Analyse the given information and answer the following questions.
A=Na=2,8,1
B=Mg=2,8,2
C=H=1
D=Cl=2,8,7
Which type of bond will be formed when A and D react ?

Options

A. Ionic bond
B. Covalent bond
C. Coordinate bond
D. None of these

Explanation

To determine the type of bond formed when A (Sodium, Na) and D (Chlorine, Cl) react, we need to analyze their electronic configurations and how they interact with each other.

### Step-by-Step Solution:

1. **Identify the Elements and Their Electronic Configurations:**
- A (Sodium, Na): Electronic configuration is 2, 8, 1.
- D (Chlorine, Cl): Electronic configuration is 2, 8, 7.

2. **Determine the Valence Electrons:**
- Sodium has 1 valence electron (the outermost shell has 1 electron).
- Chlorine has 7 valence electrons (the outermost shell has 7 electrons).

3. **Understand the Stability of the Elements:**
- Sodium (Na) wants to lose 1 electron to achieve a stable electronic configuration of 2, 8 (which is the configuration of Neon, a noble gas).
- Chlorine (Cl) wants to gain 1 electron to achieve a stable electronic configuration of 2, 8, 8 (also the configuration of Neon).

4. **Electron Transfer:**
- When sodium loses its 1 valence electron, it becomes a positively charged ion (Na⁺).
- When chlorine gains that 1 electron, it becomes a negatively charged ion (Cl⁻).

5. **Formation of Ionic Bond:**
- The transfer of an electron from sodium to chlorine results in the formation of Na⁺ and Cl⁻ ions.
- The electrostatic attraction between the positively charged sodium ion and the negatively charged chloride ion leads to the formation of an ionic bond.

### Conclusion:
The bond formed when A (Sodium) and D (Chlorine) react is an **ionic bond**.

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