Elements with their electronic configuration are given below:
Answer the following questions: 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. The most ionic compound will be formed between :

Options

A. I and IV
B. I and V
C. III and IV
D. III and V

Answer

To determine which pair of elements will form the most ionic compound, we need to analyze the given electronic configurations and identify the elements they correspond to. Ionic compounds are typically formed between metals and non-metals with a significant difference in their electronegativities.

### Step-by-Step Solution:

1. **Identify the Elements from Electronic Configurations**:
- I: `1s²2s²` → Beryllium (Be, atomic number 4)
- II: `1s²2s²2p⁶` → Neon (Ne, atomic number 10)
- III: `1s²2s²2p⁶3s²` → Magnesium (Mg, atomic number 12)
- IV: `1s²2s²2p³` → Nitrogen (N, atomic number 7)
- V: `1s²2s²2p⁵` → Fluorine (F, atomic number 9)

2. **Determine Electronegativity Values**:
- Beryllium (Be): 1.5
- Magnesium (Mg): 1.2
- Nitrogen (N): 3.0
- Fluorine (F): 4.0
- Neon (Ne): Not applicable (noble gas)

3. **Calculate Differences in Electronegativity**:
- Be and N: |1.5 - 3.0| = 1.5
- Be and F: |1.5 - 4.0| = 2.5
- Mg and N: |1.2 - 3.0| = 1.8
- Mg and F: |1.2 - 4.0| = 2.8

4. **Identify the Pair with the Largest Difference**:
- The largest difference in electronegativity is between Magnesium (Mg) and Fluorine (F), which is 2.8.

5. **Conclusion**:
- Therefore, the most ionic compound will be formed between Magnesium (Mg) and Fluorine (F).

### Final Answer:
The most ionic compound will be formed between Magnesium (Mg) and Fluorine (F).

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