Shape and polarity of the iodate ion

Question

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?

Answer 1

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).

Answer 2

For drawing structures of molecules, I always recommend following these four calculations:

1. Count the number of electrons we have!

   $7 + 3 \cdot 6 + 1 = 26$ (iodine, three oxygens and a negative charge).

2. How many electrons are required for all atoms to obtain an octet?

   $4 \cdot 8 = 32$ (four non-hydrogen atoms, each want eight)

3. How many electrons are we lacking? We need to share these, i.e. make bonds with them:

   $32 - 26 = 6$ (that is: three bonds; either three singles or a single and a double or a triple)

4. How many electrons can we distribute as lone pairs? It’s those not required for bonding:

   $26 - 6 = 20$ (giving us ten lone pairs)

Now we know that we have three bonds and that iodine is the central atom. The only way to do this is to have a single bond to each oxygen. The distribution of lone pairs is trivial since there is only one that will give every atom an octet. Assign formal charges next. Note that each oxygen appears to have seven electrons rather than six, so they get a formal minus. Iodine appears to have five rather than seven so it gets a formal two plus.

Now, consider iodine’s surroundings. There are three bonds and a lone pair. It cannot be planar since the lone pairs always want the largest possible s-character. (The theoretically largest possible s-character would be achieved with $90^\circ$ angles and iodine using only the p-orbitals for bonding. I think I remember a non-zero amount of hybridisation to relax the geometry.)

Since the molecule is non-planar, its point group must be $C_{3\mathrm{v}}$. And since $C_{3\mathrm{v}}$ does not contain an inversion centre or improper rotation, it must be polar, i.e. it must have a non-zero dipole moment.

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As an additional note, one single double bond to one of the oxygens makes no sense. There are too many electrons for that to happen. Traditionally, one considered the molecule to have two double bonds to two of the three oxygens and invoke d-orbital hybridisation for iodine. That would lead to ten bonding electrons and 16 lone pair electrons; two oxygens having given up a lone pair each for bonding. However, that view should be considered obsolete, since d-orbitals are energetically too far removed to take part in bonding for non-metal atoms.

Answer 3

In iodate ions there are three double bonds between iodine and oxygen and there is presence of one lone pair.iodine is central atom.the shape of molecule can be seen as pyramidal with corners of base triange be oxygen atoms and iodine at top.And there is no meaning for polarity for ions as they already have a charge and hence polar.