`DeltaH` in terms of 'x' `xx 10^(3)` KJ is if
`M_((g)) + 2X_((g)) rarr M_((g))^(2+) + 2X_((g))^(-)`
Given : `(I.E)_(1)` of M (g) = 705.7 kJ `mol^(-1), (I.E)_(2) " of" M(g)` = 951 kJ `mol^(-1) and (E.A)_(1)` of X(g) = `-328 kJ mol^(-1)` 'x' is

Answer

To solve the problem, we need to calculate the enthalpy change (ΔH) for the reaction given the ionization energies (IE) and electron affinity (EA) of the species involved.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ M(g) + 2X(g) \rightarrow M^{2+} + 2X^{-} \]

2. **Calculate ΔH for the Formation of M²⁺**:
To form \( M^{2+} \) from \( M \), we need to consider the ionization energies:
\[ \text{ΔH for } M \rightarrow M^{2+} + 2e^{-} = \text{IE}_1 + \text{IE}_2 \]
Given:
- \( \text{IE}_1 = 705.7 \, \text{kJ/mol} \)
- \( \text{IE}_2 = 951 \, \text{kJ/mol} \)

Therefore:
\[ \text{ΔH for } M \rightarrow M^{2+} = 705.7 + 951 = 1656.7 \, \text{kJ/mol} \]

3. **Calculate ΔH for the Formation of 2X⁻**:
To form \( 2X^{-} \) from \( 2X \), we use the electron affinity:
\[ \text{ΔH for } 2X + 2e^{-} \rightarrow 2X^{-} = 2 \times \text{EA}_1 \]
Given:
- \( \text{EA}_1 = -328 \, \text{kJ/mol} \)

Therefore:
\[ \text{ΔH for } 2X \rightarrow 2X^{-} = 2 \times (-328) = -656 \, \text{kJ/mol} \]

4. **Combine the Enthalpy Changes**:
The overall ΔH for the reaction is:
\[ \text{ΔH} = \text{ΔH for } M \rightarrow M^{2+} + \text{ΔH for } 2X \rightarrow 2X^{-} \]
Substituting the values:
\[ \text{ΔH} = 1656.7 - 656 = 1000.7 \, \text{kJ/mol} \]

5. **Relate ΔH to the Given Expression**:
The problem states that ΔH can be expressed as:
\[ \Delta H = x \times 10^3 \, \text{kJ} \]
We have calculated ΔH to be 1000.7 kJ. Therefore:
\[ 1000.7 = x \times 10^3 \]

6. **Solve for x**:
Dividing both sides by \( 10^3 \):
\[ x = \frac{1000.7}{1000} = 1.0007 \]
Rounding to one decimal place, we find:
\[ x \approx 1 \]

### Final Answer:
Thus, the value of \( x \) is approximately **1**.