on reaction with concentrated H2SO4, it suffers loss in weight and on reaction with AgNO3 solution gives a yellow precipitated which is isoluble in NH3 solution If all the ligands in the coordination sphere of the `Co(en)_2(H_2O)Cl_2 I` complex be replaced by `F^(bar)`, then the magnetic moment of the complex ion (due to spin only) will be :

Options

A. 2.8 BM
B. 5.9 BM
C. 4.9 BM
D. 1.73 BM

Explanation

To solve the problem step by step, we will analyze the information given and derive the magnetic moment of the complex ion after replacing the ligands.

### Step 1: Identify the Coordination Complex
The question states that the complex reacts with concentrated H2SO4, resulting in a loss of weight. This indicates that water is likely not part of the coordination sphere. The complex also reacts with AgNO3 to form a yellow precipitate, which suggests the presence of iodide ions (I⁻) since AgI is a yellow precipitate.

**Complex Structure:**
The complex can be represented as:
\[ \text{Co(en)}_2\text{(H}_2\text{O)}\text{Cl}_2 \]

### Step 2: Determine the Charge of the Complex
In the complex, we have:
- 2 ethylenediamine (en) ligands, which are neutral.
- 2 chloride (Cl⁻) ligands, contributing -2 to the charge.
- 1 iodide (I⁻) ion, contributing -1 to the charge.

Thus, the overall charge of the complex can be calculated as follows:
\[ \text{Charge} = 0 + 0 - 2 - 1 = -3 \]

However, since the complex is mentioned to be neutral in the context of the question, we assume the overall charge of the complex is +1.

### Step 3: Replace Ligands with F⁻
If all the ligands in the coordination sphere are replaced by F⁻, the new complex will be:
\[ \text{CoF}_6^{+1} \]

### Step 4: Determine the Oxidation State of Co
Let the oxidation state of Co be \( x \). The charge equation can be set up as follows:
\[ x + 6(-1) = +1 \]
\[ x - 6 = 1 \]
\[ x = +7 \]

### Step 5: Determine the Electron Configuration of Co
Cobalt (Co) has an atomic number of 27. The electron configuration of Co is:
\[ \text{Co: } [\text{Ar}] 3d^7 4s^2 \]

When Co is in the +7 oxidation state, it loses 2 electrons from the 4s orbital and 5 from the 3d orbital:
\[ \text{Co}^{+7}: [\text{Ar}] 3d^2 \]

### Step 6: Count Unpaired Electrons
In the \( 3d^2 \) configuration, there are 2 unpaired electrons.

### Step 7: Calculate the Magnetic Moment
The spin-only magnetic moment (\( \mu \)) is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

Substituting \( n = 2 \):
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{2 \times 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 8: Convert to Bohr Magnetons
The value \( 2\sqrt{2} \) is approximately equal to \( 2.83 \) Bohr Magnetons (BM).

### Conclusion
The magnetic moment of the complex ion after replacing all ligands with F⁻ is approximately \( 2.83 \) BM.