Elements with their electronic configuration are given below:
Answer the following questions: ltbr. I: `1s^(2)2s^(2)`
II: `1s^(2) 2s^(2) 2p^(6)`
III: `1s^(2) 2s^(2) 2p^(6)3s^(2)`
IV `1s^(2)2s^(2)2p^(3)`
V: `1s^(2)2s^(2)2p^(5)`
Q. Which of the following is the correct order of increasing size?

Options

A. `I lt III lt IV lt V`
B. `V lt IV lt III lt I`
C. `I lt IV lt V lt III`
D. `V lt IV lt I lt III`

Answer

To determine the correct order of increasing size for the given elements based on their electronic configurations, we need to analyze their positions in the periodic table.

### Step-by-Step Solution:

1. **Identify the Elements**:
- From the electronic configurations given:
- I: `1s^2 2s^2` → This corresponds to Helium (He).
- II: `1s^2 2s^2 2p^6` → This corresponds to Neon (Ne).
- III: `1s^2 2s^2 2p^6 3s^2` → This corresponds to Magnesium (Mg).
- IV: `1s^2 2s^2 2p^3` → This corresponds to Nitrogen (N).
- V: `1s^2 2s^2 2p^5` → This corresponds to Fluorine (F).

2. **Determine the Period and Group**:
- **I (He)**: 1st Period, Group 18 (Noble Gas)
- **II (Ne)**: 2nd Period, Group 18 (Noble Gas)
- **III (Mg)**: 3rd Period, Group 2 (Alkaline Earth Metal)
- **IV (N)**: 2nd Period, Group 15 (P-Block)
- **V (F)**: 2nd Period, Group 17 (Halogen)

3. **Analyze Size Trends**:
- Atomic size generally increases down a group and decreases across a period from left to right.
- Since all elements except for Magnesium (III) are in the 2nd period, we need to compare their sizes:
- In the 2nd period: N (IV) < F (V) < Ne (II) < He (I)
- He is the smallest, followed by Ne, then F, and finally N.
- Magnesium (III) is in the 3rd period and will be larger than all elements in the 2nd period.

4. **Order the Elements by Size**:
- From the analysis, we can conclude:
- III (Mg) > II (Ne) > V (F) > IV (N) > I (He)
- Therefore, the order of increasing size is:
- I (He) < IV (N) < V (F) < II (Ne) < III (Mg)

5. **Final Order**:
- The correct order of increasing size is:
- He < N < F < Ne < Mg