Q20 / 75 · JEE Main 2026 (02 April Shift 1) · Chemistry
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 $K_2CrO_4$ 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,
- A$Ca^{2+}$ and $SO_4^{2-}$
- Correct answerB$Ba^{2+}$ and $PO_4^{3-}$
- C$Mn^{2+}$ and $PO_4^{3-}$
- D$Ba^{2+}$ and $SO_4^{2-}$
Solution
An apple green colour in the flame test indicates the presence of $Ba^{2+}$ ion.
The group precipitate of $Ba^{2+}$ is $BaCO_3$, which dissolves in acetic acid to form barium acetate. Upon treatment with $K_2CrO_4$, it gives a yellow precipitate of barium chromate ($BaCrO_4$), confirming the presence of $Ba^{2+}$.
When the sodium carbonate extract is heated with conc. $HNO_3$ and ammonium molybdate, the formation of a canary yellow precipitate indicates the presence of phosphate ($PO_4^{3-}$) ion. The canary yellow precipitate is ammonium phosphomolybdate, $(NH_4)_3PO_4 \cdot 12MoO_3$.
Therefore, the cation is $Ba^{2+}$ and the anion is $PO_4^{3-}$.
Answer: $Ba^{2+}$ and $PO_4^{3-}$
The group precipitate of $Ba^{2+}$ is $BaCO_3$, which dissolves in acetic acid to form barium acetate. Upon treatment with $K_2CrO_4$, it gives a yellow precipitate of barium chromate ($BaCrO_4$), confirming the presence of $Ba^{2+}$.
When the sodium carbonate extract is heated with conc. $HNO_3$ and ammonium molybdate, the formation of a canary yellow precipitate indicates the presence of phosphate ($PO_4^{3-}$) ion. The canary yellow precipitate is ammonium phosphomolybdate, $(NH_4)_3PO_4 \cdot 12MoO_3$.
Therefore, the cation is $Ba^{2+}$ and the anion is $PO_4^{3-}$.
Answer: $Ba^{2+}$ and $PO_4^{3-}$
Hence, the correct answer is option: 'B'.