Question

In K[Cr(H\(_2\)O)\(_6\)](C\(_2\)O\(_4\))\(_2\)] .3H\(_2\)O, the spin only magnetic moment of Cr\(^3\) ion is

• A. 2.87 B.M.
• B. 3.87 B.M.
• C. 3.47 B.M.
• D. 4.89 B.M.

Hint:

The spin-only magnetic moment is calculated using the formula \( \mu = \sqrt{n(n+2)} \) where \( n \) is the number of unpaired electrons.

Answer 1

The Correct Option is B

We are given the compound \( \text{K[Cr(H}_2\text{O})_6](\text{C}_2\text{O}_4)_2] \cdot 3\text{H}_2\text{O} \) where Cr is in the +3 oxidation state. The magnetic moment is determined by the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons in the Cr\(^3\) ion. For Cr\(^3+\) (which is \( d^3 \)), there are 3 unpaired electrons, so: \[ n = 3 \] Substitute the value of \( n \) into the formula: \[ \mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} = 3.87 \, \text{B.M.} \] Thus, the spin-only magnetic moment of Cr\(^3+\) ion is 3.87 B.M.