Question

Find the value of spin only magnetic moment for chromium cation in +2 state.

• A. 3.87 BM
• B. 4.90 BM
• C. 2.84 BM
• D. 1.73 BM

Hint:

Logic Tip: You can estimate the magnetic moment almost instantly without calculating the square root! The value of $\mu$ is always $n.x$ where $n$ is the number of unpaired electrons. If $n=4$, the magnetic moment is $4.\text{something}$. Only Option B starts with a 4.

Answer 1

The Correct Option is B

Concept:
The spin-only magnetic moment ($\mu$) of a transition metal ion is calculated using the formula: $$\mu = \sqrt{n(n+2)} \text{ BM}$$ where $n$ is the number of unpaired electrons in the ion, and BM stands for Bohr Magneton. To find $n$, we must determine the electronic configuration of the specific ion.
Step 1: Write the electronic configuration of the neutral Chromium atom.
Chromium (Cr) has an atomic number of 24. Its expected electronic configuration would be $[Ar] \, 3d^4 \, 4s^2$, but due to the extra stability of half-filled subshells, an electron shifts from the $4s$ to the $3d$ orbital. Actual configuration of Cr: $[Ar] \, 3d^5 \, 4s^1$
Step 2: Determine the electronic configuration of the $Cr^{2+}$ cation.
To form the $Cr^{2+}$ cation, two electrons must be removed. Electrons are always removed from the outermost shell first (the $4s$ shell), followed by the $3d$ shell. Remove one electron from $4s$ and one from $3d$: Configuration of $Cr^{2+}$: $[Ar] \, 3d^4 \, 4s^0$
Step 3: Count the number of unpaired electrons (n).
The $3d^4$ subshell has 4 electrons. According to Hund's rule, they will occupy separate orbitals with parallel spins before pairing up. Thus, there are exactly 4 unpaired electrons: $$n = 4$$
Step 4: Calculate the spin-only magnetic moment.
Substitute $n = 4$ into the formula: $$\mu = \sqrt{4(4 + 2)}$$ $$\mu = \sqrt{4 \times 6}$$ $$\mu = \sqrt{24} \text{ BM}$$ Since $\sqrt{25} = 5$, the value of $\sqrt{24}$ must be just slightly less than 5. $$\mu \approx 4.90 \text{ BM}$$