Question

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Fe}^{2+}\) (high and low spin) b. \(\mathrm{Fe}^{3+}\) (high spin) c. \(\mathrm{Ni}^{2+}\)

Short answer

For the octahedral complex ions: a. \(\mathrm{Fe}^{2+}\) - High-spin: ``` e_g: ↑↓ ↑↓ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` Low-spin: ``` e_g: empty _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` b. \(\mathrm{Fe}^{3+}\) (high spin): ``` e_g: ↑↓ ↑ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` c. \(\mathrm{Ni}^{2+}\): ``` e_g: ↑↓ ↑↓ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ```

Step-by-step solution

Identify the metal ion's d-electron count

We need to find the number of d-electrons in the metal ions: a. \(\mathrm{Fe}^{2+}\) - Iron (Fe) is in the 3d series with an atomic number of 26. Its electronic configuration is \([Ar]3d^{6}4s^{2}\). Therefore, \(\mathrm{Fe}^{2+}\) ion has an electronic configuration of \([Ar]3d^{6}\) with 6 d-electrons. b. \(\mathrm{Fe}^{3+}\) - The \(\mathrm{Fe}^{3+}\) ion has one less electron than \(\mathrm{Fe}^{2+}\), which results in the electronic configuration \([Ar]3d^{5}\) with 5 d-electrons. c. \(\mathrm{Ni}^{2+}\) - Nickel (Ni) is in the 3d series with an atomic number of 28. Its electronic configuration is \([Ar]3d^{8}4s^{2}\). Therefore, \(\mathrm{Ni}^{2+}\) ion has an electronic configuration of \([Ar]3d^{8}\) with 8 d-electrons.

Determine Splitting Pattern in an Octahedral Field

The d-orbitals split into two sets in an octahedral field: the lower energy, triply degenerate set \(\{d_{z^2}, d_{x^2-y^2}\}\) called \(t_{2g}\), and the higher energy, doubly degenerate set \(\{d_{xy}, d_{yz}, d_{xz}\}\) called \(e_g\). The energy difference between these sets is called the crystal field splitting energy, denoted by \(\Delta_{\text{oct}}\).

Arrange the d-electrons

Now, we arrange the d-electrons into the d-orbitals according to Aufbau's principle, considering the high-spin or low-spin case: a. \(\mathrm{Fe}^{2+}\): - High-spin: Since \(\mathrm{Fe}^{2+}\) has 6 d-electrons, we will fill the 3 lower energy orbitals with 2 electrons each and the 2 higher energy orbitals with one electron each. - Low-spin: For the low-spin case, we fill the 3 lower energy orbitals with 2 electrons each, leaving the 2 higher energy orbitals unoccupied. b. \(\mathrm{Fe}^{3+}\) (high spin): With 5 d-electrons, we fill the 3 lower energy orbitals with 2 electrons each and place the fifth electron in one of the higher energy orbitals. c. \(\mathrm{Ni}^{2+}\): With 8 d-electrons, we fill all orbitals in \(t_{2g}\) set with 2 electrons each, and the remaining two electrons will be paired in two of the orbitals in the \(e_g\) set.

Draw the Orbital Splitting Diagrams

Now we can draw the d-orbital splitting diagrams for each complex ion: a. \(\mathrm{Fe}^{2+}\): High-spin: ``` e_g: ↑↓ ↑↓ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` Low-spin: ``` e_g: empty _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` b. \(\mathrm{Fe}^{3+}\) (high spin): ``` e_g: ↑↓ ↑ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ``` c. \(\mathrm{Ni}^{2+}\): ``` e_g: ↑↓ ↑↓ _________ t_{2g}: ↑↓ ↑↓ ↑↓ ```