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Oxides of Xenon

XeF6 P+other ...

Question

$X e F_{6} C o m p l e t e H y d r o l y s i s - ------------ \to P + other product$
$↓ O H^{-} / H_{2} O$
$Q$
$↓ slow disproportionation in O H^{-} / H_{2} O$
Products

Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is:

0

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2

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3

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Solution

The correct option is C $2$
$X e F_{6} + 3 H_{2} O ⟶ X e O_{3} + 3 H_{2} F_{2}$
$↓ O H^{-}$
$H X e O_{4}^{-}$
$↓ O H^{-} / H_{2} O$ (disproportionation)
$X e O_{6 (s)}^{4 -} + X e_{(g)} + H_{2} O_{(t)} + O_{2 (g)}$

Suggest Corrections

Similar questions

2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)

R a t e = k [H_{2}] [N O]^{2}

2 N O (g) ⇌ N_{2} O_{2} (g)

N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O

N_{2} O + H_{2} \to H_{2} O + N_{2}

1^{\circ}

1) E t - C O O H i . B H_{3} - ------ \to i i . H^{+} / H_{2} O

2) (C H_{3})_{2} C H - C O O H N a B H_{4} - --- \to

3) E t - C O O C H_{3} H^{+} / H_{2} O - ---- \to

4) (C H_{3})_{2} C = C H_{2} i . B H_{3}, T H F - -------- \to i i . H_{2} O_{2}, O H^{-}

5) E t - C O C l H_{2} - ------ \to P d - B a S O_{4}

6) (C H_{3})_{2} C H - O - E t H^{+} / H_{2} O - ---- \to

N a H C O_{3} + H C l \to

N a O H + Z n \to

N a C l + H_{2} O \to

M n O_{2} + H C l \to

C H_{4} + O_{2} \to

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