The spin only magnetic moment of `Ni^(2+)` in aquous solution would be

Options

A. √8
​
B. √2
C. √6
D. √15

Answer

To find the spin-only magnetic moment of \( \text{Ni}^{2+} \) in aqueous solution, we can follow these steps:

### Step 1: Determine the electronic configuration of \( \text{Ni}^{2+} \)

Nickel (Ni) has an atomic number of 28. The electronic configuration of neutral nickel is:
\[ \text{Ni}: [\text{Ar}] 3d^8 4s^2 \]

When nickel loses two electrons to form \( \text{Ni}^{2+} \), the electrons are removed first from the 4s orbital. Thus, the electronic configuration of \( \text{Ni}^{2+} \) is:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 \]

### Step 2: Identify the number of unpaired electrons in \( \text{Ni}^{2+} \)

In the \( 3d \) subshell, the distribution of electrons can be represented as follows:

- The \( 3d \) subshell can hold a maximum of 10 electrons.
- For \( 3d^8 \), the electrons will be arranged as follows:
- 5 electrons will fill the five \( d \) orbitals (each getting one electron first).
- The next 3 electrons will pair up with 3 of the already filled orbitals.

This results in the following arrangement:
- 2 orbitals have paired electrons (2 electrons each).
- 3 orbitals have 1 unpaired electron each.

Thus, the number of unpaired electrons \( N \) in \( \text{Ni}^{2+} \) is 2.

### Step 3: Use the formula for spin-only magnetic moment

The formula for calculating the spin-only magnetic moment \( \mu \) is given by:
\[ \mu = \sqrt{N(N + 2)} \]
where \( N \) is the number of unpaired electrons.

Substituting \( N = 2 \):
\[ \mu = \sqrt{2(2 + 2)} \]
\[ \mu = \sqrt{2 \times 4} \]
\[ \mu = \sqrt{8} \]

### Step 4: Conclusion

The spin-only magnetic moment of \( \text{Ni}^{2+} \) in aqueous solution is:
\[ \mu = \sqrt{8} \, \text{BM} \]

### Final Answer
The answer is \( \sqrt{8} \, \text{BM} \).

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