The correct increasing order of spin-only mag | vedclass

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Question

(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$, we calculate the number of unpaired electrons $(n)$ for each complex:$1$.

Vedclass · Accepted Answer

$C < B < D < A$

Vedclass · Answer

(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$, we calculate the number of unpaired electrons $(n)$ for each complex:$1$. $[MnBr_{4}]^{2-}$: $Mn^{2+}$ is $3d^{5}$. Since $Br^{-}$ is a weak field ligand, it forms a high-spin tetrahedral complex with $n = 5$.$2$. $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ is $3d^{9}$. It has $n = 1$ unpaired electron.$3$. $[Ni(CN)_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. Since $CN^{-}$ is a strong field ligand, it forms a square planar complex with $n = 0$.$4$. $[Ni(H_{2}O)_{6}]^{2+}$: $Ni^{2+}$ is $3d^{8}$. It forms an octahedral complex with $n = 2$ unpaired electrons.Comparing the number of unpaired electrons: $C (n=0) Thus, the increasing order of magnetic moments is $C < B < D < A$.

The correct increasing order of spin-only magnetic moment values of the complex ions $[MnBr_{4}]^{2-}$ $(A)$, $[Cu(H_{2}O)_{6}]^{2+}$ $(B)$, $[Ni(CN)_{4}]^{2-}$ $(C)$ and $[Ni(H_{2}O)_{6}]^{2+}$ $(D)$ is:

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