To solve the problem, we will analyze the given complex compound \( \text{Na}_2[\text{PtBrClI(CN)}] \) step by step to find the values of \( x \), \( y \), and \( z \).
### Step 1: Determine the Number of Geometrical Isomers (x)
1. **Identify the central atom and ligands**: The central atom is platinum (Pt), and it is coordinated by four different ligands: bromine (Br), chlorine (Cl), iodine (I), and cyanide (CN).
2. **Check for geometrical isomerism**: Since there are four different ligands attached to the platinum, we can have geometrical isomers. The possible arrangements can be determined by fixing one ligand and rotating the others.
3. **Count the geometrical isomers**:
- Fix one ligand (e.g., Br) and consider the arrangements of the other three (Cl, I, CN).
- The possible arrangements yield three distinct geometrical isomers:
1. Br, Cl, I, CN
2. Br, I, Cl, CN
3. Br, CN, Cl, I
Thus, the total number of geometrical isomers \( x = 3 \).
### Step 2: Determine the Number of Optical Isomers (y)
1. **Check for optical activity**: A compound exhibits optical isomerism if it lacks a plane of symmetry.
2. **Analyze the structure**: Since the compound is planar and has a plane of symmetry (due to the arrangement of the four different ligands around Pt), it does not exhibit optical isomerism.
Thus, the number of optical isomers \( y = 0 \).
### Step 3: Determine the Number of Ions Produced in Aqueous Solution (z)
1. **Dissociation in water**: When \( \text{Na}_2[\text{PtBrClI(CN)}] \) is dissolved in water, it dissociates into its constituent ions.
2. **Write the dissociation equation**:
\[
\text{Na}_2[\text{PtBrClI(CN)}] \rightarrow 2 \text{Na}^+ + [\text{PtBrClI(CN)}]^{2-}
\]
3. **Count the ions**:
- From the dissociation, we have 2 sodium ions \( (\text{Na}^+) \).
- The complex ion \( [\text{PtBrClI(CN)}]^{2-} \) counts as one additional ion.
Thus, the total number of ions produced \( z = 2 + 1 = 3 \).
### Final Calculation
Now, we can calculate \( x + y + z \):
\[
x + y + z = 3 + 0 + 3 = 6
\]
### Conclusion
The final answer is:
\[
\boxed{6}
\]
