Question

Predict the structures of [SbF5] 2- and [SbF4] - , in each case explain your reasoning.

Answer 1

First you have to draw a Lewis structure for the ions, then work out a VSEPR formula, and from that you can determine an appropriate geometry.

To draw a Lewis structure, start by counting up all the valence electrons available (adding or subtracting as needed to account for the ion's charge). Let's examine your first ion: [SbF5]2-.

Antimony (Sb) is part of the nitrogen family and so has 5 valence electrons. Fluorine is a halogen and has 7 valence electrons. There are five fluorine atoms, and 5 x 7 = 35. So the total number of valence electrons from the antimony and fluorine atoms is 40. But the ion also has a 2- charge, so you have to ADD 2 more valence electrons, bringing the total to 42.

Now draw the antimony atom in the center surrounded by five fluorine atoms. Satisfy the octets of all five fluorine atoms (using up 40 electrons in the process) and assign any leftover electrons (2, in this case) to the central antimony atom.

Based on this Lewis structure, we can see that the VSEPR formula for [SbF5]2- is AX5E. The electron pair geometry will be octahedral and the molecular geometry will be square pyramidal.

The ion [SbF4]- has a grand total of 34 valence electrons.

1 x Sb = 5
4 x F = 28
-1 charge = 1
Total: 34

The Lewis structure will show four flourine atoms surrounding a central antimony atom. Satisfying the octet of the peripheral fluorine atoms will leave 2 electrons, so once again the antimony atom receives a lone pair. The VSEPR formula is AX4E. The electron pair geometry is trigonal bipyramidal and the molecular geometry is distorted tetrahedral (or see-saw).