Question

Which of the following coordination entities is expected to exhibit the maximum value of magnetic moment (spin-only)? \[ (\mathrm{Mn}=25,\ \mathrm{Fe}=26,\ \mathrm{Co}=27) \]

• A. $[\mathrm{Fe(CN)_6}]^{3-}$
• B. $[\mathrm{CoF_6}]^{3-}$
• C. $[\mathrm{Mn(CN)_6}]^{3-}$
• D. $[\mathrm{Fe(H_2O)_6}]^{3+}$

Hint:

For octahedral complexes: Strong field ligands (CN$^-$, CO, NO$_2^-$) $\rightarrow$ low spin. Weak field ligands (F$^-$, H$_2$O, Cl$^-$) $\rightarrow$ high spin. More unpaired electrons imply a larger magnetic moment.

Answer 1

The Correct Option is D

Concept: The spin-only magnetic moment is given by \[ \mu=\sqrt{n(n+2)} \] where $n$ is the number of unpaired electrons. Therefore, the complex having the maximum number of unpaired electrons will exhibit the highest magnetic moment.

Step 1: Analyse $[\mathrm{Fe(CN)_6}]^{3-}$. \[ \mathrm{Fe}^{3+}:3d^5 \] CN$^-$ is a strong field ligand. Low-spin configuration: \[ t_{2g}^{5}e_g^{0} \] Number of unpaired electrons: \[ n=1 \]

Step 2: Analyse $[\mathrm{CoF_6}]^{3-}$. \[ \mathrm{Co}^{3+}:3d^6 \] F$^-$ is a weak field ligand. High-spin configuration: \[ t_{2g}^{4}e_g^{2} \] Number of unpaired electrons: \[ n=4 \]

Step 3: Analyse $[\mathrm{Mn(CN)_6}]^{3-}$. \[ \mathrm{Mn}^{3+}:3d^4 \] CN$^-$ is strong field. Low-spin configuration: \[ t_{2g}^{4} \] Number of unpaired electrons: \[ n=2 \]

Step 4: Analyse $[\mathrm{Fe(H_2O)_6}]^{3+}$. \[ \mathrm{Fe}^{3+}:3d^5 \] Water is a weak field ligand. High-spin configuration: \[ t_{2g}^{3}e_g^{2} \] Number of unpaired electrons: \[ n=5 \]

Step 5: Compare magnetic moments. Since \[ n=5 \] is maximum, the largest magnetic moment is shown by \[ [\mathrm{Fe(H_2O)_6}]^{3+}. \] Therefore, \[ \boxed{\text{Option (4)}} \]