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?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([FeF_6]^{3-}\)
The spin-only magnetic moment rises with the number of unpaired electrons, \( \mu = \sqrt{n(n+2)} \) BM. To find n, work out the oxidation state and the 3d configuration of the metal, then fill the \( t_{2g} \) and \( e_g \) sets: a strong field ligand (\( \ce{CN-} \)) pairs the electrons in \( t_{2g} \) (low spin), while weak field ligands (\( \ce{F-} \), \( \ce{Cl-} \)) leave them unpaired as far as possible (high spin).
Option (1) \( \ce{[Mn(CN)6]^3-} \): \( x + 6(-1) = -3 \), so Mn is +3. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \mathrm{3d^{4}} \). \( \ce{CN-} \) is strong field: \( t_{2g}^{4} \), 2 unpaired, \( \mu = \sqrt{8} \approx 2.83 \) BM.
Option (2) \( \ce{[FeF6]^3-} \): Fe is +3. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \mathrm{3d^{5}} \). \( \ce{F-} \) is weak field: \( t_{2g}^{3}\,e_g^{2} \), 5 unpaired, \( \mu = \sqrt{35} \approx 5.92 \) BM.
Option (3) \( \ce{[CoF6]^3-} \): Co is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing three electrons gives \( \mathrm{3d^{6}} \). \( \ce{F-} \) is weak field: \( t_{2g}^{4}\,e_g^{2} \), 4 unpaired, \( \mu = \sqrt{24} \approx 4.90 \) BM.
Option (4) \( \ce{[MnCl6]^3-} \): Mn is +3, \( \mathrm{3d^{4}} \) as in Option (1). \( \ce{Cl-} \) is weak field: \( t_{2g}^{3}\,e_g^{1} \), 4 unpaired, \( \mu \approx 4.90 \) BM.
So the highest magnetic moment, 5.92 BM, belongs to \( \ce{[FeF6]^3-} \).
Shortcut: Five unpaired electrons is the most a d subshell can hold, and it is reached only by a high spin \( \mathrm{d^{5}} \) ion. \( \ce{Fe^3+} \) with the weak field ligand \( \ce{F-} \) is the only such case among the options.