Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYEasy

In the hydrocarbon $\overset{6}{C}H_3-\overset{5}{C}H=\overset{4}{C}H-\overset{3}{C}H_2-\overset{2}{C}\equiv\overset{1}{C}H$ , the state of hybridisation of carbons 1, 3 and 5 are in the following sequence:

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

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

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

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

Step-by-Step Solution

In the given hydrocarbon: Carbon-1 is involved in a triple bond, so it is $sp$ hybridized. Carbon-3 is bonded via four single bonds, so it is $sp^3$ hybridized. Carbon-5 is involved in a double bond, so it is $sp^2$ hybridized. Therefore, the sequence of hybridization states for carbons 1, 3, and 5 is $sp$ , $sp^3$ , and $sp^2$ respectively.

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