The species with spin only magnetic moment of sqrt(24)` BM is:

Options

A. `[CoF_(3)(H_(2)O)_(3)]`
B. `[CoCl_(4)]^(2-)`
C. `[NiCl_(4)]^(2-)`
D. `[Ni(H_(2)O)_(6)]^(2+)`

Answer

To solve the problem of finding the species with a spin-only magnetic moment of \(\sqrt{24}\) BM, we will follow these steps:

### Step 1: Understand the Magnetic Moment Formula
The magnetic moment (\(\mu\)) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \text{ BM} \]
where \(n\) is the number of unpaired electrons.

### Step 2: Set Up the Equation
Given that the magnetic moment is \(\sqrt{24}\) BM, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{24} \]
Squaring both sides gives:
\[ n(n + 2) = 24 \]

### Step 3: Solve the Quadratic Equation
Rearranging the equation:
\[ n^2 + 2n - 24 = 0 \]
Now, we can use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 2\), and \(c = -24\):
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 96}}{2} \]
\[ n = \frac{-2 \pm \sqrt{100}}{2} \]
\[ n = \frac{-2 \pm 10}{2} \]
Calculating the two possible values for \(n\):
1. \(n = \frac{8}{2} = 4\)
2. \(n = \frac{-12}{2} = -6\) (not a valid solution since \(n\) cannot be negative)

Thus, \(n = 4\).

### Step 4: Identify the Coordination Complex
Now that we know there are 4 unpaired electrons, we need to find a coordination complex that corresponds to this configuration.

1. **Cobalt in +3 Oxidation State**:
- Cobalt in the +3 oxidation state has the electron configuration of \( \text{[Ar]} 4s^0 3d^6 \).
- With weak field ligands (like \( \text{F}^- \)), the electrons will remain unpaired, leading to 4 unpaired electrons.

2. **Cobalt in +2 Oxidation State**:
- Cobalt in the +2 oxidation state has the electron configuration of \( \text{[Ar]} 4s^0 3d^7 \).
- This would typically have 3 unpaired electrons with weak field ligands.

### Conclusion
The species with a spin-only magnetic moment of \(\sqrt{24}\) BM corresponds to cobalt in the +3 oxidation state with 4 unpaired electrons, likely in a complex with weak field ligands.

### Final Answer
The species with a spin-only magnetic moment of \(\sqrt{24}\) BM is a cobalt complex, specifically \( \text{CoF}_3\text{H}_2\text{O}_3 \) in the +3 oxidation state.

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