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. (Nearest Integer)

Solution

Chemistry ma’am · 3 steps

An octahedral void sits right between two anions, so walk along one cube edge and add up radii.

1. Recognise the structure. B⁻ in ccp with A⁺ in every octahedral void is the rock-salt (NaCl) structure. The radius ratio 102/181 = 0.56 lies between 0.414 and 0.732, so octahedral voids fit.
2. Look along an edge. Along one edge of the cube you meet an anion at the corner, a cation at the edge centre and an anion at the next corner. Cation and anion touch, so the edge holds two cation radii and two anion radii.

   a = 2(r₊ + r₋) = 2(102 + 181) pm

3. So. a = 2 × 283 = 566 pm, so the answer is 566.

The common slip: 283. Taking the edge as just r₊ + r₋ gives 283, but an edge spans two cation-anion contacts, so double it.

The idea to keep. In rock-salt type crystals the ions touch along the cube edge, so a = 2(r₊ + r₋). Use the anion-only ccp formula only when the anions themselves touch.

Answer from the NTA final answer key. Solution by PrepWiser.