
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
579.9k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

