Which of the following compounds is not colou | vedclass

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(B) compound is coloured if it has unpaired electrons in its $d$-orbitals,allowing for $d-d$ transitions.In $Na_2[CuCl_4]$,$Cu^{2+}$ has a $3d^9$ conf

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$Na_2[CdCl_4]$

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(B) compound is coloured if it has unpaired electrons in its $d$-orbitals,allowing for $d-d$ transitions.In $Na_2[CuCl_4]$,$Cu^{2+}$ has a $3d^9$ configuration with $1$ unpaired electron,so it is coloured.In $Na_2[CdCl_4]$,$Cd^{2+}$ has a $4d^{10}$ configuration. Since all $d$-orbitals are completely filled,no $d-d$ transition is possible,making it colourless.In $K_4[Fe(CN)_6]$,$Fe^{2+}$ is $3d^6$ (low spin),which is diamagnetic but often shows colour due to charge transfer,however,$Cd^{2+}$ complexes are characteristically colourless due to the $d^{10}$ configuration.In $K_3[Fe(CN)_6]$,$Fe^{3+}$ is $3d^5$,which is coloured.

List-$I$	List-$II$
$(a).$ $[Co(NH_3)_6][Cr(CN)_6]$	$(i).$ Linkage isomerism
$(b).$ $[Co(NH_3)_3(NO_2)_3]$	$(ii).$ Solvate isomerism
$(c).$ $[Cr(H_2O)_6]Cl_3$	$(iii).$ Coordination isomerism
$(d).$ $cis-[CrCl_2(ox)_2]^{3-}$	$(iv).$ Optical isomerism

List-$I$ (Complexes)	List-$II$ (Types)
$a$. $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$	$i$. Linkage isomerism
$b$. $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$	$ii$. Coordination isomerism
$c$. $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$	$iii$. Ionisation isomerism
$d$. $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$	$iv$. Solvate isomerism

Which of the following compounds is not coloured?

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