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Question

A large number of oxyacids of Sulphur are known. Those oxyacids with S-S linkages are in general referred to as __ acids.

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Solution

Oxyacids of Sulphur which have a Sulphur-Sulphur single bond are called thionic acids. For example, dithionic acid has the structure:

According to the VSEPR theory, what will be the hybridization and the oxidation state for the two Sulphur atoms?

Before we answer that, let us try to learn the nomenclature behind the thionic acids series.

Let us look at the following general structure structure:

Here x is any non-negative integer. If x = 0, then we will get the previous dithionic acid formula. If x = 1, total number of atoms in S – S chain is 3. Then the name of the compound for x = 1 becomes trithionic acid. In case x = 4, then there will be six atoms connected by the S – S chain. Thus, for x = 4, the name of the compound becomes hexathionic acid.

Let us take a look at the terminal Sulphur atoms. They (each atom) are connected to two different oxygen atoms via a double bonds. Also they are connected to an oxygen atom by a single bond (in other words, each terminal S atom shares an electron with a hydroxyl radical). Further, each terminal S atom is singly bonded to exactly one other Sulphur atom. We can see there are 4 ρ-bonds, two π-bonds and 0 lone pairs. So according to the VSEPR theory, the terminal Sulphur atoms are sp3 hybridized and will show tetrahedral geometry. Oxidation state of the terminal Sulphur atoms will be +5.

For x>0, each of the central non-terminal bridge S atoms will exhibit an oxidation state of 0. These S atoms will also have a tetrahedral geometry but bent shape (since each Sulphur bridge atom has two lone pairs of electrons).


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