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Binomial Coefficients

By H 2 O 2 ...

Question

By $H_{2} O_{2} / {O H}^{-}, {C r (O H)}_{3}$ changes to:

{C r O}_{3}

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{C r}_{2} O_{7}^{2 -}

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{C r O}_{4}^{2 -}

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[{C r (O H)}_{4}]^{-}

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Solution

The correct option is C ${C r O}_{4}^{2 -}$
${C r (O H)}_{3}$ is soluble in $N a O H + H_{2} O_{2}$ forming sodium chromate.

Hence, by $H_{2} O_{2} / {O H}^{-}, {C r (O H)}_{3}$ changes to ${C r} O_{4}^{2 -}$ .

$2 C r (O H)_{4}^{-} + 3 H_{2} O_{2} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + 8 H_{2} O$

Option C is correct.

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Similar questions

{C O}_{2}

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{C r O}_{4}^{2 -}

{C r}_{2} O_{7}^{2 -}

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(H_{2} {C O}_{3} ⇌ H^{+} + {H C O}_{3}^{-})

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{C r}_{2} O_{7}^{2 -}

{C r}^{3 +}

{C r}_{2} O_{7}^{2 -} = x \times

x

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