A) \[C{{H}_{3}}\overset{\centerdot }{\mathop{C}}\,{{H}_{2}}\]and \[\overset{\centerdot }{\mathop{C}}\,l\]
B) \[C{{H}_{3}}\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\]and \[C{{l}^{\text{o}-}}\]
C) \[C{{H}_{3}}\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\]and \[\overset{\centerdot }{\mathop{C}}\,l\]
D) \[C{{H}_{3}}\overset{\centerdot }{\mathop{C}}\,{{H}_{2}}\]and \[C{{l}^{\text{o}-}}\]
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. \[C{{H}_{3}}C{{H}_{2}}-\xrightarrow[fission]{Homolytic}C{{H}_{3}}\overset{\centerdot }{\mathop{C}}\,{{H}_{2}}+{{\,}^{\centerdot }}Cl\]
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