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Geometry of PF
5 molecule
The outer electronic configuration of phosphorus, the central atom, is 3s2 3p3 which may be represented as shown in fig.1.24 (a). It has three unpaired electrons in the ground state:
fig 1.24 - Formation of PF5 molecule involving sp3d Hybridization
To explain the pentavalency of phosphorus in PF5, one of the electrons in 3s orbital is promoted to the higher 3d orbital so that it has five unpaired electrons. This is called the excited state of phosphorus and is represented in fig.1.24 (b). These five orbitals hybridise to form five sp3d hybrid orbitals, which adopt trigonal bipyramidal arrangement. Each of the sp3d hybrid orbitals overlaps with 2p orbital of fluorine forming five P-F bonds as shown in fig.1.25.Thus, PF5 molecule has trigonal bipyramidal geometry. It may be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. Three bonds lie in one plane at an angle of 120o to one another as in a triangular planar arrangement. These bonds are termed as equatorial bonds. Of the remaining two bonds, one lies above and other below the equatorial plane both making an angle of 90o with the plane. These bonds are called axial bonds. It may be remembered that this geometry is not symmetrical. The axial bonds have been found to be larger than equatorial bonds. For e.g., in case of PF5 molecule each P - F axial bond = 219 pm and each P - F equatorial bond = 204 pm. As the axial bonds suffer more repulsion,the axial bond will be slightly elongated when compared to equatorial bond.




