Boron phosphide (BP) adopts the zinc blende structure. The length of the unit cell edge is \(0.457 \mathrm{nm}\). (a) Calculate the density of \(\mathrm{BP}\) in this form. (b) Boron arsenide (BAs) also forms a solid phase with the zinc blende structure. The length of the unit cell edge in this compound is $0.478 \mathrm{nm}$. What accounts for the larger unit cell length? (c) Which of the two substances has the higher density? How do you account for the difference in densities?

The zinc blende structure is a face-centered cubic (FCC) crystal lattice that consists of two interpenetrating FCC lattices with the first type of atom, say A, placed on one FCC lattice, and the second type of atom, say B, placed on the other FCC lattice. In our case, Boron Phosphide (BP) and Boron Arsenide (BAs) adopt zinc blende structures, with each Boron atom being surrounded by four phosphorus or arsenic atoms, and vice versa.

In the zinc blende structure, there are 8 corner atoms and 6 face atoms for each type of atom (A and B). Since each corner atom is shared by 8 unit cells, and each face atom is shared by 2 unit cells, the total number of whole atoms in the unit cell is: \[ \frac{8 \times 1}{8} + \frac{6 \times 1}{2} = 4 \] So, the unit cell contains 4 atoms of A and 4 atoms of B, a total of 8 atoms.

To calculate the density, we will use the formula: \[density = \frac{mass}{volume}\] where mass is calculated using the molecular weight of the compound and the number of atoms in the unit cell, and volume is calculated from the given edge length of the unit cell. (a) For Boron Phosphide: Given the unit cell edge length, \(a = 0.457×10^{-9} m\) Density = \( \frac{8 \times MolecularWeight(BP)}{Volume(unit\ cell)}\) Molecular Weight of BP = B + P = 10.8 + 30.9 = 41.7 g/mol Volume of the unit cell = \(a^3 = (0.457 \times 10^{-9})^3\) Density = \( \frac{8 \times 41.7}{(0.457 \times 10^{-9})^3} \frac{g / mol}{m^3} \times N_A \), where \(N_A\) is Avogadro's constant Density of BP: \(2.56 \times 10^3 \frac{g}{cm^3}\) (b) For Boron Arsenide: Given the unit cell edge length, \(a = 0.478×10^{-9} m\) Molecular Weight of BAs = B + As = 10.8 + 74.9 = 85.7 g/mol Volume of the unit cell = \(a^3 = (0.478 \times 10^{-9})^3\) Density = \( \frac{8 \times 85.7}{(0.478 \times 10^{-9})^3} \frac{g / mol}{m^3} \times N_A \) Density of BAs: \(5.31 \times 10^3 \frac{g}{cm^3}\)

We can now compare the densities of BP and BAs. From our calculations, the density of BP is around \(2.56 \times 10^3 \frac{g}{cm^3}\), while BAs has a higher density of about \(5.31 \times 10^3 \frac{g}{cm^3}\). The larger unit cell length in BAs is due to the larger size of the arsenic atoms compared to phosphorus atoms. Consequently, this leads to a higher density in BAs as a result of the increased molecular weight of the BAs compound. In summary, the substance with the higher density is Boron Arsenide, and the difference in densities can be attributed to the difference in atomic sizes and the molecular weight of each compound within their respective zinc blende structures.