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The oxidation number of the nitrogen atom in the ammonium ion is:
A. +3
B. +2
C. +1
D. -2
E. -3

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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Hint: Oxidation number of every atom must be known to calculate the oxidation number of an atom. We assume the atom for which the oxidation number is to be calculated as “x”.

Complete step-by-step solution:
We know that the charge of a hydrogen atom is +1 and there are four hydrogen atoms so we will take four +1 charges. Now, we need to calculate the oxidation number of Nitrogen now. So, we take the oxidation number of Nitrogen as “x”.
$\begin{array}{l}\mathop {\rm{N}}\limits^{\rm{x}} {\mathop {\rm{H}}\limits^{ + 1} _{\rm{4}}} = 1\\ \Rightarrow {\rm{x}} + 4 \times (1) = 4\\ \Rightarrow x = - 3\end{array}$
So, we can conclude that the oxidation number of nitrogen is -3. Therefore, out of the given five options, E is the correct option. Options A, B, C and D are incorrect.

Additional information:
Oxidation number refers to the number of the electrons which an atom loses or gains in order to form chemical bonds with some other atom.
Some rules of writing oxidation numbers are:
1) In a free uncombined element, the oxidation number is zero.
2) The summation of the oxidation number of all the participating atoms in a compound is always zero.
3) The summation of the oxidation number of all the participating atoms in an ion is equal to the charge of that of the ion.
4) The oxidation number of the fluorine atom in all the compounds is -1.
5) The oxidation number of the halogens in all the compound is -1
6) Hydrogen always has an oxidation number of +1 except in hydrides.
7) Oxygen has an oxidation number of -2 except that of peroxides.

Note: Oxidation number measures the loosing or the excepting of the electrons in forming chemical bonds. Oxidation number of some common atoms never changes such as hydrogen and oxygen except for some exceptions.