# How do you draw the Lewis structure for these anions? Each anion is a conjugate base of an acid. 1) IO_3^- 2) SO_3^(-2)

$1.$ $I {O}_{3}^{-}$, we have $7 + 3 \times 6 + 1 = 26$ electrons to distribute around 4 centres.........i.e. THIRTEEN electron pairs, unlucky for some. $O = {\ddot{I}}^{+} {\left(- O\right)}_{2}^{-}$ so distributes 13 electron pairs. From left to right as we face the page, there are $6$, $6$, and $2 \times 7$ VALENCE electrons associated with each atom, and this leads to our assignment of charge. Electronic structure is based on a tetrahedron, given 4 electron pairs around the central iodine atom.
$2.$ $S {O}_{3}^{2 -}$, we have $4 \times 6 + 2 = 26$ electrons to distribute around 4 centres.........i.e. THIRTEEN electron pairs again. $O = \ddot{S} {\left(- O\right)}_{2}^{-}$. There are 2 electrons on sulfur and it shares 4 electrons from the covalent bonds to oxygen. Because there are 6 valence electrons, the sulfur centre is FORMALLY neutral. On the other hand, two of the oxygen atoms have 7 electrons associated with, and therefore, these are assigned a formal negative charge. Of course the charge is distributed to the other oxygen atom by resonance.