To determine the formula of the compound, we need to analyze the given information about the lattice structure and the occupation of voids.$\\$Given:$\\$• The anions (A) form a hexagonal close packed (hcp) lattice.$\\$• The cations (C) occupy 75% of the octahedral voids.$\\$In an hcp lattice, the number of octahedral voids is equal to the number of atoms forming the lattice.$\\$Therefore, if there are $N$ anions, there are $N$ octahedral voids.$\\$Since the cations occupy 75% of the octahedral voids, the number of cations is:$\\$Number of cations $= 0.75N \\$Thus, the ratio of cations to anions is:$\\ \frac{Number \ of \ cations}{Number \ of \ anions} = \frac{0.75N}{N} = 0.75 \\$To express this ratio in the simplest whole number form, we multiply both the numerator and the denominator by 4:$\\ \frac{0.75 \times 4}{1 \times 4} = \frac{3}{4} \\$Therefore, the simplest whole number ratio of cations to anions is $C_3A_4. \\$Thus, the formula of the compound is $C_3A_4. \\$This corresponds to Option 3.