Question

Which one of the following statements is not correct?
a) Oxidation number of S in \[{{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}\] is +6
b) Oxidation number of Os in \[Os{{O}_{4}}\] is +8
c) Oxidation number of S in \[{{H}_{2}}S{{O}_{5}}\] is +8
d) Oxidation number of O in \[K{{O}_{2}}\] is -1/2

Answer 1

Hint: An oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element.


Complete answer: The oxidation number for mono-atomic ions always has the same value as the net charge corresponding to the ion.
>The hydrogen atom exhibits an oxidation state of +1 but when bonded with an element with less electronegativity than itself it exhibits an oxidation number of -1.
>Oxygen has an oxidation of -2 in most of its compounds. However, in the case of peroxides, the oxidation number corresponding to oxygen is -1.
>Group 1 elements have an oxidation state of +1 in their compounds.
>All group 2 elements exhibit an oxidation state of +2 in their compounds.
>In the compounds made up of two elements, a group 17 element has an oxidation number of -1 assigned to them.
>In the case of neutral compounds, the sum of all the oxidation numbers of the constituent atoms totals to zero and When polyatomic ions are considered, the sum of all the oxidation numbers of the atoms that constitute them equals the net charge of the polyatomic ion.
>Now, in the given compound \[{{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}\],
Let's take the oxidation state of sulphur in the given compound as x.
We know that the oxidation state of ammonium ion (\[N{{H}_{4}}\]) is +1.
\[\begin{align}
  & 2\times (1)+2x+8\times (-2)=0 \\
 & 2x-14=0 \\
 & x=+7 \\
\end{align}\]
Similarly, in \[Os{{O}_{4}}\], the oxidation state of Oxygen is -2.
Let's take the oxidation state of osmium as x.
\[\begin{align}
  & x+4\times (-2)=0 \\
 & 2x=-1 \\
 & x=-\frac{1}{2} \\
\end{align}\]
In \[{{H}_{2}}S{{O}_{5}}\], the oxidation state of oxygen is -2 and hydrogen is +1.
Let's take the oxidation state of sulphur as x.
\[\begin{align}
  & 1\times (2)+x+5\times (-2)=0 \\
 & 2+x-10=0 \\
 & x=+8 \\
\end{align}\]
Now, in \[K{{O}_{2}}\], the oxidation state of potassium is +1.
Let's take the oxidation state of Oxygen as x
\[\begin{align}
  & 1+2x=0 \\
 & 2x=-1 \\
 & x=-\frac{1}{2} \\
\end{align}\]


Therefore, from the above statements we can say that the correct option is (a).

Note: When you have to find the oxidation number of the ligands, if it is a molecule/polyatomic ion, for example, ammine (\[N{{H}_{3}}\]) is a neutral ligand so charge on it will be zero, use the standard overall charge of the species. If it is one atom, use the oxidation state rules to look up for its oxidation number.