In NaCl, the Cl^- ions are arranged in ccp arrangement where Cl^- ions are present at the corners as well as the center of the face of the cube. The Na⁺ ions are located such that each Na⁺ ion is surrounded with six chloride ions. Na⁺ ions occupy all the octahedral voids. Each chloride ion is also surrounded by six Na⁺ ions and the stoichiometry is 1:1. Thus the coordination number for both species is the same.

In CaF₂ arrangement, the Ca²⁺ ions are arranged in ccp arrangement where Ca²⁺ ions are placed at the corners as well as the center of the face of the cube. Fluoride ions occupy tetrahedral voids and in way that each ion is surrounded by four calcium ions. The coordination number of F^- is 4. There are two tetrahedral sites available for each calcium ion, and thus the F^- ions occupy all eight tetrahedral voids. However, the stoichiometry is 1:2, and the coordination number of calcium ion is 8, therefore the coordination number for these species is not the same.

In Na₂O arrangement, the O^2- ions are arranged in ccp arrangement where O^2- ions are placed at corners as well as the center of the face of the cube. Na⁺ ions occupy all tetrahedral voids surrounding O^2- ion. Each sodium ion is surrounded by 4 oxide ions and each oxide ion is surrounded by 8 sodium ions. The coordination number of Na⁺ is 4 and O^2- is 8, therefore the coordination number of these species is not the same.

In Zinc blende ZnS arrangement, the S^2- atoms are in ccp arrangement, where each sulphideion is surrounded by four zinc ions. There are eight tetrahedral voids and four sulphide ions occupy half of the tetrahedral holes. Each sulphide ion itself is surrounded by four zinc ions, thus making both of their coordination numbers 4.

Thus the correct answer is (i).