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 one is getting reduced during the formation of AD?

Options

A. A
B. D
C. Both 1 and 2
D. None of these

Explanation

To determine which element is getting reduced during the formation of AD (where A is sodium and D is chlorine), we need to analyze the oxidation states and the electron transfer involved in the formation of the ionic compound.

### Step-by-Step Solution:

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

2. **Determine the Behavior of Sodium (Na)**:
- Sodium has 1 electron in its outermost shell (valence shell).
- To achieve a stable electron configuration (octet), sodium will lose this 1 electron.
- When sodium loses 1 electron, it becomes Na⁺ (oxidation state +1).

3. **Determine the Behavior of Chlorine (Cl)**:
- Chlorine has 7 electrons in its outermost shell.
- To achieve a stable octet, chlorine needs to gain 1 electron.
- When chlorine gains 1 electron, it becomes Cl⁻ (oxidation state -1).

4. **Identify Oxidation and Reduction**:
- Oxidation is the loss of electrons, which increases the oxidation state. Here, sodium is oxidized from 0 to +1.
- Reduction is the gain of electrons, which decreases the oxidation state. Here, chlorine is reduced from 0 to -1.

5. **Conclusion**:
- In the formation of AD (NaCl), sodium (Na) is oxidized, and chlorine (Cl) is reduced.
- Therefore, the element that is getting reduced during the formation of AD is **D (Chlorine)**.

### Final Answer:
**D (Chlorine) is getting reduced during the formation of AD.**