What is the hybridization of $C$ atom in $C H_{3}$ radical?

s p^{2}

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s p^{3}

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s p

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s p^{3} d

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Solution

The correct option is A $s p^{2}$
$C$ has four electrons in its valence shell. In $C H_{3}$ radical, $C$ atom makes three sigma bonds with three $H$ atoms and have one unpaired electron. This unpaired electron does not take part in hybridization due to the low electronegativity of $H$ atom than central $C$ atom.
Therefore, the hybridization of $C$ atom in $C H_{3}$ radical will be $s p^{2}$ .

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Similar questions

C

C H_{3}

What are the hybridization states of each carbon atom(s) in the following molecules ?

(i) $C H_{3} * C H_{2} C H_{2} C H_{3}$

(ii) $C H_{2} = C^{*} H - C H = C H_{2}$

(iii) $C H_{3} * C H = C H - C H_{3}$

(iv) $H - * C \equiv C - H$

C H_{3} - O | | C * - N H_{2}

P_{2} O_{5} \to Δ C H_{3} - C * \equiv N

C H_{3} - C H = C = C H - C H_{3}

Which hybrid orbitals are used by carbon atoms in the following molecules?

CH₃–CH₃; (b) CH₃–CH=CH₂; (c) CH₃-CH₂-OH; (d) CH₃-CHO (e) CH₃COOH

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