A) \[{{C}_{2}}{{H}_{4}}\]
B) \[{{C}_{3}}{{H}_{6}}\]
C) \[{{C}_{3}}{{H}_{8}}\]
D) \[{{C}_{3}}{{H}_{4}}\]
Correct Answer: D
Solution :
Ethyne is the member of alkyne series whose general formula is \[{{C}_{n}}{{H}_{2n-2}}\] In \[{{C}_{n}}{{H}_{2n-2}}n=\] no. of atoms as 1, 2, 3... If \[n=2,\] then \[{{C}_{2}}{{H}_{2\times 2-2}}\] \[{{C}_{2}}{{H}_{2}}\,\text{Ethyne}\] If \[n=3,\]then \[{{C}_{3}}{{H}_{2\times 3-2}}\] \[{{C}_{3}}{{H}_{6-2}}\] \[{{C}_{3}}{{H}_{4}}\,\text{Propyne}\] So, the next member or homologue of ethyne is propyne \[({{C}_{3}}{{H}_{4}}).\]
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