The sum of coordination number and oxidation number of the metal M in the complex `[M(en)_(2) (C_(2)O_(4))]CI` (where en is ethylenediamine) is:

Options

A. 7
B. 8
C. 9
D. 6

Explanation

To solve the problem, let's break it down step by step:

### Step 1: Determine the Oxidation Number of Metal M

1. **Identify the ligands and their charges:**
- Ethylenediamine (en) is a neutral ligand, so it has no charge.
- Oxalate (\(C_2O_4^{2-}\)) is a bidentate ligand with a charge of -2.
- Chloride ion (\(Cl^-\)) is outside the coordination sphere and has a charge of -1.

2. **Write the formula of the complex ion:**
\[[M(en)_2(C_2O_4)]Cl\]

3. **Set up the equation for the oxidation state of M:**
Let the oxidation state of metal \(M\) be \(x\).

The sum of the charges in the complex ion should equal the charge of the complex. Since \(en\) is neutral and \(C_2O_4\) has a charge of -2:
\[ x + 2(0) + (-2) = +1 \]
(The +1 comes from the fact that the complex ion \([M(en)_2(C_2O_4)]^+\) pairs with \(Cl^-\) to form a neutral compound.)

4. **Solve for \(x\):**
\[ x - 2 = +1 \]
\[ x = +3 \]

### Step 2: Determine the Coordination Number of Metal M

1. **Identify the type of ligands and their denticity:**
- Ethylenediamine (en) is a bidentate ligand.
- Oxalate (\(C_2O_4^{2-}\)) is also a bidentate ligand.

2. **Calculate the coordination number:**
- Each ethylenediamine ligand coordinates through 2 donor atoms.
- Each oxalate ligand coordinates through 2 donor atoms.

Therefore, the coordination number is:
\[ 2(en) \times 2 + 1(C_2O_4) \times 2 = 4 + 2 = 6 \]

### Step 3: Sum of Coordination Number and Oxidation Number

1. **Add the coordination number and oxidation number:**
\[ \text{Coordination Number} + \text{Oxidation Number} = 6 + 3 = 9 \]

### Final Answer
The sum of the coordination number and oxidation number of the metal \(M\) in the complex \([M(en)_2(C_2O_4)]Cl\) is:
\[ \boxed{9} \]