An isomer of the complex `Co(en)_(2)(H_(2)O)IC l_(2)`, on reaction with concentrated `H_(2)SO_(4)`, it suffers loss in weight and on reaction with `AgNO_(3)` solution gives a yellow precipitated which is isoluble in `NH_(3)` solution.
Q. If all the ligandsin the co-ordination sphere of the above complex are replaced by `CN^(-)` ion, then the magnetic moment of the complex ion will be:

Options

A. 0.0 BM
B. 5.9 BM
C. 4.9 BM
D. 1.73 BM

Answer

To solve the problem step by step, we will analyze the given complex and its transformation when ligands are replaced.

### Step 1: Identify the original complex
The original complex given is `Co(en)₂(H₂O)ICl₂`. Here, `Co` is cobalt, `en` is ethylenediamine (a bidentate ligand), `H₂O` is water, `I` is iodine, and `Cl` is chloride.

### Step 2: Determine the coordination sphere
In the coordination sphere, we have:
- 2 ethylenediamine ligands (each occupies 2 coordination sites, total 4)
- 1 water molecule (occupies 1 coordination site)
- 2 chloride ions (each occupies 1 coordination site, total 2)

Thus, the coordination number of cobalt in this complex is 6, and the geometry is octahedral.

### Step 3: Analyze the reaction with concentrated H₂SO₄
When the complex reacts with concentrated H₂SO₄, it suffers a loss in weight. This indicates that some ligands are lost. Since water is a weakly bound ligand, it is likely that the water molecule is lost during this reaction.

### Step 4: Analyze the reaction with AgNO₃
The complex reacts with AgNO₃ to give a yellow precipitate, which is insoluble in NH₃. This indicates the presence of iodide ions (I⁻), as AgI is a yellow precipitate that is insoluble in ammonia. This suggests that the iodide is also outside the coordination sphere.

### Step 5: Formulate the new complex after ligand substitution
If all the ligands in the coordination sphere are replaced by `CN⁻` ions, the new complex will be `Co(CN)₆I`. Here, `CN⁻` is a strong field ligand and will replace all the other ligands.

### Step 6: Determine the oxidation state of cobalt
In the new complex `Co(CN)₆I`, the total charge from the six `CN⁻` ligands is -6 and from the iodine is -1. To balance the charges, cobalt must have an oxidation state of +7.

### Step 7: Determine the electron configuration of cobalt in this state
Cobalt (Co) has an atomic number of 27. Its electron configuration is:
- Neutral Co: [Ar] 4s² 3d⁷
- For Co in +7 oxidation state: It loses 7 electrons, resulting in the configuration [Ar] 3d².

### Step 8: Analyze the d-orbital filling
In the +7 oxidation state, cobalt has 2 electrons in the 3d subshell. Since `CN⁻` is a strong field ligand, it causes pairing of electrons. Thus, both electrons will occupy the same d-orbital.

### Step 9: Calculate the magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where `n` is the number of unpaired electrons. In this case, since both electrons are paired, `n = 0`.

Thus, the magnetic moment will be:
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

### Final Answer
The magnetic moment of the complex ion `Co(CN)₆I` will be **0 Bohr magnetons**.

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