`{:(A^(-) (g) rarr A^(2+)(g),,DeltaH = 1100KJ//mol),(A(g)rarrA^(2+)(g),,DeltaH = 1200 KJ//mol):}`
Electron gain enthalpy of `A` is `P xx 10^(2) KJ//mol`. What is the value of `P`?

Explanation

To solve the problem, we need to find the electron gain enthalpy of element A, given the two reactions and their enthalpy changes. Let's break down the steps:

### Step 1: Write down the given reactions and their enthalpy changes.
1. Reaction 1:
\[ A^{-} (g) \rightarrow A^{2+} (g), \Delta H_1 = 1100 \text{ kJ/mol} \]
2. Reaction 2:
\[ A (g) \rightarrow A^{2+} (g), \Delta H_2 = 1200 \text{ kJ/mol} \]

### Step 2: Understand the concept of electron gain enthalpy.
- Electron gain enthalpy is defined as the energy change when an electron is added to a gaseous atom, forming a negative ion. In this case, adding an electron to A (g) would form \( A^{-} (g) \).

### Step 3: Relate the reactions to find the electron gain enthalpy.
- We can express the electron gain enthalpy (\( \Delta H_{eg} \)) in terms of the given reactions. The relationship can be derived as follows:
\[ A (g) + e^{-} \rightarrow A^{-} (g) \]
This reaction's enthalpy change is what we need to find.

### Step 4: Use Hess's law to find the electron gain enthalpy.
- According to Hess's law, we can manipulate the given reactions to find the desired enthalpy change. We can subtract Reaction 1 from Reaction 2:
\[ \Delta H_{eg} = \Delta H_2 - \Delta H_1 \]
Substituting the values:
\[ \Delta H_{eg} = 1200 \text{ kJ/mol} - 1100 \text{ kJ/mol} = 100 \text{ kJ/mol} \]

### Step 5: Relate the electron gain enthalpy to the given format.
- The problem states that the electron gain enthalpy of A is given as \( P \times 10^{2} \text{ kJ/mol} \). Therefore, we can set up the equation:
\[ 100 \text{ kJ/mol} = P \times 10^{2} \text{ kJ/mol} \]

### Step 6: Solve for \( P \).
- Rearranging the equation gives:
\[ P = \frac{100}{100} = 1 \]

### Final Answer
Thus, the value of \( P \) is:
\[ P = 1 \]