Determine the density (in g/cm3) of an AX ionic compound in which the ionic radi

Question

Determine the density (in g/cm3) of an AX ionic compound in which the ionic radii and atomic weights of the cation and anion are given below: (answer fromt X.XX) cation: radius = 0.098 nm A = 44 g/mol anion: radius = 0.165 nm A = 53.4 g/mol Hint: Compute the cation to anion radius ratio to predict the coordination number of the ions. Remember that knowing the coordination number allows us to predict the crystal structure. Given the ratio of radii, I determined the coordination number to be 6 which means the unit cell is an octahedral. The problem is I don't know how to determine the number of atoms contained within the unit cell to calculate the mass for the density equation. Any help would be greatly appreciated.

Explanation / Answer

Given data:

Radius of cation (r) = 0.098 nm and Atomic weight of cation = 44 g/mol.

Radius of anion (R) = 0.165 nm and Atomic weight of anion = 53.4 g/mol.

Molar mass of ionic compound AX (M) = 44+53.4 = 97.4 g/mol.

Let us calculate radius ration first,

Radius ratio = (Cationic radius) / (anionic radius_

Radius ratio =r/R

Radius ration = 0.098 / 0.165

So, Radius ratio = 0.594

i.e. Co-ordination number is 6.

This implies the octahedral geometrical shape i.e. position of anions around cation i.e. octahedral unit cell for AX ionic compound.

This implies that each unit cell has 4 AX units……(Its analogous to NaCl which also has co-ordination number 6 with octahedral arrangement)

Calculation of total number of Cations (say A) and anions (say X):

Cation number (A) = (12 ions at the edge of corner x ¼ ) + (1 ion at body center x 1) = ( 12x ¼) +(1x1)= 3+1 = 4

Anion number (X) = (8 ions at corner x 1/8 ) +(6 ions at face center x ½ ) = 1+3 = 4

3) The edge length of this kind of unit cell is a = 2(r+R)

So, a = 2 (0.098+0.165 ) = 0.263 nm.

4) Density of unit cell (d) =?

We have now, Z = 4, M = 97.4 g/mol, a=0.263 nm = 0.263 x 10-7 cm, Avogadro number(N) = 6.022 x 1023 /mol

Formula for density,

d = (Z x M) /(a3 x N)

d = (4 x 97.4)/ [(0.263 x 10-7)3 x(6.022 x 1023)

d= 389.6 / 10.95

d = 35.56 g.cm-3.

Hence the density of AX is 35.56 g.cm-3.

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