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 energies (IE), we will analyze the provided values of ionization enthalpy and identify the element based on the significant jump between the ionization energies.

### Step-by-Step Solution:

1. **List the Given Ionization Energies:**
- IE1 = 5.98 eV
- IE2 = 18.82 eV
- IE3 = 28.44 eV
- IE4 = 120 eV
- IE5 = 153.8 eV

2. **Analyze the Ionization Energies:**
- The first three ionization energies (IE1, IE2, IE3) show relatively small increases:
- From IE1 to IE2: 18.82 - 5.98 = 12.84 eV
- From IE2 to IE3: 28.44 - 18.82 = 9.62 eV
- However, there is a significant jump from IE3 to IE4:
- From IE3 to IE4: 120 - 28.44 = 91.56 eV
- This large increase indicates that after removing three electrons, the atom achieves a stable electronic configuration, making it much harder to remove the fourth electron.

3. **Identify the Element:**
- The large jump between IE3 and IE4 suggests that the element likely has three valence electrons. After losing these three electrons, it reaches a stable noble gas configuration.
- Elements that typically have three valence electrons include those in Group 13 of the periodic table, such as Aluminum (Al) and Gallium (Ga).
- However, the specific values of ionization energies can help narrow it down further.

4. **Check Possible Elements:**
- **Silicon (Si)**: Configuration is 1s² 2s² 2p⁶ 3s² 3p². The jump would occur after losing 4 electrons, not 3.
- **Aluminum (Al)**: Configuration is 1s² 2s² 2p⁶ 3s² 3p¹. The jump would occur after losing 3 electrons, but the values do not match.
- **Carbon (C)**: Configuration is 1s² 2s² 2p². The jump would occur after losing 4 electrons, not 3.
- **Phosphorus (P)**: Configuration is 1s² 2s² 2p⁶ 3s² 3p³. The jump would occur after losing 5 electrons, not 3.

5. **Conclusion:**
- The element that fits this pattern, with a significant jump in ionization energy after the third electron is removed, is **Phosphorus (P)**. After removing three electrons, it achieves a stable configuration (2p⁶), making it difficult to remove the fourth electron.

### Final Answer:
The element is likely to be **Phosphorus (P)**.