Ionic radii of cation $$A^+$$ and anion $$B^-$$ are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $$B^-$$. $$A^+$$ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is _____ pm.

We are given that anion $$B^-$$ forms a cubic close packing (CCP, i.e., FCC) structure, with cation $$A^+$$ occupying all octahedral voids. The ionic radii are $$r_{A^+} = 102$$ pm and $$r_{B^-} = 181$$ pm.

In an FCC unit cell, there are 4 anions per unit cell (8 corner atoms contributing $$\frac{1}{8}$$ each and 6 face-centred atoms contributing $$\frac{1}{2}$$ each: $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$). The number of octahedral voids in an FCC unit cell equals the number of atoms, which is 4. Since $$A^+$$ occupies all octahedral voids, we have 4 cations per unit cell, giving the stoichiometry AB. This is the well-known rock salt (NaCl) structure.

In the rock salt structure, the octahedral voids are located at the edge centres and at the body centre of the cube. Consider one edge of the unit cell: at one end sits a $$B^-$$ anion (at a corner), at the midpoint of the edge sits an $$A^+$$ cation (in the octahedral void), and at the other end sits another $$B^-$$ anion (at the adjacent corner). The cation and anions touch along the edge.

The edge length $$a$$ therefore equals the distance from the centre of one anion through the cation to the centre of the next anion:

$$a = r_{B^-} + 2r_{A^+} + r_{B^-} = 2r_{A^+} + 2r_{B^-}$$

This can also be written as $$a = 2(r_{A^+} + r_{B^-})$$. The factor of 2 appears because both a cation radius and an anion radius span from the edge centre to each corner, and there are two such spans along the full edge.

Now we substitute the given values:

$$a = 2(102 + 181) = 2 \times 283 = 566 \text{ pm}$$

Hence, the correct answer is 566.