Lead Sulphide in dilute nitric acid gives:
(This question has multiple correct options)
A. Lead nitrate
B. Sulphur trioxide
C. Sulphur dioxide
D. Colloidal sulphur
Answer
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Hint: Nitric acid is a strong oxidising agent. The reactions of nitric acid are different in strong and dilute forms. Lead sulphide is salt and dilute nitric acid will act as oxidising agent. Thus sulphur is oxidised and lead will form another salt by exchange of ions with nitric acid.
Thus, 2 products out of mentioned products will be the answer for the above reaction of dilute nitric acid with Lead sulphide.
Complete step by step answer:
The nitric acid as mentioned acts differently in dilute state and concentrated state. In concentrated form, it acts as strong oxidising agent, oxidising all ions in reaction, but dilute nitric acid will act as oxidising agent and also as normal acid reaction, thus oxidising Sulphide ion of salt and reacting as normal acid with lead and forming another salt.
The reaction can be given below:
\[3PbS + 8HN{O_3} \to 3Pb{(N{O_3})_2} + 3S + 2NO + 4{H_2}O\]
Here, 3 moles Lead nitrate is produced along with 3 moles of colloidal sulphur, 2 moles of Nitrogen oxide and 4 moles of water.
So, the correct answer is Option A,D.
Additional Information:
The oxidation and reduction can be known from oxidation states of substances. Here, if oxidation state is increasing, then substances are undergoing oxidation, while if oxidation state is decreasing, then substance is undergoing reduction.
One of the major reasons for nitric acid to be a strong oxidising agent is that, nitrogen is in highest oxidation state of +5, so it cannot undergo further increase in oxidation state, so it will reduce itself, and if substance reduces itself then it is good oxidising agent.
Note: Sulphide ion of lead sulphide has oxidation state of -2, but dilute nitric acid will oxidise it to free colloidal sulphur, with oxidation state of 0. Remember, Lead ion does not undergo any oxidation or reduction, rather Nitrogen from nitrate ion undergoes reduction from -5 oxidation state to -2 oxidation state.
Thus, 2 products out of mentioned products will be the answer for the above reaction of dilute nitric acid with Lead sulphide.
Complete step by step answer:
The nitric acid as mentioned acts differently in dilute state and concentrated state. In concentrated form, it acts as strong oxidising agent, oxidising all ions in reaction, but dilute nitric acid will act as oxidising agent and also as normal acid reaction, thus oxidising Sulphide ion of salt and reacting as normal acid with lead and forming another salt.
The reaction can be given below:
\[3PbS + 8HN{O_3} \to 3Pb{(N{O_3})_2} + 3S + 2NO + 4{H_2}O\]
Here, 3 moles Lead nitrate is produced along with 3 moles of colloidal sulphur, 2 moles of Nitrogen oxide and 4 moles of water.
So, the correct answer is Option A,D.
Additional Information:
The oxidation and reduction can be known from oxidation states of substances. Here, if oxidation state is increasing, then substances are undergoing oxidation, while if oxidation state is decreasing, then substance is undergoing reduction.
One of the major reasons for nitric acid to be a strong oxidising agent is that, nitrogen is in highest oxidation state of +5, so it cannot undergo further increase in oxidation state, so it will reduce itself, and if substance reduces itself then it is good oxidising agent.
Note: Sulphide ion of lead sulphide has oxidation state of -2, but dilute nitric acid will oxidise it to free colloidal sulphur, with oxidation state of 0. Remember, Lead ion does not undergo any oxidation or reduction, rather Nitrogen from nitrate ion undergoes reduction from -5 oxidation state to -2 oxidation state.
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