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(B) In $K_3[Fe(OH)_6]$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $OH^-$ is a weak field lig

Vedclass · Accepted Answer

$5.92$ and $4.90 \ B.M.$

Vedclass · Answer

(B) In $K_3[Fe(OH)_6]$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $OH^-$ is a weak field ligand,the electrons remain unpaired. Thus,$n = 5$. The magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ B.M.$In $K_4[Fe(OH)_6]$,the oxidation state of $Fe$ is $+2$. The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$. Since $OH^-$ is a weak field ligand,the electrons remain unpaired. Thus,$n = 4$. The magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M.$Therefore,the values are $5.92 \ B.M.$ and $4.90 \ B.M.$ respectively.

The calculated spin-only magnetic moments of $K_3[Fe(OH)_6]$ and $K_4[Fe(OH)_6]$ respectively are $:$

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