Question

18 sep analyze and interpret data the table shows several ionic solids, their average bond energies in electron volts (ev), and their melting points. use the pattern indicated in the table to predict the average bond energy of calcium fluoride (caf₂), which has a melting point of 1420°c. bond energy and melting point of selected compounds material bond energy (ev/bond) melting point (°c) cesium chloride (cscl) 6.747 646 sodium chloride (nacl) 7.96 801 lithium fluoride (lif) 10.45 870 copper fluoride (cuf₂) 26.85 1360 aluminum oxide (al₂o₃) 157.46 3500

Explanation:

Step1: Observe the relationship

As the bond - energy increases, the melting point generally increases.

Step2: Determine position of CaF₂

The melting point of CaF₂ ($1420^{\circ}C$) is between that of CuF₂ ($1360^{\circ}C$) and Al₂O₃ ($3500^{\circ}C$). So its bond - energy should be between that of CuF₂ (26.85 eV/bond) and Al₂O₃ (157.46 eV/bond).

Step3: Estimate bond - energy

We can assume a linear relationship (a simple approximation). Let's consider the range of melting points from CuF₂ to Al₂O₃: $\Delta T=3500 - 1360=2140^{\circ}C$, and the range of bond - energies: $\Delta E = 157.46-26.85 = 130.61$ eV/bond. The difference in melting point between CuF₂ and CaF₂ is $\Delta T_{1}=1420 - 1360 = 60^{\circ}C$. Then the increase in bond - energy $\Delta E_{1}$ can be estimated as $\Delta E_{1}=\frac{60}{2140}\times130.61\approx3.67$ eV/bond. So the estimated bond - energy of CaF₂ is $E\approx26.85 + 3.67=30.52$ eV/bond.

Answer:

Approximately 30.52 eV/bond