Question

The spin-only magnetic moments of $[Fe(N{H}_3{)}_6{]}^{3+}$ and $[Fe{F}_6{]}^{3-}$ in $BM$ are, respectively

• A. 1.73 and 1.73
• B. 5.92 and 1.73
• C. 1.73 and 5.92
• D. 5.92 and 5.92

Answer 1

Answer: (C)

The magnetic moment of coordination compound is gives as
$\mu =\sqrt{n\left(n+2\right)}$
In case of $[Fe(N{H}_3{)}_6{]}^{3+},Fe$ is in $+3$ oxidation state thus the electronic configuration of $F{e}^{3+}$ is $3d^{5}4s^0(t_2^{3}g{e}_g^2)$, but $N{H}_3$ is a strong ligand pairing of electrons occur and thus configuration becomes $t_{2g}^5{e}_g^0$
$\Rightarrow n=1$
$\mu =\sqrt{1\left(1+2\right)}=\sqrt{3}=1.73BM$
In case of $[Fe{F}_6{]}^{3-},Fe$ is in $+3$ oxidation state, as Fe is a weak ligand, pairing will not occur thus its configuration will be
$3d^{5}4s^0\left(t_{2g}^3{e}_g^0\right)\Rightarrow n=5$
$\mu =\sqrt{5\left(5+2\right)}=\sqrt{35}=5.92BM$