If no .of oxide ions = n , then octahedral holes = n and hence ` O^(2-) " ions" = n/2`
` Ti = O = n/2 : n = 1/2 : 1 = 1: 2 ` Hence, formula of the oxide is ` TiO_(2)` . Molar mass of ` TiO _(2) = 48 +32 = 80 g mol^(-1)`
% of Ti ` = 48/80 xx 100 = 60% ` Oxidation state of Ti in `TiO_(2) = +4`