Consider `Al_(2)(OH)_(6)` compound and caculate the value of `(X+Y)+Z` Wher X=total number of `(2c-2e^(-))` bond.
Y=total number of `(3c-2e^(-))` bond.
Z=total numbe of `(3c-4e^(-))` bond

Explanation

To solve the problem, we need to analyze the compound \( \text{Al}_2(\text{OH})_6 \) and determine the values of \( X \), \( Y \), and \( Z \) based on the types of bonds present in the structure.

### Step-by-Step Solution:

1. **Identify the Structure of \( \text{Al}_2(\text{OH})_6 \)**:
- The compound consists of two aluminum (Al) atoms and six hydroxide (OH) groups.
- The hydroxide groups will coordinate with the aluminum atoms.

2. **Determine the Types of Bonds**:
- **2c-2e Bonds (X)**: These are bonds between two atoms with two electrons shared.
- **3c-2e Bonds (Y)**: These are bonds involving three centers and two electrons.
- **3c-4e Bonds (Z)**: These are bonds involving three centers and four electrons.

3. **Calculate \( Z \)**:
- In the structure, each hydroxide group (OH) can form a bond with aluminum.
- The bonds formed between Al and OH can be classified as 3c-4e bonds because the bond involves Al, O, and the other Al.
- There are 2 such 3c-4e bonds formed by the two Al atoms with the OH groups.
- Thus, \( Z = 2 \).

4. **Calculate \( Y \)**:
- Check for any 3c-2e bonds in the structure.
- In this case, there are no 3c-2e bonds present in the compound.
- Therefore, \( Y = 0 \).

5. **Calculate \( X \)**:
- Count the number of 2c-2e bonds.
- Each Al-O bond and O-H bond can be considered as a 2c-2e bond.
- There are 10 such bonds in total: 6 Al-O bonds and 4 O-H bonds.
- Thus, \( X = 10 \).

6. **Final Calculation**:
- Now, we can calculate \( (X + Y) + Z \):
\[ (X + Y) + Z = (10 + 0) + 2 = 12 \]

### Conclusion:
The value of \( (X + Y) + Z \) is \( 12 \).