In fluorite (CaF₂) structure,

C.N. of Ca⁺² = 8

C.N. of F^– = 4.

For bcc structure we have, Z = 2

Given, $$\rho $$ = 530 kg m^–3

atomic mass of Li = 6.94 mol^–1

N_A= 6.02 × 10²³ mol^–1

$$\rho $$ = 530 kg m^–3 = $${{530 \times 1000} \over {{{\left( {100} \right)}^3}}} = $$ 0.53 g cm^–3

Also, $$\rho $$ = $${{z \times At.mass} \over {{N_A} \times {a^3}}}$$

$$ \Rightarrow $$ $${a^3} = {{Z \times At.mass} \over {{N_A} \times \rho }}$$

= $${{2 \times 6.94} \over {6.02 \times {{10}^{23}} \times 0.53}}$$

= 43.5 $$ \times $$10^-10 cm³

$$ \Rightarrow $$ $$a$$ = 352 × 10^–10 cm = 352 pm

From radius ratio, $${{{r_ + }} \over {{r_ - }}} = {{0.98 \times {{10}^{ - 10}}} \over {1.81 \times {{10}^{ - 10}}}}$$ = 0.541

It lies in the range of 0.414 to 0.732 hence coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.

As Z = 4, so, structure is fcc.

Hence, r = $${a \over {2\sqrt 2 }} = {{361} \over {2\sqrt 2 }}$$ = 127.65 pm = 127 pm