Input interpretation
hypoiodite anion | structure diagram
Result
Draw the Lewis structure of hypoiodite anion. Start by drawing the overall structure of the molecule: Count the total valence electrons of the iodine (n_I, val = 7) and oxygen (n_O, val = 6) atoms, including the net charge: n_I, val + n_O, val - n_charge = 14 Calculate the number of electrons needed to completely fill the valence shells for iodine (n_I, full = 8) and oxygen (n_O, full = 8): n_I, 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: | |