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
In the formation of CD by bonding between C and D. Which is oxidised ?

Options

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

Explanation

To determine which element is oxidized in the formation of the compound CD (where C is hydrogen and D is chlorine), we need to analyze the electron configurations and the nature of the bonding between these two elements.

### Step-by-Step Solution:

1. **Identify the Elements and Their Electron Configurations**:
- C (Hydrogen, H) has the electron configuration: 1 (1 electron in its outer shell).
- D (Chlorine, Cl) has the electron configuration: 2, 8, 7 (7 electrons in its outer shell).

2. **Determine the Type of Bonding**:
- Hydrogen (C) is a non-metal and has 1 electron in its outer shell, while Chlorine (D) is also a non-metal with 7 electrons in its outer shell.
- Since both elements are non-metals, they will form a **covalent bond** by sharing electrons.

3. **Understand the Electron Sharing**:
- In a covalent bond, hydrogen will share its 1 electron with chlorine.
- Chlorine needs 1 more electron to complete its octet (8 electrons), and hydrogen needs 1 more electron to complete its duplet (2 electrons).

4. **Analyze Electronegativity**:
- Chlorine is more electronegative than hydrogen, meaning it has a stronger tendency to attract shared electrons towards itself.
- This results in a partial negative charge on chlorine and a partial positive charge on hydrogen.

5. **Determine Oxidation and Reduction**:
- Oxidation is defined as the loss of electrons or an increase in oxidation state.
- In this case, hydrogen (C) is effectively losing its electron density to chlorine (D), which means hydrogen is being oxidized.
- Conversely, chlorine (D) is gaining electron density, which means it is being reduced.

6. **Conclusion**:
- Therefore, in the formation of the compound CD, **hydrogen (C)** is oxidized.

### Final Answer:
C (Hydrogen) is oxidized in the formation of CD.

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