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Polar Molecules

HCl is prepar...

Question

$H C l$ is prepared by heating $N a C l$ with $c o n c . H_{2} S O_{4}$ . But $H I$ cannot be prepared by treating $N a I$ with $c o n c . H_{2} S O_{4}$ . Identify the reason.

H_{2} S O_{4}

is oxidised by

H I

and not oxidised by

H C l

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H I

is oxidised by

H_{2} S O_{4}

and

H C l

is not oxidised by

H I

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H I

dissociates on treatment with

H_{2} S O_{4}

and

H C l

does not dissociate

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Reaction of

H_{2} S O_{4}

with

H I

is reversible whereas with

H C l

is irreversible

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Solution

The correct option is B $H I$ is oxidised by $H_{2} S O_{4}$ and $H C l$ is not oxidised by $H I$

$H C l$ is prepared by heating $N a C l$ with $c o n c . H_{2} S O_{4}$ . But $H I$ cannot be prepared by treating $N a I$ with $c o n c . H_{2} S O_{4}$ . Because $H I$ is oxidised by $H_{2} S O_{4}$ and $H C l$ is not oxidised by $H I$ .
Hence, option $B$ is correct.

Suggest Corrections

Similar questions

H C l

is prepared by heating

N a C l

with conc.

H_{2} S O_{4}

but

H I

cannot be prepared by treating

N a I

with conc.

H_{2} S O_{4}

. Identify the reason.

Q. Assertion :Concentrated

H_{2} S O_{4}

reacts with

K C l

to give

C l_{2}

gas. Reason:

H C l

cannot be oxidised by concentrated

H_{2} S O_{4}

Q. Assertion :Concentrated

H_{2} S O_{4}

reacts with

K C l

to give

{C l}_{2}

gas Reason:

H C l

cannot be oxidised by concentrated

H_{2} S O_{4}

Q. Which of the following statements is/are correct?
Commercial

H C l

is prepared by heating

N a C l

with

H_{2} S O_{4}

as follows:

2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l

Q. In the redox titration using potassium permanganate, the acidification is usually done using

H_{2} S O_{4}

instead of

H C l

. Why?

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HCl is prepared by heating NaCl with conc.H2SO4. But HI cannot be prepared by treating NaI with conc.H2SO4. Identify the reason.

The correct option is B HI is oxidised by H2SO4 and HCl is not oxidised by HI HCl is prepared by heating NaCl with conc.H2SO4. But HI cannot be prepared by treating NaI with conc.H2SO4. Because HI is oxidised by H2SO4 and HCl is not oxidised by HI.Hence, option B is correct.

$H C l$ is prepared by heating $N a C l$ with $c o n c . H_{2} S O_{4}$ . But $H I$ cannot be prepared by treating $N a I$ with $c o n c . H_{2} S O_{4}$ . Identify the reason.