What is the hybridisation of each carbon in $H_2C = C = CH_2$?

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(N/A) The carbon atoms $1$ and $3$ are $sp^2$ hybridized because each is bonded to three atoms (two $H$ atoms and one $C$ atom) and participates in one double bond.
The central carbon atom $(C_2)$ is $sp$-hybridized because it is bonded to two carbon atoms via two double bonds.
In the structure of allene $(H_2C=C=CH_2)$,the two $\pi$ bonds are perpendicular to each other. The $2p_y$ and $2p_z$ orbitals of the central carbon atom $(C_2)$ are involved in the formation of $\pi$ bonds with the adjacent carbon atoms,leaving only the $2s$ and one $2p$ orbital for hybridization,resulting in $sp$ hybridization for $C_2$.

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Similar Questions

Which of the following statements is not correct?

The correct order of hybridisations of the central atom in $NH_3$,$[PtCl_4]^{2-}$,$PCl_5$,and $BCl_3$ is:

Match the following:
List-$I$List-$II$
$(a)$ $SF_6$$(i)$ $sp^3d^2$
$(b)$ $PCl_5$(ii) $sp^3d$
$(c)$ $XeF_4$(iii) $sp^3d^3$
$(d)$ $IF_7$(iv) $sp^3d^2$

The correct answer is:

For which of the following hybridisation is the bond angle maximum?

Which is the correct statement?
As the $s$-character of a hybrid orbital decreases:
$(I)$ The bond angle decreases
$(II)$ The bond strength increases
$(III)$ The bond length increases
$(IV)$ Size of orbitals increases

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