
The correct code for stability of oxidation states of $Sn$ and $Pb$ is :
[1] $P{b^{2 + }}\, > \,P{b^{4 + }}$ , $S{n^{2 + }}\, > \,S{n^{4 + }}$
[2] $P{b^{2 + }}\, < \,P{b^{4 + }}$ , $S{n^{2 + }}\, < \,S{n^{4 + }}$
[3] $P{b^{2 + }}\, > \,P{b^{4 + }}$, $S{n^{2 + }}\, < \,S{n^{4 + }}$
[4] $P{b^{2 + }}\, < \,P{b^{4 + }}$, $S{n^{2 + }}\, > \,S{n^{4 + }}$
[5] $S{n^{2 + }}\, < \,P{b^{2 + }}$ , $S{n^{4 + }}\,\, > \,\,P{b^{4 + }}$
[6] $S{n^{2 + }}\, < \,P{b^{2 + }}$ , $S{n^{4 + }}\,\, < \,\,P{b^{4 + }}$
A) [5] and [6]
B) [1], [3], [5], and [6]
C) [3] and [5]
D) [2] and [4]
Answer
548.7k+ views
Hint In order to answer this question, we must recall the concept of oxidation states and also we must remember the periodic properties of the groups in which the given elements lie. We will Compare the relative stabilities of the elements in their different types of oxidation stated by keeping in mind the rules which are made for concluding the correct order of stabilities in different oxidation states. By making the correct conclusion, we will select the correct options.
Complete step by step answer:
Step 1: In this step we will enlist the periodic properties of the group of carbon family:
Since we know that, on moving down the carbon family, the stability of +4 oxidation state decreases and the stability of +2 oxidation state increases.
$G{e^{2 + }}$ compounds are unstable and readily oxidised to $G{e^{4 + }}$ compounds.
$S{n^{2 + }}$ compounds are reducing agents.
$P{b^{4 + }}$ compounds act as oxidising agents and are easily reduced to $P{b^{2 + }}$ compounds.
Step 2: In this step, we will draw out the conclusion:
$P{b^{2 + }}$ oxidation state is more stable than $P{b^{4 + }}$ .
In a similar manner, $S{n^{4 + }}$ oxidation state is more stable than $S{n^{2 + }}$ .
In a similar manner, $P{b^{2 + }}$oxidation state is more stable than $S{n^{2 + }}$ .
In a similar manner, $S{n^{4 + }}$ oxidation state is more stable than $P{b^{4 + }}$ .
Therefore, option (C) is correct.
Note:Oxidation state indicates the degree of oxidation for an atom in a chemical compound; it is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states are typically represented by integers, which can be positive, negative, or zero.
Complete step by step answer:
Step 1: In this step we will enlist the periodic properties of the group of carbon family:
Since we know that, on moving down the carbon family, the stability of +4 oxidation state decreases and the stability of +2 oxidation state increases.
$G{e^{2 + }}$ compounds are unstable and readily oxidised to $G{e^{4 + }}$ compounds.
$S{n^{2 + }}$ compounds are reducing agents.
$P{b^{4 + }}$ compounds act as oxidising agents and are easily reduced to $P{b^{2 + }}$ compounds.
Step 2: In this step, we will draw out the conclusion:
$P{b^{2 + }}$ oxidation state is more stable than $P{b^{4 + }}$ .
In a similar manner, $S{n^{4 + }}$ oxidation state is more stable than $S{n^{2 + }}$ .
In a similar manner, $P{b^{2 + }}$oxidation state is more stable than $S{n^{2 + }}$ .
In a similar manner, $S{n^{4 + }}$ oxidation state is more stable than $P{b^{4 + }}$ .
Therefore, option (C) is correct.
Note:Oxidation state indicates the degree of oxidation for an atom in a chemical compound; it is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states are typically represented by integers, which can be positive, negative, or zero.
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