Question

the energy required to dissociate an ionic solid into gaseous ions (lattice energy) for the compounds naf and mgf₂ is shown in the table above. on the basis of coulomb’s law, which of the following best helps to explain the large difference between the lattice energies of naf and mgf₂?
a the solubility of mgf₂ is less than that of naf.
b the electronegativity of mg is greater than that of na.
c the mass of the mg cation is greater than that of the na cation.
d the charge of the mg cation is larger than that of the na cation.

Explanation:

Step1: Recall Coulomb's law formula

Coulomb's law formula for lattice energy \(U=\frac{kq_1q_2}{r}\), where \(k\) is a constant, \(q_1\) and \(q_2\) are the charges of ions, and \(r\) is the distance between ions.

Step2: Analyze ion charges

For \(NaF\), \(Na^+\) has a charge of \(+ 1\) and \(F^-\) has a charge of \(-1\). For \(MgF_2\), \(Mg^{2 +}\) has a charge of \(+2\) and \(F^-\) has a charge of \(-1\). The product of charges \(q_1q_2\) is larger for \(MgF_2\) (\(2\times1 = 2\)) compared to \(NaF\) (\(1\times1=1\)).

Step3: Consider ion size (distance \(r\))

The \(Mg^{2+}\) ion is smaller than \(Na^+\) (due to higher nuclear charge for the same electron shell), but the dominant factor in the large difference in lattice energy is the charge. The higher charge of \(Mg^{2+}\) (\(+2\) vs \(+1\) for \(Na^+\)) leads to a much stronger electrostatic interaction in \(MgF_2\) according to Coulomb's law.

Answer:

D. The charge of the Mg cation is larger than that of the Na cation.