JEE MAIN - Chemistry (2018 - 16th April Morning Slot - No. 20)
In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively :
sp3d, trigonal bipyramidal
sp3, tetrahedral
dsp2, square planar
d2sp3, octahedral
Explanation
Wilkinson's catalyst is [RhCl(PPh3)3]
Here central atom is Rh.
Oxidation state of Rh here is = + 1
$$ \therefore $$ Electronic configuration of Rh+ = [Kr]4d8
_16th_April_Morning_Slot_en_20_1.png)
And the complex is square planar
_16th_April_Morning_Slot_en_20_2.png)
Here central atom is Rh.
Oxidation state of Rh here is = + 1
$$ \therefore $$ Electronic configuration of Rh+ = [Kr]4d8
And the complex is square planar
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