In the complex `K_(4)[Th(C_(2)O_(4))_(2)(H_(2)O)_(2)]`. If coordination number is X and oxidation number of Th is Y. The sum of `X+Y` is ?

Answer

To solve the problem, we need to determine the coordination number (X) and the oxidation number (Y) of thorium (Th) in the complex \( K_4[Th(C_2O_4)_2(H_2O)_2] \).

### Step 1: Determine the Coordination Number (X)

1. **Identify the ligands in the complex**:
- The complex contains two types of ligands: \( C_2O_4^{2-} \) (oxalate) and \( H_2O \) (water).

2. **Determine the nature of the ligands**:
- \( C_2O_4^{2-} \) is a bidentate ligand, meaning it can bond to the central metal atom at two points. Since there are two \( C_2O_4^{2-} \) ligands, they contribute \( 2 \times 2 = 4 \) to the coordination number.
- \( H_2O \) is a monodentate ligand, meaning it can bond at only one point. Since there are two \( H_2O \) ligands, they contribute \( 1 \times 2 = 2 \) to the coordination number.

3. **Calculate the total coordination number**:
\[ X = 4 \, (\text{from } C_2O_4) + 2 \, (\text{from } H_2O) = 6 \]

### Step 2: Determine the Oxidation Number (Y)

1. **Identify the charges of the components**:
- Potassium (K) has a charge of \( +1 \). Since there are 4 potassium ions, they contribute \( +4 \) to the overall charge.
- The oxalate ligands \( C_2O_4^{2-} \) have a charge of \( -2 \). With two oxalate ligands, they contribute \( -4 \) to the overall charge.
- Water \( H_2O \) is a neutral ligand, contributing \( 0 \).

2. **Set up the equation for the overall charge**:
- The overall charge of the complex is \( 0 \) (since it is a neutral complex).
- Let \( Y \) be the oxidation number of thorium (Th). The equation can be set up as follows:
\[ +4 + Y - 4 + 0 = 0 \]

3. **Solve for Y**:
\[ Y = 0 \]

### Step 3: Calculate the Sum of X and Y

Now that we have both values:
- \( X = 6 \)
- \( Y = 0 \)

The sum \( X + Y \) is:
\[ X + Y = 6 + 0 = 6 \]

### Final Answer
The sum of \( X + Y \) is **6**.