JEE Challenger
More from Coordination Compounds
ChemEasyAdv 2013 • Paper 1

Spin Only Magnetic Moment of Transition Metal Complex Ions

Given the following complex ions denoted by PP, QQ, and RR: P=[FeF6]3−,Q=[V(H2O)6]2+,andR=[Fe(H2O)6]2+.P = [\text{FeF}_6]^{3-}, \quad Q = [\text{V}(\text{H}_2\text{O})_6]^{2+}, \quad \text{and} \quad R = [\text{Fe}(\text{H}_2\text{O})_6]^{2+}. Identify the correct sequence of these complex ions arranged in increasing order of their spin-only magnetic moments (expressed in B.M.\text{B.M.}):

Options

A

R<Q<PR < Q < P

B

Q<R<PQ < R < P

Correct
C

R<P<QR < P < Q

D

Q<P<RQ < P < R

Topics & Concepts

Coordination CompoundsCoordination CompoundsCrystal Field Theoryd-Block Elements

Step-by-Step Solution

To determine the correct order of the spin-only magnetic moments of the given complex ions, we calculate the number of unpaired electrons (nn) for each complex using Crystal Field Theory (CFT). The spin-only magnetic moment is given by the formula: μs=n(n+2) B.M.\mu_{\text{s}} = \sqrt{n(n + 2)} \text{ B.M.}

  1. For Complex P=[FeF6]3−P = [\text{FeF}_6]^{3-}:

    • The central metal atom is iron (Fe\text{Fe}, Z=26Z = 26).
    • The oxidation state of iron is +3+3: Fe3+:[Ar] 3d5\text{Fe}^{3+}: [\text{Ar}]\,3d^5
    • Fluoride ion (F−\text{F}^-) is a weak-field ligand, so Δo<P\Delta_o < P (pairing energy), leading to a high-spin complex.
    • The electronic configuration in an octahedral crystal field is: t2g3 eg2t_{2g}^3 \, e_g^2
    • Number of unpaired electrons: n=5n = 5
    • Magnetic moment: μP=5(5+2)=35≈5.92 B.M.\mu_P = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \text{ B.M.}

  2. For Complex Q=[V(H2O)6]2+Q = [\text{V}(\text{H}_2\text{O})_6]^{2+}:

    • The central metal atom is vanadium (V\text{V}, Z=23Z = 23).
    • The oxidation state of vanadium is +2+2: V2+:[Ar] 3d3\text{V}^{2+}: [\text{Ar}]\,3d^3
    • For a d3d^3 ion in an octahedral field, the electronic configuration is fixed regardless of ligand field strength: t2g3 eg0t_{2g}^3 \, e_g^0
    • Number of unpaired electrons: n=3n = 3
    • Magnetic moment: μQ=3(3+2)=15≈3.87 B.M.\mu_Q = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ B.M.}

  3. For Complex R=[Fe(H2O)6]2+R = [\text{Fe}(\text{H}_2\text{O})_6]^{2+}:

    • The central metal atom is iron (Fe\text{Fe}, Z=26Z = 26).
    • The oxidation state of iron is +2+2: Fe2+:[Ar] 3d6\text{Fe}^{2+}: [\text{Ar}]\,3d^6
    • Water (H2O\text{H}_2\text{O}) acts as a weak-field ligand with Fe2+\text{Fe}^{2+}, resulting in a high-spin complex (Δo<P\Delta_o < P).
    • The electronic configuration in an octahedral crystal field is: t2g4 eg2t_{2g}^4 \, e_g^2
    • Number of unpaired electrons: n=4n = 4
    • Magnetic moment: μR=4(4+2)=24≈4.90 B.M.\mu_R = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \text{ B.M.}

Comparing the number of unpaired electrons: nQ(3)<nR(4)<nP(5)n_Q (3) < n_R (4) < n_P (5)

Hence, the spin-only magnetic moments follow the increasing order: Q<R<PQ < R < P

Therefore, the correct option is B.

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