Question

Which of the following complex will give a white precipitate with \[BaC{{l}_{2}}(aq.)\]
A. \[[Co{{(N{{H}_{3}})}_{4}}S{{O}_{4}}]N{{O}_{2}}\]
B. \[[Cr{{(NH3)}_{5}}S{{O}_{4}}]Cl\]
C.\[~[Cr{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}\]
D. Both B and C

Answer 1

Hint: A chemical reaction that produces a precipitate when two solutions of two ionic compounds are combined is known as precipitation. Precipitation is the name for this phenomenon. In this reaction, an anion-containing solution is combined with a cation-containing solution to create an insoluble molecule.

Complete answer:According to the solubility laws, the majority of sulphate salts are soluble, with the exception of barium, lead, and calcium and the majority of chlorides salts are soluble, with the exception of mercury, silver, and calcium.
In the question, we have been asked to find the compound that will give a precipitate.

Therefore, in the first option that is option A., if we write the dissociation of \[[Co{{(N{{H}_{3}})}_{4}}S{{O}_{4}}]N{{O}_{2}}\]in solution then we’ll get:
\[[Co{{(N{{H}_{3}})}_{4}}S{{O}_{4}}]N{{O}_{2}}\to {{[Co{{(N{{H}_{3}})}_{4}}S{{O}_{4}}]}^{+}}+N{{O}_{2}}^{-}\] and if either of the produced ion will react with \[BaC{{l}_{2}}(aq.)\], we will not get any type of precipitate.

If we look at the second option, then \[{{[Cr{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]}^{+}}\] will dissociate into \[{{[Cr{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]}^{+}}\]and $C{{l}^{-}}$ions, and when they will react with \[BaC{{l}_{2}}(aq.)\]there will be no reaction and hence nothing will be given out as precipitate.

If we look at the third option is option C., then the dissociation of \[~[Cr{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}\] in the solution can be shown as:
\[~[Cr{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}\to {{[Cr{{(N{{H}_{3}})}_{5}}Cl]}^{2-}}+SO_{4}^{2-}\]here the $SO_{4}^{2-}$ion will react with the $B{{a}^{2+}}$ion which will be dissociated from \[BaC{{l}_{2}}(aq.)\] and then we will get $BaS{{O}_{4}}$which will be regarded as the precipitate.

Hence, the correct option is C.\[~[Cr{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}\]

Note: The precipitation reaction is a double displacement reaction, it should be noted. Barium sulphate precipitates because it is insoluble in water. The decreasing temperature is another factor in the precipitate's development. As a result, there is precipitation since the solubility is reduced. To find out which ions are present in the solution, we use precipitation reactions.