Q101P - Chemistry: Molecular Nature Of Matter And Change

Question

Draw orbital-energy splitting diagrams and use the spectrochemical series to show the orbital occupancy for each of the following (assuming that  is a weak-field ligand):

(a) ${\left[\mathrm{Fe}{\left(C_2{O}_4\right)}_3\right]}^{\mathbf{3}\mathbf{-}}$ (${\mathbf{C}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$creates a weaker field than ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ does.)  (b) ${\left[\mathrm{Co}{\left(\mathrm{CN}\right)}_6\right]}^{\mathbf{4}\mathbf{-}}$ (c) ${\left[{\mathrm{MnCl}}_6\right]}^{\mathbf{4}\mathbf{-}}$

Step-by-Step Solution

Answer

(a) 5 d electrons

(b) 7 d electrons

(c) 5 d electrons

Step 1: Given Information

$$\begin{gathered} \mathrm{Compounds} \mathrm{are} \\ {\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-} \\ {\left[\mathrm{Co}{\left(\mathrm{CN}\right)}_6\right]}^{4-} \\ {\left[{\mathrm{MnCl}}_6\right]}^{4-} \end{gathered}$$

Step 2: Concept Introduction

Basic Concept: The crystal field splitting energy $(∆)$ represents the difference in energy between the two higher energy $\mathbf{\left(}{\mathbf{e}}_{\mathbf{g}}\mathbf{\right)}\mathbf{d}$ orbitals and the three lower energy $\mathbf{\left(}{\mathbf{t}}_{\mathbf{2}\mathbf{g}}\mathbf{\right)}\mathbf{d}$ orbitals; there are five d orbitals that are split into two higher energy orbitals and three lower energy orbitals because of the electrostatic attractions between the metal cation and ligands' negative charge.

Spectrochemical series (from weak-field to strong-field ligands):

${\mathbf{I}}^{\mathbf{-}}\mathbf{<}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{<}{\mathbf{F}}^{\mathbf{-}}\mathbf{<}{\mathbf{OH}}^{\mathbf{-}}\mathbf{<}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{<}{\mathbf{SCN}}^{\mathbf{-}}\mathbf{<}{\mathbf{NH}}_{\mathbf{3}}\mathbf{<}\mathbf{en}\mathbf{<}{\mathbf{NO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{<}{\mathbf{CN}}^{\mathbf{-}}\mathbf{<}\mathbf{CO}$

- the weaker the field is, the lower the splitting energy is $(∆)$

- the stronger the field is, the higher the splitting energy is $(∆)$

Step 3: Solution (a)

${\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-}$

In the above compound, the oxidation state of Fe is 3+.

The electronic configuration of Fe is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^2$

Therefore the number of d electrons is 5.

The orbital splitting diagram is shown below:

Step 4: Solution (b)

${\left[\mathrm{Co}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$

In the above complex, the oxidation state of Co is 2+.

Since the electronic configuration of Co is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{7}4{\mathrm{s}}^2$

Thus the number of d electrons is 7.

The orbital splitting diagram is shown below:

Step 5: Solution (c)

${\left[{\mathrm{MnCl}}_6\right]}^{4-}$

In the above complex oxidation state of Mn is 2+.

Since the electronic configuration of Mn is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^2$.

Thus the number of d electrons is 5.

The orbital splitting diagram is shown below: