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: The spin-only magnetic moment is \( \mu = \sqrt{n(n+1)} \) BM, where \( n \) is the number of unpaired electrons. The larger the \( n \), the larger the magnetic moment.
Step 2: Write the \( d \)-electron configuration of each ion (for a 2+ ion, remove the two 4s electrons first, then adjust d):
\( Cr^{2+} \): \( 3d^{4} \) → 4 unpaired electrons.
\( Co^{2+} \): \( 3d^{7} \) → 3 unpaired electrons.
\( Fe^{2+} \): \( 3d^{6} \) → 4 unpaired electrons.
\( V^{2+} \): \( 3d^{3} \) → 3 unpaired electrons.
Step 3: The maximum number of unpaired electrons (4) is shown by \( Cr^{2+} (3d^4) \), giving \( \mu = \sqrt{4(4+1)} = \sqrt{20} = 4.9 \) BM. This is the highest among the options, so \( Cr^{2+} \) is the marked answer.
Step 4: \( Co^{2+} \) and \( V^{2+} \) have only 3 unpaired electrons \( (\mu = 3.87 \) BM\( ) \), which is smaller. Hence option (i) is correct.
\[\boxed{Cr^{2+},\ \mu = 4.9\ \text{BM}}\]