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Question

(B) substance is paramagnetic if it contains unpaired electrons $(n > 0)$. If it has no unpaired electrons $(n = 0)$,it is diamagnetic.$1$. $Cr(ClO_4)

Vedclass · Accepted Answer

$KMnO_4$

Vedclass · Answer

(B) substance is paramagnetic if it contains unpaired electrons $(n > 0)$. If it has no unpaired electrons $(n = 0)$,it is diamagnetic.$1$. $Cr(ClO_4)_3$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar]3d^3$. Here,$n = 3$,so it is paramagnetic.$2$. $KMnO_4$: $Mn$ is in $+7$ oxidation state. Electronic configuration of $Mn^{7+}$ is $[Ar]3d^0$. Here,$n = 0$,so it is diamagnetic.$3$. $TiCl_3$: $Ti$ is in $+3$ oxidation state. Electronic configuration of $Ti^{3+}$ is $[Ar]3d^1$. Here,$n = 1$,so it is paramagnetic.$4$. $VOBr_2$: $V$ is in $+4$ oxidation state. Electronic configuration of $V^{4+}$ is $[Ar]3d^1$. Here,$n = 1$,so it is paramagnetic.Therefore,$KMnO_4$ is not paramagnetic.

$A$ substance which is not paramagnetic is

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