In the following molecule,the two carbon atom | vedclass

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(C) In the $CH_3-C^* \equiv C^*-CH_3$ molecule,both $C^*$ carbon atoms are bonded to one other carbon atom via a single bond and to each other via a t

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$sp$ orbital

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(C) In the $CH_3-C^* \equiv C^*-CH_3$ molecule,both $C^*$ carbon atoms are bonded to one other carbon atom via a single bond and to each other via a triple bond.Each $C^*$ atom is attached to two atoms in total (one $C$ and one $C$),and there are no lone pairs on the carbon atoms.The steric number for each $C^*$ atom is $2 + 0 = 2$.$A$ steric number of $2$ corresponds to $sp$ hybridization.Therefore,the correct answer is Option $(C)$ $sp$ orbital.

In the following molecule,the two carbon atoms marked by asterisk $()$ possess the following type of hybridized orbitals: $CH_3-C^ \equiv C^*-CH_3$

Similar Questions

In the molecule $HC^{1} \equiv C^{2} - CH^{3} = CH^{4} - CH_{3}^{5}$,which carbon atom exhibits the lowest electronegativity?

The hybridization of carbon involved in the $C_2-C_3$ single bond in the molecule $CH \equiv C-CH=CH_2$ is

$CH_3-CH_2-CH=CH_2$ has hybridisation

Number of $sp$ hybridised carbon atoms in the given compound is:
$CH_2 = CH - C \equiv C - CH_2 - CN$

The hybridisation of the negatively charged carbon atoms in $(1)$ and $(2)$ are:
$(1)$ $CH_3^-$
$(2)$ $H_2C^- - CHO - CH_3$

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In the following molecule,the two carbon atoms marked by asterisk $(*)$ possess the following type of hybridized orbitals: $CH_3-C^* \equiv C^*-CH_3$