Question

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

Answer 1

Hint: 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.