Question

What is the distance between Na+ and Cl-ions in NaCl crystal if its density is 2.165g $cm^{3}$ and NaCl crystallizes in FCC lattice.

Answer 1

Answer:

Explanation:

NaCl crystalizes in such a way that the Cl- ions occupy the FCC lattice and all the octahedral voids are occupied by the Na+ ions. such a crystalline structure is also known as rosk salt structure which is basically shown by halides of group 1&2 elements except halides of Cesium(Cs).

each unit cell thus contains 4molecules of NaCl which is calculated as follows:

FCC lattice means atoms are present at corner and face centres. Each corner contribute 1/8th of an atom and each face centre contribute 1/2 of an atom

therefore no. of Cl- ions present per unit cell = {1/8 * no. of corners} + { 1/2 * no. of faces} = (1/8*8) + (1/2 * 6) = 1+3 = 4

In FCC or CCP type packing octahedral voids are present in the edges and body centre. Each edge contribute 1/4th of an atom and body centre contribute 1atom.

Therefore no. of Na+ ions per unit cell =

{ 1/4 * no. of edges} + { 1 at body centre} =

(1/4 * 12) + (1) = 3+1 = 4

Hence no. of NaCl molecules per unit cell = 4

using the formula M = ( d * N * a^3) / Z we can calculate the value of edge length 'a'.

in the above formula

M = molecular mass of the molecule here it is NaCl which is equal to (23+35.5) = 58.5 g

d = density of the crystal = 2.165 g cm^-3

N = avogardo's number = 6.022 * 10^ 23

Z = no. of atoms(molecule for a compound) per unit cell = 4

so using the above formula we can calculate

a= [ (M * Z) / (d * N )]^ 1/3

a = [ (58.5 * 4) / ( 2.165 * (6.022*10^ 23) )] ^ 1/3

=> a = 5.64 * 10^ -8 cm

now assuming cation anion contact along the edge we can calculate the distance between the ions by the following formula

along any edge we get 2 radius of Cl- ions present at 2 corners and from the edge centre 2 radius of Na+ ions hence

2 * ( r Cl- ) + 2 * (r Na+) = a

since distance between Cl- ion and Na+ ion is ( rCl-) + (rNa+)

hence distance between the two ions = a/2

= (5.64 * 10^-8) / 2

= 2.8203 * 10^-8 cm

= 2.8203 angstrom

1 angstrom = 10^-10 m