In which of the following complex ion the value of magnetic moment (spin only) is `sqrt(3)`B.M. and oute rd-orbitals is used in hybridization:

Options

A. `[Fe(NH_(3))_(6)]^(3+)`
B. `[Mn(CN)_(6)]^(4-)`
C. `[CuCl_(5)]^(2-)`
D. `[Co(NH_(3))_(6)]^(2+)`

Explanation

To solve the question regarding the complex ion with a magnetic moment of \(\sqrt{3}\) B.M. and the use of outer d-orbitals in hybridization, we will follow these steps:

### Step 1: Identify the Complex Ions
We need to analyze the given complex ions to determine their oxidation states, electronic configurations, and hybridization.

### Step 2: Calculate the Oxidation State
For each complex ion, we will calculate the oxidation state of the central metal ion.

1. **For CuCl\(_5^{2-}\)**:
- Let the oxidation state of Cu be \(x\).
- The equation becomes: \(x + 5(-1) = -2\).
- Solving gives \(x = +3\).

2. **For Fe(NH\(_3\))\(_6^{3+}\)**:
- Let the oxidation state of Fe be \(x\).
- The equation becomes: \(x + 6(0) = +3\).
- Solving gives \(x = +3\).

3. **For MnCl\(_6^{2-}\)**:
- Let the oxidation state of Mn be \(x\).
- The equation becomes: \(x + 6(-1) = -2\).
- Solving gives \(x = +4\).

4. **For Co(NH\(_3\))\(_6^{2+}\)**:
- Let the oxidation state of Co be \(x\).
- The equation becomes: \(x + 6(0) = +2\).
- Solving gives \(x = +2\).

### Step 3: Determine Electronic Configurations
Next, we will find the electronic configurations of the metal ions in their respective oxidation states.

1. **Cu\(^{3+}\)**:
- Ground state: \([Ar] 4s^1 3d^{10}\)
- For \(Cu^{3+}\): \([Ar] 3d^8\)

2. **Fe\(^{3+}\)**:
- Ground state: \([Ar] 4s^2 3d^6\)
- For \(Fe^{3+}\): \([Ar] 3d^5\)

3. **Mn\(^{4+}\)**:
- Ground state: \([Ar] 4s^2 3d^5\)
- For \(Mn^{4+}\): \([Ar] 3d^3\)

4. **Co\(^{2+}\)**:
- Ground state: \([Ar] 4s^2 3d^7\)
- For \(Co^{2+}\): \([Ar] 3d^7\)

### Step 4: Analyze Hybridization and Magnetic Properties
Now we will analyze the hybridization and magnetic properties of each complex.

1. **CuCl\(_5^{2-}\)**:
- Weak field ligand (Cl\(^-\)), leading to no pairing.
- Hybridization: \(sp^3d\) (outer orbital complex).
- Unpaired electrons: 2 (from \(3d^8\)).
- Magnetic moment: \(\sqrt{2(2) + 2} = \sqrt{6}\) B.M.

2. **Fe(NH\(_3\))\(_6^{3+}\)**:
- Strong field ligand (NH\(_3\)), leading to pairing.
- Hybridization: \(d^2sp^3\) (inner orbital complex).
- Unpaired electrons: 0.
- Magnetic moment: 0 B.M.

3. **MnCl\(_6^{2-}\)**:
- Weak field ligand (Cl\(^-\)), leading to no pairing.
- Hybridization: \(sp^3d^2\) (outer orbital complex).
- Unpaired electrons: 4 (from \(3d^3\)).
- Magnetic moment: \(\sqrt{4(4) + 2} = \sqrt{18}\) B.M.

4. **Co(NH\(_3\))\(_6^{2+}\)**:
- Strong field ligand (NH\(_3\)), leading to pairing.
- Hybridization: \(d^2sp^3\) (inner orbital complex).
- Unpaired electrons: 1 (from \(3d^7\)).
- Magnetic moment: \(\sqrt{3(1) + 2} = \sqrt{5}\) B.M.

### Step 5: Conclusion
Among the given options, the complex ion **CuCl\(_5^{2-}\)** is the only one that meets the criteria of having a magnetic moment of \(\sqrt{3}\) B.M. and using outer d-orbitals in hybridization.