Shape and polarity of the iodate ion I am trying to figure out the shape of the iodate ion or $\ce{IO3^-}$. No matter how many times I draw it, I keep getting iodine in the centre, and 3 oxygen atoms surrounding it, 2 with single bonds, and 1 with a double bond. This would make it a trigonal planar shape. It would mean that it is not a "bent" shape.

I went to double check on the internet, and I am now confused. On Wikipedia it says it is in fact polar and a pyramidal shape.

Why does iodine have a pair of lone electrons in its shell? Why would it not combine with one of the oxygen atoms to make a double bond?
Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).