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Variable Oxidation States

A on reaction...

Question

A on reaction with dil. $H_{2} {S O}_{4}$ gives a colourless pungent gas that can turn ${C r}_{2} O_{7}^{2 -} / H^{+}$ green. Green colour is due to formation of :

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

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{C r}^{3 +}

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{C r O}_{3}

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{C r O}_{2} {C l}_{2}

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Solution

The correct option is B ${C r}^{3 +}$
Pungent gas is ${S O}_{2}$ .
It reduces potassium dichromate to green coloured chromium sulphate $(C r_{2} (S O_{4})_{3})$
$K_{2} {C r}_{2} O_{7} + 2 H^{+} + 3 S O_{2} \to (C r_{2} (S O_{4})_{3}) (g r e e n) + 5 H_{2} O + 2 K^{+}$ . Thus, the green colour is due to formation of Cr(III) ion.

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Similar questions

H_{2} O_{2} (a q)

{C r}_{2} O_{7}^{2 -} (a q)

{C r}^{3 +} (a q)

[X] + H_{2} {S O}_{4} ⟶ [Y]

[Y] + K_{2} {C r}_{2} O_{7} + H_{2} {S O}_{4} ⟶

{S O}_{2}

K_{2} {C r}_{2} O_{7}

{C O}_{2}

H_{2} O

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

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

{C O}_{2}

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

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

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

H_{2} S O_{4}

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