Question

Calculate the sum of the spin only magnetic moments of
$K_3\left[Fe{\text{(oxalate)}}_3\right]$ and
$K_3\left[Ru{\text{(oxalate)}}_3\right]$.

Answer 1

$K_3\left[Fe{\text{(oxalate)}}_3\right]$
It is a high spin complex with $d^5$ configuration
${\mu }_{\text{spin only}}=\sqrt{n(n+2)}$
${\mu }_{\text{spin only}}=\sqrt{5(5+2)}$
${\mu }_{\text{spin only}}=\sqrt{35}=5.9BM$

$K_3\left[Ru{\text{(oxalate)}}_3\right]$
It is a low spin complex with $d^5$ configuration
${\mu }_{\text{spin only}}=\sqrt{n(n+2)}$
${\mu }_{\text{spin only}}=\sqrt{1(1+2)}$
${\mu }_{\text{spin only}}=\sqrt{3}=1.73BM$

${\mu }_{\text{Total}}=1.73+5.9=7.63BM$