which of the following compounds has the strongest coulombic attractions between its ions in the solid…

Explanation:

Step1: Recall Coulomb's law for ionic compounds

The strength of Coulombic attractions between ions in an ionic - solid is given by (F = k\frac{q_1q_2}{r^2}), where (k) is a constant, (q_1) and (q_2) are the charges of the ions, and (r) is the distance between the ions. The greater the product of the ion - charges ((q_1q_2)) and the smaller the ionic radius ((r)), the stronger the Coulombic attraction.

Step2: Analyze the ion - charges in the given compounds

In (MgO), (Mg^{2 +}) and (O^{2-}) have charges of (+ 2) and (-2) respectively, so (q_1q_2=4). In (CaO), (Ca^{2 +}) and (O^{2-}) also have (q_1q_2 = 4). In (MgCl_2), (Mg^{2+}) and (Cl^-) have (q_1q_2=2). In (CaCl_2), (Ca^{2+}) and (Cl^-) have (q_1q_2 = 2). So, (MgO) and (CaO) have a greater product of ion - charges compared to (MgCl_2) and (CaCl_2).

Step3: Compare the ionic radii

The ionic radius of (Mg^{2+}) is smaller than that of (Ca^{2+}) (as (Mg) is above (Ca) in the periodic table and ionic radius increases down a group). The ionic radius of (O^{2-}) is smaller than that of (Cl^-). Since the distance between the ions ((r)) is smaller in (MgO) (due to the smaller ionic radii of (Mg^{2+}) and (O^{2-})) and the product of the ion - charges is the same ((q_1q_2 = 4)) for (MgO) and (CaO), the Coulombic attraction is stronger in (MgO).

Answer:

A. (MgO)