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.
Fe(II) octahedral complexes can show different 'spin only' magnetic moments for arrangements in the presence of weak and strong ligands. The 'spin only' magnetic moments of these complexes with weak and strong ligands, respectively, are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.90 BM and 0 BM
In an octahedral complex the five d orbitals split into a lower set \( t_{2g} \) (three orbitals) and an upper set \( e_g \) (two orbitals), separated by the splitting energy \( \Delta_o \). With a weak field ligand, \( \Delta_o \) is smaller than the pairing energy, so electrons enter the \( e_g \) orbitals singly before pairing (high spin). With a strong field ligand, \( \Delta_o \) is larger than the pairing energy, so electrons pair up in \( t_{2g} \) first (low spin). The spin-only magnetic moment is \( \mu = \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons.
Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Fe^2+} \): \( \mathrm{3d^{6}} \).
With a weak field ligand: \( t_{2g}^{4}\,e_g^{2} \), four unpaired electrons:
\( \displaystyle \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \mathrm{BM} \)
With a strong field ligand: \( t_{2g}^{6}\,e_g^{0} \), all six electrons paired:
\( \displaystyle \mu = \sqrt{0(0+2)} = 0\ \mathrm{BM} \)
So the values are 4.90 BM and 0 BM, respectively (as in \( \ce{[Fe(H2O)6]^2+} \) and \( \ce{[Fe(CN)6]^4-} \)).