If a solid `A^(o+)B^(ɵ)` having `ZnS` Structure is heated so that the ions along two of the axis passing throgh the face centre particles are lost and bivalent ion `(Z)` enters herre to maintain the electrical neutrality, so that the new formula unit becomes `A_(x)B_(y)Z_(c)`, report the value of `x + y + c`.

Explanation

To solve the problem, we need to analyze the changes in the solid structure when it is heated and how the introduction of the bivalent ion \( Z^{2+} \) affects the overall composition. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Structure
The solid \( A^{+}B^{-} \) has a ZnS structure. In this structure:
- Each unit cell contains 4 \( A^{+} \) ions and 4 \( B^{-} \) ions.
- The \( B^{-} \) ions are located at the face centers and corners of the unit cell.

### Step 2: Identify the Loss of Ions
When the solid is heated, ions along two of the axes passing through the face center particles are lost. This means:
- 2 \( B^{-} \) ions are removed from the face centers (1 from each of the 2 face centers).

### Step 3: Calculate Remaining Ions
After losing 2 \( B^{-} \) ions:
- The number of \( B^{-} \) ions left = Initial \( B^{-} \) ions - Lost \( B^{-} \) ions
- Remaining \( B^{-} \) ions = \( 4 - 2 = 2 \)

### Step 4: Introduce the Bivalent Ion
To maintain electrical neutrality, a bivalent ion \( Z^{2+} \) enters the structure. Since 2 \( B^{-} \) ions were lost, we need 1 \( Z^{2+} \) ion to balance the charge:
- The relationship is \( 2B^{-} = 1Z^{2+} \).

### Step 5: Write the New Formula
The new formula unit is \( A_{x}B_{y}Z_{c} \). From our calculations:
- \( x = 4 \) (number of \( A^{+} \) ions remains the same)
- \( y = 2 \) (remaining \( B^{-} \) ions)
- \( c = 1 \) (number of \( Z^{2+} \) ions)

### Step 6: Calculate \( x + y + c \)
Now, we sum the values:
\[ x + y + c = 4 + 2 + 1 = 7 \]

### Final Answer
The value of \( x + y + c \) is \( 7 \).

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