Question

Which gives ppt with $ {K_2}Cr{O_4} $
(A) $ H{g_2}^{2 + } $ , $ P{b^{2 + }} $ , $ A{g^ + } $ , $ B{a^{2 + }} $
(B) $ P{b_{{2^{}}}}^ + $ , $ A{g^ + } $ , $ B{a^{2 + }} $
(C) $ A{g^ + } $ , $ B{a^{2 + }} $
(D) $ P{b_{{2^{}}}}^ + $ , $ B{a^{2 + }} $

Answer 1

Hint :Potassium chromate is a strong oxidizing agent. In the option we are provided with different cations. Depending on their reactions with the Potassium chromate each of them gives precipitate of different color.

Complete Step By Step Answer:
In a precipitation reaction the insoluble salt is settled at the bottom of the beaker due to the presence of two solutions containing soluble salts. This reaction helps in determining the presence of various ions in solution.
Let’s take the given ions of lead, barium, silver and mercury one by one and sort their reaction with the Potassium Chromate.
The reaction of lead ion with Potassium Chromate is
 $ P{b^{2 + }} + {K_2}Cr{O_4} \to PbCr{O_4} + {K^ + } $
In the above reaction Lead Chromate is formed giving a yellow precipitate.
Similarly Potassium Chromate reacts with Barium ion to give Barium Chromate. The precipitate formed is also yellow in colour. Given below is the reaction
 $ B{a^{2 + }} + {K_2}Cr{O_4} \to BaCr{O_4} + {K^ + } $
In the same manner Silver ions react with Potassium Chromate to form black red precipitate. The reaction of the silver ion with Potassium Chromate is as follow:
 $ A{g^ + } + {K_2}Cr{O_4} \to A{g_2}Cr{O_4} + {K^ + } $
Mercury ion reacts with Potassium Chromate is as follow:
 $ H{g_{{2^{}}}}^{2 + } + {K_2}Cr{O_4} \to HgCr{O_4} + 4{K^ + } $
This above reaction gives a black precipitate due to formation of mercury chromate.
Therefore the answer is option (A).

Note :
In all of the above reactions chromate is formed by displacing one of the ions with another. This type of reaction where two compounds react and cation as well as the anion of the two reactants switch places or one displaces the other forming a new compound or product is called double displacement reaction. For example the ion $ Ag $ switches position with $ K $ in the above mentioned reaction.