A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with K$$_2$$CrO$$_4$$ to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. HNO$$_3$$ and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively,

In salt analysis we first identify the cation.

Flame test : The solid moistened with conc. HCl gives an apple-green colour in the non-luminous flame. Among common cations only $$Ba^{2+}$$ imparts an apple-green colour (Ca$$^{2+}$$ gives brick-red, Sr$$^{2+}$$ gives crimson, Mn$$^{2+}$$ gives no characteristic colour). Hence the cation is very likely $$Ba^{2+}$$.

Confirmatory test for Ba2+ : The “group precipitate” (carbonate of the Group V cations Ba, Sr, Ca obtained with $$\left(NH_4\right)_2CO_3$$) is dissolved in dilute acetic acid and the solution is treated with $$K_2CrO_4$$. $$Ba^{2+}+CrO_4^{2-}\;\longrightarrow\;BaCrO_4\downarrow$$ Yellow precipitate of $$BaCrO_4$$ confirms the presence of $$Ba^{2+}$$ (SrCrO4 is light yellow but dissolves in acetic acid; CaCrO4 is white). Thus the cation is definitely $$Ba^{2+}$$.

Now test for the anion.

Test of the sodium-carbonate extract : The extract is boiled with conc. $$HNO_3$$ to oxidise/removal of interfering anions and then treated with ammonium molybdate solution. Formation of a canary-yellow precipitate indicates $$\left(NH_4\right)_3\left[PO_4\cdot12MoO_3\right]$$, the characteristic test for the phosphate ion $$PO_4^{3-}$$.

Thus, the salt contains the cation-anion pair $$Ba^{2+}$$ and $$PO_4^{3-}$$.

Checking the options, only Option B matches this pair.

Answer : Option B which is: $$Ba^{2+}$$ and $$PO_4^{3-}$$