Identify the ion (hydrated in solution) which | vedclass

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(C) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $C

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$Co^{3+} (0)$

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(C) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$. (Correct)$2$. For $Cu^{2+}$ $(3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$. (Correct)$3$. For $Co^{3+}$ (hydrated,i.e.,$[Co(H_2O)_6]^{3+}$): $H_2O$ is a weak field ligand. The configuration is $t_{2g}^4 e_g^2$,so $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$. The given value $0$ is incorrect.$4$. For $Fe^{2+}$ $(3d^6)$: In aqueous solution,$[Fe(H_2O)_6]^{2+}$ has $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$. (Correct)Thus,$Co^{3+}$ is not correctly matched.

Identify the ion (hydrated in solution) which is not correctly matched with its spin only magnetic moment (in $BM$) given in brackets.

Similar Questions

Identify the homoleptic complex(es) that is/are low spin.
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$(D) \ [Co(NH_3)_6]^{3+}$
$(E) \ [Cr(H_2O)_6]^{2+}$
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