Metals have few electrons in their valence shell while non-metals generally have more electrons in their valence shell. Metallic character is closely related to atomic radius and ionisation enthalpy. Metallic character increases from top to bottom in a group and decreases from let to right in a period of periodic table. metallic character is inversely related to electronegativity of element.
Q. `3 N_0//2` atoms of `X_((g))` are converted it into `X_((g))^(+)` by energy `E_(1)`, 2 `N_0//3` atoms of `X_((g))` are converted it into `X_((g))^(-)` by energy `E_(2)`. hence, ionisation potential and electron affinity of `X_((g))` are: (`N_(0)`=Avogadro's number)

Options

A. `(2E_(1))/(3N_(0)),(2E_(2))/(3N_(0))`
B. `(2E_(1))/(3N_(0)),(3E_(2))/(2N_(0))`
C. `(3E_(1))/(2N_(0)),(3E_(2))/(2N_(0))`
D. `(3E_(1))/(2N_(0)),(2E_(2))/(3N_(0))`

Answer

To solve the problem, we need to determine the ionization potential and electron affinity of the element \(X\) based on the given information about the energy required for ionization and electron gain.

### Step 1: Understand the Ionization Process
We are given that \( \frac{3N_0}{2} \) atoms of \(X_{(g)}\) are converted into \(X_{(g)}^+\) by energy \(E_1\). The ionization potential (IP) is defined as the energy required to remove an electron from one mole of gaseous atoms to form cations.

**Calculation of Ionization Potential:**
- Energy \(E_1\) corresponds to \( \frac{3N_0}{2} \) atoms.
- To find the energy required for one atom to become \(X^+\), we divide \(E_1\) by the number of atoms:

\[ \text{IP} = \frac{E_1}{\frac{3N_0}{2}} = \frac{2E_1}{3N_0} \]

### Step 2: Understand the Electron Affinity Process
Next, we are given that \( \frac{2N_0}{3} \) atoms of \(X_{(g)}\) are converted into \(X_{(g)}^-\) by energy \(E_2\). The electron affinity (EA) is defined as the energy released when one mole of gaseous atoms gains an electron to form anions.

**Calculation of Electron Affinity:**
- Energy \(E_2\) corresponds to \( \frac{2N_0}{3} \) atoms.
- To find the energy released for one atom to become \(X^-\), we divide \(E_2\) by the number of atoms:

\[ \text{EA} = \frac{E_2}{\frac{2N_0}{3}} = \frac{3E_2}{2N_0} \]

### Step 3: Final Results
We have derived the formulas for both ionization potential and electron affinity:
- Ionization Potential (IP):

\[ \text{IP} = \frac{2E_1}{3N_0} \]

- Electron Affinity (EA):

\[ \text{EA} = \frac{3E_2}{2N_0} \]

### Conclusion
Thus, the ionization potential and electron affinity of \(X_{(g)}\) are:

- Ionization Potential: \( \frac{2E_1}{3N_0} \)
- Electron Affinity: \( \frac{3E_2}{2N_0} \)