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 from the element M based on the given ionization energy (IE) values, we can follow these steps:

### Step 1: Analyze the Ionization Energy Values
The first four ionization energy values provided are:
- 1st IE = 6.11 eV
- 2nd IE = 11.87 eV
- 3rd IE = 50.91 eV
- 4th IE = 67.27 eV

### Step 2: Identify the Trend in Ionization Energies
We observe that the first and second ionization energies are relatively close (6.11 eV and 11.87 eV). However, there is a significant jump between the second and third ionization energies (from 11.87 eV to 50.91 eV). This large increase indicates that removing the third electron requires significantly more energy.

### Step 3: Interpret the Large Jump in Ionization Energy
The large difference between the second and third ionization energies suggests that after removing two electrons, the atom achieves a more stable electronic configuration. This stability implies that the element is likely to form a +2 oxidation state (M²⁺) when it loses two electrons.

### Step 4: Conclusion on the Ion Formed
Since the third ionization energy is much higher, it indicates that removing a third electron from the M²⁺ ion is much more difficult. Therefore, the most stable ion that would be formed from element M is M²⁺.

### Final Answer
The ion which would be formed is M²⁺.

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