Animated Solution for Chemistry - d and f-Block Elements: In mild alkaline medium, thiosulphate ion is oxidised by ${\text{MnO}}_4^{-}$ to "A". The oxidation state of sulphur in "A" is ......... .

Imagine you are standing on a microscopic battlefield where electrons are the ultimate currency. On one side, we have the thiosulphate ion (${\text{S}}_2{\text{O}}_3^{2-}$), a species eager to give away its electrons. On the other side stands the mighty permanganate ion (${\text{MnO}}_4^{-}$), one of the most powerful and notorious oxidising agents in all of chemistry.

But in redox chemistry, the environment dictates the rules of engagement. The problem explicitly states that this reaction takes place in a mild alkaline medium. This is the critical piece of intelligence that determines the fate of our reactants.

When permanganate operates in a mild alkaline or neutral medium, it doesn't go all the way down to ${\text{Mn}}^{2+}$ like it does in acidic conditions. Instead, it accepts exactly three electrons and precipitates out as the brown solid, manganese dioxide (${\text{MnO}}_2$).

Simultaneously, the thiosulphate ion is subjected to a brutal oxidation. The permanganate strips it of its electrons, forcing the sulphur atoms to their highest stable oxidation state. The thiosulphate is completely transformed into the sulphate ion (${\text{SO}}_4^{2-}$).

Therefore, our mystery product "A" is the sulphate ion.

Now that we have unmasked Product A as ${\text{SO}}_4^{2-}$, our mission is to find the oxidation state of its central sulphur atom.

To do this, we rely on a fundamental law of redox chemistry: The sum of the oxidation states of all atoms in a polyatomic ion must perfectly equal the net charge of that ion.

Let's break down the sulphate ion. It consists of one central sulphur atom and four surrounding oxygen atoms. The entire structure carries a net charge of $-2$.

We know that oxygen, being highly electronegative, almost always demands an oxidation state of $-2$ (except in rare cases like peroxides). Let's assign the unknown oxidation state of sulphur as the variable $x$.

Setting up our algebraic equation based on our fundamental law:

Now, we simply execute the math. Four oxygen atoms, each contributing $-2$, gives us a total of $-8$.

Moving the $-8$ across the equals sign:

And there is our final, elegant answer. The oxidation state of sulphur in the sulphate ion is +6. This makes perfect chemical sense, as sulphur is in Group 16 and has exactly 6 valence electrons to lose!