Question

In the coordination compound ${\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left[}\mathbf{Fe}\mathbf{\left(}{\mathbf{OH}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{F}}_{\mathbf{5}}\mathbf{\right]}$, the$\mathbf{Fe}$ is octahedrally coordinated.

(a) Based on the fact that$\mathbf{F}$ is a weak-field ligand, predict whether this compound is diamagnetic or paramagnetic. If it is paramagnetic, tell how many unpaired electrons it has.

(b) By comparison with other complexes reviewed in this chapter, discuss the likely color of this compound.

(c) Determine the$\mathit{d}$-electron configuration of the iron in this compound.

(d) Name this compound.

Short answer

1. The complex ion ${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}$ has 5 unpaired electrons. And it is a paramagnetic compound.
2. We can except that the ${\left({\text{NH}}_4\right)}_2\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]$ complex is only lightly colored.
3. Theelectron configuration for the ion is ${\left({\text{t}}_{2\text{g}}\right)}^3{\left({\text{e}}_{\text{g}}\right)}^2$.
4. The name of the complex is ammonium aquapentafluoroiron (III).

Step-by-step solution

Step 1:

(a)

Given complex ${\left({\text{NH}}_4\right)}_2\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]$ has ${\left({\text{NH}}_4\right)}^{+}$cation and${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}$ as anion. In the${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{-}$ anion, the charge on each of the six ${\text{F}}^{-}$ligands is $-1$. And water is a neutral ligand. Therefore, the oxidation state of the Fe must be $+3$, because$1\times 0$ (for the water)$+5\times -1$ (for$\mathrm{F}$)$+3$ (for Fe) equals the required$-2$ .
$\begin{array}{c}{\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}\\ \uparrow \uparrow \uparrow \end{array}$

O.S.: $+3-0-1$( for each NCS)

Thus iron in${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}$ is ${\text{d}}^5$and${\text{F}}^{-}$ is a weak-field ligand. Therefore, the $\mathrm{d}$electron configuration for the ion is,

Thus, the complex ion has 5 unpaired electrons. And it is a paramagnetic compound.

Step 2:

(b)

The anion ${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}$is a high spin ${\text{d}}^5$complex like ${\left[\text{Fe}{\left({\text{OH}}_2\right)}_6\right]}^3$so this complex is probably a pale violet. Here, the cation${\left({\text{NH}}_4\right)}^{+}$ is colorless.
In other words the transition metal ions which have completely filled $\text{d}$-orbitals ( $3{\text{d}}^{10},4{\text{d}}^{10}$or $5{\text{d}}^{10}$) such as${\text{Zn}}^{2+},{\text{Cd}}^{2+},{\text{Hg}}^{2+},{\text{Cu}}^{+},{\text{Ag}}^{+}$ and${\text{Au}}^{+}$ ions are colorless as the excitation of electron or electrons is not possible within-orbitals. The transition metal ions which have completely empty$\mathrm{d}$-orbitals such as ${\text{Se}}^{3+},{\text{Ti}}^{4+}$, etc., are also colorless because they cannot undergo transitions. High-spin${\text{d}}^5$ complexes can only undergo $\text{d}-\text{d}$spin-forbidden transitions. Although the energies of these transitions do occur in the visible region, their probability is so low that high-spin ${\text{d}}^5$ complexes are invariably only lightly colored. Whereas Electronic transitions are allowed for low spin${\mathrm{d}}^5$ complexes and are highly colored as expected.
Therefore, we can except that the ${\left({\text{NH}}_4\right)}_2\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]$complex is only lightly colored.

Step 3:

(c)

Thus iron in ${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{-}$is${\text{d}}^5$ and${\text{F}}^{-}$ is a weak-field ligand. Therefore, the$\mathrm{d}$ electron configuration for the ion is,

High-spin

Weak-field ligand

5 Unpaired electrons

The $\mathrm{d}$electron configuration for the ion is${\left({\text{t}}_{2\text{g}}\right)}^3{\left({\text{e}}_{\text{g}}\right)}^2$ .

Step 4:

(d)

Given complex is ${\left({\text{NH}}_4\right)}_2\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]$

${\text{NH}}_4^{+}$: ammonium

${\left[\text{Fe}\left({\text{OH}}_2\right){\text{F}}_5\right]}^{2-}$;

$\left.\begin{array}{r}{\text{OH}}_2:\mathrm{aqua}\\ {\text{F}}^{-}:\mathrm{fluoro}\end{array}\right\}$ alphabetical

Charge on iron is

$\begin{array}{rr}\text{Fe}+0+5(-1)& =-2\\ \text{Fe}& =+3\end{array}$

Thus the name of the complex is ammonium aquapentafluoroiron (III).