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B
⊕CH3
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C
CH4
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D
Both (B) and (C)
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Solution
The correct option is D∙CH3 Alkyl free radicals are either planar or pyramidal in structure, they have sp2 hybridisation with impaired e− located primarily in p−orbital.
In CH3, it has sp2 hybridisation with two e− on each and one empty p−orbital.
In CH4, every p−orbital is involved in hybridisation making it sp3.