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Chapter 12: Problem 61

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?

Short Answer

Expert verified
In summary, Boron Arsenide (BAs) has a higher density of about \(5.31 \times 10^3 \frac{g}{cm^3}\) compared to Boron Phosphide's (BP) \(2.56 \times 10^3 \frac{g}{cm^3}\). The larger unit cell length in BAs is due to the larger size of arsenic atoms compared to phosphorus atoms, which also leads to a higher molecular weight for BAs. The difference in densities between the two substances can be attributed to the difference in atomic sizes and molecular weights within their respective zinc blende structures.

Step by step solution

01

Understand the zinc blende structure

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.
02

Calculate the number of atoms in the unit cell for both BP and BAs

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.
03

Calculate the density of BP and BAs

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, \(a = 0.457脳10^{-9} m\) 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, \(a = 0.478脳10^{-9} m\) 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: \(5.31 \times 10^3 \frac{g}{cm^3}\)
04

Compare the densities and account for the differences

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Calculation
Density is a crucial property used in chemistry and material science to express how much mass is present in a given volume of a substance. In this context, we focus on calculating the density of compounds that form a zinc blende structure like Boron Phosphide (BP) and Boron Arsenide (BAs). Here is a simple way to calculate density using the formula:
  • Density = \( \frac{\text{mass}}{\text{volume}} \)
This formula helps relate the molecular weight of a compound to the volume of its unit cell.
First, determine the mass present in a unit cell. This can be done by calculating the total molecular weight of the atoms in the unit cell. Each zinc blende unit cell contains 8 atoms. Knowing the molecular weights of individual atoms helps provide this total mass.
Next, compute the volume of the unit cell, which is given by the cube of the length of its edge, \( a^3 \). With both mass and volume, you can easily substitute these values into the density formula to find the density of each compound.
Crystal Lattice
Understanding crystal lattices helps us see how atoms are arranged in a crystalline solid. The zinc blende structure is a specific type of crystal lattice that is prevalent in many compounds, including BP and BAs. It can be visualized as two interpenetrating face-centered cubic (FCC) lattices.
In a zinc blende crystal lattice, one type of atom occupies the lattice points of the FCC structure. The other type fits into the tetraherron gaps formed between the FCC atoms. For BP and BAs:
  • Boron atoms take the positions of one FCC sub-lattice.
  • Phosphorus or arsenide fill the voids in this lattice.
This highly symmetric arrangement is why the zinc blende structure exhibits many important mechanical and electronic properties, making them valuable for various technological applications.
Molecular Weight
Molecular weight is a critical parameter in determining the physical properties of compounds such as BP and BAs. It is defined as the sum of the weights of the atoms in a molecule and is usually expressed in grams per mole (g/mol).
For Boron Phosphide, we combine the atomic weights of boron (10.8 g/mol) and phosphorus (30.9 g/mol), resulting in a total molecular weight of 41.7 g/mol.
Boron Arsenide's molecular weight is calculated similarly. With boron at 10.8 g/mol and arsenic at 74.9 g/mol, we get a total molecular weight of 85.7 g/mol.
  • The heavier molecular weight of BAs leads to a greater mass in its unit cell compared to BP.
This variance in molecular weight directly influences the density calculations, as heavier compounds will generally have higher densities if the atomic packing factor is similar.
Atomic Size
Atomic size is essential in determining how atoms arrange themselves in a solid and ultimately affect properties like density. In our context of zinc blende structures, comparing atomic sizes between phosphorus and arsenic provides insight into why unit cell lengths vary and why BAs has a higher density than BP.
Arsenic atoms are larger than phosphorus atoms, which explains why BAs forms a larger unit cell \((0.478 \ \text{nm})\) compared to BP \((0.457 \ \text{nm})\). Despite the larger unit cell:
  • The mass increase from the heavier arsenic atoms outweighs the volume increase, resulting in higher density for BAs.
Understanding the effects of atomic size, therefore, allows us to comprehend and compare the physical properties between substances that have similar structures but different constituent atoms.

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Most popular questions from this chapter

Which type (or types) of crystalline solid is characterized by each of the following? (a) High mobility of electrons throughout the solid; (b) softness, relatively low melting point; (c) high melting point and poor electrical conductivity; \((\mathbf{d})\) network of covalent bonds.

One method to synthesize ionic solids is by the heating of two reactants at high temperatures. Consider the reaction of \(\mathrm{MgO}\) with \(\mathrm{TiO}_{2}\) to form \(\mathrm{MgTiO}_{3}\). Determine the amount of each of the two reactants to prepare \(3.250 \mathrm{~g} \mathrm{MgTiO}_{3}\), assuming the reaction goes to completion. (a) Write a balanced chemical reaction. (b) Calculate the formula weight of \(\mathrm{MgTiO}_{3}\). (c) Determine the moles of \(\mathrm{MgTiO}_{3}\). (d) Determine moles and mass (g) of MgO required. (e) Determine moles and mass (g) of \(\mathrm{TiO}_{2}\) required.

Arrange the following metals in increasing order of expected melting point: La, W, Ta, Hf. Explain this trend in melting points.

Which of these statements about alloys and intermetallic compounds is false? (a) Bronze is an example of an alloy. (b) "Alloy" is just another word for "a chemical compound of fixed composition that is made of two or more metals." (c) Intermetallics are compounds of two or more metals that have a definite composition and are not considered alloys. (d) If you mix two metals together and, at the atomic level, they separate into two or more different compositional phases, you have created a heterogeneous alloy. (e) Alloys can be formed even if the atoms that comprise them are rather different in size.

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