Which one of the following is correct order of increase of size?

Options

A. `Mg lt Na^(+) lt F^(-) lt Al`
B. `Na^(+) lt Al lt Mg lt F^(-)`
C. `Na^(+) lt F^(-) lt Al lt Mg`
D. `Na^(+) lt F^(-) lt Mg lt Al`

Explanation

To determine the correct order of increase of size among the given species (Na⁺, Mg, Al, and F⁻), we will analyze the electronic configurations and the trends in ionic and atomic sizes.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations:**
- Sodium (Na) has an atomic number of 11:
- Electronic configuration: 1s² 2s² 2p⁶ 3s¹
- Sodium ion (Na⁺) loses one electron:
- Electronic configuration: 1s² 2s² 2p⁶ (10 electrons)
- Magnesium (Mg) has an atomic number of 12:
- Electronic configuration: 1s² 2s² 2p⁶ 3s²
- Magnesium does not lose or gain electrons, so it has 12 electrons.
- Aluminum (Al) has an atomic number of 13:
- Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p¹
- Aluminum has 13 electrons.
- Fluoride ion (F⁻) has an atomic number of 9:
- Electronic configuration: 1s² 2s² 2p⁶ (10 electrons, after gaining one electron).

2. **Determine Isoelectronic Species:**
- Both Na⁺ and F⁻ are isoelectronic with 10 electrons.
- Isoelectronic species have the same number of electrons but different nuclear charges.
- In isoelectronic species, the size of the anion (F⁻) is larger than that of the cation (Na⁺) because the effective nuclear charge is lower in the anion, leading to a larger size.

3. **Compare Neutral Atoms:**
- Now we compare the sizes of the neutral atoms: Mg and Al.
- Magnesium (Mg) has 12 electrons, while Aluminum (Al) has 13 electrons.
- As we move from left to right across a period, the size decreases due to increasing effective nuclear charge. Therefore, Mg is larger than Al.

4. **Final Comparison:**
- From the above analysis, we have:
- F⁻ > Na⁺ (both are isoelectronic)
- Mg > Al (size decreases from Mg to Al)
- Thus, the order of size from largest to smallest is:
- F⁻ > Na⁺ > Mg > Al

5. **Conclusion:**
- The correct order of increase of size is:
- Al < Mg < Na⁺ < F⁻