The decreasing order of acidic character of the following is:
$I . C H_{3} S H$ $I I . C H_{3} O H$ $I I I . H_{2} O$ $I V . E t O H$

I > II > III > IV

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IV > III > II > I

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I > III > II > IV

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III > I > II > IV

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Solution

The correct option is A I > II > III > IV
Thiols (RSH) are stronger acid than alcohols.
Methanol $(p K_{a} = 15.5)$ is a stronger acid than $H_{2} O (p K_{a} = 15.7)$ , but other alcohols are weaker than $H_{2} O$ . So, the correct order of the acidic character is represented by the option A.

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C H_{3} S H

C H_{3} O H

H_{2} O

i . B C l_{3}

i i . A l C l_{3}

i i i . G a C l_{3}

i v . I n C l_{3}

P h O^{⊖}

C H_{3} S^{⊖}

⊖ O H

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Order of decreasing acidity of the following is

(I) HCOOH

(II) ${C H}_{3} C O O H$

(III) ${C l}_{2} C H C O O H$

(IV) ${C F}_{3} C O O H$

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