The state of hybridisation of C_2, C_3, C_5 a | vedclass

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Question

(D) The correct option is $(d)$.In the given hydrocarbon structure:$1.$ $C_2$ is involved in a triple bond $(C \equiv C)$,so it is $sp$ hybridized.$2.

Vedclass · Accepted Answer

$sp, sp^3, sp^2$ and $sp^3$

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(D) The correct option is $(d)$.In the given hydrocarbon structure:$1.$ $C_2$ is involved in a triple bond $(C \equiv C)$,so it is $sp$ hybridized.$2.$ $C_3$ is bonded to four other atoms ($C_2, C_4, H$,and the $C$ of the $CH_3$ group) via single bonds,so it is $sp^3$ hybridized.$3.$ $C_5$ is involved in a double bond $(C=C)$,so it is $sp^2$ hybridized.$4.$ $C_6$ is bonded to four other atoms ($C_5, C_7$,and the $C$ atoms of two $CH_3$ groups) via single bonds,so it is $sp^3$ hybridized.Thus,the sequence for $C_2, C_3, C_5$ and $C_6$ is $sp, sp^3, sp^2, sp^3$.

The state of hybridisation of $C_2, C_3, C_5$ and $C_6$ of the hydrocarbon $CH_3-C_6(CH_3)_2-C_5H=C_4H-C_3H(CH_3)-C_2 \equiv C_1H$ is in the following sequence:

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