Question

On the addition of a solution containing $Cr{{O}_{4}}^{2-}$ ions to the solution of $B{{a}^{2+}},S{{r}^{2+}},C{{a}^{2+}}$ ions, the precipitate obtained first will be of:
(A) $CaCr{{O}_{4}}$
(B) $SrCr{{O}_{4}}$
(C) $BaCr{{O}_{4}}$
(D) A mixture of all the three

Answer 1

Hint: The amount of a solute that will dissolve in the solvent at a specific temperature and pressure is known as its solubility. The solute may be a solid, liquid, or gas. The solvent is either a solid or a liquid.

Complete Step by Step Solution:
All the three ions are the ions of group 2, which are alkaline earth metals. The alkaline earth metals contain two electrons in their valence shell. The compound that will be least soluble will precipitate out first. As we move down the group 2, the size of the cation increases.

The size of the cation of group 2 is given in the order: $B{{e}^{2+}} < M{{g}^{2+}} < C{{a}^{2+}} < S{{r}^{2+}} < B{{a}^{2+}} < R{{a}^{2+}}$.
With an increase in size, the hydration energy decreases. Also, the lattice energy decreases, but the decrease in lattice energy is small as compared to the increase in the size of cations. The decrease in hydration energy is more significant as compared to the decrease in lattice energy. Therefore, as we move down the group, the decrease in hydration energy results in a decrease in the solubility of compounds of alkaline earth metals.

As the size of $B{{a}^{2+}}$is largest among the given ions, hence, it will precipitate out first.
On the addition of a solution containing $Cr{{O}_{4}}^{2-}$ ions to the solution of $B{{a}^{2+}},S{{r}^{2+}},C{{a}^{2+}}$ions, the precipitate obtained first will be of $BaCr{{O}_{4}}$.
Correct Option: (C) $BaCr{{O}_{4}}$.

Note: The alkaline earth metals are so named because they are basic in nature and are found in the earth’s crust as metal oxides. These have low ionisation energy, low electronegativity and low electron affinity. These are good conductors of electricity.