Question

Which of the following options are correct regarding the complex $\left[Fe\right(CN{)}_6{]}^{3-}$ and $\left[Fe\right(CN{)}_6{]}^{4-}$:

• A. number of unpaired electrons are $1$ and $0$ respectively.
• B. number of unpaired electrons are $0$ and $1$ respectively.
• C. magnetic moment are $\sqrt{3}BM$ and $0$ respectively.
• D. magnetic moment are $0$ and $\sqrt{3}BM$ respectively.

Answer 1

Answer: (A), (C)

number of unpaired electrons are $1$ and $0$ respectively.
magnetic moment are $\sqrt{3}BM$ and $0$ respectively.

This coordination compound has iron as the central transition metal and $6$ cyanides as monodentate ligands.

Oxidation number of $Fe$ in $\left[Fe\right(CN{)}_6{]}^{3-}=x+(-6)=-3\Rightarrow x=+3$

Oxidation number of $Fe$ in $\left[Fe\right(CN{)}_6{]}^{4-}=y+(-6)=-2\Rightarrow y=+2$

Electronic configuration of $Fe=\left[Ar\right]3d^6,4s^2$

In $F{e}^{3+}=\left[Ar\right]3d^5$

$F{e}^{2+}=\left[Ar\right]3d^6$

Since cyanide is a strong field ligand, it will be a low spin complex. The splitting energy is greater than the pairing energy for electrons and therefore, $5$ valence electrons of $F{e}^{3+}$ will pair up and the $1$ of these electron will remain unpaired and $6$ valance electrons of $F{e}^{2+}$ will pair up and there will be no unpaired electrons.

The magnetic moment for $F{e}^{3+}=\sqrt{n(n+2}=\sqrt{1(1+2)}=\sqrt{3}$

The magnetic moment for $F{e}^{3+}=\sqrt{n(n+2}=\sqrt{0(0+2)}=\sqrt{0}$

$F{e}^{3+}$ is paramagnetic while $F{e}^{2+}$ is diamagnetic.