`Delta_(f)H^(ɵ)` of hypothetical MX is `-150 kJ mol^(-1)` and for `MX_(2)` is `-600 kJ mol^(-1)`. The enthalpy of disproportionation of MX is `=-100 x kJ mol^(-1)`.
Find the value of x.

Explanation

To find the value of \( x \) in the given problem, we will follow these steps:

### Step 1: Understand the Disproportionation Reaction
The disproportionation reaction of \( MX \) can be written as:
\[ 2 \, MX \rightarrow M + MX_2 \]
This means that two moles of \( MX \) decompose into one mole of \( M \) and one mole of \( MX_2 \).

### Step 2: Write the Enthalpy Change for the Disproportionation Reaction
The enthalpy change for the disproportionation reaction is given as:
\[ \Delta H_{disproportionation} = -100x \, \text{kJ/mol} \]

### Step 3: Use the Enthalpy of Formation Values
We know the enthalpy of formation for \( MX \) and \( MX_2 \):
- For \( MX \):
\[ \Delta_f H^\circ (MX) = -150 \, \text{kJ/mol} \]
- For \( MX_2 \):
\[ \Delta_f H^\circ (MX_2) = -600 \, \text{kJ/mol} \]

### Step 4: Calculate the Enthalpy Change for the Reaction
The enthalpy change for the reaction can also be calculated using the enthalpies of formation:
\[ \Delta H_{reaction} = \Delta H_f (products) - \Delta H_f (reactants) \]
Substituting the values:
\[ \Delta H_{reaction} = \Delta H_f (M) + \Delta H_f (MX_2) - 2 \Delta H_f (MX) \]
Since the enthalpy of formation for \( M \) is not given, we will denote it as \( \Delta H_f (M) \).

### Step 5: Set Up the Equation
Substituting the known values:
\[ \Delta H_{reaction} = \Delta H_f (M) - 600 - 2(-150) \]
This simplifies to:
\[ \Delta H_{reaction} = \Delta H_f (M) - 600 + 300 \]
\[ \Delta H_{reaction} = \Delta H_f (M) - 300 \]

### Step 6: Equate the Two Expressions for Enthalpy Change
Now, we can set the two expressions for the enthalpy change equal to each other:
\[ -100x = \Delta H_f (M) - 300 \]

### Step 7: Solve for \( x \)
To find \( x \), we need to express \( \Delta H_f (M) \). However, we can rearrange the equation:
\[ \Delta H_f (M) = -100x + 300 \]
Now, we need to find a relationship to eliminate \( \Delta H_f (M) \). We know that the enthalpy of formation for \( M \) should be consistent with the other values.

### Step 8: Assume \( \Delta H_f (M) = 0 \) for simplicity
If we assume \( \Delta H_f (M) = 0 \) (which is a common assumption for elements in their standard state), we can substitute this into the equation:
\[ 0 = -100x + 300 \]
Solving for \( x \):
\[ 100x = 300 \]
\[ x = 3 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{3} \]