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. 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 the most ionic compound that can be formed from the given electronic configurations, we need to analyze the elements based on their ability to lose or gain electrons. Here's a step-by-step solution:

### Step 1: Identify the Elements
From the electronic configurations provided, we can identify the elements:

- **I:** `1s²2s²` → This is Beryllium (Be), a Group 2 element (alkaline earth metal).
- **II:** `1s²2s²2p⁶` → This is Neon (Ne), a noble gas.
- **III:** `1s²2s²2p⁶3s²` → This is Magnesium (Mg), also a Group 2 element.
- **IV:** `1s²2s²2p³` → This is Nitrogen (N), a Group 15 element (non-metal).
- **V:** `1s²2s²2p⁵` → This is Fluorine (F), a Group 17 element (halogen).

### Step 2: Determine Ionic Character
To form an ionic compound, we need a metal that can easily lose electrons (forming a cation) and a non-metal that can easily gain electrons (forming an anion).

- **Metals:**
- Be (from I) and Mg (from III) can lose 2 electrons to form Be²⁺ and Mg²⁺.

- **Non-metals:**
- N (from IV) can gain 3 electrons to form N³⁻.
- F (from V) can gain 1 electron to form F⁻.

### Step 3: Assess Ionic Bond Formation
The most ionic character in a compound is typically observed when:
1. The metal has a low ionization energy (easily loses electrons).
2. The non-metal has a high electron affinity (easily gains electrons).

- **Mg (from III)** can lose 2 electrons easily.
- **F (from V)** can gain 1 electron easily.

### Step 4: Form the Ionic Compound
The most ionic compound will be formed between:
- **Mg (III)** and **F (V)**, resulting in the compound MgF₂.

### Conclusion
Thus, the most ionic compound will be formed between the element with configuration III (Magnesium, Mg) and the element with configuration V (Fluorine, F).

### Final Answer
The most ionic compound will be formed between **III (Mg) and V (F)**.

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