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) \[BrC{{H}_{2}}COOH>CIC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]\[>C{{H}_{3}}COOH\]
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 almost proportionately with the increase in electronegativity of the halogen present. So, the correct order is: |
| FCH2COOH > ClCH2COOH> BrCH2COOH> CH3COOH |
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