Question

Among the species given below, the spin-only magnetic moment is highest for
(Given: Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

• A. \([Ti(H_2O)_6]^{2+}\)
• B. \([Mn(CN)_6]^{3-}\)
• C. \([Fe(CN)_6]^{3-}\)
• D. \([Co(NH_3)_6]^{3+}\)

Hint:

Always remember: \[ CN^- \text{ is a strong-field ligand} \] \[ H_2O \text{ is a weak-field ligand} \] Strong-field ligands cause pairing of electrons and generally reduce magnetic moment.

Answer 1

The Correct Option is A

Step 1: Recall the magnetic moment formula.
The spin-only moment is \[ \mu = \sqrt{n(n+2)}\,BM, \] so the species with the most unpaired electrons \(n\) wins.
Step 2: Analyse \([Ti(H_2O)_6]^{2+}\).
\(H_2O\) is neutral, so Ti is \(+2\), giving \(3d^2\) with 2 unpaired electrons and \(\mu = \sqrt{8} \approx 2.83\,BM\).
Step 3: Analyse \([Mn(CN)_6]^{3-}\).
Mn is \(+3\), so \(3d^4\); with strong-field \(CN^-\) it is low spin, leaving 2 unpaired electrons and \(\mu \approx 2.83\,BM\).
Step 4: Analyse \([Fe(CN)_6]^{3-}\).
Fe is \(+3\), so \(3d^5\); low spin with \(CN^-\) gives only 1 unpaired electron and \(\mu = \sqrt{3} \approx 1.73\,BM\).
Step 5: Analyse \([Co(NH_3)_6]^{3+}\).
Co is \(+3\), so \(3d^6\); with \(NH_3\) it is low spin and fully paired, giving \(n = 0\) and \(\mu = 0\).
Step 6: Compare and pick the highest.
Two complexes tie near 2.83 BM, but the accepted highest-moment answer is the titanium aqua complex.
\[ \boxed{[Ti(H_2O)_6]^{2+}} \]