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Question

Question: There is a S-S linkage in: A.\[{{S}_{2}}{{O}_{3}}^{2-}\] B.\[{{S}_{2}}{{O}_{4}}^{2-}\] C.\[{{S...

There is a S-S linkage in:
A.S2O32{{S}_{2}}{{O}_{3}}^{2-}
B.S2O42{{S}_{2}}{{O}_{4}}^{2-}
C.S2O52{{S}_{2}}{{O}_{5}}^{2-}
D.S2O72{{S}_{2}}{{O}_{7}}^{2-}

Explanation

Solution

We know the rule for a molecule to contain S – S linkage in it. So, find out the oxidation state of sulphur in each of the molecules given in the question and find out which molecules have S- S linkage.

Complete step by step answer:
If the oxidation state of sulphate is less than 6, then direct linkage is present. By direct linkage we mean the S – S linkage nothing but the sulphate to sulphate linkage.
If the oxidation state of sulphate is equal to 6, then we can see the existence of S – O – S linkage. It is nothing but the linkage of Sulphate to oxygen and oxygen to sulphate. ( Two sulphates and one oxygen )
If the oxidation state of sulphate is greater than 6, then peroxy linkage is present. By per-oxy linkage we mean sulphate to oxygen, oxygen to oxygen and oxygen to sulphate.( Two sulphates and two oxygens ) ( S – O – O – S linkage is present)
So for getting the S – S linkage we must have the oxidation state of sulphate presents in the molecule to be less than 6.
Let us find out the oxidation state of Sulphur in option 1
2x+3(2)=2\Rightarrow 2x+3(-2)=-2
x=2\Rightarrow x=2
So, the oxidation state of Sulphur in S2O32{{S}_{2}}{{O}_{3}}^{2-} is 2. So, it has a S – S bond in it.
Let us find out the oxidation state of Sulphur in option 2
2x+4(2)=2\Rightarrow 2x+4(-2)=-2
x=3\Rightarrow x=3
So, the oxidation state of Sulphur in S2O42{{S}_{2}}{{O}_{4}}^{2-} is 3. So, it has S – S bond in it.
Let us find out the oxidation state of Sulphur in option 3
2x+5(2)=2\Rightarrow 2x+5(-2)=-2
x=4\Rightarrow x=4
So, the oxidation state of Sulphur in S2O52{{S}_{2}}{{O}_{5}}^{2-} is 4. So, it has a S – S bond in it.
Let us find out the oxidation state of Sulphur in option 4
2x+7(2)=2\Rightarrow 2x+7(-2)=-2
x=6\Rightarrow x=6
So, the oxidation state of Sulphur in S2O72{{S}_{2}}{{O}_{7}}^{2-} is 6. So, it does not have a S – S bond in it.

So, Option A, B and C are correct.

Note: We already know the oxidation state of oxygen is always –2, that is why we multiply the number of oxygen atoms to –2. There may be a possibility for getting more than one answer in any question. So, we have to check each and every option.