Input interpretation
hypoiodous acid
Chemical names and formulas
formula | HOI Hill formula | HIO name | hypoiodous acid alternate names | hydroxidoiodine | iodanol mass fractions | H (hydrogen) 0.7% | I (iodine) 88.2% | O (oxygen) 11.1%
Lewis structure
Draw the Lewis structure of hypoiodous acid. Start by drawing the overall structure of the molecule: Count the total valence electrons of the hydrogen (n_H, val = 1), iodine (n_I, val = 7), and oxygen (n_O, val = 6) atoms: n_H, val + n_I, val + n_O, val = 14 Calculate the number of electrons needed to completely fill the valence shells for hydrogen (n_H, full = 2), iodine (n_I, full = 8), and oxygen (n_O, full = 8): n_H, full + n_I, full + n_O, full = 18 Subtracting these two numbers shows that 18 - 14 = 4 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 2 bonds and hence 4 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 14 - 4 = 10 electrons left to draw: Answer: | |
3D structure
3D structure
Basic properties
molar mass | 143.911 g/mol
Units
Chemical identifiers
CAS number | 14332-21-9 PubChem CID number | 123340 SMILES identifier | OI