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Preparation of Carboxylic Acids from Hydrocarbons

When Br2 is...

Question

When $B r_{2}$ is treated with aqueous solutions of $N a F, N a C l, N a I$ separately:

F_{2}, C l_{2}

and

I_{2}

are liberated

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only

F_{2}

and

C l_{2}

are liberated

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only

I_{2}

is liberated

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only

C l_{2}

is liberated

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Solution

The correct option is C only $I_{2}$ is liberated
Within a group, the element above can displace the element below in a redox reaction.

When bromine gas is passed through solutions of $N a F, N a C l, N a I$ , bromine cannot displace fluorine and chlorine compound but it can displace iodine compound

$2 N a I + {B r}_{2} ⟶ 2 N a B r + I_{2}$

Iodine gas is released.

Hence, option $C$ is correct.

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Similar questions

B r_{2}

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

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} :

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} :

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

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

x

y

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

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

F_{2}, C l_{2}, B r_{2}, I_{2}

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

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