Predict the hybridisation and geometry of the following complexes.
Text Solution
Verified by Experts(a) [NiBr 4 ]; (b) [AuCl 4 ]
Complex | Hybridization | Geometry | |
(a) | [NiBr4]2– | sp3 | Tetrahedral |
(b) | [AuCl4]– | dsp2 | square planar |
(c) | [Pt(NH3)4]2+ | dsp2 | square planar |
Sol. In the paramagnetic and tetrahedral complex [NiBr 4 ] 2– , the nickel is in +2 oxidation state and the ion has the electronic configuration 3d 8 . The hybridisation scheme is as shown below.
Ni 2+ , [Ar]3d 8 
[NiBr 4 ] 2– 
sp 3 hybrid orbitals & tetrahedral.
In [AuCl 4 ] – the gold is in +3 oxidation state and 5d 8 configuration has higher CFSE. It is square planar and diamagnetic.
In [Pt(NH 3 ) 4 ] 2+ the platinum is in +2 oxidation state and 5d 8 configuration has higher CFSE. It is square planar and diamagnetic.
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