Which p and d orbitals of central atoms are involved in the hybridisation of the following compounds
(a) `[Pt CI_(4)]^(2-)`
(b) `AsF_(3)`
(c ) `IF_(3)`
(d) `XeO_(2)F_(2)`
(e) `XeOF_(4)`
(f) `XeF_(6)`
(g) `IF_(7)` .

Explanation

For explanation
(a) Coordination number of `Pt^(2+)` is 4 Hybridisation at `Pt^(2+)` is `dsp^(2)` (square planar) The orbitals involved in hybridisation are
`dx^(2) -y^(2) sp^(2)` or `dx^(2) - y^(2) + s + px + py`
(b) Hybridisation as As atom `= (1)/(2) (V +M)`
` = (1)/(2) (5 +5) = 5 = sp^(3) d`
(Trigonal bipyramid)
The orbitals involved in hybridisation are `sp^(3) dz^(2)`
(c) Hybrisdisation at I atom ` = (1)/(2) (7 + 3) =5 = sp^(3) d`
Geometry is trigonal bipyramid (Tbp) Due to 2 lp on I atom. it has `T` shape The orbitals involved in hybridisation are `sp^(3) dz^(2)`
(d) Hydridisation at `X` atom ` = (1)/(2) (8 + 0 +2) = 5 = sp^(3) d`
Geometry is Tbp but due to 1 lp it has see-saw shape the orbitals involved in hybridisation are `sp^(3) dz^(2)`
(e) Hybridisation at `Xe` atom ` = (1)/(2) (8 + 0 + 4) = 6 = sp^(3) d^(2)`
Geometry is octahedral `(OH)` but due to 1 lp it has square pyramid shape The orbitals involved in hybrisation are `sp^(3), d_(x2 -y^(2),d_(z2)` or `(s+p_(x)+p_(y)+p_(z)+d_(x2 -y^(2))+d_(z2))`
(f) Hybridisation at `Xe` atom `=(1)/(2) (8 +6) = 7 = sp^(3) d^(3)`
Geometry is pentagonal bipyramid (pbp) but due to 1 (lp) it has distorted octahedral shape The orbitals involved in hybridisation are `sp^(3)` , dxy, dyz,dxz or `(s + p_(x) + p_(y) + p_(z) + dxy + dyz + dxz)`
Hybridisation at I atom ` = (1)/(2) (7 +7) = 7 = sp^(3) d^(3)`
Geometry and shape is pentagonal bipyramid (pbp) since I atoms do not have any `lp' s` The orbitals involved in hybridisation are `sp^(3)` dxy dyz dxz or `( s + p_(x) + p_(y) + p_(z) + dxy + dyz + dxz)` .