Question

Which statement from following is correct for homolytic fission?

• A. In this single electron of shared pair moves to one of bonded atom and other to other atom,
• B. In this electron deficient and electron rich species are formed.
• C. In this electronegative atom pulls both electrons of shared pair.
• D. In this carbon free radical formed has $\text{sp}^3$ hybridisation.

Hint:

The prefix "homo-" means equal. Therefore, "homolytic" fission means a perfectly equal and fair split of the bonding pair of electrons—one electron goes to each partner, represented visually by single-barbed "fishhook" arrows.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to identify the correct statement describing the process of homolytic bond fission (homolysis) from the options provided.

Step 2: Key Formula or Approach:
Homolytic cleavage involves the symmetrical breaking of a covalent bond where each participating atom retains exactly one electron from the shared electron pair, leading to the formation of neutral free radicals: $$\text{A}\mathrel{-}\text{B} \xrightarrow{h\nu} \text{A}^{\bullet} + \text{B}^{\bullet}$$

Step 3: Detailed Explanation:
Let's analyze the validity of each statement: 1.

Option (A): Correct. Symmetrical cleavage dictates that one electron of the shared pair moves to one bonded atom and the remaining electron moves to the other atom.
2.

Option (B): Incorrect. This describes heterolytic fission, which yields ions (cations and anions) that are electron-deficient and electron-rich. Homolysis yields neutral free radicals.
3.

Option (C): Incorrect. An electronegative atom pulling both electrons describes heterolytic fission.
4.

Option (D): Incorrect. Simple alkyl free radicals (like the methyl radical) generally assume a planar geometry with $\text{sp}^2$ hybridization, not $\text{sp}^3$, containing the unpaired electron in an unhybridized p-orbital. Thus, this is not a generally correct statement.

Step 4: Final Answer:
The correct statement is option (A).