Modern methods of structural analysis using X-rays provide valuable information about the three dime — Inorganic Chemistry — Solid State Chemistry Question
Ionic and metallic structures
Modern methods of structural analysis using X-rays provide valuable information about the three dimensional arrangement of atoms, molecules or ions in a given crystal structure.
Crystal structure of rock salt (NaCl) is given below.
[VISUAL]
What is the type of crystal lattice presented in the diagram?
Model Answer
The lattice of NaCl consist of interpenetrating fcc lattices of Na+ and Cl–.
What is the coordination number of a sodium ion in this structure?
Model Answer
The co-ordination number of sodium is six since, it is surrounded by six nearest chloride ions.
What is the number of formula units of NaCl per unit cell?
Model Answer
For NaCl, the number of Na+ ions is: twelve at the edge centres shared equally by four unit cells thereby effectively contributing 12 × 1/4 = 3 Na+ ions per unit cell and one at body center. Thus, a total of 3 + 1 = 4 Na+ ions per unit cell. Number of Cl– ions is: six at the center of the faces shared equally by two unit cells, thereby effectively contributing 6 × 1/2 = 3 Cl– ions per unit cell and eight at the corners of the unit cell shared equally by eight unit cells thereby effectively contributing 8 × 1/8 = 1 Cl– ion per unit cell. Thus, a total of 3 + 1 = 4 Cl– ions per unit cell. Hence, the number of formula units of NaCl per unit cell = 4 Na+ + 4 Cl– = 4 NaCl.
Calculate the r_Na / r_Cl limiting radius ratio for this structure. Why is the array of chloride ions slightly expanded, with the nearest Cl-Cl distance being 400 pm, compared to the close packed value of 362 pm?
Model Answer
The face diagonal of the cube is equal to √2 times 'a' the lattice constant for NaCl. The anions/anions touch each other along the face diagonal. The anion/cations touch each other along the cell edge. Thus, a = 2 (r_Na + r_Cl) ... (1). Face diagonal √2 a = 4 r_Cl ... (2). Substituting for 'a' from (1) into (2) we get: √2 * 2 (r_Na + r_Cl) = 4 r_Cl from which, the limiting radius ratio r_Na / r_Cl = 0.414. The chloride ion array is expanded to make the octahedral holes large enough to accommodate the sodium ions since, the r_Na / r_Cl ratio of 0.564 is larger than the ideal limiting value of 0.414 for octahedral six coordination number.
What happens when the cation radius in the structure shown above is progressively increased till the cation/anion radius ratio reaches a value of 0.732?
Model Answer
As the cation radius is progressively increased, the anions will no longer touch each other and the structure becomes progressively less stable. There is insufficient room for more anions till the cation / anion radius ratio equals 0.732 when, eight anions can just be grouped around the cation resulting in a cubic eight coordination number as in CsCl.
What is the range of cation/anion radius ratio for which the structure like that of NaCl is stable?
Model Answer
Generally, the fcc structure with a six coordination number is stable in the cation/anion radius ratio range 0.414 to 0.732. That is, if 0.414 < r+/r- < 0.732 then, the resulting ionic structure will generally be NaCl type fcc.
The Cu-K X-ray (λ = 154pm) reflection from (200) planes of sodium chloride crystal is observed at 15.8°. Given that the radius of the chloride ion is 181 pm, calculate:
i. the separation between adjacent 200 planes of NaCl.
Model Answer
Bragg's law states λ = 2 d_hkl sin θ. 154 pm = 2 × d_200 sin(15.8°). d_200 = 154 pm / (2 × sin(15.8°)) = 154 pm / (2 × 0.2722) = 283 pm. Thus, the separation between the (200) planes of NaCl is 283 pm.
the length of the unit cell edge (lattice constant) of NaCl.
Model Answer
Length of the unit cell edge, a = d_100 = 2 × d_200. a = 2 × 283 pm = 566 pm.
the radius of the sodium ion.
Model Answer
Since it is an fcc lattice, cell edge, a = 2 (r_Na+ + r_Cl–). radius of sodium ion r_Na+ = a/2 – r_Cl– = (566 / 2) – 181 = 102 pm. In the alternative form: 2 × r_Na+ = a - 2 × r_Cl– = 566 – 362 = 204 pm, which yields r_Na+ = 102 pm.
The diagram of a cubic close packing (ccp) and a hexagonal close packing (hcp) lattice arrangement (assuming rigid sphere model) is given below.
[VISUAL]
Describe the difference between the ccp and hcp lattice arrangements.
Model Answer
The difference in an hcp and a ccp arrangement is as follows:
The two 'A' layers in a hcp arrangement are oriented in the same direction making the packing of successive layers ABAB.. and the pattern repeats after the second layer whereas, they are oriented in the opposite direction in a ccp arrangement resulting in a ABCABC… packing pattern which repeats after the third layer.
The unit cell in a ccp arrangement is based on a cubic lattice whereas in a hcp arrangement it is based on a hexagonal lattice.
Calculate the packing fraction for a ccp arrangement.
Model Answer
Packing fraction = (volume occupied by 4 atoms) / (volume of unit cell). Let 'a' be the length of the unit cell edge. Since it is an fcc lattice, face diagonal = √2 a = 4 r ... (1). Volume of the unit cell = a^3. Volume occupied by 4 atoms = 4 × (4/3) π r^3 ... (2). Substituting for 'a' from (1) into (2), we get: Packing fraction = (16/3 π r^3) / (4 r / √2)^3 = π / (3√2) ≈ 0.74. Thus, the packing fraction in a ccp arrangement = 0.74.
Will the coordination number, and the packing fraction in a hcp arrangement be the same as that in a ccp arrangement?
Model Answer
The coordination number (12) and the packing fraction (0.74) remain the same in a hcp as in a ccp arrangement.
Nickel (at.wt. 58.69) crystallizes in the ccp structure. X-ray diffraction studies indicate that its unit cell edge length is 352.4 pm. Given that the density of Nickel is 8.902 g cm^-3, calculate:
i. the radius of the nickel atom.
Model Answer
For an fcc, face diagonal = √2 a = 4 r_Ni where a = lattice constant and r_Ni = radius of the nickel atom. r_Ni = (√2 × a) / 4 = (√2 × 352.4 pm) / 4 = 124.6 pm.
the volume of the unit cell.
Model Answer
Volume of unit cell = a^3 = (3.524 Å)^3 = 43.76 Å^3 (or 4.376 × 10^–23 cm^3).
the Avogadro number.
Model Answer
Density of nickel ρ_Ni = Z × M / (V × N_A). Number of Ni atoms per unit cell, Z = 4 for an fcc lattice. Avogadro constant N_A = Z × M / (V × ρ_Ni) = (4 × 58.69 g mol^–1) / (8.902 g cm^–3 × 43.76 × 10^–24 cm^3) = 6.02×10^23 mol^–1.