The `IE_1 , IE_2 , IE_3` and `IE_4` of an element are 5.7, 11.03, 42.89, 57 eV respectively. The element is likely to be

Options

A. P
B. Sr
C. F
D. Cs

Answer

To determine which element corresponds to the given ionization energies (IE1 = 5.7 eV, IE2 = 11.03 eV, IE3 = 42.89 eV, IE4 = 57 eV), we will analyze the differences between these ionization energies and their implications regarding the electronic configuration of the element.

### Step-by-Step Solution:

1. **List the Ionization Energies**:
- IE1 = 5.7 eV
- IE2 = 11.03 eV
- IE3 = 42.89 eV
- IE4 = 57 eV

2. **Calculate the Differences Between Successive Ionization Energies**:
- Difference between IE1 and IE2:
\[ \Delta IE_{1-2} = IE2 - IE1 = 11.03 - 5.7 = 5.33 \text{ eV} \]
- Difference between IE2 and IE3:
\[ \Delta IE_{2-3} = IE3 - IE2 = 42.89 - 11.03 = 31.86 \text{ eV} \]
- Difference between IE3 and IE4:
\[ \Delta IE_{3-4} = IE4 - IE3 = 57 - 42.89 = 14.11 \text{ eV} \]

3. **Analyze the Differences**:
- The first difference (5.33 eV) is relatively small.
- The second difference (31.86 eV) is significantly larger, indicating that removing the third electron requires much more energy than the previous two. This suggests that after removing the first two electrons, we are likely disrupting a stable electron configuration.
- The third difference (14.11 eV) is smaller than the second but still indicates a higher energy requirement compared to the first difference.

4. **Identify the Element**:
- The large jump in ionization energy from IE2 to IE3 suggests that the element has two electrons in its outer shell that can be removed relatively easily, but the third electron is from a more stable, inner shell configuration.
- This pattern is characteristic of elements in Group 15 (the nitrogen group), where elements like phosphorus (atomic number 15) have the configuration of [Ne] 3s² 3p³. Removing two electrons from the 3s and 3p orbitals is easier, but removing a third electron would require breaking into a stable filled shell.

5. **Conclusion**:
- Based on the analysis, the element with the given ionization energies is likely **Phosphorus (P)**.