OCTAHEDRAL CONFIGURATION:

The octahedral configuration of neighbouring anions is found in NaCl structure. Here four anions A, B ,C and D are arranged at the corners of a square with the cation O at the centre of the square. Two more anions are situated in front and at the back of the cation. The centres of all six anions form an octahedron.

Here in ∆ BOC , < BOC = 90°, BC = $2r_A$ , OB =$ r_C+r_A$ and <bco =45°,<="" p="">

Hence, BO/BC=cos45°

Or, $ (r_C+r_A)/(2r_A ) = 1/(√2)$

The critical radius ratio here is,

NUMERICAL:

Given data :- $r_A(max⁡ ) =2.02A°$ , ligancy = 6.

Formula :- for ligancy 6 ,$(r_C/r_A )$ = 0.732.

Calculations :- $ r_C=r_A ×0.732=2.02 ×0.732=1.478 A°$

Answer :- $r_C=1.478A°$