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Q5.29:

Amongst the following ions which one has the highest magnetic moment value?

Solution :

Given Variables & Initial Setup

To determine the ion with the highest magnetic moment among a group of transition metal ions, we must utilize the Spin-Only Magnetic Moment formula derived from the number of unpaired electrons ($n$):

$\mu_{eff} = \sqrt{n(n+2)} \, \text{B.M.}$

[Per the Quantum Mechanical Model of the atom, the magnetic moment increases monotonically with the number of unpaired electrons ($n$).]

Step 1: Analysis of Electronic Configurations

Let us consider a standard set of common transition metal ions encountered in coordination chemistry exercises (e.g., $[Cr(H_2O)_6]^{3+}$, $[Fe(H_2O)_6]^{2+}$, $[Co(H_2O)_6]^{2+}$, and $[Ni(H_2O)_6]^{2+}$):

Ion Atomic Number ($Z$) Electronic Configuration Number of Unpaired Electrons ($n$)
$Cr^{3+}$ 24 $[Ar] 3d^3$ 3
$Fe^{2+}$ 26 $[Ar] 3d^6$ 4
$Co^{2+}$ 27 $[Ar] 3d^7$ 3
$Ni^{2+}$ 28 $[Ar] 3d^8$ 2

Step 2: Calculating Magnetic Moments

Applying the spin-only formula for the ions identified:

  • For $Cr^{3+}$ ($n=3$): $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.}$
  • For $Fe^{2+}$ ($n=4$): $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.}$
  • For $Co^{2+}$ ($n=3$): $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.}$
  • For $Ni^{2+}$ ($n=2$): $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, \text{B.M.}$

Step 3: Visual Representation of $d$-Orbital Occupancy ($Fe^{2+}$ case)

Fe²+ [Ar] 3d⁶ configuration: 4 unpaired electrons lead to the highest spin-only magnetic moment.

Analytical Conclusion

The magnetic moment is directly proportional to the number of unpaired electrons. Among the transition metal ions typically compared, the $Fe^{2+}$ ion possesses the maximum number of unpaired electrons ($n=4$), resulting in the highest calculated magnetic moment of $4.90 \, \text{B.M.}$

Final Solution: The ion with the highest magnetic moment is $Fe^{2+}$ ($4.90 \, \text{B.M.}$), provided it is in a high-spin configuration.


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