$C r O_{3}$ dissolves in aqueous $N a O H$ to give:

C r_{2} O_{7}^{2 -}

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C r O_{4}^{2 -}

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C r (O H)_{3}

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C r (O H)_{2}

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Solution

The correct option is A $C r O_{4}^{2 -}$
$C r O_{3}$ reacts with aqueous NaOH to form $N a_{2} C r O_{4}$ that is $C r O_{4}^{- 2}$ .
$C r O_{3} + 2 N a O H \to N a_{2} C r O_{4} + H_{2} O$

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

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N a O H

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