Question

The expected spin-only magnetic moments(BM) for $\left[Fe\right(CN{)}_6{]}^{4-}$ and $[Fe{F}_6{]}^{3-}$ respectively are:

• A. $1.73$ and $1.73$
• B. $1.73$ and $5.92$
• C. $0.0$ and $1.73$
• D. $0.0$ and $5.92$

Answer 1

The correct option is D $0.0$ and $5.92$

In, $\left[Fe\right(CN{)}_6{]}^{4-}$ due to strong ligand pairing of electron happens and because of pairing it have zero unpaired electron so magnetic moment is zero.

But in $[Fe{F}_6{]}^{3-}$ fluoride is a weak ligand and because of that,

here $F{e}^{3+}$ has 5 unpaired electron and it have $\sqrt{n(n+2)}=5.92$ BM, magnetic moment.