Question

The density of an ionic compound (molecular weight $=58.5$ g/mol) is $2.165$ kg m${}^{-3}$ and the edge length of unit cell is $562$ pm, then the closest distance between $A^{\oplus }$ and $B^{⊝}$ and the number of atoms per unit cell is:

• A. $281$ pm, $4$
• B. $562$ pm, $2$
• C. $562$ pm, $4$
• D. $281$ pm, $2$

Answer 1

The correct option is A $281$ pm, $4$

Density $=\frac{z\times \frac{M}{1000}}{N_A\times (a\times {10}^{-12}{)}^3}$

$2.165=\frac{z\times \frac{58.5}{1000}}{6\times {10}^{23}\times (562\times {10}^{-12}{)}^3}$
$z=\frac{2.165\times 6\times {10}^{23}\times (177\times {10}^6\times {10}^{-36})\times 1000}{58.5}$

$z=4$
Closest distance between the $A^{+}$ and $B^{-}$ is equal to $\frac{a}{2}$.