The first, second, third and fourth, ionization potential values of an element are `8.4, 25.15, 37.92` and `256.3eV` respectively. The element is

Options

A. `M^(+3) `
B. `M^(+2) `
C. `M^(+4 )`
D. `M^(+1)` and `M^(+2)`

Answer

To determine the element based on the given ionization potential values, we can follow these steps:

### Step 1: Understand Ionization Energy
Ionization energy (or ionization potential) is the energy required to remove an electron from an isolated gaseous atom. The first ionization energy (IE1) is the energy needed to remove the first electron, the second ionization energy (IE2) is for the second electron, and so on.

### Step 2: Analyze the Given Ionization Energies
The ionization energies provided are:
- IE1 = 8.4 eV
- IE2 = 25.15 eV
- IE3 = 37.92 eV
- IE4 = 256.3 eV

### Step 3: Identify Patterns in the Ionization Energies
Notice that the first three ionization energies (IE1, IE2, and IE3) have relatively small increases between them:
- Difference between IE1 and IE2: 25.15 - 8.4 = 16.75 eV
- Difference between IE2 and IE3: 37.92 - 25.15 = 12.77 eV

However, the jump from IE3 to IE4 is significant:
- Difference between IE3 and IE4: 256.3 - 37.92 = 218.38 eV

### Step 4: Interpret the Large Jump in Ionization Energy
The large increase in energy required to remove the fourth electron (from 37.92 eV to 256.3 eV) suggests that the element is likely to have a stable oxidation state after losing three electrons. This indicates that the element likely forms a stable M³⁺ ion.

### Step 5: Determine the Element
Elements that typically form stable M³⁺ ions include aluminum (Al), gallium (Ga), and indium (In). Among these, aluminum is the most common and well-known element that forms a stable +3 oxidation state.

### Conclusion
Based on the analysis of the ionization energies, the element is **Aluminum (Al)**.