JEE Advance - Chemistry (2004 - No. 4)

A compound AB has rock salt type structure. The formula weight of AB is 6.023 Y amu, and the closest A - B distance is Y1/3 nm, where Y is an arbitrary number
(a) Find the density of the lattice
(b) If the density of lattice is found to be 20 kg m-3, then predict the type of defect
Density = 5.0 kg/m³, Defect = Metal excess defect
Density = 10.0 kg/m³, Defect = Schottky defect
Density = 5.0 kg/m³, Defect = Frenkel defect
Density = 2.5 kg/m³, Defect = Metal deficiency defect
Density = 5.0 kg/m³, Defect = Metal deficiency or excess defect

Explanation

AB has a rock salt (NaCl) structure. This type of crystal structure possesses fcc unit cell and contains four formula units per unit cell, i.e., Z = 4.

In case of a rock salt structure, the edge-length (a) of the unit cell = 2 $$\times$$ (radius of cation + radius of anion)

Therefore, the edge-length (a) of the unit cell of AB crystal = 2 $$\times$$ Y1/3 nm = 2Y1/3 $$\times$$ 10$$-$$9 m.

We know, $$\rho = {{Z \times M} \over {N \times {a^3}}}$$

Given : M = 6.022 Y g mol$$-$$1 = 6.022 $$\times$$ 10$$-$$3 Y kg mol$$-$$1

$$\therefore$$ $$\rho = {{4 \times 6.022 \times {{10}^{ - 3}}Y} \over {6.022 \times {{10}^{23}} \times {{(2{Y^{1/3}} \times {{10}^{ - 9}})}^3}}} = 5.0$$ kg m$$-$$3

(1) Density of the crystal = 5.0 kg m$$-$$3

(2) The observed density (= 20 kg m$$-$$3) is higher than that of the calculated density. This indicates that the crystal structure of AB is likely to have non-stoichiometric defect in the form of metal excess or metal deficiency defect or to have impurity defect in the form of substitutional impurity defect or interstitial impurity defect.

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