Question

What is the oxidation number of $N$ in $NH_4^+$ ?

Answer 1

To solve the given question, we should have knowledge about oxidation numbers and how to calculate it. Oxidation Number is the number of electrons gained or lost or shared to form a compound from an element.

Complete step by step answer:
Some general rules that follow oxidation number are that the oxidation number of a free element is always $0$, the oxidation number of a Group $1$ element in a compound is $+1$ , a Group $2$ element in a compound is $+2$ and the sum of oxidation numbers of all of the atoms in a compound is of the total charge of the molecule.
Step-1 :
The Nitrogen atom has different oxidation numbers with different atoms to form a different number of compounds. Oxidation state of Nitrogen generally is $-3$.
Step-2 :
If we look at any other compound like nitrate which has a formula of $NO_3^-$ . The oxidation state of nitrogen can be calculated as the Oxidation number of $N+3 \times$ Oxidation number of oxygen = Total charge on the molecule.
$x+3(-2)=-1$
$x=+5$
Step-3 :
Now, if we look forward to the given compound as a whole is not neutral, in fact, it contains a charge of a positive one. So, the oxidation number of nitrogen here can be calculated as :
$x+4=+1$
$x=-3$

Note:
The nitrogen gas has a molecular formula of which is neutral, so, the charge is zero in this case. The atomic number of Nitrogen is $7$ with a mass of $14amu$ .
While calculating the oxidation number of an atom in a compound, make sure to determine the charge of the whole compound as it is the most important step while calculating the oxidation number.