Question

Kcl has same structure as NaCl. If the volume of the unit cell is 247•7 cubic angstroms, the distance between centres of k+ and Cl- ions is

Answer 1

Answer:

The distance between centers $K^+ \&Cl^-$ ion is 2.2185 $\AA$.

Explanation;

Volume of KCl structure = 247.7 $\AA^3$

KCL and NaCl has FCC crystal structure with value of Z equal to 4.

Let the distance between centers $K^+ \&Cl^-$ be r.

a = Edge length of the crystal

Volume of the crystal =$247.7\AA^3=a^3$

$a=6.275 \AA$

In F.C.C cubic lattice

$r=\frac{a}{2\sqrt{2}}=\frac{6.275 \AA}{2\times \sqrt{2}}=2.2185 \AA$

The distance between centers $K^+ \&Cl^-$ ion is 2.2185 $\AA$.