xenon hexafluoride

xenon hexafluoride

formula | F_6Xe_1 Hill formula | F_6Xe name | xenon hexafluoride IUPAC name | hexafluoroxenon mass fractions | F (fluorine) 46.5% | Xe (xenon) 53.5%

Draw the Lewis structure of xenon hexafluoride. Start by drawing the overall structure of the molecule: Count the total valence electrons of the fluorine (n_F, val = 7) and xenon (n_Xe, val = 8) atoms: 6 n_F, val + n_Xe, val = 50 Calculate the number of electrons needed to completely fill the valence shells for fluorine (n_F, full = 8) and xenon (n_Xe, full = 8): 6 n_F, full + n_Xe, full = 56 Subtracting these two numbers shows that 56 - 50 = 6 bonding electrons are needed, which are already accounted for in the structure. Note that the valence shell of xenon has been expanded to 6 bonds. After accounting for the expanded valence, there are 6 bonds and hence 12 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 50 - 12 = 38 electrons left to draw: Answer: | |

molar mass | 245.283 g/mol

PubChem CID number | 139546 SMILES identifier | F[Xe](F)(F)(F)(F)F InChI identifier | InChI=1S/F6Xe/c1-7(2, 3, 4, 5)6