An ionic compound $AB$ crystallises in b.c.c lattice. If its unit cell edge length is 412 pm and radius of $B^{-}$ is 181 pm, the radius of $A^{+}$ is : 

For b.c.c lattice,

$2\left(r_{{\mathrm{A}}^{+}}+r_{{\mathrm{B}}^{-}}\right)=a\sqrt{3}$

$\begin{array}{l}\therefore r_{A^{+}}+r_{{\mathrm{B}}^{-}}=\frac{a\sqrt{3}}{2}=\frac{412\times 1.732}{2}=356.8\mathrm{pm}\\ r_{{\mathrm{A}}^{+}}+181\mathrm{pm}=356.8\mathrm{pm}\\ \therefore r_{{\mathrm{A}}^{+}}=356.8\mathrm{pm}-181\mathrm{pm}=\mathbf{175}.8\mathrm{pm}.\end{array}$