The hybridization and the oxidation number of the central metal ion in Wilkinson's catalyst is:

d s p^{2}, + 2

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s p^{3}, + 1

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d s p^{2}, + 4

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d s p^{2}, + 1

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Solution

The correct option is B $d s p^{2}, + 1$
$P P h_{3}$ is a neutral ligard. $C l^{-}$ is a uninegatively charged ligand. Hence Rh is in +1 state with $d s p^{2}$ hybridization

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The correct order of hybridization of the central atom in the following species $N H_{3}, [P t C l_{4}]^{2 -}$ , $P C l_{5}$ and $B C l_{3}$ is

[N i C l_{4}]^{2 -}, [N i (C N)_{4}]^{2 -} & [N i (C O)_{4}]

The correct order of hybridisation of the central atom in the following species $N H_{3}$ , ${[P t C l_{4}]}^{2 -}$ , $P C l_{5}$ and $B C l_{3}$

[N i (C O)_{4}]

[N i (C N)_{4}]^{2 -}

d s p^{2}

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