Dinitrogen pentoxide ${{N}_{2}}{{O}_{5}}$, a colourless solid, prepared by:
(A)- heating $N{{H}_{4}}N{{O}_{2}}$ with an excess of oxygen
(B)- dehydrating $HN{{O}_{3}}$ with $CaO$
(C)- dehydrating $HN{{O}_{3}}$ with ${{P}_{4}}{{O}_{10}}$
(D)- heating a mixture of $HN{{O}_{2}}$ and $Ca{{(N{{O}_{3}})}_{2}}$

Hint: Dinitrogen pentoxide is covalently bonded and is quite unstable. Since, it is a reaction of nitric acid without water, giving ${{N}_{2}}{{O}_{5}}$ as a product, thus known as nitric anhydride.
\[\text{2 }HN{{O}_{3}}~\to \text{ }{{N}_{2}}{{O}_{5}}+\text{ }{{H}_{2}}O~\].

Complete step by step answer:
The colourless solid of ${{N}_{2}}{{O}_{5}}$, is one of the binary nitrogen oxides compound that contains only nitrogen and oxygen bonds. Since, it is reaction of nitric acid without water, giving water and ${{N}_{2}}{{O}_{5}}$ as products, thus known as anhydride.
It consists of nitronium ion, $N{{O}_{2}}^{+}$ (linear shape)and nitrate anions, $N{{O}_{3}}^{-}$ (trigona planar in shape), having oxidation state $(+5)$ for both nitrogen centers. In its gas phase, the two ions are covalently bonded to form dinitrogen pentoxide.

Dinitrogen pentoxide is prepared by dehydrating nitric acid, $HN{{O}_{3}}$ with phosphorus oxide ${{P}_{4}}{{O}_{10}}$ .
\[{{P}_{4}}{{O}_{10}}~+\text{ 4 }HN{{O}_{3}}~\to \text{ 2}{{N}_{2}}{{O}_{5}}+\text{ }HP{{O}_{3}}~\]
Therefore, it is found that out of all the four options, dinitrogen pentoxide is prepared by option (C)- dehydrating $HN{{O}_{3}}$ with ${{P}_{4}}{{O}_{10}}$ . \[{{\text{P}}_{\text{4}}}{{\text{O}}_{\text{10}}}\]
So, the correct answer is “Option C”.

Additional Information:
 It is a strong oxidizing agent, as it oxidises iodine to ${{I}_{2}}{{O}_{5}}$.
One of major path of formation of ${{N}_{2}}{{O}_{5}}$ is through reaction of nitric oxide with ozone, ${{O}_{3}}$.
\[N{{O}_{2}}+\,\,{{O}_{3}}\to N{{O}_{3}}+\,\,{{O}_{2}}\]
\[N{{O}_{2}}+N{{O}_{3}}\rightleftharpoons {{N}_{2}}{{O}_{5}}\]

This process is further effective in the removal of reactive chlorine from a catalytic ozone destroying cycle by forming unreactive chlorine nitrate, $ClON{{O}_{2}}$.
It also acts as an unreactive reservoir, thereby reducing the concentration of the $N{{O}_{X}}$ species temporarily.

Note: For in case of dehydrating nitric acid, $HN{{O}_{3}}$ with calcium oxide, $CaO$ ,we get calcium nitrate and water.
\[\underset{\text{base}}{\mathop{CaO}}\,~+~\underset{\text{acid}}{\mathop{2HN{{O}_{3}}}}\,~\to ~~\underset{\text{acid}}{\mathop{Ca{{(N{{O}_{3}})}_{2}}}}\,~+\underset{\text{water}}{\mathop{~2{{H}_{2}}O}}\,\]
This is an example of double-displacement reaction as well as neutralisation reaction.
- The heating of ammonium nitrite, $N{{H}_{4}}N{{O}_{2}}$ with an excess of oxygen, decomposes to give nitrogen gas and water.
\[N{{H}_{4}}N{{O}_{2}}(s)~\to ~{{N}_{2}}(g)+2{{H}_{2}}O(l)\]
- The heating of mixture of nitrous acid, $HN{{O}_{2}}$ and calcium nitrate, $Ca{{(N{{O}_{3}})}_{2}}$ gives calcium nitrite and nitric acid.
\[Ca{{\left( N{{O}_{3}} \right)}_{2}}~+~2HN{{O}_{2}}~\to ~Ca{{\left( N{{O}_{2}} \right)}_{2}}~+~2HN{{O}_{3}}\]