Question

I can't seem to figure out what the hybridization of PF6(-) is. could someone explain how to find it correctly for me?

Answer 1

Hybridisation of central phosphorous atom in PF6(-) can be find out by considering its electronic configuration. Phosphorous contains 15 electrons and its electronic configuration can be written as follows:

1s2   2s2   2p6 3s2    3p3

Thus, it has five valence electrons (3s2    3p3 ). Out of these two electrons in 3s orbital are paired. Therefore, it has only three unpaired electrons in 3p orbital which can form three bonds with another atom.

However, phosphorous forms six bonds with six fluorine atoms in PF6(-). This can be explained by hybridization of orbitals of central phosphorous atom. Out of two paired electrons in 3s orbital one gets promoted to one of empty 3d orbital, taking total count of unpaired electrons to five. Since, central phosphorous has one negative charge, it gets one extra electron from fluoride ion into its another empty 3d orbital.

Therefore, total unpaired electron in phosphorous atom becomes six (one is 3s, three in 3p and two in 3d). These six orbitals undergo hybridization to produce six sp3d2 hybrid orbitals containing one unpaired electron each which forms bond with six fluorine atoms.

Therefore, hybridization of central phosphorous atom in PF6(-) is sp3d2