Answer
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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
>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.
Similarly, the oxidation state of Oxygen is -2.
\[\begin{align}
& 2\times (1)+2x+8\times (-2)=0 \\
& 2x-14=0 \\
& x=+7 \\
\end{align}\]
>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
The oxidation state of 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).
Additional Information:
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.
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.
Complete answer
>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.
Similarly, the oxidation state of Oxygen is -2.
\[\begin{align}
& 2\times (1)+2x+8\times (-2)=0 \\
& 2x-14=0 \\
& x=+7 \\
\end{align}\]
>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
The oxidation state of 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).
Additional Information:
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.
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.
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