Question

The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are 
A. $ Cr^{2+} $ 
B. $ Fe^{2+} $ 
C. $ Fe^{3+} $ 
D. $ Co^{2+} $ 
E. $ Mn^{2+} $ 
Choose the correct answer from the options given below

• A. A, C and E only
• B. B and E only
• C. B and E only
• D. A, B and E only

Hint:

Use the spin-only magnetic moment formula to calculate the number of unpaired electrons. Then, determine the electronic configurations of the given metal ions and identify those with 4 unpaired electrons.

Answer 1

The Correct Option is D

To identify metal ions with a calculated spin-only magnetic moment of 4.9 Bohr Magneton (B.M.), the formula for spin-only magnetic moment is applied:

\(\mu = \sqrt{n(n+2)} \, \text{B.M.}\)

Here, \( n \) represents the number of unpaired electrons. A magnetic moment of 4.9 B.M. corresponds to \( n = 4 \) unpaired electrons, confirmed by the following calculation:

\[\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \, \text{B.M.}\]

The analysis of each metal ion is as follows:

• \( \text{Cr}^{2+} \): Chromium (atomic number 24) has an electron configuration of \( [\text{Ar}] \, 3d^5 \, 4s^1 \). Upon forming \( \text{Cr}^{2+} \), it loses two electrons (one from 4s and one from 3d), resulting in \( [\text{Ar}] \, 3d^4 \). This configuration contains 4 unpaired electrons.
• \( \text{Fe}^{2+} \): Iron (atomic number 26) has an electron configuration of \( [\text{Ar}] \, 3d^6 \, 4s^2 \). When forming \( \text{Fe}^{2+} \), it loses two electrons from the 4s orbital, yielding \( [\text{Ar}] \, 3d^6 \). This configuration has 4 unpaired electrons.
• \( \text{Fe}^{3+} \): The formation of \( \text{Fe}^{3+} \) involves losing three electrons (two from 4s and one from 3d), resulting in \( [\text{Ar}] \, 3d^5 \). This configuration has 5 unpaired electrons.
• \( \text{Co}^{2+} \): Cobalt (atomic number 27) has an electron configuration of \( [\text{Ar}] \, 3d^7 \, 4s^2 \). For \( \text{Co}^{2+} \), two electrons are lost from the 4s orbital, resulting in \( [\text{Ar}] \, 3d^7 \). This configuration contains 3 unpaired electrons.
• \( \text{Mn}^{2+} \): Manganese (atomic number 25) has an electron configuration of \( [\text{Ar}] \, 3d^5 \, 4s^2 \). The loss of two electrons from the 4s orbital forms \( \text{Mn}^{2+} \), yielding \( [\text{Ar}] \, 3d^5 \). This configuration has 5 unpaired electrons.

Based on this analysis, \( \text{Cr}^{2+} \), \( \text{Fe}^{2+} \), and \( \text{Mn}^{2+} \) exhibit 4 unpaired electrons, thus possessing a spin-only magnetic moment approximately equal to 4.9 B.M.

Conclusion: The correct answer is option A, B, and E only.