Step 1: Identify the nitrogen-containing compound

The question asks about the reaction between nitric acid ($HNO_3$) and phosphorus pentoxide ($P_4O_{10}$). Phosphorus pentoxide is a very strong dehydrating agent. It reacts with nitric acid by removing a molecule of water from two molecules of nitric acid.

The reaction is the dehydration of nitric acid to form its anhydride, dinitrogen pentoxide ($N_2O_5$).

$2HNO_3 \xrightarrow{P_4O_{10}} N_2O_5 + H_2O$

The overall balanced chemical equation is:

$4HNO_3 + P_4O_{10} \rightarrow 4HPO_3 + 2N_2O_5$

The nitrogen-containing compound produced is dinitrogen pentoxide ($N_2O_5$).

Step 2: Evaluate the properties of $N_2O_5$

Now we need to check the given options for $N_2O_5$.

Option A: can also be prepared by reaction of $P_4$ and $HNO_3$.

The reaction between white phosphorus ($P_4$) and concentrated nitric acid ($HNO_3$) is a redox reaction. Nitric acid is a strong oxidizing agent and oxidizes phosphorus to phosphoric acid ($H_3PO_4$), while it gets reduced itself.

$P_4 + 20HNO_3(\text{conc.}) \rightarrow 4H_3PO_4 + 20NO_2 + 4H_2O$

The nitrogen-containing compound produced in this reaction is nitrogen dioxide ($NO_2$), not dinitrogen pentoxide ($N_2O_5$). Therefore, statement A is incorrect.

Option B: is diamagnetic.

A compound is diamagnetic if all its electrons are paired. We can determine this by counting the total number of valence electrons in the $N_2O_5$ molecule.

• Valence electrons from 2 Nitrogen atoms = $2 \times 5 = 10$
• Valence electrons from 5 Oxygen atoms = $5 \times 6 = 30$
• Total valence electrons = $10 + 30 = 40$

Since the total number of valence electrons is an even number (40), and in the molecular orbital configuration of $N_2O_5$ all electrons are paired, the molecule is diamagnetic. Molecules with an odd number of total electrons are always paramagnetic. Those with an even number are typically diamagnetic, with notable exceptions like $O_2$. $N_2O_5$ is not an exception. Therefore, statement B is correct.

Option C: contains one N$-$N bond.

Let's examine the structure of the $N_2O_5$ molecule in its covalent (gaseous) state. The structure consists of two nitro groups ($-NO_2$) linked by an oxygen atom.

The structure is $O_2N-O-NO_2$. It contains an $N-O-N$ linkage, not a direct $N-N$ bond. In the solid state, it exists as an ionic salt, nitronium nitrate $([NO_2]^+[NO_3]^-)$, which also does not contain an $N-N$ bond. Therefore, statement C is incorrect.

Option D: reacts with Na metal producing a brown gas.

$N_2O_5$ is a strong oxidizing agent, as nitrogen is in its highest oxidation state (+5). Sodium (Na) is a strong reducing agent (an alkali metal). They will react in a redox reaction.

The reaction is:

$N_2O_5 + Na \rightarrow NaNO_3 + NO_2$

In this reaction, one nitrogen atom in $N_2O_5$ remains in the +5 oxidation state (in $NaNO_3$), while the other is reduced to the +4 oxidation state (in $NO_2$). Sodium is oxidized from 0 to +1. The product, nitrogen dioxide ($NO_2$), is a reddish-brown gas. Therefore, the reaction produces a brown gas. Statement D is correct.

Conclusion

Based on the analysis, the correct statements are B and D.