A) Propene
B) Propyne
C) Propadiene
D) None of these
Correct Answer: C
Solution :
\[\underset{\text{Propadiene}}{\mathop{\underset{s{{p}^{2}}}{\mathop{C{{H}_{3}}}}\,=\underset{sp}{\mathop{{{C}_{{}}}}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,}}\,\] Propadiene has both sp and \[s{{p}^{2}}\]-hybridized carbon atoms.
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