Question

Barium chloride on reacting with ammonium sulfate forms barium sulfate and ammonium chloride. Which of the following correctly represents the type of the reaction involved?
(i) Displacement reaction
(ii) Precipitation reaction
(iii) Combination reaction
(iv) Double displacement reaction
A) (i) only
B) (ii) only
C) (iv) only
D) (ii) and (iv)

Answer 1

Hint: The barium sulfate is a white colour sparingly soluble salt. When introduced into the water it is seen at the bottom of the beaker as a precipitate. The barium chloride and ammonium sulfate undergo the reaction such that the cation or anions of the reactant interchange with each other to form a double displaced product.

Complete step by step solution:
A precipitation reaction is defined as the reaction which proceeds with the formation of insoluble salt when the two solutions containing the soluble salt have interacted with each other. This insoluble salt is obtained at the bottom of the beaker and is also called as the precipitate. These reactions help in determining the presence of various ions in the solution.
In a precipitation reaction, the two solutions of different salts combine and the cation and anion of the solution combine in a solution to form an insoluble salt which then precipitates out from the solution.
The barium chloride $\text{ BaC}{{\text{l}}_{\text{2}}}$ reacts with the ammonium sulfate $\text{ (N}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$.the reaction proceeds with the formation of barium sulfate $\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$and ammonium chloride$\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }$. They $\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$have large lattice energy and thus less solubility. The lattice energy depends on the radius and charge of the ion since the barium and sulfate both have a significantly large radius they are not normally soluble in water. The solubility product of the salt is less and it remains as the precipitate. Therefore, this reaction is the precipitation reaction.
$\begin{matrix}
\text{BaC}{{\text{l}}_{\text{2}}} & \text{+} & {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} & \to & \text{BaS}{{\text{O}}_{\text{4}}}\text{(}\downarrow \text{)} & \text{+} & \text{2N}{{\text{H}}_{\text{4}}}\text{Cl} \\
\text{(barium Chloride)} & {} & \text{(Ammonium Sulphate)} & {} & \text{(Barium sulphate)} & {} & \text{(Ammonium chloride)} \\
\end{matrix}$
The $\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$is a white colour precipitate which can be seen at the bottom of the beaker.
The type of reactions in which the two compounds react by the exchange of ions between them to form a new compound is called the double displacement reaction. In a double replacement reaction, the negative ion is exchanged by the positive ion to form a new pair of ions. A general reaction for the double displacement is as shown below:
${{\text{A}}^{\text{+}}}{{\text{B}}^{-}}\text{+}{{\text{C}}^{\text{+}}}{{\text{D}}^{-}}\text{ }\to \text{ }{{\text{A}}^{\text{+}}}{{\text{D}}^{-}}\text{+}{{\text{C}}^{\text{+}}}{{\text{B}}^{-}}$
Here, A forms a new compound with the D and B forms a new compound with C.
$\begin{matrix}
\text{BaC}{{\text{l}}_{\text{2}}} & \text{+} & {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} & \to & \text{BaS}{{\text{O}}_{\text{4}}}\text{(}\downarrow \text{)} & \text{+} & \text{2N}{{\text{H}}_{\text{4}}}\text{Cl} \\
\text{(barium Chloride)} & {} & \text{(Ammonium Sulphate)} & {} & \text{(Barium sulphate)} & {} & \text{(Ammonium chloride)} \\
\end{matrix}$
The barium chloride $\text{ BaC}{{\text{l}}_{\text{2}}}$ reacts with the ammonium sulfate $\text{ (N}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$.the reaction proceeds with the formation of barium sulfate $\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$and ammonium chloride$\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }$. If we look at the products closely we observe that the cation in barium chloride which is $\text{B}{{\text{a}}^{+}}$ reacting with the anion of ammonium sulfate which is $\text{SO}_{\text{4}}^{\text{2}-}$ and result in the formation of a new compound$\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$. Similarly the chloride of barium chloride reacts with the ammonium ion to form$\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }$. Since there is a double displacement of ions in the reaction. This is a double displacement reaction.
Therefore, the reaction of $\text{ BaC}{{\text{l}}_{\text{2}}}$ with the$\text{ (N}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$to give the $\text{ BaS}{{\text{O}}_{\text{4}}}\text{ }$and $\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }$is precipitation and double displacement reaction.

Hence, (D) is the correct option.

Note: Note that in double displacement reactions either cation are swipe or anion are swipe. You cannot swipe both as you end up getting the same substance with which you started.