Question

the charges and sizes of the ions in an ionic compound affect the strength of the electrostatic intera thus the strength of the lattice energy of the ionic compound. arrange the compounds according to lattice energies based on the relative ion charges and sizes. highest lattice energy lowest lattice energy answer bank naf mgo mgf₂ kcl

Explanation:

Step1: Recall the formula for lattice energy

The lattice energy \(U\) is given by the formula \(U=\frac{kq_1q_2}{r}\), where \(k\) is a constant, \(q_1\) and \(q_2\) are the charges of the ions, and \(r\) is the distance between the ions (related to the ionic radii). Lattice energy is directly proportional to the product of ionic charges and inversely proportional to the ionic radii.

Step2: Analyze the ionic charges

• For \(MgO\): \(Mg^{2 +}\) and \(O^{2-}\), so \(q_1q_2=(+ 2)\times(-2)=-4\)
• For \(MgF_2\): \(Mg^{2+}\) and \(F^-\), so \(q_1q_2=(+2)\times(-1)\times2=- 2\)
• For \(NaF\): \(Na^+\) and \(F^-\), so \(q_1q_2=(+1)\times(-1)=-1\)
• For \(KCl\): \(K^+\) and \(Cl^-\), so \(q_1q_2=(+1)\times(-1)=-1\)

Since \(MgO\) has the highest product of ionic charges (\(\vert q_1q_2\vert = 4\)), it has a strong tendency to have a high lattice energy compared to the others based on charge alone.

Step3: Analyze the ionic radii (when charges are similar)

• For \(NaF\) and \(KCl\):

The ionic radius of \(K^+\) (\(138\space pm\)) is larger than \(Na^+\) (\(102\space pm\)), and the ionic radius of \(Cl^-\) (\(181\space pm\)) is larger than \(F^-\) (\(119\space pm\)). Using \(U=\frac{kq_1q_2}{r}\), since \(q_1q_2\) is the same (\(-1\)) for \(NaF\) and \(KCl\), but \(r\) (sum of cation and anion radii) is larger for \(KCl\) (\(r = r_{K^+}+r_{Cl^-}\)) than for \(NaF\) (\(r=r_{Na^+}+r_{F^-}\)), so \(U_{NaF}>U_{KCl}\)

• For \(MgF_2\) and \(NaF\):

\(Mg^{2 +}\) has a smaller ionic radius (\(72\space pm\)) than \(Na^+\) (\(102\space pm\)), and \(q_1q_2\) for \(MgF_2\) (\(-2\)) has a higher magnitude than \(NaF\) (\(-1\)). So \(U_{MgF_2}>U_{NaF}\)

Answer:

Highest lattice energy: \(MgO\), then \(MgF_2\), then \(NaF\), Lowest lattice energy: \(KCl\)