First, second, third and fourth IE values in eV of M are 6.11, 11.87, 50.91, 67.27 respectively. The ion which would be formed is:

Options

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

Answer

To determine the ion that would be formed by element M based on its ionization energy (IE) values, we can follow these steps:

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

### Step 2: Analyze Given Ionization Energies
The given ionization energies for element M are:
- IE1 = 6.11 eV
- IE2 = 11.87 eV
- IE3 = 50.91 eV
- IE4 = 67.27 eV

### Step 3: Identify Stability of Ions
To determine which ion is most stable, we look for a significant jump in the ionization energy values. A large increase in ionization energy indicates that the ion formed before this jump is more stable.

### Step 4: Compare Ionization Energies
- The difference between IE1 (6.11 eV) and IE2 (11.87 eV) is relatively small:
- Difference = 11.87 - 6.11 = 5.76 eV
- The difference between IE2 (11.87 eV) and IE3 (50.91 eV) is very large:
- Difference = 50.91 - 11.87 = 39.04 eV
- The difference between IE3 (50.91 eV) and IE4 (67.27 eV) is also significant, but not as drastic:
- Difference = 67.27 - 50.91 = 16.36 eV

### Step 5: Determine the Stable Ion
The large jump in ionization energy from IE2 to IE3 indicates that removing the third electron from M2+ requires significantly more energy, suggesting that M2+ is a stable ion. Therefore, M is likely to form a +2 ion.

### Conclusion
Based on the analysis of the ionization energies, the ion that would be formed by element M is:
**M²⁺**