Input interpretation
iodine pentafluoride
Chemical names and formulas
formula | F_5I_1 Hill formula | F_5I name | iodine pentafluoride IUPAC name | pentafluoro-$l^{5}-iodane mass fractions | F (fluorine) 42.8% | I (iodine) 57.2%
Lewis structure
Draw the Lewis structure of iodine pentafluoride. Start by drawing the overall structure of the molecule: Count the total valence electrons of the fluorine (n_F, val = 7) and iodine (n_I, val = 7) atoms: 5 n_F, val + n_I, val = 42 Calculate the number of electrons needed to completely fill the valence shells for fluorine (n_F, full = 8) and iodine (n_I, full = 8): 5 n_F, full + n_I, full = 48 Subtracting these two numbers shows that 48 - 42 = 6 bonding electrons are needed, which are already accounted for in the structure. Note that the valence shell of iodine has been expanded to 5 bonds. After accounting for the expanded valence, there are 5 bonds and hence 10 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 42 - 10 = 32 electrons left to draw: Answer: | |
Basic properties
molar mass | 221.89649 g/mol
Units
Chemical identifiers
PubChem CID number | 522683 SMILES identifier | FI(F)(F)(F)F InChI identifier | InChI=1S/F5I/c1-6(2, 3, 4)5