Which of the following complex ions has a mag | vedclass

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(A) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n 2)} \ BM$,where $n$ is the number of unpaired electrons.For $[Cr(H_

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$[Mn(H_2O)_6]^{4+}$

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(A) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n 2)} \ BM$,where $n$ is the number of unpaired electrons.For $[Cr(H_2O)_6]^{3 }$,the oxidation state of $Cr$ is $ 3$. The electronic configuration of $Cr^{3 }$ is $[Ar]3d^3$,which has $n = 3$ unpaired electrons.For $[Mn(H_2O)_6]^{4 }$,the oxidation state of $Mn$ is $ 4$. The electronic configuration of $Mn^{4 }$ is $[Ar]3d^3$,which has $n = 3$ unpaired electrons.For $[Mn(H_2O)_6]^{3 }$,the oxidation state of $Mn$ is $ 3$. The electronic configuration of $Mn^{3 }$ is $[Ar]3d^4$,which has $n = 4$ unpaired electrons.For $[Fe(H_2O)_6]^{3 }$,the oxidation state of $Fe$ is $ 3$. The electronic configuration of $Fe^{3 }$ is $[Ar]3d^5$,which has $n = 5$ unpaired electrons.For $[Cu(NH_3)_4]^{2 }$,the oxidation state of $Cu$ is $ 2$. The electronic configuration of $Cu^{2 }$ is $[Ar]3d^9$,which has $n = 1$ unpaired electron.Since $[Cr(H_2O)_6]^{3 }$ and $[Mn(H_2O)_6]^{4 }$ both have $n = 3$,they have the same magnetic moment.

Which of the following complex ions has a magnetic moment same as $[Cr(H_2O)_6]^{3+}$?

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