Question

The calculated spin only magnetic moment of $Cr^{2+}$ ion is :

• A. 3.87 BM
• B. 4.90 BM
• C. 5.92 BM
• D. 2.84 BM

Answer 1

The Correct Option is B

To solve the problem of calculating the spin-only magnetic moment for the \(Cr^{2+}\) ion, we can follow these steps:

Firstly, identify the electronic configuration of the chromium (Cr) atom. Neutral Cr has an atomic number of 24, which gives it the electronic configuration of \([Ar] \, 3d^5 \, 4s^1\).

The ion \(Cr^{2+}\) results from removing two electrons from the neutral Cr atom. These electrons are removed from the outermost shell first, which in this case are the 4s and then the 3d orbitals.

Thus, the electronic configuration of \(Cr^{2+}\) is \([Ar] \, 3d^4\).

The magnetic moment due to spin is calculated using the formula:

\(\mu = \sqrt{n(n+2)} \, BM\)

where \(n\) is the number of unpaired electrons.

For a \(3d^4\) configuration, there are 4 electrons in the 3d subshell. Arranging these according to Hund’s rule in the five 3d orbitals:

The electron configuration is represented as \(\uparrow \, \uparrow \, \uparrow \, \uparrow\) (4 unpaired electrons).

Thus, \(n = 4\) unpaired electrons.

Substitute \(n = 4\) into the formula:

\(\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \, BM\)

This confirms the calculated spin-only magnetic moment for \(Cr^{2+}\) is 4.90 BM.

The correct answer is: 4.90 BM.