The first I.E. of Na, Mg, Al and Si are in the order

Options

A. `Na lt Mg lt Al lt Si`
B. `Na lt Al lt Mg lt Si`
C. `Na lt Al lt Si lt Mg`
D. `Na gt Mg gt Al gt Si`

Answer

To determine the first ionization energy (I.E.) of sodium (Na), magnesium (Mg), aluminum (Al), and silicon (Si), we need to analyze the electronic configurations and the trends in ionization energy across a period in the periodic table.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations:**
- Sodium (Na): 1s² 2s² 2p⁶ 3s¹
- Magnesium (Mg): 1s² 2s² 2p⁶ 3s²
- Aluminum (Al): 1s² 2s² 2p⁶ 3s² 3p¹
- Silicon (Si): 1s² 2s² 2p⁶ 3s² 3p²

2. **Understand the Trend in Ionization Energy:**
- Ionization energy generally increases across a period from left to right due to increasing nuclear charge, which holds the electrons more tightly.
- However, there are exceptions due to the stability of filled and half-filled subshells.

3. **Compare the Ionization Energies:**
- As we move from Na to Mg, we see an increase in ionization energy because Mg has a higher nuclear charge and a more stable electron configuration (fully filled 3s subshell).
- When we move from Mg to Al, we expect an increase in I.E. but Al has one more electron in the 3p orbital. Removing an electron from a p orbital (in Al) requires less energy than removing an electron from a fully filled s orbital (in Mg).
- Moving from Al to Si, the I.E. increases again because Si has a higher nuclear charge and an additional electron in the p orbital.

4. **Establish the Order of Ionization Energies:**
- Based on the above analysis, the order of first ionization energies is:
- Na < Al < Mg < Si

### Final Order:
The first ionization energy of Na, Mg, Al, and Si is in the order:
**Na < Al < Mg < Si**