Arrange the following hydroxy compounds in the order of increasing acid strength:

$(a)$ $C r O_{2} (O H)_{2}$
$(b)$ $C r (O H)_{2}$
$(c)$ $C r (O H)_{3}$

C r (O H)_{2} < C r O_{2} (O H)_{2} < C r (O H)_{3}

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C r (O H)_{2} < C r (O H)_{3} < C r O_{2} (O H)_{2}

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C r (O H)_{3} < C r O_{2} (O H)_{2} < C r (O H)_{2}

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C r (O H)_{3} < C r (O H)_{2} < C r O_{2} (O H)_{2}

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Solution

The correct option is A $C r (O H)_{2} < C r (O H)_{3} < C r O_{2} (O H)_{2}$
As the oxidation state of the central element of the compound increases, its acidic strength also increases. The oxidation states of $C r$ in $C r (O H)_{2}, C r (O H)_{3}$ and $C r O_{2} (O H)_{2}$ are $+ 6, + 3$ and $+ 2$ respectively. Hence, the correct order is as follows:
$C r (O H)_{2} < C r (O H)_{3} < C r O_{2} (O H)_{2}$
Hence, option B is the correct answer.

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$C r O_{3}$ dissolves in aqueous NaOH to give

C r (O H)_{2}

2.0 \times 10^{- 16}

25^{\circ} C

p H

C r^{2 +}

C r (O H)_{2}

2 \times 10^{- 4} M C r^{2 +}

$C r O_{3}$ dissoloves in aqueous NaOH to give :

C r (O H)_{3} + (O H)^{-} + {I O_{3}}^{-} \to {C r O_{4}}^{2 -} + H_{2} O + I^{-}

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