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(C) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$. $Cr^{+6} = [Ar] 3d^0$. In $(NH_4)_2[TiCl_6]$,the oxidation state of $Ti$ is $+4$. $Ti^{+4} =

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$VOSO_4$

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(C) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$. $Cr^{+6} = [Ar] 3d^0$. In $(NH_4)_2[TiCl_6]$,the oxidation state of $Ti$ is $+4$. $Ti^{+4} = [Ar] 3d^0$. In $VOSO_4$,the oxidation state of $V$ is $+4$. $V^{+4} = [Ar] 3d^1$. In $K_3[Cu(CN)_4]$,the oxidation state of $Cu$ is $+1$. $Cu^{+1} = [Ar] 3d^{10}$. In $VOSO_4$,the $V^{+4}$ configuration is $d^1$,which contains one unpaired $d$-electron,making it paramagnetic and coloured. All other compounds listed have no unpaired $d$-electrons,making them diamagnetic in nature.

Among the following,the compound that is both paramagnetic and coloured is

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