Input interpretation
hypobromite anion
Lewis structure
Draw the Lewis structure of hypobromite anion. Start by drawing the overall structure of the molecule: Count the total valence electrons of the bromine (n_Br, val = 7) and oxygen (n_O, val = 6) atoms, including the net charge: n_Br, val + n_O, val - n_charge = 14 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8) and oxygen (n_O, full = 8): n_Br, full + n_O, full = 16 Subtracting these two numbers shows that 16 - 14 = 2 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There is 1 bond and hence 2 bonding electrons in the diagram. Fill in the remaining unbonded electrons on each atom. In total, there remain 14 - 2 = 12 electrons left to draw. Lastly, fill in the formal charges: Answer: | |
General properties
formula | (BrO)^- net ionic charge | -1 alternate names | oxobromate | oxidobromate | hypobromite | hypobromite(1-)
Other properties
ion class | anions | oxoanions | polyatomic ions
Thermodynamic properties
molar free energy of formation Δ_fG° | aqueous | -33.4 kJ/mol (kilojoules per mole) molar heat of formation Δ_fH° | aqueous | -94.1 kJ/mol (kilojoules per mole) molar entropy S° | aqueous | 42 J/(mol K) (joules per mole kelvin)