Why does $C{H_4}$ not undergo a nitration reaction?

Hint: Here, we need to recall substitution reactions of alkanes where H atom of alkanes is substituted by another group.

Complete step by step answer:
Substitution reaction: The hydrogen atoms attached to carbon are replaced by the other atoms or compounds to make a new product. For example, halogenation. Methane reacts with bromine in the presence of light to replace H-atom by Br to make bromomethane.$C{H_4}{\text{ + B}}{{\text{r}}_2}{\text{ }}\xrightarrow{{h\nu }}{\text{ C}}{{\text{H}}_3}Br{\text{ + HBr}}$ . Other substitution reactions of alkanes are sulfonation and nitration.
Nitration: It is the substitution reaction in which a hydrogen atom of alkane is replaced by $ - N{O_2}$ group. Lower members of alkanes like methane do not undergo nitration at ordinary temperatures. Alkanes are highly stable and exhibit inertness. Their less reactivity or inertness can be explained as:
All the C-C & C-H bonds being stronger sigma bonds in alkanes and not influenced by the acids, alkalis, oxidants under the ordinary conditions.
The C-H (weak polar) bonds and C-C (completely non polar) bonds in alkanes are basically non polar because of the small electronegativity difference in C (2.6) and H (2.1). Therefore,the polar species (electrophiles or nucleophiles) are not able to attack the bonds under the ordinary conditions. Thus, because of the less reactive nature alkanes don't undergo nitration easily.
However, with fuming $HNO$_3 in the vapour phase at 667-793 K under pressure, methane undergoes nitration. This is called vapour phase nitration. By the help of this process higher as well as lower alkanes can be converted into the nitroalkanes.$C{H_3} - H{\text{ + HON}}{{\text{O}}_2}{\text{ }}\xrightarrow{{723{\text{ K}}}}{\text{ C}}{{\text{H}}_3} - N{O_2}{\text{ + }}{{\text{H}}_2}O$

Note:
The same reasons are true for sulphonation of methane. The reactivity and temperature conditions methane does not undergo nitration easily.