NaFMgO

Boiling Point (°C)
16953600







$N a ^{ + }$$Mg ^{2+}$$F ^{-}$$Cl ^{-}$$O ^{2-}$

Ionic Radius (pm)
7672133181140






Based on the data in the tables above, which of the following statements provides the best prediction for the boiling point of NaCl?


ANaCl will have a lower boiling point than NaF because the coulombic attractions are weaker in NaCl than in NaF.
BNaCl will have a boiling point between that of NaF and MgO because the covalent character of the bonds in NaCl is intermediate between that of MgO and NaF.
CNaCl will have a higher boiling point than MgO because the ions are spaced farther apart in NaCl.
DNaCl will have a higher boiling point than MgO because the energy required to transfer electrons from the anion to the cation is larger in NaCl than in MgO.

Verified Answer :

  • The boiling point of an ionic compound is influenced by the lattice energy, which depends on the ionic charges and the interionic distance.
  • \(\text{NaCl}\) and \(\text{NaF}\) have the same charge product (\(1\)), but \(\text{NaCl}\) has a larger interionic distance (\(257 \text{ pm}\) vs \(209 \text{ pm}\)).
  • Larger distance reduces the Coulombic attraction, leading to a weaker ionic bond and a lower boiling point.
  • Therefore, \(\text{NaCl}\) has a lower boiling point than \(\text{NaF}\), which is option (A).
  • The final answer is \(\boxed{\text{A}}\).

Explanation

  1. Understanding the Physical Principles of Ionic Boiling Points

To predict the boiling point of \(\text{NaCl}\), we analyze the factors that determine the strength of ionic bonds, specifically Coulombic attraction and lattice energy. Coulombic force between two ions is directly proportional to the product of their charges (\(q_1 \times q_2\)) and inversely proportional to the square of the distance between their centers (\(r^2\)), where \(r\) is the sum of the ionic radii (\(r = r_{\text{cation}} + r_{\text{anion}}\)). Consequently, the lattice energy, which correlates with the boiling point, is approximately proportional to \(\frac{q_1 \times q_2}{r}\).

  1. Analyzing Charge and Distance for NaF, MgO, and NaCl

We calculate the relative interionic distances and analyze the charges for each compound:

  • For \(\text{NaF}\): Charges are \(\text{Na}^+\) (\(+1\)) and \(\text{F}^-\) (\(-1\)). Product of charges is \(|(+1) \times (-1)| = 1\). The distance is \(r = 76 \text{ pm} + 133 \text{ pm} = 209 \text{ pm}\).
  • For \(\text{MgO}\): Charges are \(\text{Mg}^{2+}\) (\(+2\)) and \(\text{O}^{2-}\) (\(-2\)). Product of charges is \(|(+2) \times (-2)| = 4\). The distance is \(r = 72 \text{ pm} + 140 \text{ pm} = 212 \text{ pm}\).
  • For \(\text{NaCl}\): Charges are \(\text{Na}^+\) (\(+1\)) and \(\text{Cl}^-\) (\(-1\)). Product of charges is \(|(+1) \times (-1)| = 1\). The distance is \(r = 76 \text{ pm} + 181 \text{ pm} = 257 \text{ pm}\).
  1. Comparing NaCl to NaF and Conclusion

Now we compare \(\text{NaCl}\) with \(\text{NaF}\). Both have the same charge product (\(1\)). However, the interionic distance for \(\text{NaCl}\) (\(257 \text{ pm}\)) is significantly larger than the interionic distance for \(\text{NaF}\) (\(209 \text{ pm}\)). According to the relationship \(\text{Lattice Energy} \propto \frac{q_1 \times q_2}{r}\), since the charge product is the same and the distance \(r\) is larger in \(\text{NaCl}\), the attractive forces are weaker in \(\text{NaCl}\) than in \(\text{NaF}\). Weaker attractive forces result in a lower boiling point. Thus, \(\text{NaCl}\) must have a lower boiling point than \(\text{NaF}\). This aligns with option (A).

Examples

In chemistry, predicting the properties of materials is essential for material design. For instance, if you are selecting a material for a high-temperature application, you look for compounds with high lattice energies, which generally come from ions with high charges and small ionic radii. By understanding these principles, scientists can predict whether a material like NaCl (common salt) will melt or boil at a lower temperature than a ceramic like MgO or a fluoride like NaF.