The $C r O_{3}$ on reaction with $H C l$ and $N a O H (a q)$ gives, respectively :

C r O_{2} C l_{2}

C r O_{4}^{2 -}

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C r {(O H)}_{2}

C r O_{4}^{2 -}

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C l_{2}

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

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C l_{2}

C r {(O H)}_{3}

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Solution

The correct option is D $C r O_{2} C l_{2}$ , $C r O_{4}^{2 -}$
The $C r O_{3}$ on reaction with $H C l$ and $N a O H (a q)$ gives $C r O_{2} C l_{2}$ and $C r O_{4}^{2 -}$ respectively.
The reactions are as follows:
$C r O_{3} + 2 H C l \to C r O_{2} C l_{2} + H_{2} O$
$C r O_{3} + 2 O H^{-} \to C r O_{4}^{2 -} + H_{2} O$

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Similar questions

$C r O_{3}$ 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^{-}

C r O_{4}^{2 -} (a q) + S^{2 -} (a q) ⟶ C r {(O H)}_{3} (s) + S (s)

C O_{2}

C r O_{4}^{2 -}

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

C r O_{4}^{2 -} (a q) C O_{2} - - \to C r_{2} O_{7}^{2 -}

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

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