
The pair of compounds that can exist together is:
(A) $FeC{l_3},SnC{l_2}$
(B) $HgC{l_2},SnC{l_2}$
(C) $FeC{l_2},SnC{l_2}$
(D) $FeC{l_3},KI$
Answer
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Hint: Find which salt is oxidizing agent and which one is reducing agent. Both oxidizing agent and reducing agent cannot remain together as there will be a redox reaction between them.
Complete answer:
Here, we are given the pairs of compounds. We need to find out the pair that can exist together. That means there should be no reaction between them when in contact. Let’s find out which pair can exist together.
- We can see that all the metals present in the compounds can exist in more than one oxidation state and this makes them undergo redox reactions.
- So, in redox reactions, an oxidising agent and reducing agent react together to give a reaction. So, a pair that involves both these agents cannot exist together.
A) In this pair Fe in $FeC{l_3}$ is in +3 oxidation state. So, this is an oxidizing agent. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation. So, they both cannot exist together as there will be a reaction.
B) $HgC{l_2}$ is an oxidizing agent because Hg is in +2 oxidation state. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation. So, they both cannot exist together as there will be a reaction.
C) $FeC{l_2}$ is a reducing agent as Fe is in +2 oxidation state and it can undergo oxidation. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation only. Thus, there will be no reaction if they are together.
D) In this pair Fe in $FeC{l_3}$ is in +3 oxidation state. So, this is an oxidizing agent. KI is a reducing agent as iodine ion can undergo oxidation. So, there will be a reaction.
Therefore, the correct answer is (C).
Note:
Do not get confused between oxidising agent and reducing agent. Oxidising agent oxidizes other compounds and itself undergoes reduction. Reducing agent reduces other compounds and itself undergoes oxidation.
Complete answer:
Here, we are given the pairs of compounds. We need to find out the pair that can exist together. That means there should be no reaction between them when in contact. Let’s find out which pair can exist together.
- We can see that all the metals present in the compounds can exist in more than one oxidation state and this makes them undergo redox reactions.
- So, in redox reactions, an oxidising agent and reducing agent react together to give a reaction. So, a pair that involves both these agents cannot exist together.
A) In this pair Fe in $FeC{l_3}$ is in +3 oxidation state. So, this is an oxidizing agent. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation. So, they both cannot exist together as there will be a reaction.
B) $HgC{l_2}$ is an oxidizing agent because Hg is in +2 oxidation state. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation. So, they both cannot exist together as there will be a reaction.
C) $FeC{l_2}$ is a reducing agent as Fe is in +2 oxidation state and it can undergo oxidation. $SnC{l_2}$ is a reducing agent because Sn is in +2 oxidation state and it can undergo oxidation only. Thus, there will be no reaction if they are together.
D) In this pair Fe in $FeC{l_3}$ is in +3 oxidation state. So, this is an oxidizing agent. KI is a reducing agent as iodine ion can undergo oxidation. So, there will be a reaction.
Therefore, the correct answer is (C).
Note:
Do not get confused between oxidising agent and reducing agent. Oxidising agent oxidizes other compounds and itself undergoes reduction. Reducing agent reduces other compounds and itself undergoes oxidation.
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