For the following questions enter the correct numerical value (in decimal -notation , truncated `//` rounded - off to the second decimal place e.g., 6.25 , 7.00 -0.33,- 30,30.27 , - 127 .30 ) using the mouse and the on- screen virtual numeric keypad in the place designated to enter the answer.
The amount of energy released when `1 xx 10^(10)` atoms of chlorine in vapour state are converted to `C1^(-)` ions according to the equation,
`C1 (g) + e^(-)` to `C1^(-) ` ( g) " is `57.86 xx 10^(-1) J`
Calculate the electron gain enthalpy of chlorine atom in terms of eV per atom.

Explanation

To solve the problem of calculating the electron gain enthalpy of a chlorine atom in terms of eV per atom, we will follow these steps:

### Step 1: Understand the Given Information
We know that the energy released when \( 1 \times 10^{10} \) atoms of chlorine are converted to \( Cl^- \) ions is given as \( -57.86 \times 10^{-1} \) J.

### Step 2: Calculate the Total Energy Released for One Mole
Since the energy provided is for \( 1 \times 10^{10} \) atoms, we need to find out how much energy is released for one mole of chlorine atoms.

1 mole of any substance contains \( 6.02 \times 10^{23} \) atoms (Avogadro's number).

Using a unitary method, we can find the energy released for one mole:
\[ \text{Energy per mole} = \left( \frac{-57.86 \times 10^{-1} \text{ J}}{1 \times 10^{10} \text{ atoms}} \right) \times 6.02 \times 10^{23} \text{ atoms} \]

### Step 3: Calculate the Energy Released
Now, substituting the values:
\[ \text{Energy per mole} = \left( -5.786 \text{ J} \right) \times 6.02 \times 10^{23} \text{ atoms} \]
Calculating this gives:
\[ \text{Energy per mole} \approx -348.49 \text{ kJ/mol} \]

### Step 4: Convert Energy from kJ/mol to J/mol
Since we want the energy in Joules per mole:
\[ -348.49 \text{ kJ/mol} = -348490 \text{ J/mol} \]

### Step 5: Convert Joules to Electron Volts
We know that \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). To convert the energy from J/mol to eV/atom, we use:
\[ \text{Energy in eV} = \frac{-348490 \text{ J/mol}}{6.02 \times 10^{23} \text{ atoms/mol}} \times \frac{1 \text{ eV}}{1.602 \times 10^{-19} \text{ J}} \]

### Step 6: Calculate the Final Value
Calculating this gives:
\[ \text{Energy in eV} \approx \frac{-348490}{6.02 \times 10^{23}} \times \frac{1}{1.602 \times 10^{-19}} \approx -3.61 \text{ eV/atom} \]

### Final Answer
Thus, the electron gain enthalpy of a chlorine atom is approximately:
\[ \boxed{-3.61} \]