Referring to the following reactions the missing products A, B, C, and D respectively are

${\mathrm{NH}}_4\mathrm{Cl }\left(\mathrm{aq}\right) + {\mathrm{NaNO}}_{\mathrm{2 }}\left(\mathrm{aq}\right) \to \left[\mathrm{A}\right] + {\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right)+ \mathrm{NaCl }\left(\mathrm{aq}\right)$

${\left({\text{NH}}_{\text{4}}\right)}_2{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}} \stackrel{\Delta }{\to } \left[\text{B}\right] + {\text{H}}_2\text{O} \left(\text{l}\right)+ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}} \left(\text{s}\right)$

$\mathrm{Cu} + {\mathrm{HNO}}_3 \left(\mathrm{dilute}\right) \to \mathrm{Cu}({\mathrm{NO}}_3{)}_2 + [\mathrm{C}] + {\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

$\mathrm{Cu} + {\mathrm{HNO}}_3 \left(\mathrm{concentrate}\right) \to \mathrm{Cu}({\mathrm{NO}}_3{)}_2 + [\mathrm{D}] + {\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

• Correct answer
  A

  ${\text{N}}_{\text{2}},{\text{N}}_2,\text{NO},{\text{NO}}_{\text{2}}$
• B

  ${\text{N}}_2,{\text{NH}}_{\text{3}},{\text{N}}_{\text{2}},\text{NO}$
• C

  ${\text{N}}_2,{\text{N}}_{\text{2}},{\text{NO}}_2,{\text{NO}}_{\text{2}}$
• D

  ${\text{N}}_2,{\text{NH}}_{\text{3}},{\text{NO}}_2,{\text{N}}_2{\text{O}}_{\text{4}}$

Hence, the correct answer is option: 'A'.