The size of the following species increases in the order

Options

A. `Mg^(2+)ltNa^(+)ltF^(-)`
B. `F^(-)ltNa^(+)ltMg^(2+)`
C. `MgltF^(-)ltNa^(+)`
D. `Na^(+)ltF^(-)ltMg^(2+)`

Answer

To determine the order of increasing size for the species Mg²⁺, Na⁺, and F⁻, we can follow these steps:

### Step 1: Identify the species and their electron configurations
- **Mg²⁺**: Magnesium loses two electrons, resulting in an electron configuration of 1s². It has 10 electrons.
- **Na⁺**: Sodium loses one electron, resulting in an electron configuration of 1s². It also has 10 electrons.
- **F⁻**: Fluoride gains one electron, resulting in an electron configuration of 1s² 2s² 2p⁶. It has 10 electrons as well.

### Step 2: Recognize that these species are isoelectronic
All three species (Mg²⁺, Na⁺, and F⁻) have the same number of electrons (10 electrons), making them isoelectronic.

### Step 3: Understand the effect of nuclear charge on size
Isoelectronic species are compared based on their nuclear charge (the number of protons in the nucleus):
- **Mg²⁺**: 12 protons
- **Na⁺**: 11 protons
- **F⁻**: 9 protons

### Step 4: Determine the relative sizes
The size of the ions decreases with increasing nuclear charge because a higher positive charge pulls the electrons closer to the nucleus:
- **Mg²⁺** has the highest nuclear charge (12 protons) and will be the smallest.
- **Na⁺** has a nuclear charge of 11 protons and will be larger than Mg²⁺ but smaller than F⁻.
- **F⁻** has the lowest nuclear charge (9 protons) and will be the largest.

### Step 5: Write the order of increasing size
Thus, the order of increasing size is:
**Mg²⁺ < Na⁺ < F⁻**

### Final Answer
The size of the following species increases in the order: **Mg²⁺ < Na⁺ < F⁻**.

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