Question

{"given":"Magnetic oxide of cobalt crystallizes with cobalt atoms occupying one-eighth of tetrahedral holes and one-half of octahedral holes in a closest packed array of oxide ions. This compound is used in recording tapes.","key_observation":"In a closest packed structure, the number of tetrahedral holes is twice the number of atoms in the lattice, while the number of octahedral holes equals the number of atoms. To find the formula, we need to calculate the ratio of cobalt to oxygen atoms based on the occupation of these holes by cobalt atoms in the oxide ion lattice.","option_analysis":[{"label":"(A)","text":"$Co_2O_3$","verdict":"incorrect","explanation":"This formula gives a Co:O ratio of 2:3, which doesn't match the calculated ratio from the given hole occupancy pattern of cobalt atoms in the oxide lattice."},{"label":"(B)","text":"$Co_5O_8$","verdict":"incorrect","explanation":"This formula gives a Co:O ratio of 5:8, which doesn't correspond to the stoichiometry derived from one-eighth tetrahedral and one-half octahedral hole occupation by cobalt atoms."},{"label":"(C)","text":"$CoO$","verdict":"incorrect","explanation":"This formula represents a 1:1 ratio of cobalt to oxygen, which is inconsistent with the given occupation pattern of tetrahedral and octahedral holes in the closest packed oxide structure."},{"label":"(D)","text":"$Co_3O_4$","verdict":"correct","explanation":"Let n be the number of oxide ions. Tetrahedral holes = 2n, octahedral holes = n. Cobalt atoms = (1/8)(2n) + (1/2)(n) = n/4 + n/2 = 3n/4. The Co:O ratio = 3n/4:n = 3:4, giving the formula $Co_3O_4$."}],"answer":"(D)","formula_steps":[]}