How many chlorine atoms can you ionize in the process `Cl rarr Cl^+ + e`, by the energy liberated from the following process ?
`Cl + e^- rarr Cl^(-) for 6 xx 10^23 "atoms"`
Given electron affinity ofm `Cl = 3.61 eV`, and `I P` of `Cl = 17.422 eV`.

Options

A. Positive
B. Negative
C. Zero
D. Unpredictable

Answer

To solve the problem, we need to determine how many chlorine atoms can be ionized using the energy released from the process of chlorine gaining an electron.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Electron affinity of Cl (EA) = 3.61 eV
- Ionization potential of Cl (IP) = 17.422 eV
- Number of chlorine atoms = \(6 \times 10^{23}\)

2. **Calculate the Total Energy Released:**
The energy released when \(6 \times 10^{23}\) chlorine atoms gain electrons can be calculated using the formula:
\[ \text{Total Energy Released} = \text{Number of Atoms} \times \text{Electron Affinity} \]
Substituting the values:
\[ \text{Total Energy Released} = 6 \times 10^{23} \times 3.61 \, \text{eV} \]
\[ = 21.66 \times 10^{23} \, \text{eV} \]

3. **Set Up the Equation for Ionization:**
Let \(Z\) be the number of chlorine atoms that can be ionized. The energy required to ionize \(Z\) chlorine atoms is given by:
\[ \text{Energy Required for Ionization} = Z \times \text{Ionization Potential} \]
Substituting the values:
\[ \text{Energy Required for Ionization} = Z \times 17.422 \, \text{eV} \]

4. **Equate the Total Energy Released to the Energy Required for Ionization:**
Since the energy released from the first process is used to ionize the chlorine atoms, we can set up the equation:
\[ 21.66 \times 10^{23} \, \text{eV} = Z \times 17.422 \, \text{eV} \]

5. **Solve for \(Z\):**
Rearranging the equation gives:
\[ Z = \frac{21.66 \times 10^{23}}{17.422} \]
Performing the division:
\[ Z \approx 1.243 \times 10^{23} \]

### Conclusion:
The number of chlorine atoms that can be ionized is approximately \(1.243 \times 10^{23}\).