Which of the following compounds has a magnet | vedclass

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(D) The magnetic moment depends on the number of unpaired electrons $(n)$. The formula is $\mu = \sqrt{n(n+2)} \ BM$.For $[Cr(H_{2}O)_{6}]^{3+}$: $Cr$

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$[Mn(H_{2}O)_{6}]^{4+}$

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(D) The magnetic moment depends on the number of unpaired electrons $(n)$. The formula is $\mu = \sqrt{n(n+2)} \ BM$.For $[Cr(H_{2}O)_{6}]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$. Number of unpaired electrons $(n)$ = $3$.Now, let us check the number of unpaired electrons for each option:$A$) $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ is $[Ar] 3d^{9}$, $n = 1$.$B$) $[Mn(H_{2}O)_{6}]^{3+}$: $Mn^{3+}$ is $[Ar] 3d^{4}$, $n = 4$.$C$) $[Fe(H_{2}O)_{6}]^{3+}$: $Fe^{3+}$ is $[Ar] 3d^{5}$, $n = 5$.$D$) $[Mn(H_{2}O)_{6}]^{4+}$: $Mn^{4+}$ is $[Ar] 3d^{3}$, $n = 3$.Since $[Mn(H_{2}O)_{6}]^{4+}$ has $3$ unpaired electrons, it has the same magnetic moment as $[Cr(H_{2}O)_{6}]^{3+}$.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

Which of the following compounds has a magnetic moment identical to $[Cr(H_{2}O)_{6}]^{3+}$?

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