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Oxidation state of S in \[{H_2}{S_2}{O_8}\] is:A. 6B. 7C. +8D. 0

Last updated date: 23rd May 2024
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Hint:The oxidation state is also termed as oxidation number. \${H_2}{S_2}{O_8}\$ is a neutral compound, so the sum of all the oxidation states of atoms present in it will be equal to zero.

Complete step by step solution:

The oxidation state is defined as the overall electrons lost or gain by an atom in order to form a chemical bond with the other atom resulting in the formation of the molecule.

The given formula is \${H_2}{S_2}{O_8}\$. The chemical formula of \${H_2}{S_2}{O_8}\$ is peroxydisulfuric acid. It is an inorganic colourless solid and the most powerful peroxyacid oxidant.

The structure of peroxydisulphuric acid is shown below.

Peroxydisulfuric acid

Here, the two groups are attached by two oxygen atoms known as peroxy linkage (\$O - O\$ ). The compounds which possess peroxy linkage are known as peroxy compounds.

As we need to find the oxidation state of sulphur, let's denote it to “x”.

For any neutral atom, the sum total of the oxidation states of the atoms involved is equal to zero.

The oxidation state of oxygen involved in peroxy linkage is \$ - 1\$ and all the other oxygen has \$ - 2\$ oxidation state. The hydrogen has \$ + 1\$ oxidation state.

The oxidation state of sulphur is calculated as shown below.

\$ \Rightarrow 2( + 1) + 2(x) + 2( - 1) + 6( - 2) = 0\$

\$ \Rightarrow 2 + 2x - 2 - 12 = 0\$

\$ \Rightarrow 2 + 2x - 14 = 0\$

\$ \Rightarrow 2x - 12 = 0\$

\$ \Rightarrow 2x = 12\$

\$ \Rightarrow x =  + 6\$

Therefore, the oxidation state of sulphur is \$ + 6\$ . Hence, option (A) is correct.

Note: The oxidation state of oxygen is usually \$ - 2\$ in all the compounds but when oxygen is present as peroxy linkage then the oxidation state of oxygen is \$ - 1\$. Also, the oxidation state of hydrogen is \$ + 1\$ but when it is attached to a less electronegative element, it is in \$ - 1\$ oxidation state.