In this example problem, we will draw the Lewis structure for xenon hexafluoride. Our first step is to sum the valence electrons. We have one xenon atom with eight valence electrons, and we have six fluorine atoms with seven valence electrons each. Our charge is zero, so we have a total of 50 valence electrons. Our second step is to draw the structure with only single bonds. Xenon is the central atom to which the six fluorine atoms are attached. The third step is to satisfy the octet and duet rules by adding lone pairs of electrons. If we run out of electrons, we can create double and triple bonds. If we have extra electrons, we can add them to the central atom if the atom is in the third row or below in the periodic table. So, we begin by satisfying the octet rule for all of the fluorine atoms, which are the peripheral atoms. If we count up all of the electrons that we've used so far, we'll have used 48 electrons. So, we have two of our 50 valence electrons left over. These electrons will add to xenon. The fourth step is to use formal charge to evaluate the Lewis structure. The formal charge for an atom is equal to the number of valence electrons in a neutral atom of that element minus the number of bonds plus the number of lone pair electrons. For each of our seven fluorine atoms, the formal charge of fluorine is equal to seven valence electrons in a neutral atom minus one bond plus six lone pair electrons, which is equal to zero. For our xenon, xenon normally has eight valence electrons minus six bonds and two lone pair electrons, which equals zero. We know this is a good Lewis structure for xenon hexafluoride because formal charges are minimized.