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The hybridization of carbon atoms in ${\text{C - C}}$ single bond of is:
(A) ${\text{s}}{{\text{p}}^{\text{3}}}{\text{ - s}}{{\text{p}}^{\text{3}}}$
(B) ${\text{s}}{{\text{p}}^2}{\text{ - s}}{{\text{p}}^{\text{3}}}$
(C) ${\text{sp - s}}{{\text{p}}^2}$
(D) ${\text{s}}{{\text{p}}^{\text{3}}}{\text{ - sp}}$

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Hint: (1)Hybridization refers to the intermixing of two or more atomic orbitals which are of nearly the same energies to give rise to the formation of new orbitals called hybrid orbitals.
(2) For the carbon atom, hybridization takes place in three ways: ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridization, ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridization and sp hybridization.
Complete step-by-step answer: The given compound is .
We need to find out the hybridization of the carbon atoms in the carbon-carbon single bond of the given compound.
When the 2s and the three 2p orbitals of carbon, viz. the 2px, 2py and 2pz orbitals get intermixed, four new orbitals called the ${\text{s}}{{\text{p}}^{\text{3}}}$ hybrid orbitals will be formed. An ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridIzed orbital has s-character 25% and p-character 75%.
In ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridization, the 2s orbital and two of the three 2p orbitals of carbon get intermixed, while the third 2p orbital remains unchanged. A ${\text{s}}{{\text{p}}^{\text{2}}}$ hybrid orbital has 33% s-character and 66% p-character. In sp hybridization, the 2s orbital and one of the three 2p orbitals of carbon get intermixed, while the remaining two 2p orbitals are left unchanged.
Let us consider the carbon atom of the ${\text{C - C}}$ single bond which is attached to the triple bond. Since carbon has 4 valence electrons and double or triple bonds don’t take part in hybridization, thus, this carbon is forming two bond pairs, viz., one with the other triple bonded carbon and the other with the single bonded carbon $\left( {{\text{C - C}}} \right)$. Also, there are no lone pairs on this carbon. So it is sp hybridized.
Now, let us consider the carbon atom of the ${\text{C - C}}$ single bond which is attached to the double bond. Since carbon has 4 valence electrons and double or triple bonds don’t take part in hybridization, thus, this carbon is forming three bond pairs, viz., one ${\text{C - H}}$ bond pair, one ${\text{C - C}}$ bond pair and one ${\text{C = C}}{{\text{H}}_2}$ bond pair. Also, there are no lone pairs. So this carbon is ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridized.

Out of the given options, the option (c) is correct.

Note: A carbon atom bonded to other atoms by two sigma and two pi bonds (e.g. alkynes) is always sp hybridized. A carbon atom bonded to other atoms by three sigma and one pi bond will always be ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridized.