A) \[BrC{{H}_{2}}COOH\]
B) \[ClC{{H}_{2}}COOH\]
C) \[FC{{H}_{2}}COOH\]
D) \[IC{{H}_{2}}COOH\]
Correct Answer: C
Solution :
Greater the electron withdrawing nature of the halogen atom, stronger will be the acid. Hence, \[FC{{H}_{2}}COOH\] is most acidic due to the most electron withdrawing power of the fluorine atom.
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