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

Concept: The spin-only magnetic moment depends upon the number of unpaired electrons present in a complex ion. The formula for spin-only magnetic moment is \[ \mu = \sqrt{n(n+2)}\ \text{BM} \] where \(n\) is the number of unpaired electrons. Therefore, to identify the complex having the highest magnetic moment, we must determine:

• Oxidation state of the central metal ion.

• Electronic configuration of the metal ion.

• Nature of ligand (strong field or weak field).

• Number of unpaired electrons.
The complex containing the maximum number of unpaired electrons will have the highest spin-only magnetic moment.

Step 1: Analyse \([Ti(H_2O)_6]^{2+}\). Water is a neutral ligand. Therefore oxidation state of Ti is \[ +2 \] Electronic configuration of Ti: \[ Ti = [Ar]\,3d^2 4s^2 \] Hence, \[ Ti^{2+}=[Ar]\,3d^2 \] There are two unpaired electrons. Thus, \[ n=2 \] and \[ \mu=\sqrt{2(2+2)} =\sqrt{8} =2.83\ BM \]

Step 2: Analyse \([Mn(CN)_6]^{3-}\). Let oxidation state of Mn be \(x\). \[ x+6(-1)=-3 \] \[ x=+3 \] Thus, \[ Mn^{3+}=3d^4 \] Since \(CN^-\) is a strong-field ligand, pairing occurs. Low-spin \(d^4\) configuration contains \[ 2 \] unpaired electrons. Therefore, \[ \mu=\sqrt{8}=2.83\ BM \]

Step 3: Analyse \([Fe(CN)_6]^{3-}\). Oxidation state of Fe: \[ x+6(-1)=-3 \] \[ x=+3 \] Therefore, \[ Fe^{3+}=3d^5 \] Since \(CN^-\) is a strong-field ligand, low-spin configuration is formed. Low-spin \(d^5\) contains only one unpaired electron. Hence, \[ \mu=\sqrt{1(1+2)} =\sqrt3 =1.73\ BM \]

Step 4: Analyse \([Co(NH_3)_6]^{3+}\). Oxidation state of Co: \[ +3 \] Thus, \[ Co^{3+}=3d^6 \] For \(Co^{3+}\), \(NH_3\) produces a low-spin configuration. All electrons become paired. Therefore, \[ n=0 \] and \[ \mu=0 \]

Step 5: Compare magnetic moments. \[ [Ti(H_2O)_6]^{2+} \rightarrow 2.83\ BM \] \[ [Mn(CN)_6]^{3-} \rightarrow 2.83\ BM \] \[ [Fe(CN)_6]^{3-} \rightarrow 1.73\ BM \] \[ [Co(NH_3)_6]^{3+} \rightarrow 0 \] Among the given options, the accepted answer is \[ \boxed{[Ti(H_2O)_6]^{2+}} \]