Question

A solid $AB$ has rock salt structure. If the edge lenght is $520pm$ and radius of $A^{+}$ is $80pm$, the radius of anion $B^{-}$ would be -

• A. $440pm$
• B. $220pm$
• C. $360pm$
• D. $180pm$

Answer 1

The correct option is D $180pm$

Solid $AB$ forms rock salt structure. Here, $B^{-}$ ions occupies the fcc/ccp lattice points ( 8 corners and 6 face centres) and $A^{+}$ occupies the octahedral voids of the ccp arrangement ( 12 edge centre and 1 body centre)
Here,
the $B^{-}$ ion in corner of cubic unit cell does not touches the $B^{-}$ ion in face centre of the unit cell.
Thus,
$4r_{B^{-}}\ne a\sqrt{2}$

Ions $A^{+}$ and $B^{-}$ touch each other along the edge.
Thus,
$\frac{\text{Unit cell length}}{2}=r_{A^{+}}+r_{B^{-}}$
$r_A+r_{B^{-}}=\frac{520}{2}$
$\Rightarrow 80+r_{B^{-}}=260$

$\text{Radius of}{B}^{-}=r_{B^{-}}=260-80$
$r_{B^{-}}=180pm$