Question

Among the oxides of Nitrogen, the highest and lowest oxidation states for Nitrogen are +5 and +1 respectively – True or False?

• A. True
• B. False

Answer 1

The correct option is A True

Nitrogen forms a number of oxides:
$N_{2}O$ is nitrous oxide, $NO$ is nitric oxide,$N_2{O}_3$ is dinitrogen trioxide, $N{O}_2$ is nitrogen dioxide and $N_2{O}_5$ is dinitrogen pentoxide. Between Nitrogen and Oxygen, which is more electronegative? It is oxygen.
Let us briefly recall the rules for allocating oxidation numbers:

1. The oxidation number of an element in its free state is zero — for example, $H_2$, $S_8$, $P_4$ etc.
2. In the case monoatomic ions, it is the same as the charge on the ion, for example: $C{a}^{2+}=+2$ and $N^{3-}=-3$
3. For a neutral compound, the sum of all oxidation numbers is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the overall charge on the ion.
4. The first two groups - alkali metal (IA family) and alkaline earth metal (IIA family) generally tend to have +1 and +2 oxidation states respectively.
5. For most compounds, the oxidation number of oxygen is generally –2. There is, however, an exception in the case of peroxides ($H_2{O}_2$, $N{a}_2{O}_2$). Usually, in cases where an - O – O bond is involved, oxygen attains an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.
6. Fluorine is the most electronegative element in the periodic table. It almost always has the oxidation number as –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they’re in combination with an oxygen, fluorine or between themselves (halogens).
7. Usually, the oxidation state of hydrogen in a compound is +1. As an exception they tend to have –1 oxidation states in binary metal hydrides like NaH LiH etc.

Let the oxidation state of the central Nitrogen atom be ‘x’. Let us apply the usual rules from above:
Hydrogen has an oxidation number of +1 while oxygen has -2. Also, the sum of oxidation states of all the atoms in a neutral atom is 0.
Hence,
2x + (-2) = 0;
Solving, x = +1
Now try that for the remaining oxides. You will see that Nitrogen is in its +5 oxidation state in $N_2{O}_5$.