Sulphuryl chloride $(S O_{2} C l_{2})$ reacts with white phosphorus $(P_{4})$ to give:

Name: Sulphuryl chloride (SO2Cl2) reacts with white phosphorus (P4) to give:
Uploaded: 2020-01-06
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Question

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Answer 1

Sulphuryl chloride ( $S O_{2} C l_{2}$ ) reacts with white phosphorus ( $P_{4}$ ) to give phosphorous pentachloride and sulphur dioxide.

$10 S O_{2} C l_{2} + P_{4} \to 4 P C l_{5} + 10 S O_{2}$

Hence, option A is correct.

Answer 2

Sulphuryl chloride (

S O_{2} C l_{2}

) reacts with white phosphorus (

P_{4}

) to give phosphorous pentachloride and sulphur dioxide.

10 S O_{2} C l_{2} + P_{4} \to 4 P C l_{5} + 10 S O_{2}

Hence, option A is correct.

Sulphuryl chloride $(S O_{2} C l_{2})$ reacts with white phosphorus $(P_{4})$ to give:

$P C l_{5}, S O_{2}$

$O P C l_{3}, S O C l_{2}$

$P C l_{5}, S O_{2}, S_{2} C l_{2}$

$O P C l_{3}, S O_{2}, S_{2} C l_{2}$

Sulphuryl chloride ( $S O_{2} C l_{2}$ ) reacts with white phosphorus ( $P_{4}$ ) to give phosphorous pentachloride and sulphur dioxide.

$10 S O_{2} C l_{2} + P_{4} \to 4 P C l_{5} + 10 S O_{2}$

Hence, option A is correct.

Hydrogen sulphide is passed through acidified ferric chloride solution.

$S n C l_{2}$ gives white precipitate with $H g C l_{2}$ which later on turns grey. Explain why?

Iodine and hypo react to produce :

A black sulphide when treated with ozone becomes white. The white compound is :

When $H_{2} S$ gas is passed into $S O_{2}$ in aqueous solution, colloidal yellowish white turbidity appears.

Calculate the weight of this turbidity if $11.2 L$ of each gas reacts at $N T P$ .

Revise with Concepts

Phosphorus and its Allotropes

Sulphuryl chloride (SO2​Cl2​) reacts with white phosphorus (P4​) to give:

PCl5​,SO2​

OPCl3​,SOCl2​

PCl5​,SO2​,S2​Cl2​

OPCl3​,SO2​,S2​Cl2​

Sulphuryl chloride (SO2​Cl2​) reacts with white phosphorus (P4​) to give phosphorous pentachloride and sulphur dioxide. 10SO2​Cl2​+P4​→4PCl5​+10SO2​ Hence, option A is correct.

Hydrogen sulphide is passed through acidified ferric chloride solution.

SnCl2​ gives white precipitate with HgCl2​ which later on turns grey. Explain why?

Iodine and hypo react to produce :

A black sulphide when treated with ozone becomes white. The white compound is :

When H2​S gas is passed into SO2​ in aqueous solution, colloidal yellowish white turbidity appears. Calculate the weight of this turbidity if 11.2 L of each gas reacts at NTP.

Revise with Concepts

Phosphorus and its Allotropes

Sulphuryl chloride $(S O_{2} C l_{2})$ reacts with white phosphorus $(P_{4})$ to give:

$P C l_{5}, S O_{2}$

$O P C l_{3}, S O C l_{2}$

$P C l_{5}, S O_{2}, S_{2} C l_{2}$

$O P C l_{3}, S O_{2}, S_{2} C l_{2}$

Sulphuryl chloride ( $S O_{2} C l_{2}$ ) reacts with white phosphorus ( $P_{4}$ ) to give phosphorous pentachloride and sulphur dioxide.

$10 S O_{2} C l_{2} + P_{4} \to 4 P C l_{5} + 10 S O_{2}$

Hence, option A is correct.

$S n C l_{2}$ gives white precipitate with $H g C l_{2}$ which later on turns grey. Explain why?

When $H_{2} S$ gas is passed into $S O_{2}$ in aqueous solution, colloidal yellowish white turbidity appears.

Calculate the weight of this turbidity if $11.2 L$ of each gas reacts at $N T P$ .