The values of `IE_1 , IE_2 , IE_3, IE_4` and `IE_5` of an element are 5.98, 18.82, 28.44, 120 and 153.8 eV respectively. The element is likely to be

Options

A. Si
B. AI
C. C
D. P

Answer

To determine which element corresponds to the given ionization energy values (IE1 = 5.98 eV, IE2 = 18.82 eV, IE3 = 28.44 eV, IE4 = 120 eV, IE5 = 153.8 eV), we will analyze the pattern of these ionization energies.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
Ionization energy (IE) is the energy required to remove an electron from an atom in its gaseous state. The first ionization energy (IE1) is the energy required to remove the first electron, the second ionization energy (IE2) is for the second electron, and so on.

2. **Identifying the Trend in Ionization Energies**:
- The first four ionization energies (IE1, IE2, IE3) show a gradual increase in energy:
- IE1 = 5.98 eV
- IE2 = 18.82 eV
- IE3 = 28.44 eV
- However, there is a significant jump between IE3 and IE4:
- IE4 = 120 eV (a large increase compared to the previous values).

3. **Interpreting the Large Jump**:
- The large jump in ionization energy from IE3 to IE4 indicates that the fourth electron is being removed from a much more stable inner electron shell, rather than from the outer shell. This suggests that the first three electrons are valence electrons, while the fourth electron is from a filled inner shell.

4. **Determining the Element**:
- We need to identify an element that has three valence electrons and a stable inner shell configuration.
- The elements in the periodic table with three valence electrons are typically found in Group 13 (e.g., Aluminum).

5. **Electron Configuration**:
- Let's check the electron configurations of the possible elements:
- **Silicon (Si)**: Atomic number 14 → 1s² 2s² 2p⁶ 3s² 3p² (4 valence electrons)
- **Aluminum (Al)**: Atomic number 13 → 1s² 2s² 2p⁶ 3s² 3p¹ (3 valence electrons)
- **Carbon (C)**: Atomic number 6 → 1s² 2s² 2p² (4 valence electrons)
- **Phosphorus (P)**: Atomic number 15 → 1s² 2s² 2p⁶ 3s² 3p³ (5 valence electrons)

6. **Conclusion**:
- The element that matches the criteria of having three valence electrons and a significant jump in ionization energy after the removal of the third electron is **Aluminum (Al)**. The fourth ionization energy corresponds to removing an electron from a stable inner shell.

### Final Answer:
The element is likely to be **Aluminum (Al)**.