The hybridization of P is PO3−4 is the same as of:
I in ICl+2
Number of hybrid orbitals of P in PO3−4=12[5+0+3]=4(sp3)
No. of hybrid orbitals of N in NO−3=12[5+0+1]=3(sp2)
No. of hybrid orbitals of I+ in ICl+2(5s25p2x5py5Pz:(sp3)2(sp3)2(sp3)1(sp3)1
- atom in excited state: 5s25p45d1z25d0z2−y2
I - atom is sp3d2 hybridised state in ICI−4 ∶
(sp3d2)2(sp3d2)2(sp3d2)1(sp3d2)1(sp3d2)1(sp3d2)1
-Two lone pairs occupy the axial positions of octahedron and the shape of ICl−4 becomes square planar.