Input interpretation
sulfuryl bromide fluoride
Chemical names and formulas
formula | SO_2BrF Hill formula | BrFO_2S name | sulfuryl bromide fluoride mass fractions | Br (bromine) 49% | F (fluorine) 11.7% | O (oxygen) 19.6% | S (sulfur) 19.7%
Lewis structure
Draw the Lewis structure of sulfuryl bromide fluoride. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), fluorine (n_F, val = 7), oxygen (n_O, val = 6), and sulfur (n_S, val = 6) atoms: n_Br, val + n_F, val + 2 n_O, val + n_S, val = 32 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), fluorine (n_F, full = 8), oxygen (n_O, full = 8), and sulfur (n_S, full = 8): n_Br, full + n_F, full + 2 n_O, full + n_S, full = 40 Subtracting these two numbers shows that 40 - 32 = 8 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.58 (sulfur), 2.96 (bromine), 3.44 (oxygen), and 3.98 (fluorine). Because the electronegativity of sulfur is smaller than the electronegativity of oxygen, expand the valence shell of sulfur to 6 bonds. Therefore we add a total of 2 bonds to the diagram: Answer: | |
Basic properties
molar mass | 162.96 g/mol melting point | -86 °C boiling point | 41 °C solubility in water | reacts
Units
Chemical identifiers
CAS number | 13536-61-3 SMILES identifier | BrS(=O)(=O)F