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Anomalous Behaviour of Fluorine

In the follow...

Question

In the following reaction,
$X e F_{2} + B r O_{3}^{-} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F$ :

X e F_{2}

is an oxidising agent

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X e F_{2}

is an reducing agent

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B r O_{3}^{-}

is oxidised to

B r O_{4}^{-}

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X e F_{2}

is reduced to HF

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Solution

The correct options are
A $X e F_{2}$ is an oxidising agent
C $B r O_{3}^{-}$ is oxidised to $B r O_{4}^{-}$
$X e F_{2} + B r O_{3}^{-} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F$
+2 +5 +7 0
as in this reaction $X e F_{2}$ gains electron so $X e F_{2}$ is an oxidising agent and reduced.
Here, Br looses electron so $B r O_{3}^{-}$ is oxidised to $B r O_{4}^{-}$ .

Suggest Corrections

Similar questions

B r O_{3}^{-}

B r O_{4}^{-}

X e F_{2}

X e F_{2} m_{1} + B r O_{3}^{-} m_{2} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

m_{1}

m_{2}

X e F_{2}

B r O_{3}^{-}

X e F_{2}

B r O_{3}^{-}

X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F

e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}

In $X e F_{2},$ Xe is ................. hybridised

M g + 2 H^{+} \to M g^{2 +} + H_{2}

M g

H^{+}

M g

H^{+}

Question 10

Hydrogen peroxide is .......

(a) an oxidising agent

(b) a reducing agent

(d) neither oxidising nor reducing agent

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