Q20 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry

The correct order of spin only magnetic moments among the following is: [Given: Atomic numbers: $\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ ]

  • Correct answer
    A
    $\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
  • B
    $\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}$
  • C
    $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}$
  • D
    $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}$

Solution

For $\left[\mathrm{MnCl}_4\right]^{2-}$: Mn is in $+2$ oxidation state ($d^5$ configuration). Since $\mathrm{Cl}^{-}$ is a weak field ligand, the electrons remain unpaired. Number of unpaired electrons $n = 5$. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.

For $\left[\mathrm{CoCl}_4\right]^{2-}$: Co is in $+2$ oxidation state ($d^7$ configuration). Since $\mathrm{Cl}^{-}$ is a weak field ligand, the electrons remain unpaired. Number of unpaired electrons $n = 3$. Magnetic moment $\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87$ BM.

For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Fe is in $+2$ oxidation state ($d^6$ configuration). Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing of electrons. All $6$ electrons pair up in the $t_{2g}$ orbitals. Number of unpaired electrons $n = 0$. Magnetic moment $\mu = 0$ BM.

Comparing the values: $5.92 > 3.87 > 0$. Therefore, the order is $\left[\mathrm{MnCl}_4\right]^{2-} > \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$.

Answer: $\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$

Hence, the correct answer is option: 'A'.

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