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

K_{2} C r_{2} O_{7}

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{(N H_{4})}_{2} [T i C l_{6}]

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V O S O_{4}

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K_{3} [C u {(C N)}_{4}]

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Solution

The correct option is D $V O S O_{4}$
In $K_{2} C r_{2} O_{7}$ , the oxidation state of $C r$ is +6.
${C r}^{^{+} 6} = [A r] 3 d^{0}$

In ${(N H_{4})}_{2} [T i C l_{6}]$ , the oxidation state of $T i$ is +4.
${T i}^{+ 4} = [A r] 3 d^{0}$

In $V O S O_{4}$ , the oxidation state of $V$ is +4.
$V^{+ 4} = [A r] 3 d^{1}$

In $K_{3} [C u (C N)_{4}]$ , the oxidation state of $C u$ is +1.
${C u}^{+ 1} = [A r] {3 d}^{10}$

In $V O S O_{4}$ , $V^{+ 4}$ configuration is $d^{1}$ , so it has one unpaired d-electron so it is paramagnetic and colored. All other compounds don't have any unpaired d electrons, so they are diamagnetic in nature.

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