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−
B.S2O42−
C.S2O52−
D.S2O72−
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
⇒x=2
So, the oxidation state of Sulphur in S2O32− 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
⇒x=3
So, the oxidation state of Sulphur in S2O42− 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
⇒x=4
So, the oxidation state of Sulphur in S2O52− 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
⇒x=6
So, the oxidation state of Sulphur in S2O72− 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.