Draw the structure, name the shape and show bond angles of the molecules XeF4 and SbF4-. In your answer explain why each structure is different, despite both having a central atom, surrounded by 4 fluorine atoms.

Xe is in group 8, the noble gases. Therefore it has 8 electrons in its outer shell. Fluorine atoms have 7 electrons, yet one is an unpaired electron, so this is the electron in fluorine that gets involved in bonding. There are 4 of these, so around the xenon atom, there are (8+4=) 12 electrons, so 6 electron pairs; and as there are 4 fluorine atoms bonding, there are 4 bond pairs, and 2 lone pairs of electrons. Therefore this is an octahedral shape, with 90 degree angles between each bond and lone pair of electrons. Due to VSEPR (valence shell electron pair repulsion) theory, the most energetically stable form of XeF4 will form, with the lone pairs as far away from each other as possible, hence forming the shape seen. Sb is in group 5, so it has 5 electrons in its outer shell. There are 4 Sb-F bonds, and also a negative charge on the central antinomy atom, so there are in total (5+4+1=) 10 electrons. Therefore there are 5 pairs. This suggests it is a trigonal bipyramedal shape, (seesaw) typically with bond angles of 120 and 90 degrees. However as there are 4 bond pairs, and 1 lone pair, then the lone pair will distort this perfect geometry, forming angles between Sb-F bonds to be less than 90 and 120 degrees. The lone pair forms on the axial part of the molecule as there is minimum repulsion. It is due to VSEPR theory that both structures are different. The number of bond pairs and lone pairs combine to give a shape, while the ratio of both determine any varience from the regular geometry.