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 determine which metal ions have a calculated spin-only magnetic moment value of 4.9 Bohr Magneton (B.M.), we will use the formula for spin-only magnetic moment:

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

where \( n \) is the number of unpaired electrons. A magnetic moment of 4.9 B.M. corresponds to \( n = 4 \) unpaired electrons, as calculated below:

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

Now, let's analyze each metal ion:

• \( \text{Cr}^{2+} \): Chromium has an atomic number of 24. The electron configuration of \( \text{Cr} \) is \( [\text{Ar}] \, 3d^5 \, 4s^1 \). For \( \text{Cr}^{2+} \), it loses 2 electrons: one from the 4s orbital and one from the 3d orbital, resulting in \( [\text{Ar}] \, 3d^4 \). This configuration has 4 unpaired electrons.
• \( \text{Fe}^{2+} \): Iron has an atomic number of 26. The electron configuration of \( \text{Fe} \) is \( [\text{Ar}] \, 3d^6 \, 4s^2 \). For \( \text{Fe}^{2+} \), it loses 2 electrons, both from the 4s orbital, resulting in \( [\text{Ar}] \, 3d^6 \). This configuration has 4 unpaired electrons.
• \( \text{Fe}^{3+} \): For \( \text{Fe}^{3+} \), it loses 3 electrons (2 from the 4s and 1 from the 3d), resulting in \( [\text{Ar}] \, 3d^5 \). This configuration has 5 unpaired electrons.
• \( \text{Co}^{2+} \): Cobalt has an atomic number of 27. The electron configuration of \( \text{Co} \) is \( [\text{Ar}] \, 3d^7 \, 4s^2 \). For \( \text{Co}^{2+} \), it loses 2 electrons, both from the 4s orbital, resulting in \( [\text{Ar}] \, 3d^7 \). This configuration has 3 unpaired electrons.
• \( \text{Mn}^{2+} \): Manganese has an atomic number of 25. The electron configuration of \( \text{Mn} \) is \( [\text{Ar}] \, 3d^5 \, 4s^2 \). For \( \text{Mn}^{2+} \), it loses 2 electrons, both from the 4s orbital, resulting in \( [\text{Ar}] \, 3d^5 \). This configuration has 5 unpaired electrons.

From this analysis, the ions \( \text{Cr}^{2+} \), \( \text{Fe}^{2+} \), and \( \text{Mn}^{2+} \) have 4 unpaired electrons and hence, a spin-only magnetic moment close to 4.9 B.M.

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

Answer 2

Given magnetic moment = 4.9 B.M.
We know, M.M = \( \sqrt{n(n+2)} \) B.M. 
Where, \( n = \) Number of unpaired electrons (\( e^- \)) 
\( 4.9 = \sqrt{n(n+2)} \) We get \( n = 4 \)  
(A) \( \mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4 \) (4 unpaired \( e^- \))  
(B) \( \mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6 \) (4 unpaired \( e^- \))  
(C) \( \mathrm{Fe}^{3+} = [\mathrm{Ar}]\,3d^5 \) (5 unpaired \( e^- \)) 
(D) \( \mathrm{Co}^{2+} = [\mathrm{Ar}]\,3d^7 \) (3 unpaired \( e^- \))  
(E) \( \mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5 \) (5 unpaired \( e^- \))