Read the passage carefully and answer the Questions.

In the formation of coordination complexes, if the inner d orbital (n-1d) is used in hybridisation, the complex, is called an inner orbital or low spin or spin paired complex. And if it uses outer d orbital (nd) in hybridisation (like $sp^3d^2$), it is called outer orbital or high spin or spin free complex. The degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the complex. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For coordination complexes, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, $μ = \sqrt{n (n+2)}$ where $n$ is the number of unpaired electrons and $μ$ is the magnetic moment in units of Bohr magneton (BM). The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine.

Which of the following complex has the highest magnetic moment?

$[FeF_6]^{3-}$

The correct answer is Option (2) → $[FeF_6]^{3-}$

Magnetic moment depends on the number of unpaired electrons.

Spin-only magnetic moment formula:

$\mu = \sqrt{n(n + 2)} \text{ BM}$

where $n = $ number of unpaired electrons.

First determine the oxidation state and electronic configuration.

(A) $[Mn(CN)_6]^{3-}$

$CN^-$ is a strong field ligand causing low-spin configuration.

Oxidation state of $Mn$:

$Mn + 6(-1) = -3$

$Mn = +3$

$Mn^{3+} \rightarrow 3d^4$

Strong field $\rightarrow$ low spin $d^4$

Unpaired electrons $= \mathbf{2}$

(B) $[FeF_6]^{3-}$

$F^-$ is a weak field ligand, producing high-spin complex.

Oxidation state:

$Fe + 6(-1) = -3$

$Fe = +3$

$Fe^{3+} \rightarrow 3d^5$

High-spin $d^5 \rightarrow \mathbf{5}$ unpaired electrons

(C) $[CoF_6]^{3-}$

$F^-$ weak field ligand $\rightarrow$ high-spin complex.

Oxidation state:

$Co + 6(-1) = -3$

$Co = +3$

$Co^{3+} \rightarrow 3d^6$

High-spin $d^6 \rightarrow \mathbf{4}$ unpaired electrons

(D) $[MnCl_6]^{3-}$

$Cl^-$ weak field ligand $\rightarrow$ high-spin complex.

$Mn^{3+} \rightarrow 3d^4$

High-spin $d^4 \rightarrow$ 4 unpaired electrons

Number of Unpaired Electrons

• $[Mn(CN)_6]^{3-} \rightarrow 2$
• $[FeF_6]^{3-} \rightarrow 5$
• $[CoF_6]^{3-} \rightarrow 4$
• $[MnCl_6]^{3-} \rightarrow 4$

Highest Magnetic Moment

Maximum unpaired electrons $= \mathbf{5}$

Thus the highest magnetic moment is for:

$[FeF_6]^{3-}$