Question:
The spin only magnetic moment of the element having highest third ionization enthalpy among Ti, V, Cr, Mn and Fe in its \(+3\) state is (BM):
The spin only magnetic moment of the element having highest third ionization enthalpy among Ti, V, Cr, Mn and Fe in its \(+3\) state is (BM):
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A half-filled \(d^5\) configuration is exceptionally stable; therefore \(Mn^{2+}\) shows the highest third ionization enthalpy among the given elements.
Updated On: Jun 17, 2026
- 3.87
- 4.90
- 5.92
- 2.84
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The Correct Option is B
Solution and Explanation
Concept:
The highest third ionization enthalpy corresponds to the element whose \(M^{2+}\) configuration is exceptionally stable.
Step 1: Identify the element Electronic configurations: \[ Mn=[Ar]3d^5 4s^2 \] After removal of two electrons: \[ Mn^{2+}=[Ar]3d^5 \] This is a half-filled and highly stable configuration. Therefore removal of the third electron requires maximum energy. Hence Mn has the highest third ionization enthalpy.
Step 2: Configuration of \(Mn^{3+}\) \[ Mn^{3+}=[Ar]3d^4 \] Number of unpaired electrons: \[ n=4 \]
Step 3: Calculate magnetic moment Spin-only formula: \[ \mu=\sqrt{n(n+2)} \] \[ \mu=\sqrt{4(4+2)} \] \[ \mu=\sqrt{24} \] \[ \mu=4.90\ \text{BM} \] \[ \boxed{4.90\ \text{BM}} \]
Step 1: Identify the element Electronic configurations: \[ Mn=[Ar]3d^5 4s^2 \] After removal of two electrons: \[ Mn^{2+}=[Ar]3d^5 \] This is a half-filled and highly stable configuration. Therefore removal of the third electron requires maximum energy. Hence Mn has the highest third ionization enthalpy.
Step 2: Configuration of \(Mn^{3+}\) \[ Mn^{3+}=[Ar]3d^4 \] Number of unpaired electrons: \[ n=4 \]
Step 3: Calculate magnetic moment Spin-only formula: \[ \mu=\sqrt{n(n+2)} \] \[ \mu=\sqrt{4(4+2)} \] \[ \mu=\sqrt{24} \] \[ \mu=4.90\ \text{BM} \] \[ \boxed{4.90\ \text{BM}} \]
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