The calculated spin only magnetic moment of Cr\(^{2+}\) ion is
Show Hint
You can quickly estimate the magnetic moment. The value \(\sqrt{n(n+2)}\) is always slightly less than n+1. For n=4, the value is just under 5. For n=3, it's just under 4 (3.87). This can help you select the correct option without a calculator.
Step 1: Determine the electronic configuration of Cr\(^{2+}\).
The atomic number of Chromium (Cr) is 24.
Its neutral electronic configuration is exceptional: [Ar] 3d\(^5\) 4s\(^1\).
To form the Cr\(^{2+}\) ion, we remove two electrons. The first electron is removed from the outermost shell (4s), and the second is removed from the next shell (3d).
So, the electronic configuration of Cr\(^{2+}\) is [Ar] 3d\(^4\).
Step 2: Determine the number of unpaired electrons (n).
The 3d subshell has 5 orbitals. According to Hund's rule, the 4 electrons in the 3d subshell will occupy separate orbitals with parallel spins.
There are 4 unpaired electrons. So, n = 4. Step 3: Key Formula for Spin-only Magnetic Moment.
The spin-only magnetic moment (\(\mu\)) is calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \]
where 'n' is the number of unpaired electrons. The unit is Bohr Magneton (BM). Step 4: Calculate the magnetic moment.
Substitute n = 4 into the formula:
\[ \mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \]
Now, we estimate the value: \(\sqrt{16} = 4\) and \(\sqrt{25} = 5\). So \(\sqrt{24}\) is just below 5.
Calculating it more precisely: \(\sqrt{24} \approx 4.8989...\) BM. Step 5: Final Answer.
The calculated value is approximately 4.90 BM.
