Question

The correct set of oxidation numbers of nitrogen in ammonium nitrate is:
A. -3,+3
B. -1,+1
C. +1,-1
D. -3,+5

Answer 1

Hint: To answer this question, we should know that oxidation number or oxidation state is the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom.

Step by step answer:
We should know that when an atom participates in an oxidation-reduction reaction is assigned an oxidation number that reflects its ability to acquire, donate, or share electrons. We should know that oxidation states are represented by integers which may be positive, zero, or negative. We should note that in some cases, the average oxidation state of an element is a fraction, such as $${\displaystyle \frac{8}{3}}$$ for iron in magnetite$$\left(F{e}_3{O}_4\right)$$. One thing to note also that the increase in oxidation state of an atom, by a chemical reaction, is known as an oxidation; and a decrease in oxidation state is known as a reduction.

We should know some rules to find the correct oxidation number of nitrogen in ammonium nitrate. First of all we should know that in ammonium nitrate the chemical formula is $$N{H}_{4}N{O}_3$$. Now, we notice that it has two nitrogen atoms, so we will find the oxidation number of Nitrogen in this compound by separately knowing oxidation state in cation and anion part of this compound.

So, we know that ammonium nitrate chemical compound is $$N{H}_{4}N{O}_3$$​, so we can write this compound as $$N{H}_4^{+}andN{O_3}^{-}.$$
So, we will find oxidation number for $$N{H_4}^{+},$$
The first step is to let the oxidation state of N=x
We know that oxidation state of H=+1
Sum of total oxidation state of all atoms = Overall charge on the compound.

$$\begin{array}{rl}& (x)+(1\times 4)=+1\\ & (x)=1-4=-3\end{array}$$
Oxidation state of N is -3.

Now, we will find the oxidation state of N in $$N{O_3}^{-}$$. We should let the oxidation of N as y.
To answer this, we should know that oxidation state of O=−2
Applying the above formula, we will get:
$$\begin{array}{rl}& (y)+3\times (-2)=-1\\ & y=-1+6\\ & y=+5\end{array}$$
So, the oxidation state of N in $$N{O_3}^{-}$$- is +5.
So, now we can say that oxidation state of N in $$N{H}_{4}N{O}_3$$​= −3, +5.

So, the correct option is D.

Note: We should know some important points about the oxidation state of elemental atoms. We should know that the oxidation state of elemental atoms such as sodium, magnesium, iron is zero and also the net oxidation state of neutral molecules such as oxygen, chlorine, water, ammonia, methane, and potassium permanganate is zero. We should note that the oxidation state of charged ions is equal to the net charge of the ion. Alkali metals have an oxidation state of +1, alkali earth metals have +2 oxidation number, oxidation state of boron family is +3. And it is important to note that the oxidation number of hydrogen in proton $$(H^{+})$$ is +1, and in hydride is -1. Oxidation number of oxygen in oxide ion $$\left(O^{2-}\right)$$ is -2, and in peroxide ion is -1.