Question

Which of the following ions has highest magnetic moment?

• A. \( Cr^{2+} \)
• B. \( Co^{2+} \)
• C. \( Fe^{2+} \)
• D. \( V^{2+} \)

Hint:

Count unpaired \( 3d \) electrons of each 2+ ion and use \( \mu=\sqrt{n(n+1)} \); the d4 ion wins.

Answer 1

The Correct Option is A

Step 1: Magnetic moment grows with the count of unpaired electrons, so the task is simply to find which ion carries the most unpaired \( 3d \) electrons.
Step 2: Fill the \( 3d \) subshell by Hund's rule for each dipositive ion. \( V^{2+} \) is \( d^3 \) (three unpaired). \( Cr^{2+} \) is \( d^4 \) (four unpaired: \( t_{2g}^{3}e_g^{1} \) high spin). \( Fe^{2+} \) is \( d^6 \) (four unpaired). \( Co^{2+} \) is \( d^7 \) (three unpaired).
Step 3: Convert to numbers: \( d^3 \) and \( d^7 \) give 3 unpaired electrons; \( d^4 \) gives the peak of 4 for the left half of the series. \( Cr^{2+} \) therefore reaches the largest unpaired count of 4.
Step 4: Using \( \mu=\sqrt{n(n+1)} \), four unpaired electrons give \( 4.9 \) BM versus \( 3.87 \) BM for three. So \( Cr^{2+} \) has the highest magnetic moment and option (i) is chosen.