Assertion $A$ : Aniline on nitration yields ortho, meta and para nitro derivatives of aniline.
Reason $R$: Nitrating mixture is a strong acidic mixture.
In the light of the above statements, choose the correct answer from the options given below

$1.$ The compound ${X}$ is
$(A)$ $\mathrm{NaNO}_3$ $(B)$ $\mathrm{NaCl}$ $(C)$ $\mathrm{Na}_2 \mathrm{SO}_4$ $(D)$ $\mathrm{Na}_2 \mathrm{~S}$
$2.$ The compound ${Y}$ is
$(A)$ $\mathrm{MgCl}_2$ $(B)$ $\mathrm{FeCl}_2$ $(C)$ $\mathrm{FeCl}_3$ $(D)$ $\mathrm{ZnCl}_2$
$3.$ The compound ${Z}$ is
$(A)$ $\left.\mathrm{Mg}_2 \mid \mathrm{Fe}(\mathrm{CN})_6\right]$ $(B)$ $\operatorname{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(D)$ $\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
Give the answer question $1,2$ and $3.$
The structure of $C$ would be
$STATEMENT-2$: The colour of the compound formed in the reaction of aniline with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol is due to the extended conjugation.