Step 1: Understanding the Question:
We are asked to find the spin-only magnetic moment ($\mu$) of a Nickel ion with an atomic number of 28 holding a $+2$ charge ($\text{Ni}^{2+}$).
Step 2: Key Formula or Approach:
The spin-only magnetic moment is calculated using the formula:
$$\mu = \sqrt{n(n + 2)}\ \text{BM}$$
Where $n$ is the total number of unpaired d-orbital electrons and BM stands for Bohr Magnetons.
Step 3: Detailed Explanation:
First, write down the ground-state electron configuration of a neutral Nickel atom ($Z = 28$):
$$\text{Ni} = [\text{Ar}]\, 3\text{d}^8\, 4\text{s}^2$$
To form the $\text{Ni}^{2+}$ cation, the atom loses its 2 outermost valence electrons from the $4\text{s}$ subshell:
$$\text{Ni}^{2+} = [\text{Ar}]\, 3\text{d}^8$$
Now, let's distribute these 8 electrons across the 5 degenerate d-orbitals following Hund's Rule:
The first 5 electrons occupy the five orbitals individually with parallel spins ($\uparrow$, $\uparrow$, $\uparrow$, $\uparrow$, $\uparrow$).
The remaining 3 electrons pair up in the first three orbitals ($\uparrow\downarrow$, $\uparrow\downarrow$, $\uparrow\downarrow$, $\uparrow$, $\uparrow$).
Counting the remaining single occupants shows exactly $n = 2$ unpaired electrons.
Substitute $n = 2$ into our magnetic moment formula:
$$\mu = \sqrt{2(2 + 2)} = \sqrt{2 \times 4} = \sqrt{8}\ \text{BM}$$
Since we know $\sqrt{9} = 3.0$, the value of $\sqrt{8}$ must be slightly less than 3, which calculates out to approximately $2.83\ \text{BM}$. This lines up with option (C).
Step 4: Final Answer:
The spin-only magnetic moment of $\text{Ni}^{2+}$ is 2.8 BM, corresponding to option (C).