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Question

Identify the anion present in the compound Y is treated with Silver nitrate solution a white precipitate is obtained which is soluble in excess of Ammonium hydroxide solution?


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Solution

The Anion present is Chloride ion, Cl-.

The compound may be Sodium chloride.

When sodium chloride reacts with silver nitrate to form a white precipitate of silver chloride and sodium nitrate.

NaCl (aq) + AgNO3 (aq) ⇢ AgCl(s) + NaNO3(aq)

On ionization NaCl which gives Na+ and Cl- ions

Similarly AgNO3 becomes Ag+ and NO3- ions.

On reaction between these two components (NaCl and AgNO3) Cation Na+ combines with NO3- anion and anion Cl- combines with Ag+ cation.

i.e the anion present in the given compound Y (NaCl) is Cl-.


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