Atomic radius of Li is 1.23A⁰ and ionic radius of Li⁺ is 0.76A⁰. The percentage of volume occupied by 2s electron is $\mathrm{\_\_\_\_\_}$
 

Volume of Li = $\frac{4}{3}{\mathrm{\pi r}}^3 =\frac{4}{3}\times 3.14\times {(1.23)}^3 =7.79{\mathrm{A}}^0$ (1s²2s¹)

Volume of Li = $\frac{4}{3}{\mathrm{\pi r}}^3 =\frac{4}{3}\times 3.14\times {(0.76)}^3 =1.83{\mathrm{A}}^0$ (1s²2s¹)

volume occupied by 2s electron = 7.79-1.83 = 5.96 A⁰

% volume occupied by 2s electron  = $\frac{5.96}{7.79}\times 100 =0.765 \times 100 =76.5$