Looking for classes? Ksquare Career Institute, Bengaluru →
AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ions in the lattice is
A
300 pm
B
335 pm
C
250 pm
D
200 pm
Detailed Solution
In a body-centred cubic (CsCl type) arrangement one kind of ion is at the corners and the oppositely charged ion is at the body centre.
The ions touch along the body diagonal, whose length is $\sqrt{3}a$.
The body diagonal contains corner ion – centre ion – corner ion: $2(r^+ + r^-) = \sqrt{3}a$
Distance between oppositely charged ions: $r^+ + r^- = \frac{\sqrt{3}a}{2} = \frac{1.732\times387}{2}$
$= 335.1$ pm $\approx 335$ pm
The ions touch along the body diagonal, whose length is $\sqrt{3}a$.
The body diagonal contains corner ion – centre ion – corner ion: $2(r^+ + r^-) = \sqrt{3}a$
Distance between oppositely charged ions: $r^+ + r^- = \frac{\sqrt{3}a}{2} = \frac{1.732\times387}{2}$
$= 335.1$ pm $\approx 335$ pm