The pair in which both species have the same | vedclass

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Question

(B) To have the same magnetic moment,the species must have the same number of unpaired electrons $(n)$.$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$.

Vedclass · Accepted Answer

$[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$

Vedclass · Answer

(B) To have the same magnetic moment,the species must have the same number of unpaired electrons $(n)$.$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. It has $4$ unpaired electrons.$2$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. In a tetrahedral field,it has $3$ unpaired electrons.$3$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. It has $4$ unpaired electrons.$4$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. It has $5$ unpaired electrons.Comparing the number of unpaired electrons,$[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ both have $4$ unpaired electrons. Therefore,they have the same magnetic moment.

The pair in which both species have the same magnetic moment (spin-only value) is:

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