Step 1: Understanding the Concept:
Magnetic moments in coordination compounds result from the angular momentum of electrons.
For transition metals of the first series (3d series), the orbital angular momentum is "quenched" by the surrounding ligand field.
This means we can calculate the magnetic moment using only the spin of the unpaired electrons.
This is known as the "spin-only" magnetic moment.
Key Formula or Approach:
The formula for the spin-only magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n + 2)} \text{ BM} \]
where:
\( n \) is the number of unpaired electrons.
BM stands for Bohr Magneton, the unit of magnetic moment.
Step 2: Detailed Explanation:
The question specifies that the complex has \( n = 3 \) unpaired electrons.
Substitute the value of \( n \) into the spin-only formula:
\[ \mu = \sqrt{3(3 + 2)} \]
\[ \mu = \sqrt{3 \times 5} \]
\[ \mu = \sqrt{15} \]
Now, let's find the approximate value of \(\sqrt{15}\).
Since \( \sqrt{9} = 3 \) and \( \sqrt{16} = 4 \), the value of \( \sqrt{15} \) must be between 3 and 4, very close to 4.
By calculation, \( \sqrt{15} \approx 3.8729 \).
Looking at the options provided:
(A) 1.73 BM corresponds to \( n = 1 \) (\(\sqrt{3}\)).
(B) 2.84 BM corresponds to \( n = 2 \) (\(\sqrt{8}\)).
(C) 3.87 BM corresponds to \( n = 3 \) (\(\sqrt{15}\)).
(D) 4.90 BM corresponds to \( n = 4 \) (\(\sqrt{24}\)).
Therefore, the value 3.87 BM is the correct match for 3 unpaired electrons.
In chemical terms, this would be seen in complexes like high-spin \(Cr^{3+}\) (\(d^3\)) or high-spin \(Co^{2+}\) (\(d^7\)).
Step 3: Final Answer:
The magnetic moment for \( n = 3 \) is 3.87 BM.