Boron Phosphide, BP, is a semiconductor, and a hard, abrasion resistant material. It is made by reacting Boron tribromide and phosphorus tribromide in a hydrogen atmosphere at high temperature (>750 degrees C)

(a) Write a Balanced chemical equation for the synthesis of BP

(b) Boron Phosphide crystallizes in zinc blende structure, formed from boron atoms in a face-centered cubic lattice and phosphorus atoms in tetrahedral holes. How many tetrahedralholes are filled with P atoms in each unit cell?

(c) The length of a unit cell of BP is 478pm. What is the density of the solid in g/cm?

(d) Calculate the closest distance between a B and a P atom in the unit cell?
(Assume the B atoms do not touch along the cell edge. The B atoms in the facestouch the B atoms at the corners of the unit cell)

(e.) Determine the radii of B and P, Compared with published values, is the solid ionic ormolecular?

Birke....

(a) The balanced chemical equation for the synthesis of boron phosphide (BP) is as follows:

3BBr3 + 2PBr3 + 4H2 → 2BP + 12HBr

(b) Boron phosphide crystallizes in a zinc blende structure, where boron atoms form a face-centered cubic lattice and phosphorus atoms occupy tetrahedral holes. In a zinc blende structure, each tetrahedral hole is occupied by one atom. Therefore, in each unit cell of BP, there are 4 tetrahedral holes, all filled with phosphorus (P) atoms.

(c) To find the density of the solid (BP) in g/cm³, we need to know the molar mass of BP and the volume of the unit cell.

The molar mass of BP can be calculated by adding the atomic masses of boron (B) and phosphorus (P).

Molar mass of BP = (atomic mass of B) + (atomic mass of P)
= (10.81 g/mol) + (30.97 g/mol)
= 41.78 g/mol

The volume of a unit cell can be determined using the length of a unit cell provided (478 pm). However, we need to convert picometers (pm) to centimeters (cm) before calculating the volume.

1 pm = 1 × 10^-10 cm

Length of unit cell in cm = 478 pm × (1 × 10^-10 cm/pm)
= 4.78 × 10^-8 cm

The volume of a unit cell can be calculated using the formula:

Volume of a cube = (length of side)³

Volume of unit cell = (4.78 × 10^-8 cm)³
= 103.65 × 10^-24 cm³

Now, we can calculate the density by dividing the molar mass by the volume of the unit cell:

Density = (Molar mass of BP) / (Volume of unit cell)
= (41.78 g/mol) / (103.65 × 10^-24 cm³)
≈ 4.03 g/cm³ (approximately)

Therefore, the density of boron phosphide (BP) is approximately 4.03 g/cm³.

(d) The closest distance between a B (boron) and a P (phosphorus) atom in the unit cell can be determined using the length of a unit cell.

In a zinc blende structure, the closest distance between two atoms is along the body diagonal of the unit cell.

The body diagonal of a cube can be found using the formula:

Body Diagonal = √(3 × (length of side)²)

Body Diagonal = √(3 × (4.78 × 10^-8 cm)²)
≈ 8.27 × 10^-8 cm (approximately)

Therefore, the closest distance between a B and a P atom in the unit cell is approximately 8.27 × 10^-8 cm.

(e) To determine the radii of B and P and assess whether the solid is ionic or molecular, we can compare the observed values with the published ionic radii of boron and phosphorus.

Based on experimental data and theoretical calculations, the ionic radii of boron (B³⁺) and phosphorus (P³⁻) are approximately:

- Boron (B³⁺) ionic radius: 27 pm
- Phosphorus (P³⁻) ionic radius: 220 pm

If the observed values for B and P radii are close to these ionic radii, then the solid is more likely to be ionic in nature.

However, without specific information about the observed values or further calculations, it is not possible to determine the exact radii of B and P in comparison to the published values or conclusively determine whether the solid is ionic or molecular.

(a) The balanced chemical equation for the synthesis of Boron Phosphide (BP) can be written as follows:

3BBr3 + 2PBr3 + 6H2 → 2BP + 9HBr

This equation represents the reaction of Boron tribromide (BBr3) and phosphorus tribromide (PBr3) in the presence of hydrogen gas (H2) at high temperature (>750 degrees C) to produce Boron Phosphide (BP) and hydrogen bromide gas (HBr).

(b) Boron Phosphide crystallizes in a zinc blende structure, where boron atoms occupy the face-centered cubic (fcc) lattice, and phosphorus atoms occupy the tetrahedral holes. In the zinc blende structure, each unit cell contains four atoms, i.e., two boron atoms and two phosphorus atoms. Since each phosphorus atom occupies a tetrahedral hole, there are two tetrahedral holes filled with P atoms in each unit cell.

(c) To calculate the density of the solid, we need to know the molar mass of Boron Phosphide (BP). The molar mass of BP is calculated as follows:

Molar mass of BP = (Atomic mass of boron) + (Atomic mass of phosphorus)

Using the atomic masses from the periodic table, we find that the molar mass of boron is approximately 10.81 g/mol, and the molar mass of phosphorus is approximately 30.97 g/mol.

Therefore, the molar mass of Boron Phosphide (BP) ≈ 10.81 g/mol + 30.97 g/mol = 41.78 g/mol

Given that the length of a unit cell of BP is 478 pm (picometers), we need to convert it to centimeters (cm):

1 pm = 1 × 10^(-10) cm

Therefore, 478 pm = 478 × 10^(-10) cm = 4.78 × 10^(-8) cm

Now, we can calculate the density of the solid using the formula:

Density = (Molar mass of BP) / (Volume of the unit cell)

Assuming the unit cell is a cube, the volume of the unit cell is given by:

Volume = (Length of the unit cell)^3

Volume = (4.78 × 10^(-8) cm)^3

Finally, we can plug in the values to obtain the density of Boron Phosphide (BP) in g/cm³.

(d) The closest distance between a B and a P atom in the unit cell can be determined by considering the arrangement of atoms in the zinc blende structure.

In the zinc blende structure, the boron atoms in the face-centered cubic (fcc) lattice touch the boron atoms at the corners of the unit cell. Hence, the distance between the B atoms in the unit cell can be calculated using the length of the unit cell (4.78 × 10^(-8) cm) and the diagonal of the cube formula:

Diagonal = (Length of the unit cell) × √3

Calculating the diagonal:

Diagonal = (4.78 × 10^(-8) cm) × √3

Now, we need to consider the positions of the phosphorus atoms in the tetrahedral holes. The phosphorus atoms are located at the center of each edge of the cube.

The distance between a B and a P atom can be calculated by multiplying the diagonal length by 1/4, since the phosphorus atoms are located at the center of each edge:

Distance between B and P atoms = (Diagonal length) × (1/4)

(e) To determine whether the solid is ionic or molecular, we can compare the radii of the boron (B) and phosphorus (P) atoms. If the solid is ionic, the difference in radii between the cation (B) and anion (P) should be substantial. If the solid is molecular, the radii should be relatively similar.

To access the published values for the atomic radii of boron (B) and phosphorus (P), you can refer to reliable sources such as periodic tables or reputable scientific databases. By comparing the experimental radii with the calculated distances between B and P atoms in the unit cell, you can determine whether the solid is ionic or molecular based on the difference in radii.

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