A) 1
B) 2
C) 7
D) 4
Correct Answer: D
Solution :
\[\frac{{{r}_{{{H}_{2}}}}}{{{r}_{HC}}}=\sqrt{\frac{{{M}_{HC}}}{{{M}_{{{H}_{2}}}}}}\Rightarrow 3\sqrt{3}=\,\,\sqrt{\frac{{{M}_{HC}}}{2}}\] \[{{\operatorname{M}}_{HC}}={{({{3}^{3}})}^{2}}\times 2=54\] \[\therefore \,\,\,\,{{C}_{n}}{{H}_{2n-2}},\,\,12\times n+(2n-2)=54\,\,or\,\,n=4\]
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