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The state of hybridization of $C_2$, $C_3$, $C_5$ and $C_6$ of the hydrocarbon
is in the following sequence
is in the following sequence
A
sp, $sp^2$, $sp^3$ and $sp^2$
B
sp, $sp^3$, $sp^2$ and $sp^3$
C
$sp^3$, $sp^2$, $sp^2$ and sp
D
sp, $sp^2$, $sp^2$ and $sp^3$
Detailed Solution
The hybridisation of a carbon atom follows from the number of σ bonds it forms: 4 σ bonds means $sp^3$, 3 σ bonds means $sp^2$ and 2 σ bonds means sp.
$C_2$ is part of the triple bond: 2 σ bonds (and 2 π bonds), so sp.
$C_3$ is the CH carbon carrying a methyl group, with four single bonds: 4 σ bonds, so $sp^3$.
$C_5$ is part of the double bond: 3 σ bonds (and 1 π bond), so $sp^2$.
$C_6$ is the carbon bonded to three methyl groups and $C_5$, with four single bonds: 4 σ bonds, so $sp^3$.
Sequence: sp, $sp^3$, $sp^2$ and $sp^3$
$C_2$ is part of the triple bond: 2 σ bonds (and 2 π bonds), so sp.
$C_3$ is the CH carbon carrying a methyl group, with four single bonds: 4 σ bonds, so $sp^3$.
$C_5$ is part of the double bond: 3 σ bonds (and 1 π bond), so $sp^2$.
$C_6$ is the carbon bonded to three methyl groups and $C_5$, with four single bonds: 4 σ bonds, so $sp^3$.
Sequence: sp, $sp^3$, $sp^2$ and $sp^3$