The edge length of a face centered cubic cell of an ionic substance is $$508$$ pm. If the radius of the cation is $$110$$ pm, the radius of the anion is

The data given are typical of an NaCl-type lattice.

In the NaCl structure the anions (Cl−) occupy the lattice points of a face-centred cubic (fcc) cell, while the cations (Na+) sit in all the octahedral voids (edge centres and body centre).

Consider one edge of the cube. Along this edge we encounter: corner anion → edge-centre cation → next corner anion. Hence the sequence of touching spheres is

anion   +   cation   +   anion.

The distance between the centres of the corner anion and the edge-centre cation equals half the edge length.$$ \frac{a}{2} $$. Because the two ions are in contact along this line, the separation of their centres also equals the sum of their radii: $$r_{-}+r_{+}$$.

Therefore $$ r_{-}+r_{+}= \frac{a}{2}\qquad -(1)$$

The edge length is given as $$a=508\text{ pm}$$ and the cation radius is $$r_{+}=110\text{ pm}$$. Substituting in (1):

$$ r_{-}= \frac{508}{2}-110 = 254-110 = 144\text{ pm}$$

Hence the radius of the anion is $$144\text{ pm}$$.

Option D which is: 144 pm