Question

Which of the following transition metal has the highest magnetic moment?

• A. $Sc^{3+}$
• B. $Ti^{3+}$
• C. $Cr^{2+}$
• D. $Fe^{2+}$
• E. $Mn^{2+}$

Hint:

$d^5$ configuration in a weak field always yields the maximum spin.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The magnetic moment of a transition metal ion (in the 'spin-only' approximation) is determined by the number of unpaired electrons in its d-orbitals. A higher number of unpaired electrons results in a higher magnetic moment.
Step 2: Key Formula or Approach:
1. Write the electron configuration of the neutral atom. 2. Determine the electron configuration of the ion by removing electrons, first from the outermost s-orbital, then from the d-orbital. 3. Count the number of unpaired electrons (n) in the d-orbitals. 4. The spin-only magnetic moment (\( \mu \)) is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \text{ Bohr Magnetons (BM)} \] Since the formula is a monotonically increasing function of n, the species with the most unpaired electrons will have the highest magnetic moment. Step 3: Detailed Explanation:
Let's find the number of unpaired electrons (n) for each ion: - (A) Sc\(^{3+}\): Scandium (Sc, Z=21) has configuration [Ar] 3d\(^1\) 4s\(^2\). To form Sc\(^{3+}\), it loses all 3 valence electrons. The configuration is [Ar] 3d\(^0\). Number of unpaired electrons, \( n = 0 \). - (B) Ti\(^{3+}\): Titanium (Ti, Z=22) has configuration [Ar] 3d\(^2\) 4s\(^2\). To form Ti\(^{3+}\), it loses two 4s electrons and one 3d electron. The configuration is [Ar] 3d\(^1\). Number of unpaired electrons, \( n = 1 \). - (C) Cr\(^{2+}\): Chromium (Cr, Z=24) has an exceptional configuration [Ar] 3d\(^5\) 4s\(^1\). To form Cr\(^{2+}\), it loses one 4s and one 3d electron. The configuration is [Ar] 3d\(^4\). Number of unpaired electrons, \( n = 4 \). - (D) Fe\(^{2+}\): Iron (Fe, Z=26) has configuration [Ar] 3d\(^6\) 4s\(^2\). To form Fe\(^{2+}\), it loses two 4s electrons. The configuration is [Ar] 3d\(^6\). The 3d\(^6\) configuration has 4 unpaired electrons (and one paired). Number of unpaired electrons, \( n = 4 \). - (E) Mn\(^{2+}\): Manganese (Mn, Z=25) has configuration [Ar] 3d\(^5\) 4s\(^2\). To form Mn\(^{2+}\), it loses two 4s electrons. The configuration is [Ar] 3d\(^5\). This is a half-filled d-subshell, with all 5 electrons unpaired. Number of unpaired electrons, \( n = 5 \). Comparing the number of unpaired electrons: Sc\(^{3+}\) (n=0), Ti\(^{3+}\) (n=1), Cr\(^{2+}\) (n=4), Fe\(^{2+}\) (n=4), Mn\(^{2+}\) (n=5). The maximum number of unpaired electrons is 5, found in Mn\(^{2+}\).
Step 4: Final Answer:
Since Mn\(^{2+}\) has the highest number of unpaired electrons (n=5), it will have the highest magnetic moment.