The values of `IE_(1),IE_(2),IE_(3),IE_(4)` and `IE_(5)` of an element are 7.1, 14.3, 34.5, 46.8 and 162.2 eV respectively. The element is likely to be

Options

A. Na
B. Si
C. F
D. Ca

Explanation

To determine the element based on the provided ionization enthalpy values, we can follow these steps:

### Step 1: List the Ionization Energies
The given ionization energies (IE) are:
- \( IE_1 = 7.1 \, \text{eV} \)
- \( IE_2 = 14.3 \, \text{eV} \)
- \( IE_3 = 34.5 \, \text{eV} \)
- \( IE_4 = 46.8 \, \text{eV} \)
- \( IE_5 = 162.2 \, \text{eV} \)

### Step 2: Calculate the Differences Between Consecutive Ionization Energies
Next, we calculate the differences between consecutive ionization energies:
- \( IE_2 - IE_1 = 14.3 - 7.1 = 7.2 \, \text{eV} \)
- \( IE_3 - IE_2 = 34.5 - 14.3 = 20.2 \, \text{eV} \)
- \( IE_4 - IE_3 = 46.8 - 34.5 = 12.3 \, \text{eV} \)
- \( IE_5 - IE_4 = 162.2 - 46.8 = 115.4 \, \text{eV} \)

### Step 3: Analyze the Differences
From the calculated differences, we observe:
- The first three differences are relatively small.
- However, the difference between \( IE_4 \) and \( IE_5 \) is significantly larger (115.4 eV).

### Step 4: Identify the Reason for the Large Jump
The large jump in ionization energy from \( IE_4 \) to \( IE_5 \) suggests that removing the fifth electron requires a lot more energy. This indicates that after removing four electrons, the element achieves a noble gas configuration, making it much more stable.

### Step 5: Determine the Valence Electrons
Since the element has a large increase in ionization energy after removing four electrons, it likely has four electrons in its outer shell.

### Step 6: Identify the Element
Elements with four valence electrons include:
- Carbon (C)
- Silicon (Si)
- Germanium (Ge)

Given that silicon is in the same period as the other elements and has four valence electrons, we can conclude that the element is likely to be **Silicon (Si)**.

### Final Answer
The element is likely to be **Silicon (Si)**.

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