A) \[FC{{H}_{2}}COOH\,>\,ClC{{H}_{2}}COOH>BrC{{H}_{2}}COOH\]
B) \[ClC{{H}_{2}}COOH>\,BrC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]
C) \[BrC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]
D) \[ClC{{H}_{2}}COOH>FC{{H}_{2}}COOH>BrC{{H}_{2}}COOH\]
Correct Answer: A
Solution :
Acidity decreases with the decrease in electronegativity of halogen i.e., \[FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>BrC{{H}_{2}}COOH\]
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