QUESTIONS & ANSWERS (Defects in Crystals )
Q1.Titanium monoxide has a rock-salt structure. X-ray diffraction data show that the length of one edge of the cubic unit cell for TiO with a 1:1 ratio of Ti to O is 4.18Å, and the density as determined by volume and mass measurements is 4.92 g cm–3. Do the data indicate that defects are present? If so, are they vacancy or interstitial defects? [Ti = 47.88 u]
Solution: The presence of vacancies (Schottky defects) at the Ti and O sites should be reflected in a lower measured density than that calculated from the size of the unit cell and the assumption that every Ti and O site is occupied. Interstitial (Frenkel) defects would give little of any difference between the measured and theoretical densities. There are four formula units per unit all so theoretical density is
d = 4 (47.88+16) / 6.023 × 1023 (4.18 × 1023) = 4.81 g cm–3
This is significantly greater than the measured density. The crystal must, therefore, contain numerous vacancies. Because the overall composition of the solid is TiO, there must be equal number of vacancies on cation and anion sites.
Q2. Find out the ratio of the mole-fraction of the Frenkel’s defect in NaCl crystal at 4000K temperature. The amount of energy needed to form Frenkel’s defects and Schottky defects are respectively 2 eV and 4 eV.
Given that 1ev = 1.6 × 10–19V And k = 1.23 × 10–23
Solution: Let the Frenkel defect be n in the ionic crystal of N ions within interstitial space.
Since, n = vN × Ni e–E/2kT
Since for this crystal
Ni = 2N
? n = v2 × N × e–E/kT
? n/N = v2e–2×1.6 × 10–19 / 2×1.28 ×10–23 × 4000 = v2 × e–2.8985
x1 = = 7.79 × 10–2
Now for Schottky defect
Let n defects be present in N-ions of the crystal
? mole fraction = n/N = e–E/2kT
n = e–4×1.6×10–19 / 2×1.38 × 10–23 × 4000 = e–5.797
x2 = 3.03 × 10–3
So, the ratio of mole-fraction of Frenkel’s and Schottky defects are
= x1 / x2 = 7.79 × 10–2 / 3.03 × 10–3 = 25.71 / 1
Q3. Titanium crystallizes in a face centered cubic lattice. It reacts with C or H interstitially, by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes, but carbon occupies octahedral holes.
(a) Predict the formula of titanium hydride and titanium carbide formed by saturating the titanium lattice with either “foreign” element.
(b) What is the maximum ratio of “foreign” atom radius to host atom radius that can be tolerated in a tetrahedral hole without causing a strain in the host lattice?
Solution: (a) There are 4 – atoms per unit cell and 8-tetrahedral sites per unit cell.
So, the ratio of atoms to tetrahedral sites = 1:2
So, the formula would be TiH2
For carbide
Since carbon occupies octahedral holes
So, ratio of octahedral hole to atom = 1:1
The formula of carbide is TiC
(b) Since for tetrahedral hole
The limiting ratios radius
r+ / r– = 0.225 – 0.414
here r+ = foreign atom radius
r– = host atom radius
Without causing a strain in the host lattice i.e., r+ / r– = 0.225 (minimum value)