A) \[C{{H}_{3}}\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\] and \[\overset{\bullet }{\mathop{Cl}}\,\]
B) \[C{{H}_{3}}\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\]and \[C{{l}^{O-}}\]
C) \[C{{H}_{3}}\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\] and \[\overset{\bullet }{\mathop{Cl}}\,\]
D) \[C{{H}_{3}}\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]and \[C{{l}^{O-}}\]
Correct Answer: A
Solution :
Key Idea In homolysis, the covalent bond is broken in such a way that each resulting species known as free radical. \[C{{H}_{3}}C{{H}_{2}}-Cl\xrightarrow[fission]{Homolytic}C{{H}_{3}}\overset{\bullet }{\mathop{C}}\,{{H}_{2}}{{+}^{\bullet }}Cl\]
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