The first, second and third ionisation potentials (E1, E2, and E3) for an element are 7eV, 12.5eV and 142.3ev respectively. The most stable oxidation state of the element will be

Options

A. `+1`
B. `+2`
C. `+3`
D. `+4`

Answer

To determine the most stable oxidation state of the element based on its ionization potentials (E1, E2, and E3), we can follow these steps:

### Step 1: Understand Ionization Energies
The first ionization potential (E1) is the energy required to remove the first electron from a neutral atom. The second ionization potential (E2) is the energy required to remove the second electron from a singly charged ion, and the third ionization potential (E3) is the energy required to remove the third electron from a doubly charged ion.

### Step 2: Analyze the Given Ionization Potentials
- E1 = 7 eV (removing the first electron)
- E2 = 12.5 eV (removing the second electron)
- E3 = 142.3 eV (removing the third electron)

### Step 3: Compare Ionization Energies
The significant jump in ionization energy from E2 to E3 indicates that removing the third electron is much more difficult than removing the first two. This suggests that after losing two electrons, the remaining electrons are more stable and likely part of a complete shell.

### Step 4: Determine the Most Stable Oxidation State
Since the first two ionization energies are relatively low compared to the third, we can conclude that the most stable oxidation state of the element is +2. This is because the element can easily lose two electrons to achieve a stable electronic configuration, while losing a third electron requires a significantly higher amount of energy, indicating that the atom is more stable in the +2 state.

### Conclusion
The most stable oxidation state of the element is +2.

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