
What is the oxidation number of $O$ in the ion $O{H^ - }$?
Answer
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Hint: When an atom loses or gains an electron in order to form a chemical bond, then the charge acquired by the atom is known as its oxidation number. It is sometimes referred to as oxidation state. The atoms may have more than one oxidation state depending upon the type of atom they are bonded with.
Complete answer:
Some basic rules to assign oxidation number to an element are as follows:
1. The oxidation number of a free element is always considered as zero.
2. The oxidation number of a mono-atomic ion is always equal to the charge present on the ion.
3. For a given molecule $A - B$, the cleavage of a bond takes place in such a way that negative charge is assigned to more electronegative element as ${A^ + }$ and ${B^ - }$, where B is more electronegative than A.
4. The oxidation number of hydrogen atom is always $ + 1$ with an exception of hydrides. In hydrides like $NaH$, it is bonded to a less electronegative element and thus exists in $ - 1$ oxidation state.
5. The oxidation number of the oxygen atom is always $ - 2$ with an exception of peroxides. In peroxides, it exists in its $ - 1$ oxidation state.
6. The oxidation state of group 1 elements i.e., alkali metal in a compound is $ + 1$.
7. The oxidation state of group 2 elements i.e., alkaline earth metal in a compound is $ + 2$.
8. The oxidation state of group 17 elements i.e., halogen in a compound is $ - 1$.
9. The sum of the oxidation number of each atom in a neutral compound is always equal to zero.
10. The sum of the oxidation number of each atom in a polyatomic ion is equal to the overall charge present on the ion.
Now, as per question we are given a polyatomic ion i.e., hydroxide ion which is chemically represented as $O{H^ - }$. Assuming the oxidation number of the oxygen atom be $x$.
We know that the hydrogen atom is generally present in its $ + 1$ oxidation state. Therefore, the value of $x$ can be calculated as follows:
$x + 1 = - 1$
$ \Rightarrow x = - 1 - 1$
$ \Rightarrow x = - 2$
Hence, the oxidation number of $O$ in the $O{H^ - }$ ion is $ - 2$.
Note:
It is important to note that if an atom undergo increase in the oxidation state during a chemical reaction, then it is known as oxidation and it involves loss of electrons during the process whereas if a decrease in the oxidation number of the atom is observed, then it is known as reduction and it involves gain of electron during the process.
Complete answer:
Some basic rules to assign oxidation number to an element are as follows:
1. The oxidation number of a free element is always considered as zero.
2. The oxidation number of a mono-atomic ion is always equal to the charge present on the ion.
3. For a given molecule $A - B$, the cleavage of a bond takes place in such a way that negative charge is assigned to more electronegative element as ${A^ + }$ and ${B^ - }$, where B is more electronegative than A.
4. The oxidation number of hydrogen atom is always $ + 1$ with an exception of hydrides. In hydrides like $NaH$, it is bonded to a less electronegative element and thus exists in $ - 1$ oxidation state.
5. The oxidation number of the oxygen atom is always $ - 2$ with an exception of peroxides. In peroxides, it exists in its $ - 1$ oxidation state.
6. The oxidation state of group 1 elements i.e., alkali metal in a compound is $ + 1$.
7. The oxidation state of group 2 elements i.e., alkaline earth metal in a compound is $ + 2$.
8. The oxidation state of group 17 elements i.e., halogen in a compound is $ - 1$.
9. The sum of the oxidation number of each atom in a neutral compound is always equal to zero.
10. The sum of the oxidation number of each atom in a polyatomic ion is equal to the overall charge present on the ion.
Now, as per question we are given a polyatomic ion i.e., hydroxide ion which is chemically represented as $O{H^ - }$. Assuming the oxidation number of the oxygen atom be $x$.
We know that the hydrogen atom is generally present in its $ + 1$ oxidation state. Therefore, the value of $x$ can be calculated as follows:
$x + 1 = - 1$
$ \Rightarrow x = - 1 - 1$
$ \Rightarrow x = - 2$
Hence, the oxidation number of $O$ in the $O{H^ - }$ ion is $ - 2$.
Note:
It is important to note that if an atom undergo increase in the oxidation state during a chemical reaction, then it is known as oxidation and it involves loss of electrons during the process whereas if a decrease in the oxidation number of the atom is observed, then it is known as reduction and it involves gain of electron during the process.
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