xenon tetrafluoride

xenon tetrafluoride

formula | F_4Xe_1 Hill formula | F_4Xe name | xenon tetrafluoride IUPAC name | tetrafluoroxenon mass fractions | F (fluorine) 36.7% | Xe (xenon) 63.3%

Draw the Lewis structure of xenon tetrafluoride. 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: 4 n_F, val + n_Xe, val = 36 Calculate the number of electrons needed to completely fill the valence shells for fluorine (n_F, full = 8) and xenon (n_Xe, full = 8): 4 n_F, full + n_Xe, full = 40 Subtracting these two numbers shows that 40 - 36 = 4 bonding electrons are needed, which are already accounted for in the structure. Note that the valence shell of xenon has been expanded. After accounting for the expanded valence, there are 4 bonds and hence 8 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 36 - 8 = 28 electrons left to draw: Answer: | |

molar mass | 207.287 g/mol

PubChem CID number | 123324 SMILES identifier | F[Xe](F)(F)F InChI identifier | InChI=1S/F4Xe/c1-5(2, 3)4