The symmetry requirement makes things even simpler. If we choose the H-F bond as the z axis, then the p and p orbitals have the wrong symmetry and cannot combine with the Is orbital of hydrogen. Electrons in these two orbitals remain as lone pairs on the fluorine atom. Only the 2p, orbital has the proper symmetry, and it combines with the hydrogen is to form a bonding MO and an antibonding MO with s symmetry: 'and . The bonding MO is lower in energy than either the 1s of H or the 2p, of F, and the antibonding MO is higher than either. These bonding and antibonding MO's and the 2p lone pairs, are represented in the center of the energy-level diagram at the left. Ten electrons are available to the HIP molecule, nine from F and one from H. Of these, two will fill the fluorine is atomic orbital and two more the 2s, not shown on the energy-level diagram. The next two electrons fill the bonding o-~ MO and provide the attraction that holds the molecule together. The remaining four electrons occupy the 2p, and 2p, orbitals on fluorine, as lone pairs. The simple Lewis diagram of this molecule is accurate.