$$BaC{l_2} + {K_2}Cr{O_4} \to BaCr{O_4} + 2KCl.BaCr{O_4}$$
is insoluble in acetic acid and $$Ba$$  gives apple green colour in flame test.

$$\eqalign{ & C{l_2} + {H_2}O \to HCl + HClO \cr & HCl + AgN{O_3} \to \mathop {AgCl \downarrow }\limits_{{\text{White ppt}}} + HN{O_3} \cr & Mg + 2HCl \to MgC{l_2} + {H_2} \uparrow \cr & {\text{Hence}}\,X\,{\text{is}}\,C{l_2}\,{\text{and}}\,Y\,{\text{is}}\,{H_2} \cr} $$

If $$N{a_2}C{O_3}$$   is used in place of $${\left( {N{H_4}} \right)_2}C{O_3}.$$    It will precipitate group $$V$$ radicals as well as magnesium radicals. The reason for this is the high ionization of $$N{a_2}C{O_3}$$   in water into $$N{a^ + }$$  and $$CO_3^{2 - }.$$  Now the higher concentration of $$CO_3^{2 - }$$  is available which exceeds the solubility product of group $$V$$ radicals as well as that of magnesium radicals.

$$SC{N^ - } + F{e^{3 + }} \to \mathop {Fe{{\left( {SCN} \right)}_3}}\limits_{{\text{blood red ppt}}{\text{.}}} $$

Precipitate of $$Zn{\left( {OH} \right)_2}$$  formed at initial stage dissolves in excess of $$N{H_4}OH$$  due to the formation of tetrammine $$Zn\left( {II} \right)$$  complex.
$$Z{n^{2 + }} + 2N{H_4}OH \to Zn{\left( {OH} \right)_2} \downarrow + 2NH_4^ + $$
$$Zn{\left( {OH} \right)_2} + 4NH_4^ + \to {\left[ {Zn{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }} + 2{H_2}O + 2{H^ + }$$

When the sodium fusion extract is added with $$FeC{l_3}$$  and then the resulting solution is acidified with dilute hydrochloric acid, the appearance of Prussian blue colouration confirms the presence of nitrogen in the organic compound.
$$Na + C + N \to NaCN$$
$$FeS{O_4} + 2NaCN \to $$     $$Fe{\left( {CN} \right)_2} + N{a_2}S{O_4}$$
$$Fe{\left( {CN} \right)_2} + 4NaCN \to $$     $$N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$
$$3N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + 4FeC{l_3} \to $$      $$\mathop {F{e_4}{{\left[ {Fe\left( {C{N_6}} \right)} \right]}_3}}\limits_{{\text{Prussian blue}}} + 12NaCl$$

\[\begin{align} & N{{a}_{2}}S{{O}_{3}}+2HCl\left( dil \right)\to 2NaCl+{{H}_{2}}O+S{{O}_{2}} \\ & {{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+\underset{\text{Green}}{\mathop{C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}}}\,+{{H}_{2}}O \\ \end{align}\]

Only $$F{e^{3 + }}\,ion$$   give blood red colouration with $$SC{N^ - }\,ions.$$
$$F{e^{3 + }} + SC{N^ - } \to \mathop {{{\left[ {Fe\left( {SCN} \right)} \right]}^{2 + }}}\limits_{{\text{(dark red)}}} $$

NOTE:
Only group $$I$$ cations are precipitated by dil. $$HCl$$
$$\mathop {SO_3^{2 - }}\limits_{\left( X \right)} + {H_2}S{O_4} \to \mathop {S{O_2}}\limits_{\left( Y \right)} + {H_2}O + SO_4^{2 - }$$
$$3S{O_2} + {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to {K_2}S{O_4} + \mathop {C{r_2}{{\left( {S{O_4}} \right)}_3}}\limits_{\left( {{\text{green}}\,\,{\text{colour}}\,{\text{solution}}} \right)} + {H_2}O$$
Only $$PbC{l_2}\,{\text{and}}\,H{g_2}C{l_2}$$    will precipitate as $$P{b^{2 + }}\,{\text{and}}\,Hg_2^{2 + }$$   as first group basic radicals and their solubility product is less than the other radicals.
NOTE: dil. $$HCl$$  is the first group reagent.