Question
The compound 1, 2-butadiene has
A.
only $$sp$$ hybridized carbon atoms
B.
only $$s{p^2}$$ hybridized carbon atoms
C.
both $$sp$$ and $$s{p^2}$$ hybridized carbon atoms
D.
$$sp$$ , $$s{p^2}$$ and $$s{p^3}$$ hybridized carbon atoms
Answer :
$$sp$$ , $$s{p^2}$$ and $$s{p^3}$$ hybridized carbon atoms
Solution :
$$\eqalign{
& \mathop {C{H_3}}\limits^4 - \mathop {CH}\limits^3 = \mathop C\limits^2 = \mathop {C{H_2}}\limits^1 \cr
& {\text{Hybridisation in}}\,{C_1} = s{p^2},{C_2} = sp,{C_3} = s{p^2},{C_4} = s{p^3}. \cr} $$