The spin-only magnetic moment [in units of Bohr magneton, `(mu_(B)` of `Ni^(2+)` in aqueous solution would be (atomic number of `Ni=28)`

Options

A. `2.84`
B. `4.90`
C. `0`
D. `1.73`

Explanation

To find the spin-only magnetic moment of Ni²⁺ in aqueous solution, we can follow these steps:

### Step 1: Determine the electronic configuration of Ni
The atomic number of Nickel (Ni) is 28. The electronic configuration of Ni is:
\[ \text{Ni: } [\text{Ar}] 3d^8 4s^2 \]

### Step 2: Determine the electronic configuration of Ni²⁺
When Nickel loses two electrons to form Ni²⁺, the electrons are removed from the 4s orbital first. Therefore, the electronic configuration of Ni²⁺ is:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 4s^0 \]

### Step 3: Identify the number of unpaired electrons
In the 3d subshell of Ni²⁺, we have 8 electrons. The electron filling in the 3d subshell follows Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up. The filling of the 3d orbitals can be represented as follows:
- 3d: ↑↓ ↑↓ ↑↓ ↑ ↑

From this configuration, we can see that there are 2 unpaired electrons in the 3d subshell.

### Step 4: Calculate the spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (μ) in units of Bohr magneton (μ_B) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

In this case, \( n = 2 \):
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{2 \times 4} = \sqrt{8} \]

### Step 5: Simplify the expression
Calculating the square root:
\[ \mu = \sqrt{8} = 2.83 \, \mu_B \]

### Step 6: Round to appropriate significant figures
Rounding to two decimal places, we get:
\[ \mu \approx 2.84 \, \mu_B \]

### Final Answer
The spin-only magnetic moment of Ni²⁺ in aqueous solution is approximately **2.84 Bohr magnetons**.

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