Question

$N{H}_{4}Cl$ crystallizes in a body-centered cubic type lattice with a unit cell edge length of $387$ pm. The distance (in pm) between the oppositively charged ions in the lattice is :

• A. $335.1$
• B. $83.77$
• C. $274.46$
• D. $137.23$

Answer 1

Answer: (A)

$335.1$

In a body-centered cubic lattice, oppositely charged ions touch each other along the cross - diagonal of the cube.

Hence, we can write,

$2r^{+}+2r^{–}=\sqrt{3}a{r}^{+}+r^{–}=\frac{\sqrt{3}a}{2}{r}^{+}+r^{–}=\frac{\sqrt{3}}{2}\times 387pm=335.15pm$

Option A is correct.