
Which of the following is an example of comproportionation reaction?
(This question has multiple correct options)
A. \[Pb + Pb{O_2} + 2{H_2}S{O_4} \to PbS{O_4} + 2{H_2}O\]
B. \[15Se + SeC{l_4} + 4AlC{l_3} \to 2S{e_8}{[AlC{l_4}]_2}\]
C. \[2{H_2}S + S{O_2} \to 2S + 2{H_2}O\]
D. \[2A{g^{2 + }} + Ag \to 2A{g^ + }\]
Answer
486k+ views
Hint: To solve this question, we need to first check if the reactants have one common element in them. If they do, we need to check the oxidation states of the common element and compare it to the conditions of a comproportionation reaction.
Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Comproportionation reactions are those chemical reactions which involve 2 reactants containing the same element, but in different oxidation states, to form a product involving the same element which has a different oxidation state. Comproportionation reactions are also known as disproportionation reactions.
Let us now discuss the reactions given to us:
1. \[Pb + Pb{O_2} + 2{H_2}S{O_4} \to PbS{O_4} + 2{H_2}O\]
In the reactants, the common element present is Pb. The oxidation state of lead in Pb is Zero while in \[Pb{O_2}\] is +4. In the products, we can see that lead has formed a compound with sulphate ion and has an oxidation state of +2.
2. \[15Se + SeC{l_4} + 4AlC{l_3} \to 2S{e_8}{[AlC{l_4}]_2}\]
In the reactants, the common element present is selenium. The oxidation state of selenium in Se is zero while in \[SeC{l_4}\] is +4. In the products, Se has formed the compound \[S{e_8}{[AlC{l_4}]_2}\] where its oxidation state is +2.
3. \[2{H_2}S + S{O_2} \to 2S + 2{H_2}O\]
In the reactants, the common element present is sulphur. The oxidation state of sulphur in hydrogen sulphide is (-2) while in sulphur dioxide it is (+4). In the products, the oxidation state of S is zero.
4. \[2A{g^{2 + }} + Ag \to 2A{g^ + }\]
In the reactants, the common element present is silver. The oxidation of silver in \[A{g^{2 + }}\] is +2 while in Ag is zero. In the products, the oxidation state of \[A{g^ + }\] is (+1)
Hence, we can see that in all the reactions above, all of them satisfy the conditions for being called as a comproportionation reaction.
Hence, Options A, B, C and D are the correct options
Note: Comproportionation reactions cannot be carried out with the elements that exhibit only a single oxidation state. These elements include metals like sodium, caesium, magnesium, potassium, calcium, etc.
Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Comproportionation reactions are those chemical reactions which involve 2 reactants containing the same element, but in different oxidation states, to form a product involving the same element which has a different oxidation state. Comproportionation reactions are also known as disproportionation reactions.
Let us now discuss the reactions given to us:
1. \[Pb + Pb{O_2} + 2{H_2}S{O_4} \to PbS{O_4} + 2{H_2}O\]
In the reactants, the common element present is Pb. The oxidation state of lead in Pb is Zero while in \[Pb{O_2}\] is +4. In the products, we can see that lead has formed a compound with sulphate ion and has an oxidation state of +2.
2. \[15Se + SeC{l_4} + 4AlC{l_3} \to 2S{e_8}{[AlC{l_4}]_2}\]
In the reactants, the common element present is selenium. The oxidation state of selenium in Se is zero while in \[SeC{l_4}\] is +4. In the products, Se has formed the compound \[S{e_8}{[AlC{l_4}]_2}\] where its oxidation state is +2.
3. \[2{H_2}S + S{O_2} \to 2S + 2{H_2}O\]
In the reactants, the common element present is sulphur. The oxidation state of sulphur in hydrogen sulphide is (-2) while in sulphur dioxide it is (+4). In the products, the oxidation state of S is zero.
4. \[2A{g^{2 + }} + Ag \to 2A{g^ + }\]
In the reactants, the common element present is silver. The oxidation of silver in \[A{g^{2 + }}\] is +2 while in Ag is zero. In the products, the oxidation state of \[A{g^ + }\] is (+1)
Hence, we can see that in all the reactions above, all of them satisfy the conditions for being called as a comproportionation reaction.
Hence, Options A, B, C and D are the correct options
Note: Comproportionation reactions cannot be carried out with the elements that exhibit only a single oxidation state. These elements include metals like sodium, caesium, magnesium, potassium, calcium, etc.
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