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