The compound 1, 2-butadiene has:
A. Only sp hybridised carbon atoms
B. Only \[s{p^2}\] hybridised carbon atoms
C. Both sp and \[s{p^2}\] hybridised carbon atoms
D. \[sp\], \[s{p^2}\] and \[s{p^3}\] hybridised carbon atoms

Hint: We know that the phenomenon of mixing of orbitals of the same atom with slight difference in energies so as to redistribute the energies and give new orbitals of equivalent energy and shape is termed as hybridization.

Complete step-by-step solution:
Here we first find the hybridisation of each carbon individually either it is \[s{p^2}\], \[s{p^3}\] or \[sp\].
\[s{p^3}\] hybridization uses four \[s{p^3}\] hybridized atomic orbitals. So, there must be the presence of four groups of electrons.
\[s{p^2}\] hybridization uses three \[s{p^2}\] hybridized atomic orbitals. So, there must be the presence of three groups of electrons.
\[sp\] hybridization uses two \[sp\] hybridized atomic orbitals. So, there must be the presence of three groups of electrons.
Let’s come to the question. The structure of 1, 2-butadiene is as follows:

,the 1st carbon is \[s{p^2}\] hybridized as three electrons groups present, 2nd carbon is \[sp\] hybridised (two electron groups) , 3rd carbon is \[s{p^2}\] hybridized (three electrons groups) and 4th carbon is \[s{p^3}\] hybridized because of presence of four electron groups.
Therefore, in 1, 2-butadiene, \[s{p^2}\], \[sp\] and \[s{p^3}\] carbons are present. Hence, D is the correct option.

Note: Hybridisation of carbon atom can be found by counting the number of electron groups surrounding the carbon atom. If four groups present such as in case of \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\] the hybridization is \[s{p^3}\]. If three electron groups are present, hybridization is \[s{p^2}\] and if two electron groups present, hybridization is \[sp\].