The magnetic moment (in BM) of `Zn^(2+)` ion according to spin-only formula is

Options

A. zero
B. `1.73`
C. `2.84`
D. `3.87`

Explanation

To find the magnetic moment of the `Zn^(2+)` ion using the spin-only formula, we will follow these steps:

### Step 1: Determine the electron configuration of Zinc (Zn)
Zinc has an atomic number of 30. The electron configuration of neutral Zinc is:
\[ \text{Zn: } [\text{Ar}] 3d^{10} 4s^2 \]

### Step 2: Determine the electron configuration of `Zn^(2+)`
To find the configuration of `Zn^(2+)`, we need to remove 2 electrons from the neutral Zinc. The electrons are removed first from the 4s orbital, followed by the 3d orbital. Thus, the electron configuration for `Zn^(2+)` is:
\[ \text{Zn}^{2+}: [\text{Ar}] 3d^{10} \]

### Step 3: Identify the number of unpaired electrons
In the `Zn^(2+)` ion, the 3d subshell is fully filled with 10 electrons. The filling of the 3d subshell can be represented as:
- 5 pairs of electrons (since each d orbital can hold 2 electrons).

Since all electrons are paired, there are no unpaired electrons in `Zn^(2+)`. Therefore, the number of unpaired electrons \( n \) is:
\[ n = 0 \]

### Step 4: Use the spin-only formula to calculate the magnetic moment
The spin-only formula for magnetic moment (\( \mu \)) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
Substituting \( n = 0 \):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

### Conclusion
The magnetic moment of `Zn^(2+)` ion is:
\[ \mu = 0 \text{ BM} \]

### Final Answer
The magnetic moment of `Zn^(2+)` ion according to the spin-only formula is **0 BM**.

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