A) \[N{{H}_{4}}\]
B) \[B{{a}^{2+}}\]
C) \[C{{a}^{2+}}\]
D) \[N{{a}^{+}}\]
Correct Answer: B
Solution :
\[B{{a}^{2+}}\]ion gives a yellow precipitate of barium chromate with potassium chromate. \[{{K}_{2}}Cr{{O}_{4}}+B{{a}^{2+}}\xrightarrow{{}}BaCr{{O}_{4}}+2{{K}^{+}}\]
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