The structure of sodium chloride, NaCl, is one of the basic crystal structure types of ionic compoun — Inorganic Chemistry — Solid State Chemistry Question
Structures in the solid state
The structure of sodium chloride, NaCl, is one of the basic crystal structure types of ionic compounds. Its face-centred cubic unit cell is shown in Figure 1. [VISUAL]
The lattice constant of NaCl is a = 5.64 Å and the radius of the sodium(I) ion is r(Na+) = 1.16 Å.
Calculate the ionic radius of a chloride ion, r(Cl–).
Model Answer
It is obvious from the picture that:
a(NaCl) = 2 × r(Na+) + 2 × r(Cl−), and therefore:
r(Cl–) = ½ × (5.64 − 2 × 1.16) Å = 1.66 Å.
Potassium chloride, KCl, crystallizes in the same crystal structure type. The density of solid KCl is ρ(KCl) = 1.98 g cm–3.
Calculate the ionic radius of a potassium ion, r(K+).
Model Answer
The density of KCl is:
ρ(KCl) = m / V = [4 × M(KCl)] / [NA × a(KCl)³], and therefore:
a(KCl) = {[4 × M(KCl)] / [NA × ρ(KCl)]}⅓ = [(4 × 74.55) / (6.022 × 10²³ × 1.98)]⅓ cm = 6.30 × 10⁻⁸ cm = 6.30 Å
r(K+) = ½ × Å = ½ × (6.30 − 2 × 1.66) Å = 1.49 Å.
The structure of ionic compounds can be estimated using relative sizes of the cation and anion, as r+/r− ratio determines what kind of cavity found in the anionic lattice could be occupied by the cation.
The ionic radius of a lithium ion is r(Li+) = 0.90 Å. Estimate whether LiCl adopts the same crystal structure type as NaCl or not.
Model Answer
The ratio of ionic radii of Li+ to Cl− is:
r(Li+) / r(Cl−) = 0.90 / 1.66 = 0.54. It is higher than the relative size of the octahedral cavity (0.41), which is a critical value for an ion to occupy this cavity. Thus, occupying this cavity by Li+ ion will result in a stable arrangement and LiCl should crystallize in the NaCl type of structure.
(Taking into account the smaller size of Li+ compared to Na+, it is not necessary to consider the upper limit of an ion size for a stable arrangement. However, for completeness, one can assume that the relative size of a cation with respect to an anion higher than 0.73 should lead to the change of the coordination sphere and enforce a cubic coordination environment and the structure type of CsCl. In fact, the ratio for KCl is somewhat above this limiting value, but KCl still adopts the structural type of NaCl as stated above.)
Some ionic compounds of divalent ions also crystallize in the crystal structure type of NaCl, for example galena, PbS. Its lattice constant is a = 5.94 Å.
Calculate the density of galena.
Model Answer
ρ(PbS) = m / V = [4 × M(PbS)] / [NA × a(PbS)³] = (4 × 239.3) / (6.022 × 10²³ × 5.94³) g Å⁻³ = 7.58 g cm⁻³
Since silver(I) ions can be substituted for lead(II) ions in the structure of PbS, galena is a very important silver ore. To ensure electro-neutrality of the crystal, the decrease in the overall positive charge is compensated by the vacancies of sulphide anions. The composition of such a phase can be expressed by a general formula Pb₁₋ₓAgₓSy.
Derive the value of y as a function of x.
Model Answer
Due to total electro-neutrality, one can derive: 2(1 − x) + 1x = 2y, and thus: y = 1 − ½x
A general formula of silver-containing galena is thus Pb₁₋ₓAgₓS₁₋½ₓ.
A sample of silver-containing galena, in which a part of the lead(II) ions are substituted by silver(I) ions and the decrease in charge is compensated by the vacancies of sulphide ions, has a density of 7.21 g cm⁻³. The lattice constant of this sample is a = 5.88 Å.
Calculate the value of the stoichiometric coefficient x.
Model Answer
ρ(Pb₁₋ₓAgₓS₁₋½ₓ) = m / V = [4 × M(Pb₁₋ₓAgₓS₁₋½ₓ)] / [NA × a(Pb₁₋ₓAgₓS₁₋½ₓ)³], and therefore:
M(Pb₁₋ₓAgₓS₁₋½ₓ) = ρ(Pb₁₋ₓAgₓS₁₋½ₓ) × NA × a(Pb₁₋ₓAgₓS₁₋½ₓ)³ / 4 = [7.21 × 6.022 × 10²³ × (5.88 × 10⁻⁸)³] / 4 g mol⁻¹ = 220.7 g mol⁻¹, and thus:
207.2(1 − x) + 107.9x + 32.1(1 − ½x) = 220.7, and x = 0.16
Zinc blende (sphalerite, ZnS) crystallizes in a different crystal structure type, which is closely related to the structure of diamond. Both types of structures are shown in Figure 2. [VISUAL]
How many formula units (ZnS) are there in the unit cell of sphalerite?
Model Answer
Four.
Heavier elements of group IV (i.e. group 14), silicon and germanium, also adopt the structure of diamond. The radius of elemental germanium is r(Ge) = 1.23 Å.
Calculate the density of solid germanium.
Model Answer
According to the crystal structure type, the bond distance of Ge–Ge corresponds to one fourth of the body diagonal of the unit cell and, thus, atomic radius is equal to one eighth of the body diagonal. Thus, the lattice parameter a(Ge) is:
a(Ge) = 8 × r(Ge) / √3 = 8 × 1.23 / √3 Å = 5.68 Å,
giving the density:
ρ(Ge) = m / V = [8 × M(Ge)] / [NA × a(Ge)³] = (8 × 72.6) / (6.022 × 10²³ × 5.68³) g Å⁻³ = 5.26 g cm⁻³
(Alternatively, substituting the a(Ge) by the term 8 × r(Ge)/√3 in the latter equation gives the following formula:
ρ(Ge) = m / V = [3√3 × M(Ge)] / [64 × NA × r(Ge)³] = 5.26 g cm⁻³
without the need to calculate the lattice parameter. However, the similarity of the lattice constant of Ge with that of the isoelectronic GaAs is emphasized by the previous approach.)
Germanium is a semiconductor similar to silicon. It is used in electro-technology and similarly to silicon, it is also very fragile. Therefore, more flexible isoelectronic gallium arsenide, GaAs, is used in some practical applications. This compound belongs to semi-conductors of III–V type (compounds of elements from groups III and V, i.e. groups 13 and 15, respectively) and adopts the structure of sphalerite. The lattice constants of Ge and GaAs are very similar, and a(GaAs) = 5.65 Å. An analogous compound GaP also adopts the structure of sphalerite, but has a smaller unit cell with a(GaP) = 5.45 Å.
Calculate the difference between the radii of P and As in the respective compounds with gallium (GaP versus GaAs).
Model Answer
According to the crystal structure type, the bond distance of Ga–As (and Ga–P) corresponds to one fourth of the body diagonal of the unit cell. Thus:
d(Ga–As) = (5.65 × √3) / 4 Å = 2.45 Å
d(Ga–P) = (5.45 × √3) / 4 Å = 2.36 Å
The radius of phosphorus in these types of compounds is 0.09 Å smaller than the radius of arsenic.