Question

Consider an ionic solid MX with \(\mathrm{NaCl}\) structure. Construct a new structure \((\mathbf{Z})\) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions \((\mathbf{X})\) except the central one (ii) Replace all the face centered cations ( \(\mathbf{M}\) ) by anions \((\mathbf{X})\) (iii) Remove all the corner cations \((\mathbf{M})\) (iv) Replace the central anion \((\mathbf{X})\) with cation \((\mathbf{M})\) The value of \(\left(\frac{\text { number of anions }}{\text { number of cations }}\right)\) in \(\mathbf{Z}\) is

Short answer

\(\left(\frac{\text{number of anions}}{\text{number of cations}}\right)\) in structure \(\mathbf{Z}\) is 4.

Step-by-step solution

- Visualize the original MX structure

First, recall that the NaCl structure has chloride anions (Cl-) at the corners and face centers of the cube, with sodium cations (Na+) positioned at the center of the edges and in the center of the cube. In the MX structure, M represents the cations and X represents the anions.

- Apply instruction (i)

Remove all anions (X) except the central one. Initially, there is one anion at each of the 8 corners (each shared by 8 unit cells, contributing 1/8 of an anion to the unit cell) and 6 face-centered anions (each shared by 2 unit cells, contributing 1/2 of an anion to the unit cell), summing up to 4 whole anions per unit cell. After removing all but the central anion, there is now 1 anion left in the unit cell.

- Apply instruction (ii)

Replace all face centered cations (M) with anions (X). Since there are 6 face-centered positions each originally occupied by an M (which is shared with the neighboring cell, so each contributes 1/2 to the unit cell), this step adds 3 more anions to the unit cell.

- Apply instruction (iii)

Remove all corner cations (M). Initially, there are 8 corner positions with cations, each shared by 8 unit cells contributing 1/8 of a cation to the unit cell, summing up to 1 whole cation per unit cell. These are now removed entirely.

- Apply instruction (iv)

Replace the central anion (X) with a cation (M). After replacing the central anion with a cation, the total number of anions remains 4 (3 from face centered and 1 we added in step 2 and then subtracted in this step), while the number of cations becomes 1.

- Calculate the ratio \(\left(\frac{\text{number of anions}}{\text{number of cations}}\right)\) in structure Z

From the previous steps, we have 4 anions and 1 cation. The ratio is therefore \(\left(\frac{\text{number of anions}}{\text{number of cations}}\right) = \frac{4}{1} = 4\).

Key concepts

NaCl Structure

The structure of Sodium Chloride (NaCl) is one of the classic examples of ionic solid structures taught in chemistry and material science. This structure features a highly ordered, repeating pattern known as a crystal lattice. In the NaCl structure, chloride ions (Cl-) form a cubic array, occupying the corners and the centers of each face of a cube. These anions essentially shape the lattice structure around which sodium ions (Na+) find their positions at the centers of the edges and exactly in the center of the cube.

This arrangement allows each ion, both cation and anion, to be surrounded by ions of the opposite charge, explaining why NaCl and similar ionic solids are such effective conductors of electricity when dissolved in water; the ions are free to move. Such geometric arrangement is crucial in understanding many physical properties of an ionic solid, including its density, melting point, and solubility.

Because ions in the NaCl structure are held together by strong electrostatic forces -- the attraction between opposite charges -- the resulting compound is characterized by high melting and boiling points, and a hard, brittle nature. The NaCl lattice serves as a foundational model for understanding more complex ionic structures.

Unit Cell Construction

When studying crystalline materials, the unit cell is the smallest division of the crystal lattice that still reflects the overall structure of the entire solid. In the case of the NaCl structure, the unit cell is a cube, but within that cube, the arrangement of ions is carefully balanced to reflect the crystal as a whole. Constructing a unit cell for an ionic solid involves several steps:

First, identify the positions of the ions in the lattice. For NaCl, as mentioned previously, the Cl- ions reside at the corners and face-centers, whereas the Na+ ions are at edge-centers and the cube center. Each position within this unit cell is a point in space where an ion is repeated throughout the entire lattice when the unit cells are stacked in three dimensions. Constructing a unit cell with these positions in mind allows us to make accurate predictions and calculations about the properties of the material.

Moreover, by changing the positions of the ions according to certain rules or instructions as in the exercise, one can visualize how altering the structure can lead to different properties. These manipulations help students to understand how different crystal structures can be formed and the creativity required in material science to engineer new materials with desired characteristics.

Cation-Anion Ratio

The ratio of cations to anions in an ionic solid is a key factor in determining its chemical formula and properties. In the NaCl structure, the ratio is 1:1, with each Na+ ion being surrounded by six Cl- ions, and vice versa. This ratio is essential in maintaining electrical neutrality within the crystal lattice, as the charges must balance.

In the exercise, transformations were made to the original NaCl-inspired structure of an ionic solid MX, resulting in changes to this crucial ratio. The sequential instructions of anion and cation removal and replacement have profound effects on the resulting structure, which would in real-life scenarios greatly alter the properties of the compound.

A different cation-anion ratio will affect not just the chemical stoichiometry but also the physical structure: lattice parameters may change, possibly resulting in different crystal systems or alterations in conductivity and mechanical strength. Such exercises reinforce the connection between the microscopic arrangement of ions and the macroscopic physical and chemical behavior of materials. Understanding this relationship is vital for anyone studying chemistry, materials science, or related fields.