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:

• 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.

• 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$.

• Atomic packing factor(APF)

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

Hence, APF=$\frac{2π}{3}\frac{(r_C^3+ r_A^3)}{(rC+rA)^3} $

• Void space.

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

NUMERICAL:-

Given Data :- $ r_C=0.98A° , r_A=1.81A°$

Formula :- APF = $ \frac{2π}{3}.\frac{(r_C^3+ r_A^3)}{((r_C+r_A )^3 )}.$

Calculations :- APF = $\frac{2π}{3}.\frac{(0.98^3+1.81^3)}{((0.98+1.81)^3 )}$ = 0.66.

Answer :- APF = 0.66