Step 1: Understanding the Concept:
The magnetic properties of transition metal complexes and ions are primarily due to the presence of unpaired electrons in their \(d\)-orbitals. Each electron has a magnetic dipole moment stemming from two sources: its orbital motion around the nucleus and its spin around its own axis. For many first-row transition metal ions, the "orbital contribution" is "quenched" (cancelled out) by the surrounding ligands or crystal field, meaning the magnetic moment can be accurately calculated using only the spin component. This is known as the "spin-only" magnetic moment. The magnitude of this moment is measured in units called Bohr Magnetons (BM). As the number of unpaired electrons increases, the total magnetic moment of the ion increases predictably.
Step 2: Key Formula or Approach:
The spin-only magnetic moment (\(\mu\)) is given by the formula:
\[ \mu = \sqrt{n(n+2)} \]
Where:
- \(\mu\) = Magnetic moment in Bohr Magnetons (BM)
- \(n\) = Number of unpaired electrons
Step 3: Detailed Explanation:
We are given that the magnetic moment \(\mu\) is 3.87 BM. We need to find the integer value of \(n\).
Method 1: Mathematical Calculation
\[ 3.87 = \sqrt{n(n+2)} \]
Squaring both sides to remove the square root:
\[ (3.87)^{2} = n(n+2) \]
\[ 14.9769 = n^{2} + 2n \]
Rearranging into a quadratic equation:
\[ n^{2} + 2n - 14.9769 = 0 \]
Since \(n\) must be a whole number, we can look for integers where \(n(n+2)\) is close to 15.
If \(n = 3\), then \(3(3+2) = 3 \times 5 = 15\).
The square root of 15 is approximately 3.872. This perfectly matches the given value.
Method 2: Electronic Configuration Check
Chromium (Cr) has an atomic number of 24. Its ground-state configuration is \([Ar] 3d^{5} 4s^{1}\).
To form the Cr\(^{3+}\) ion, we must remove 3 electrons. Electrons are removed first from the \(4s\) orbital and then from the \(3d\) orbital:
- Remove 1 from \(4s\) \(\rightarrow\) \([Ar] 3d^{5}\)
- Remove 2 from \(3d\) \(\rightarrow\) \([Ar] 3d^{3}\)
In the \(3d^{3}\) configuration, there are three electrons in the \(d\) subshell. According to Hund’s Rule of maximum multiplicity, these three electrons will occupy three separate \(d\) orbitals with parallel spins to minimize repulsion. Therefore, the number of unpaired electrons (\(n\)) is 3.
Both the mathematical formula and the chemistry of the atom confirm that 3 unpaired electrons result in a magnetic moment of 3.87 BM.
Step 4: Final Answer:
The number of unpaired electrons present in the Cr\(^{3+}\) ion is 3.
Thus, option (3) is correct.