A) \[C{{H}_{3}}I-66{{\,}^{o}}C,42{{\,}^{o}}C\]
B) \[C{{H}_{3}}Br-94{{\,}^{o}}C,3{{\,}^{o}}C\]
C) \[{{C}_{2}}{{H}_{5}}Cl-{{139}^{o}}C,12{{\,}^{o}}C\]
D) \[C{{H}_{3}}F-115{{\,}^{o}}C,78{{\,}^{o}}C\]
Correct Answer: A
Solution :
\[C{{H}_{3}}Br(B.P.=3{{\,}^{o}}C)\]\[C{{H}_{3}}F(B.P.=-78{{\,}^{o}}C)\]and \[{{C}_{2}}{{H}_{5}}Cl(B.P.=12{{\,}^{o}}C)\] are all gases at room temperature as their boiling point is much less than room temperature i.e., they boil (vaporise) at a temperature much less than room temperature. But \[\text{C}{{\text{H}}_{\text{3}}}\]I has\[B.P.={{42}^{o}}\] which is above the room temperature so, it is liquid at room temperature and solidify at around - \[66{{\,}^{o}}C\].
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