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Reaction with Metals

Which one wil...

Question

Which one will liberate $B r_{2}$ from KBr ?

A

I_{2}

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B

S O_{2}

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C

H I

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D

C l_{2}

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Solution

The correct option is B $C l_{2}$
$2 K B r + C l_{2} \to 2 K C l + B r_{2}$
The reaction is based on the fact that $C l_{2}$ has greater oxidizing power than $B r_{2}$ and therefore can oxidize Br from KBr.

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Similar questions

The correct order of the oxidizing power of Cl₂, Br₂, I₂ and O₂ is:

{C l}_{2}

{B r}_{2}

K I

I_{2}

{C C l}_{4}

{C l}_{2}

K B r

{B r}_{2}

(orange-yellow in

{C C l}_{4}

{C l}_{2}

would change mixture of

K I a n d K B r

into violet layer in

{C C l}_{4}

. Reason: Oxidising power is in order

I_{2} < {B r}_{2} < {C l}_{2} < F_{2}

.

Q. Match the following

	Column-1		Column-2
A	Acidic reactions	p	${C l}_{2} + 2 K B r \to 2 K C l + {B r}_{2}$
B	Displacement reactions	q	$C a + {C l}_{2} \to {C a C l}_{2}$
C	Oxidising properties	r	${C l}_{2} + H_{2} O \to H O C l + H C l$
D	Addition reactions	s	$2 K I \to {C l}_{2} + 2 K C l + I_{2}$
E	Reaction with metals	t	${S O}_{2} + {C l}_{2} \to {S O}_{2} {C l}_{2}$

Q. Consider the following orders –
(1)

F_{2} > C l_{2} > B r_{2} > I_{2} :

boiling point
(2)

F_{2} > C l_{2} > B r_{2} > I_{2} :

oxidizing nature
(3)

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

Acidic nature
(8)

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

Oxidizing nature
Then calculate

(x^{2} + y^{2})

x

is correct order and

y

is incorrect order.

B r_{2}

is treated with aq soln of

N a F, N a C l & N a I

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