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(B) The reaction of $HNO_3$ with $P_4O_{10}$ is a dehydration reaction: $2HNO_3 + P_4O_{10} \rightarrow N_2O_5 + H_4P_4O_{12}$.The product is $N_2O_5$

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$B, D$

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(B) The reaction of $HNO_3$ with $P_4O_{10}$ is a dehydration reaction: $2HNO_3 + P_4O_{10} \rightarrow N_2O_5 + H_4P_4O_{12}$.The product is $N_2O_5$ (dinitrogen pentoxide).$(A)$ $N_2O_5$ cannot be prepared by the reaction of $P_4$ and $HNO_3$ (which produces $H_3PO_4$ and $NO_2$).$(B)$ $N_2O_5$ is diamagnetic because all electrons are paired.$(C)$ $N_2O_5$ has the structure $O_2N-O-NO_2$,which contains an $N-O-N$ bond,not an $N-N$ bond.$(D)$ $N_2O_5$ reacts with $Na$ metal to produce $NaNO_3$ and $NO_2$ (a brown gas): $Na + N_2O_5 \rightarrow NaNO_3 + NO_2 \uparrow$.Thus,statements $(B)$ and $(D)$ are correct.

The nitrogen-containing compound produced in the reaction of $HNO_3$ with $P_4O_{10}$:
$(A)$ can also be prepared by the reaction of $P_4$ and $HNO_3$
$(B)$ is diamagnetic
$(C)$ contains one $N-N$ bond
$(D)$ reacts with $Na$ metal producing a brown gas

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The nitrogen-containing compound produced in the reaction of $HNO_3$ with $P_4O_{10}$:$(A)$ can also be prepared by the reaction of $P_4$ and $HNO_3$$(B)$ is diamagnetic$(C)$ contains one $N-N$ bond$(D)$ reacts with $Na$ metal producing a brown gas