JEE Main 2013 — Chemistry Physical Chemistry

2013mcqmedium

Values of dissociation constant, $K_a$ are given as follows : $\begin{array}{lc}\text { Acid } & \boldsymbol{K}_{\boldsymbol{a}} \\ \mathrm{HCN} & 6.2 \times 10^{-10} \\ \mathrm{HF} & 7.2 \times 10^{-4} \\ \mathrm{HNO}_2 & 4.0 \times 10^{-4}\end{array}$ Correct order of increasing base strength of the base $\mathrm{CN}^{-}, \mathrm{F}^{-}$ and $\mathrm{NO}_2^{-}$ will be :

Official previous-year question

Held on 22 Apr 2013 · Verified 6 Jul 2026.

Options

A
$\mathrm{F}^{-} < \mathrm{CN}^{-} < \mathrm{NO}_2^{-}$
B
$\mathrm{NO}_2^{-} < \mathrm{CN}^{-} < \mathrm{F}^{-}$
C
$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$
D
$\mathrm{NO}_2^{-} < \mathrm{F}^{-} < \mathrm{CN}^{-}$

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Solution

Higher the value of $\mathrm{K}_a$ lower will be the value of $\mathrm{pK}_{\mathrm{a}}$ i.e. higher will be the acidic nature. Further since $\mathrm{CN}^{-}, \mathrm{F}^{-}$ and $\mathrm{NO}_2^{-}$ are conjugate base of the acids $\mathrm{HCN}, \mathrm{HF}$ and $\mathrm{HNO}_2$ respectively hence the correct order of base strength will be $\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$ $(\because$ stronger the acid weaker will be its conjugate base)

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Options

$\mathrm{F}^{-} < \mathrm{CN}^{-} < \mathrm{NO}_2^{-}$

$\mathrm{NO}_2^{-} < \mathrm{CN}^{-} < \mathrm{F}^{-}$

$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$

$\mathrm{NO}_2^{-} < \mathrm{F}^{-} < \mathrm{CN}^{-}$

Solution

JEE Main 2013 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2013 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry