(CN)6​]

The complex ion is [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−.

Let oxidation state of Fe be xxx: x+6(−1)=−4⇒x=+2x+6(-1)=-4 \Rightarrow x=+2x+6(−1)=−4⇒x=+2

So Fe2+Fe^{2+}Fe2+ is 3d63d^63d6.

CN−CN^-CN− is a strong field ligand, so octahedral low-spin: d6⇒t2g6eg0d^6 \Rightarrow t_{2g}^6 e_g^0d6⇒t2g6​eg0​

Number of unpaired electrons: n=0n=0n=0

Thus, μII=0\mu_{II}=0μII​=0


  1. Complex (III): Na3[Fe(C2O4)3]Na_3[Fe(C_2O_4)_3]Na3​[Fe(C2​O4​)3​] with Δo>P\Delta_o>PΔo​>P

The complex ion is [Fe(C2O4)3]3−[Fe(C_2O_4)_3]^{3-}[Fe(C2​O4​)3​]3−.

Each oxalate ligand C2O42−C_2O_4^{2-}C2​O42−​ has charge −2-2−2. Let oxidation state of Fe be xxx: x+3(−2)=−3⇒x=+3x+3(-2)=-3 \Rightarrow x=+3x+3(−2)=−3⇒x=+3

So Fe3+Fe^{3+}Fe3+ is 3d53d^53d5.

Given Δo>P\Delta_o>PΔo​>P, it is low spin octahedral: d5⇒t2g5eg0d^5 \Rightarrow t_{2g}^5 e_g^0d5⇒t2g5​eg0​

Number of unpaired electrons: n=1n=1n=1

Thus, μIII=1(1+2)=3\mu_{III} = \sqrt{1(1+2)}=\sqrt{3}μIII​=1(1+2)​=3​


  1. Complex (IV): (Et4N)2[CoCl4](Et_4N)_2[CoCl_4](Et4​N)2​[CoCl4​]

The complex ion is [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2−.

Let oxidation state of Co be xxx: x+4(−1)=−2⇒x=+2x+4(-1)=-2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2

So Co2+Co^{2+}Co2+ is 3d73d^73d7.

Cl−Cl^-Cl− is a weak field ligand, and tetrahedral complexes are generally high spin.

For tetrahedral d7d^7d7 high spin, number of unpaired electrons: n=3n=3n=3

Thus, μIV=3(3+2)=15\mu_{IV} = \sqrt{3(3+2)} = \sqrt{15}μIV​=3(3+2)​=15​


  1. Compare magnetic moments
  • (I): n=4n=4n=4, μ=24\mu=\sqrt{24}μ=24​
  • (IV): n=3n=3n=3, μ=15\mu=\sqrt{15}μ=15​
  • (III): n=1n=1n=1, μ=3\mu=\sqrt{3}μ=3​
  • (II): n=0n=0n=0, μ=0\mu=0μ=0

Therefore, (I)>(IV)>(III)>(II)\boxed{(I)>(IV)>(III)>(II)}(I)>(IV)>(III)>(II)​


  1. Match with options

This corresponds to Option D.


  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.

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