A) \[C{{H}_{3}}COOH>BrC{{H}_{2}}COOH>\]\[ClC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]
B) \[FC{{H}_{2}}COOH>C{{H}_{3}}COOH>\]\[BrC{{H}_{2}}COOH>ClC{{H}_{2}}COOH\]
C) \[BrC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>\]\[FC{{H}_{2}}COOH>C{{H}_{3}}COOH\]
D) \[FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>\]\[BrC{{H}_{2}}COOH>C{{H}_{3}}COOH\]
Correct Answer: D
Solution :
The acidity of halogenated acid increases with increase in electronegativity of the halogen present. Hence, the correct order of the acidity is as: \[FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>BrC{{H}_{2}}COOH>\] \[C{{H}_{3}}COOH\]
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