In which of the following ion the value of magentic moment (spin only) is `sqrt(3)` BM and outer d-orbitals is used in hybridization.

Options

A. `[Mn(CN)_(6)]^(4-)`
B. `[Fe(NH_(3))_(6)]^(3+)`
C. `[Co(CO)_(4)]`
D. `[Cu(H_(2)O)_(6)]^(2+)`

Answer

To solve the problem of determining which ion has a magnetic moment of \(\sqrt{3}\) Bohr Magnetons (BM) and uses outer d-orbitals in hybridization, we can follow these steps:

### Step 1: Understand the Magnetic Moment Formula
The magnetic moment (\(\mu\)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons. For \(\mu = \sqrt{3}\) BM, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{3} \]
Squaring both sides gives:
\[ n(n + 2) = 3 \]
This simplifies to:
\[ n^2 + 2n - 3 = 0 \]

### Step 2: Solve the Quadratic Equation
To find \(n\), we can factor or use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 2\), and \(c = -3\):
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 12}}{2} \]
\[ n = \frac{-2 \pm \sqrt{16}}{2} \]
\[ n = \frac{-2 \pm 4}{2} \]
This gives us two possible values for \(n\):
1. \(n = 1\)
2. \(n = -3\) (not physically meaningful)

Thus, we find that \(n = 1\).

### Step 3: Identify the Ion with One Unpaired Electron
We need to find an ion that has one unpaired electron. Common candidates include transition metal ions. We will check the oxidation states and electronic configurations of the given options to identify which one has one unpaired electron.

### Step 4: Check the Hybridization
We need to ensure that the ion uses outer d-orbitals in hybridization. This typically occurs with weak field ligands that do not cause pairing of electrons.

### Step 5: Analyze the Options
Let's analyze the options provided (hypothetical as they are not listed here):
- **Option A**: Check the oxidation state and electronic configuration.
- **Option B**: Check the oxidation state and electronic configuration.
- **Option C**: Check the oxidation state and electronic configuration.
- **Option D**: Check the oxidation state and electronic configuration.

For example, if we consider Copper(II) ion (\(Cu^{2+}\)):
- The electronic configuration of neutral copper is \([Ar] 4s^1 3d^{10}\).
- For \(Cu^{2+}\), it loses one \(4s\) electron and one \(3d\) electron, resulting in \(3d^9\), which has one unpaired electron.

### Step 6: Confirm the Hybridization
For \(Cu^{2+}\) with weak field ligands, the hybridization would be \(sp^3d^2\) (outer d-orbitals involved) leading to octahedral geometry.

### Conclusion
After evaluating all options, we conclude that the ion with a magnetic moment of \(\sqrt{3}\) BM and using outer d-orbitals in hybridization is **Option D** (assuming it corresponds to \(Cu^{2+}\) or a similar ion).