# Question #c7810

Jul 21, 2017

$\text{IO"_2"F"_2^"-}$ has a see-saw shape.

#### Explanation:

Step 1. Draw a skeleton Lewis structure I which every atom has an octet. Step 3. Count electrons

The skeleton structure has 32 electrons.

We have available

$\text{1I + 2F + 2O + 1e"^"-" = 7 + 14 +12 + 1 = }$ 34 electrons

There is an extra pair of electrons.

They go on the central $\text{I}$ atom.  Step 4. Minimize formal charges

Put a double bond to one of the $\text{O}$ atoms. Step 5. Determine the electron geometry

There are four bonded atoms and a lone pair.

This is an $\text{AX"_4"E}$ molecule.

The electron geometry is trigonal bipyramidal. The bulky lone pair and the oxygen atoms go in the equatorial positions, and the fluorine atoms go in the axial positions.

Step 6. Determine the molecular geometry

The molecular geometry ignores the lone pair.

The molecular geometry is see-saw. So, the structure of $\text{IO"_2"F"_2^"-}$ is Note: The two $\text{O}$ atoms are equivalent because we could have put the double bond to the other $\text{O}$ atom.