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Q.
The spin-only magnetic moments of $[Fe(NH_3)_6]^{3+}$ and $[FeF_6]^{3-}$ in $BM$ are, respectively
KVPYKVPY 2013Coordination Compounds
Solution:
The magnetic moment of coordination compound is gives as
$\mu=\sqrt{n\left(n+2\right)}$
In case of $[Fe(NH_3)_6]^{3+}, Fe$ is in $+3$ oxidation state thus the electronic configuration of $Fe^{3+}$ is $3d^{5} 4s^{0} (t^3_2g e^2_g)$, but $NH_3$ is a strong ligand pairing of electrons occur and thus configuration
becomes $t ^{5}_{2g} e^{0}_{g}$
$\Rightarrow n=1$
$\mu=\sqrt{1\left(1+2\right)}=\sqrt{3}=1.73 \,BM$
In case of $[FeF_6]^{3-}, Fe$ is in $+3$ oxidation state, as Fe is a weak ligand, pairing will
not occur thus its configuration will be
$3d^{5} 4s^{0} \left(t^{3}_{2g} e^{0}_{g}\right) \Rightarrow n=5$
$\mu=\sqrt{5\left(5+2\right)}=\sqrt{35}=5.92\,BM$