Question

MnO has either the \(\mathrm{NaCl}\) type structure or the \(\mathrm{CsCl}\) type structure (see Exercise 67). The edge length of the \(\mathrm{MnO}\) unit cell is \(4.47 \times 10^{-8} \mathrm{~cm}\) and the density of \(\mathrm{MnO}\) is \(5.28 \mathrm{~g} / \mathrm{cm}^{3}\) a. Does \(\mathrm{Mn} \mathrm{O}\) crystallize in the \(\mathrm{NaCl}\) or the \(\mathrm{CsCl}\) type structure? b. Assuming that the ionic radius of oxygen is \(140 . \mathrm{pm}\), estimate the ionic radius of manganese.

Short answer

a. MnO crystallizes in the NaCl type structure, as its calculated density is closer to the given density of MnO. b. The estimated ionic radius of manganese is \(618\ \text{pm}\).

Step-by-step solution

Calculate the number of formula units per unit cell for NaCl and CsCl type structures

To do this, we need to remember that the NaCl structure has a face-centered cubic arrangement of atoms, while the CsCl structure has a simple cubic arrangement. In the NaCl structure, there are 4 formula units per unit cell, while in the CsCl structure, there is 1 formula unit per unit cell.

Calculate the mass of one formula unit of MnO

Given the atomic weights of Mn and O: Mn = 54.94 g/mol and O = 16.00 g/mol, the molar mass of MnO is: MnO = 54.94 g/mol + 16.00 g/mol = 70.94 g/mol Now, let's convert this to the mass of one formula unit: Mass of one formula unit of MnO = 70.94 g/mol /(6.022 x 10²³) = 1.178 x 10⁻²² g

Calculate density for both NaCl and CsCl type structures

To calculate the density, we need to find the mass of the formula units in the unit cell and divide it by the volume of the unit cell. For NaCl type structure: Number of formula units = 4 Mass in the unit cell = 4 × 1.178 × 10⁻²² g = 4.712 × 10⁻²² g Edge length (a) = \(4.47 × 10^{-8}\ \text{cm}\) Volume of unit cell = (4.47 x 10⁻⁸)³ cm³ = 8.931 x 10⁻²³ cm³ Density (NaCl type) = (4.712 x 10⁻²² g) / (8.931 x 10⁻²³ cm³) = 5.276 g/cm³ For CsCl type structure: Number of formula units = 1 Mass in the unit cell = 1 × 1.178 × 10⁻²² g = 1.178 × 10⁻²² g Volume of unit cell remains the same. Density (CsCl type) = (1.178 x 10⁻²² g) / (8.931 x 10⁻²³ cm³) = 1.319 g/cm³ Comparing the two calculated densities with the given density of MnO, it's clear that the NaCl type structure has a closer density. a. So, MnO crystallizes in the NaCl type structure.

Estimate the ionic radius of manganese

In the NaCl type structure, the ions are in contact along the face diagonal. Therefore, the face diagonal length is equal to the sum of the ionic radii of Mn and O ions. Since the structure is face-centered cubic, the face diagonal is \(\sqrt{2}\) times the edge length, so: Face diagonal length = \(\sqrt{2}\) × \(4.47 × 10^{-8}\ \text{cm}\) = \(6.32 × 10^{-8}\ \text{cm}\) Given the ionic radius of oxygen (O) as \(1.40 × 10^{-10}\ \text{cm}\), we can solve for the ionic radius of manganese (Mn): Face diagonal length = Ionic radius of Mn + Ionic radius of O Ionic radius of Mn = Face diagonal length - Ionic radius of O Ionic radius of Mn = \(6.32 × 10^{-8}\ \text{cm}\) - \(1.40 × 10^{-10}\ \text{cm}\) = \(6.18 × 10^{-8}\ \text{cm}\) Converting to picometers: Ionic radius of Mn = \(6.18 × 10^{-8}\ \text{cm}\) × 10¹º pm/cm = \(618\ \text{pm}\) b. The estimated ionic radius of manganese is \(618\ \text{pm}\).

Key concepts

NaCl type structure

The NaCl type structure, also known as the rock-salt structure, is prevalent in many binary compounds like table salt and metal oxides. It is characterized by a face-centered cubic (fcc) lattice arrangement. Here, each ion is surrounded by six ions of the opposite charge. This leads to a coordination number of 6. The unit cell of NaCl is composed of a cubic arrangement where there are 4 formula units per unit cell.

This structural arrangement results in an efficient packing of ions and significantly influences the physical properties, such as density. In our context, MnO adopts this NaCl structure, which affects its density calculation as shown in the solution. The near match to the given density helps confirm this structure, distinguishing it from other potential arrangements like the CsCl type.

CsCl type structure

In contrast, the CsCl type structure is a unique arrangement where the ions form a simple cubic structure. This is characterized by a body-centered cubic (bcc) lattice. The primary feature of the CsCl structure is that each ion is directly surrounded by 8 neighboring ions of the opposite charge, giving it a coordination number of 8.

This structural arrangement results in only 1 formula unit per unit cell. Such a setup can lead to significantly different physical properties, such as density. As seen in the solution, the calculated density for MnO does not match well with this structure, which suggests that MnO does not crystallize in the CsCl type. The larger coordination number often impacts the lattice energy and stability of the compound.

density calculation

Calculating the density of a crystallized compound involves some key steps. Density (\(\rho\)) is expressed as mass per unit volume. Knowing the type of crystal structure can guide this calculation:

• Identify the number of formula units within the unit cell.
• Calculate the total mass of those units using known atomic masses, as seen with MnO's 70.94 g/mol molar mass.
• Calculate the unit cell's volume using the cube of the edge length.
• Divide the total mass by this volume to yield the density.

As demonstrated, the step showing calculations for NaCl and CsCl structures helps decide which structure aligns with the given density of MnO. The match in density with the NaCl structure indicates the compound's likely configuration.

ionic radius estimation

Estimating the ionic radius within a crystal structure can give insight into the spatial arrangement of ions. In a NaCl type structure, this helps determine the sizes of the ions packed into the lattice. When calculating ionic radii, the contact between ions along diagonals is considered:

• Start by finding the face diagonal using geometry (for fcc: \(\sqrt{2}\) times the edge length).
• The face diagonal equals the sum of the ionic radii of the metal and non-metal in the lattice.
• By isolating the unknown radius, such as Mn's, from the diagonal's length, you arrive at its specific measure.

In our solution, it shows how using the known radius of oxygen and the crystallography principles, the radius of manganese is estimated effectively. This understanding is crucial in material science for predicting and analyzing compound behaviors.