Question

which of the following compounds has the strongest coulombic attractions between its ions in the solid state? a mgo b cao c mgcl2 d cacl2

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. \(MgO\)