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. Magnesium
B. Silicon
C. Sodium
D. Aluminium

Answer

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

### Step 1: Understand Ionization Potential
Ionization potential (IP) is the energy required to remove an electron from an atom. The first ionization potential is the energy needed to remove the first electron, the second IP is for the second electron, and so on. A significant jump in ionization potential values indicates that the electron being removed is from a more stable electron configuration.

### Step 2: Analyze Given Ionization Potentials
The given ionization potentials are:
- First IP: 8.4 eV
- Second IP: 25.15 eV
- Third IP: 37.92 eV
- Fourth IP: 256.3 eV

Notice that there is a large jump from the third to the fourth ionization potential (from 37.92 eV to 256.3 eV). This suggests that after removing three electrons, the atom reaches a stable electronic configuration, and removing the fourth electron requires significantly more energy.

### Step 3: Consider Possible Elements
We need to find an element whose electronic configuration supports this pattern of ionization potentials. The large jump indicates that after removing three electrons, the element achieves a noble gas configuration.

### Step 4: Evaluate Options
1. **Magnesium (Mg)** - Atomic number 12: Configuration is 1s² 2s² 2p⁶ 3s². Removing two electrons gives a stable configuration (Neon). Hence, this option is incorrect.

2. **Silicon (Si)** - Atomic number 14: Configuration is 1s² 2s² 2p⁶ 3s² 3p². Removing four electrons would not yield a stable noble gas configuration. Hence, this option is incorrect.

3. **Sodium (Na)** - Atomic number 11: Configuration is 1s² 2s² 2p⁶ 3s¹. Removing one electron gives a stable configuration (Neon). Hence, this option is incorrect.

4. **Aluminum (Al)** - Atomic number 13: Configuration is 1s² 2s² 2p⁶ 3s² 3p¹. Removing three electrons gives a stable configuration (Neon), and removing the fourth electron requires a significant amount of energy, matching the large jump in IP values. Hence, this option is correct.

### Conclusion
The element with the given ionization potentials is **Aluminum (Al)**.

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