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 Answer
May 27, 2017

Answer:

Well, count your valence electrons and your charges up first.........

Explanation:

#1.# #IO_3^-#, we have #7+3xx6+1=26# electrons to distribute around 4 centres.........i.e. THIRTEEN electron pairs, unlucky for some. #O=ddotI^(+)(-O)_2^-# so distributes 13 electron pairs. From left to right as we face the page, there are #6#, #6#, and #2xx7# 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.# #SO_3^(2-)#, we have #4xx6+2=26# electrons to distribute around 4 centres.........i.e. THIRTEEN electron pairs again. #O=ddotS(-O)_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.

In each example, the ELECTRONIC geometry around the ipso atom is tetrahedral; the atomic geometry is trigonal pyramidal. Do you appreciate this distinction?