The colorless species is

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(C) To determine the color of the species,we look at the oxidation state of Vanadium $(V)$ and the presence of unpaired $d$-electrons.$1$. In $VCl_3$,

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$Na_3VO_4$

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(C) To determine the color of the species,we look at the oxidation state of Vanadium $(V)$ and the presence of unpaired $d$-electrons.$1$. In $VCl_3$,$V$ is in the $+3$ oxidation state ($3d^2$ configuration). It has unpaired electrons and is colored.$2$. In $VOSO_4$,$V$ is in the $+4$ oxidation state ($3d^1$ configuration). It has one unpaired electron and is colored.$3$. In $Na_3VO_4$,$V$ is in the $+5$ oxidation state ($3d^0$ configuration). There are no $d$-electrons available for $d-d$ transitions,making it colorless.$4$. In $[V(H_2O)_6]SO_4 \cdot H_2O$,$V$ is in the $+2$ oxidation state ($3d^3$ configuration). It has unpaired electrons and is colored.Therefore,the colorless species is $Na_3VO_4$.

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