A) \[NH_{4}^{+}\]
B) \[B{{a}^{2+}}\]
C) \[C{{a}^{2+}}\]
D) \[N{{a}^{+}}\]
Correct Answer: B
Solution :
\[{{K}_{2}}Cr{{O}_{4}}+B{{a}^{2+}}\xrightarrow{\,}\,\underset{Yellow\,\,ppt}{\mathop{BaCr{{O}_{4}}}}\,\downarrow +2{{K}^{+}}\]
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