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

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 \( e^- \))