A dopant for an n-type semiconductor must have 5 valence electrons. N, P, As, and Sb all have this property (they all lie in the 5th column of the periodic table). B, however, is element 5. It lies in the 3rd column of the periodic and so has 3 valence electrons instead of 5. This makes B a p-type dopant instead of an n-type dopant. Therefore, answer (D) is correct.