Like H2S, NH3 looks like a symmetric molecule with A=4.832 and D 2 =0.9999. This implies that the volume of N 2sp 3 hybridized orbitals is much larger than that of the H1s orbital; thus, the tetrahedral symmetry remains major. However, even though a tetrahedron gives n=8, it doesn't satisfy A. Hence, considering the packing of molecules, the immigration of H among the four vortexes of a tetrahedron is a kind of freedom, provided that immigration is forbidden in liquid because another NH3 occupies the vacant orbital via the hydrogen bond. By this means, n=11 is obtained for NH3. Such an approach is also for H2O. Form M.P. to B.P. of H2O, A=5.9334 corresponds to n≥12, while above B.P., A=5.1168. As shown in Fig. , since there are two spare sp 3 hybridized orbitals in H2O, two H atoms can immigrate between two spare orbitals. It produces 6 additional operations. Therefore, n = 14 for H2O.