Identify the group of compounds that is in the same homologous series.

C_{2} H_{4}, C_{3} H_{6}, C_{4} H_{8}

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C_{2} H_{6}, C_{5} H_{12}, C_{7} H_{16}

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C_{2} H_{2}, C_{4} H_{6}, C_{6} H_{10}

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C_{2} H_{4}, C_{4} H_{6}, C_{6} H_{14}

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Solution

The correct options are
A $C_{2} H_{4}, C_{3} H_{6}, C_{4} H_{8}$
B $C_{2} H_{6}, C_{5} H_{12}, C_{7} H_{16}$
C $C_{2} H_{2}, C_{4} H_{6}, C_{6} H_{10}$
Elements in the same homolgous series differs by $C_{n} H_{2 n}$ group(where n = 1,2,3,4,...).
Considering the elements, $C_{2} H_{4}, C_{3} H_{6}, C_{4} H_{8}$
The first 2 elements differs by $C H_{4}$ ie., $C_{n} H_{2 n}, w h e r e n = 2$ .
The 2nd and 3rd elements differs by $C H_{4}$ ie., $C_{n} H_{2 n}, w h e r e n = 2$
So, $C H_{2}, C_{3} H_{6}, C_{4} H_{8}$ are of the same homologous series.
Considering the elements, $C_{2} H_{6}, C_{5} H_{1} 2, C_{7} H_{16}$
The first 2 elements differs by $C_{3} H_{6}$ ie., $C_{n} H_{2 n}, w h e r e n = 3$ .
The 2nd and 3rd elements differs by $C_{2} H_{4}$ ie., $C_{n} H_{2 n}, w h e r e n = 2$
So, $C_{2} H_{6}, C_{5} H_{1} 2, C_{7} H_{16}$ are of the same homologous series.
Considering the elements, $C_{2} H_{2}, C_{4} H_{6}, C_{6} H_{10}$
The first 2 elements differs by $C_{2} H_{4}$ ie., $C_{n} H_{2 n}, w h e r e n = 2$ .
The 2nd and 3rd elements differs by $C_{2} H_{4}$ ie., $C_{n} H_{2 n}, w h e r e n = 2$
So, $C_{2} H_{2}, C_{4} H_{6}, C_{6} H_{10}$ are of the same homologous series.
Considering the elements, $C_{2} H_{4}, C_{4} H_{6}, C_{6} H_{14}$
The first 2 elements differs by $C_{2} H_{2}$ which doesn't satisfy $C_{n} H_{2 n}$ .
The 2nd and 3rd elements differs by $C_{2} H_{8}$ which doesn't satisfy $C_{n} H_{2 n}$ .
So, $C_{2} H_{4}, C_{4} H_{6}, C_{6} H_{14}$ are not from the same homologous series.

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