NON – STOICHIOMETRIC DEFECTS  

If as a result of imperfection, the ratio of number of cation to anion becomes different from that indicated by the ideal chemical formula, the defects are called non – stoichiometric defects. These defects arise either due to excess of metal atoms or non metal atom or presence of impurities / foreign particle.

(a) Metal excess defects due to anion vacancies: -

A compound may have excess metal ion if a negative ion is absent from its lattice site leaving a hole which is occupied by electron to maintain neutrality.

The holes that are occupied by electrons are called ‘F’ centres (or colour centres) and are responsible for the colour of the compound and many interesting properties.

                                   2425_Metal excess and metal excess defect.JPG    

(ii) Metal excess defects due to interstitial cations  

Metal excess may also be caused by an extra cation (positive ion) present in an interstitial site. Electrical neutrality is maintained by presence of an electron in another interstitial site. This defect is similar to Frenkel defect and is found in crystals having Frenkel defects.   

(iii) Metal deficiency due to cation vacancies  

The non-stoichiometric compounds may have metal deficiency due to the absence of a metal from its lattice site. The charge is balanced by an adjacent ion having higher positive charge. This type of defects are generally shown by compounds of transition elements.  

(c) Point defects due to the presence of foreign atoms  

These defects arise when foreign atoms are present at the lattice site (in place of host atoms) or at the vacant interstitial sites. In the former case, we get substitutional solid solutions. The formation of former depends upon the electronic structure of impurity while that of later on the size of impurity.   

Illustration 46. 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) = 4.81g 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.  

Illustration 47. 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 = √N × Ni    e–E/2kT 

                          Since for this crystal  

                        Ni = 2N 

                          ∴ n = √2 × N × e–E/kT 

                          ∴ n/N = √2e–2×1.6 × 10–19 / 2×1.28 ×10–23 × 4000 =  √2 × 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 

Illustration 48. 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)

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