NaCl STRUCTURE:-

This is an ionic structure in which the $Na^+$ ions and $Cl^- $ ions are alternately arranged. It is a combination of two FCC sublattice one made up of $Na^+$ ions and the other of $Cl^-$ ions as if one sublattice is translated through the other along the cube edges.

NaCl unit cell with $Na^+$ ions occupying the regular FCC lattice points with $Cl^-$ ions positioned at alternate points. A face of this unit cell is shown.

Another NaCl unit cell can be considered with the positions of $Na^+$ and $Cl^-$ ions interchanged. The face of such a unit cell is shown.

NaCl UNIT CELL PARAMETER:

1. Total number of molecule / unit cells

Calculation for $Na^+$ = Here $Na^+$ forms a FCC structure. Hence total number of $Na^+$ ions = 4

Calculation for 〖Cl〗^- = There are 12〖Cl〗^- ions at the edges. Every edge lattice points is shared by four neighbouring unit cell. Hence every edge lattice point carries ¼ of an atom. There is one whole 〖Cl〗^- ion at the centre of the structure. Hence ,

$ Total number of Cl^- ions = (12 ×1/4) + 1 = 4.$

Since there are 4 $Na^+$ ions and four $Cl^-$ ions in a NaCl unit cell , there are four NaCl molecule present in a unit cell.

Hence number of molecule / unit cell = 4.

1. Atomic Radius (r)

Since NaCl is an ionic structure and cations are smaller than anions it is assumed that radius of cation =$ r_C$ and the radius of an anion =$ r_A.$

1. Atomic packing factor(APF)

   $ APF = ((4 ×4/3 πr_C^3 ) ×( 4×4/3 πr_A^3))/a^3 \hspace{1cm} it is found that a=2r_C+2r_A$

Hence, $ APF=(2π/3)(r_C^3+ r_A^3)/(rC+rA)^3 $

Void space.

This is given by $ [1- (2π/3)(r_C^3+ r_A^3)/(rC+rA)^3 ]$