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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 hybridization of carbons 1, 3 and 5 are in the following sequence :
$\overset{6}{C}H_3-\overset{5}{C}H=\overset{4}{C}H-\overset{3}{C}H_2-\overset{2}{C}\equiv\overset{1}{C}H$
The state of hybridization of carbons 1, 3 and 5 are in the following sequence :
A
$sp$, $sp^2$, $sp^3$
B
$sp^3$, $sp^2$, $sp$
C
$sp^2$, $sp$, $sp^3$
D
$sp$, $sp^3$, $sp^2$
Detailed Solution
The hybridisation of a carbon atom is decided by the number of sigma bonds it forms: 4 sigma bonds means $sp^3$, 3 sigma bonds (one double bond) means $sp^2$, and 2 sigma bonds (a triple bond) means $sp$.
Hybridisation of the carbons from C6 to C1: $sp^3$, $sp^2$, $sp^2$, $sp^3$, $sp$, $sp$
C1 is part of the triple bond ($C\equiv CH$), so it is $sp$ hybridised.
C3 is a $CH_2$ group with four single bonds, so it is $sp^3$ hybridised.
C5 is part of the double bond ($CH=CH$), so it is $sp^2$ hybridised.
The state of hybridization of carbons 1, 3 and 5 is $sp$, $sp^3$ and $sp^2$ respectively.
Hybridisation of the carbons from C6 to C1: $sp^3$, $sp^2$, $sp^2$, $sp^3$, $sp$, $sp$
C1 is part of the triple bond ($C\equiv CH$), so it is $sp$ hybridised.
C3 is a $CH_2$ group with four single bonds, so it is $sp^3$ hybridised.
C5 is part of the double bond ($CH=CH$), so it is $sp^2$ hybridised.
The state of hybridization of carbons 1, 3 and 5 is $sp$, $sp^3$ and $sp^2$ respectively.