The ionization energies for B, TI and In are x, y and z kcal `mol^(-1)` respectively. Choose the correct relationship between them. 

Options

A. `z gt x = y `
B. `x gt y lt z`
C. `x gt y ~~ z`
D. `x  lt y gt z`

Answer

To determine the correct relationship between the ionization energies of Boron (B), Tellurium (Te), and Indium (In), we need to analyze the periodic trends in ionization energy.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
- Ionization energy is the energy required to remove an electron from an isolated gaseous atom.
- It is influenced by the effective nuclear charge (Z_eff) and the size of the atom.

2. **Periodic Trends**:
- As we move across a period from left to right, ionization energy generally increases due to increasing nuclear charge and decreasing atomic radius.
- As we move down a group, ionization energy generally decreases because the outer electrons are further from the nucleus and experience greater shielding.

3. **Analyzing the Elements**:
- Boron (B) is in Group 13 and has a relatively small atomic size and high ionization energy compared to elements in lower periods.
- Indium (In) is also in Group 13 but is located lower than Boron, meaning it has more electron shells and a larger atomic radius, resulting in lower ionization energy.
- Tellurium (Te) is in Group 16 and is located below oxygen and sulfur. It has a larger atomic size compared to Boron but is higher than Indium in terms of ionization energy due to its position in the periodic table.

4. **Comparing Ionization Energies**:
- Based on the periodic trends:
- Boron (B) will have the highest ionization energy (x).
- Tellurium (Te) will have a lower ionization energy than Boron but higher than Indium (y).
- Indium (In) will have the lowest ionization energy (z).

5. **Establishing the Relationship**:
- Therefore, we can establish the relationship as follows:
- x (Boron) > y (Tellurium) > z (Indium).
- This means that the correct relationship is x > y > z.

### Final Answer:
The correct relationship between the ionization energies of Boron (x), Tellurium (y), and Indium (z) is:
**x > y > z**