Question

The complex with spin-only magnetic moment of 4.9 B.M. is

• A. $\left[Fe\right(CN{)}_6{]}^{3+}$
• B. $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$
• C. $\left[Fe\right(CN{)}_6{]}^{4-}$
• D. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

Answer 1

The correct option is D $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

$4\cdot 9Bm$ corresponds to 4 unpaired electrons which is the case with $F{e}^{2+}$ in $[FeC{H}_{2}o16{]}^{2+}\cdot (3d^4)$.

In this coordination compound, Iron has the oxidation state of +2.

So, in this cation form, there are six valence electrons.

Water is a weak ligand.

So, these electrons do not get paired up in the iron d- subshell.

So, the value comes out to be 4.9 BM.