$[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic whereas $[Fe(CN)_6]^{3-}$ is weakly paramagnetic. Explain.
(N/A) In both $[Fe(H_2O)_6]^{3+}$ and $[Fe(CN)_6]^{3-}$,$Fe$ exists in the $+3$ oxidation state,i.e.,in $d^5$ configuration.
Since $CN^{-}$ is a strong field ligand,it causes the pairing of unpaired electrons. Therefore,there is only one unpaired electron left in the $d$-orbital.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{1(1+2)}$
$= \sqrt{3}$
$= 1.732 \, BM$
On the other hand,$H_2O$ is a weak field ligand. Therefore,it cannot cause the pairing of electrons. This means that the number of unpaired electrons is $5$.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{5(5+2)}$
$= \sqrt{35}$
$\simeq 5.92 \, BM$
Thus,it is evident that $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic (with $5$ unpaired electrons),while $[Fe(CN)_6]^{3-}$ is weakly paramagnetic (with $1$ unpaired electron).
Since $CN^{-}$ is a strong field ligand,it causes the pairing of unpaired electrons. Therefore,there is only one unpaired electron left in the $d$-orbital.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{1(1+2)}$
$= \sqrt{3}$
$= 1.732 \, BM$
On the other hand,$H_2O$ is a weak field ligand. Therefore,it cannot cause the pairing of electrons. This means that the number of unpaired electrons is $5$.
Therefore,
$\mu = \sqrt{n(n+2)}$
$= \sqrt{5(5+2)}$
$= \sqrt{35}$
$\simeq 5.92 \, BM$
Thus,it is evident that $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic (with $5$ unpaired electrons),while $[Fe(CN)_6]^{3-}$ is weakly paramagnetic (with $1$ unpaired electron).
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