Arrange the following in the decreasing order of hydrogen bonding.
$C H_{4}, C H_{3} - O H, C H_{3} - F, C H_{3} - N H_{2}$

C H_{3}, - O H > C H_{3} - N H_{2} > C H_{3} - F > C H_{4}

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C H_{3}, - O H > C H_{3} - F > C H_{3} - N H_{2} > C H_{4}

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C H_{3}, - F > C H_{3} - O H > C H_{3} - N H_{2} > C H_{4}

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C H_{4}, > C H_{3} - F > C H_{3} - N H_{2} > C H_{3} - O H

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Solution

The correct option is C $C H_{3}, - F > C H_{3} - O H > C H_{3} - N H_{2} > C H_{4}$
Due to greater electronegativity ,In $C H_{3} - F$ ,Florine has greater delta negative charge on it ,so,it shows greater hydrogen bonding than $C H_{3} - O H; C H_{3} - N H_{3}$ .hence the correct order is $C H_{3}, - F > C H_{3} - O H > C H_{3} - N H_{2} > C H_{4}$

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