Here we need to determine the FIF bond angle for the following ion if 4 minus. So iodine is in Group 7A, so it has seven valence electrons. It will use four of them to connect to the fluorines. Fluorines are also in Group 7A, so they have 7 valence electrons, and they also only make single bonds.
We've used four out of iodine 7 valence electrons and we're going to say here it has three more that are not being used. But remember -1 charge means we'd gain an outside electron, which would give to the iodine. Ince it has a charge, you put it in brackets with the charge on the outside.
Now here we would have what we have, two loan payers and four bonding groups or surrounding elements connected to iodine. Here we'd say that the bond angles here and here from this particular shape would be approximately 90�. So that would be our bond angle for IF4-.
