Question

Name the type of following chemical reaction
$Pb{O_2} + S{O_2} \to PbS{O_4}$
A.Combination
B.Displacement
C.Double displacement
D.Neutralization

Answer 1

Hint:Redox reactions are chemical reactions where both reduction and oxidation take place. Oxidation is the addition of oxygen to a substance or removal of hydrogen from a substance. The reduction is the addition of hydrogen to a substance or removal of oxygen from a substance. There are mainly four types of redox reactions.

Complete step by step answer:
Now let’s discuss the Redox reactions.
In redox reactions both oxidation and reduction occur.
Oxidation is the addition of oxygen to a substance or removal of hydrogen from a substance.
As a result of oxidation, there is an increase in the oxidation number of the element in the substance.
The reduction is the addition of hydrogen to a substance or removal of oxygen from a substance.
As a result of the reduction, there is a decrease in the oxidation number of the element in the substance.
Therefore reactions that involve a change in the oxidation number of the reacting species are called redox reactions.
Now let’s consider the reaction given in the question.
$Pb{O_2} + S{O_2} \to PbS{O_4}$.
The above reaction is an example of a redox reaction.
We know when there is an increase in the oxidation state of the compound, it is said that the compound undergoes oxidation.
Here in the above chemical reaction sulfur in $S{O_2}$ undergo oxidation
In $S{O_2}$ sulfur is in $ + 4$ oxidation state but when it comes to the product side sulfur loses two electrons and thus in the product the oxidation state of sulfur is $ + 6$ oxidation state.
So here there is an increase in the oxidation number of the element sulfur.
Thus here oxidation takes place.
But when we come to $Pb{O_2}$ the oxidation state of lead is also in $ + 4$. In the product $PbS{O_4}$ oxidation state of $Pb$ is in $ + 2$. That is, $Pb$ in the $Pb{O_2}$ gains two electrons in $PbS{O_4}$. Thus here reduction occurs. There is a decrease in the oxidation number of $Pb$.
Therefore the reaction $Pb{O_2} + S{O_2} \to PbS{O_4}$ is an oxidation-reduction reaction or we can say it is a redox reaction since both reduction and oxidation takes place in the reaction.
Therefore $Pb{O_2}$ undergo reduction and $S{O_2}$ undergo oxidation.
There are mainly four types of redox reaction:
Combination reactions
Decomposition reactions
Displacement reactions
Disproportionation reactions
Here the reaction $Pb{O_2} + S{O_2} \to PbS{O_4}$ is a combination reaction.
Combination reactions are redox reactions where two or more simple substances combine to form a more complex product. The reactants are elements or compounds but the product is always a compound.
So $Pb{O_2} + S{O_2} \to PbS{O_4}$ is a combination reaction in which two simple compounds react to give a complex compound.
In displacement reactions, the atom in a compound is replaced by an atom of another element. Here in the reaction given in the question, no such replacements take place.
Therefore the answer is option (A) combination.



Note:
An increase in the oxidation number of the element indicates oxidation and a decrease in the oxidation number of the element indicates reduction. If an element gains electrons, there is a decrease in oxidation number and the process is reduction. If an element in a compound loses electrons, there is an increase in the oxidation number and it is called oxidation.