Which of the following complexes has its hybr | vedclass

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(C) Let us analyze each complex:$1$. $AuCl_4^-$: $Au^{3+}$ is a $5d^8$ ion. It forms a square planar complex with $dsp^2$ hybridisation.$2$. $[Co(oxal

Vedclass · Accepted Answer

$RhCl(PPh_3)_3 - dsp^2$

Vedclass · Answer

(C) Let us analyze each complex:$1$. $AuCl_4^-$: $Au^{3+}$ is a $5d^8$ ion. It forms a square planar complex with $dsp^2$ hybridisation.$2$. $[Co(oxalato)_3]^{3-}$: $Co^{3+}$ is a $3d^6$ ion. Oxalate is a strong field ligand,leading to inner orbital complex formation with $d^2sp^3$ hybridisation.$3$. $RhCl(PPh_3)_3$: This is Wilkinson's catalyst. $Rh^+$ is a $4d^8$ ion,forming a square planar complex with $dsp^2$ hybridisation.$4$. $[Fe(NH_3)_6]^{2+}$: $Fe^{2+}$ is a $3d^6$ ion. $NH_3$ acts as a strong field ligand for $Fe^{2+}$,resulting in $d^2sp^3$ hybridisation. However,option $C$ is the most standard representation of $dsp^2$ for $Rh(I)$ square planar complexes.

List-$I$ (Tetrahedral Complex)	List-$II$ (Electronic configuration)
$A$. $TiCl_4$	$I$. $e^2, t_2^0$
$B$. $[FeO_4]^{2-}$	$II$. $e^4, t_2^3$
$C$. $[FeCl_4]^{-}$	$III$. $e^0, t_2^0$
$D$. $[CoCl_4]^{2-}$	$IV$. $e^2, t_2^3$

Which of the following complexes has its hybridisation correctly matched?

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