Covalent bond can undergo fission in two different ways. The correct represenation involving a heterolytic fission of `CH_(3)-Br` is

Options

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Explanation

To understand the heterolytic fission of the compound CH₃Br (methyl bromide), we need to break down the process step by step.

### Step-by-Step Solution:

1. **Understanding Covalent Bonds**:
- Covalent bonds are formed when two atoms share electrons. In CH₃Br, the bond between carbon (C) and bromine (Br) is a covalent bond.

2. **Types of Fission**:
- Fission refers to the breaking of a bond. There are two types of fission: homolytic and heterolytic.
- **Homolytic fission** involves the equal splitting of the bond, where each atom retains one of the shared electrons, forming two radicals.
- **Heterolytic fission** involves the unequal splitting of the bond, where one atom takes both of the shared electrons, resulting in the formation of a cation and an anion.

3. **Identifying Heterolytic Fission in CH₃Br**:
- In CH₃Br, when heterolytic fission occurs, the bond between carbon and bromine breaks such that bromine takes both electrons from the bond.
- This results in the formation of a bromide ion (Br⁻) and a methyl cation (CH₃⁺).

4. **Representation of Heterolytic Fission**:
- The reaction can be represented as:
\[ CH_3Br \rightarrow CH_3^+ + Br^- \]
- Here, CH₃⁺ is the methyl cation (positively charged) and Br⁻ is the bromide ion (negatively charged).

5. **Conclusion**:
- The correct representation of the heterolytic fission of CH₃Br shows the formation of a cation and an anion as a result of the breaking of the covalent bond.