Question

The metal ion (in gaseous state) with lowest spin-only magnetic moment value is

• A. V²⁺
• B. Ni²⁺
• C. Cr²⁺
• D. Fe²⁺

Answer 1

The Correct Option is A

To determine which metal ion has the lowest spin-only magnetic moment, we need to calculate the magnetic moment for each option. The formula for the spin-only magnetic moment (\mu_{so}) is given by:

\mu_{so} = \sqrt{n(n+2)} \, \mu_B

where n is the number of unpaired electrons and \mu_B is the Bohr magneton.

1. V²⁺: The electron configuration is [Ar] 3d³. Hence, there are 3 unpaired electrons. We calculate as follows:
   \mu_{so} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \mu_B
2. Ni²⁺: The electron configuration is [Ar] 3d⁸. Hence, there are 2 unpaired electrons. Calculate:
   \mu_{so} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, \mu_B
3. Cr²⁺: The electron configuration is [Ar] 3d⁴. Hence, there are 4 unpaired electrons. Calculate:
   \mu_{so} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \mu_B
4. Fe²⁺: The electron configuration is [Ar] 3d⁶. Hence, there are 4 unpaired electrons. Calculate:
   \mu_{so} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \mu_B

From the calculations, Ni²⁺ has a spin-only magnetic moment of approximately 2.83 \mu_B, which is the lowest among the given options. Therefore, the correct answer is Ni²⁺.