Question

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\), (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), (a high-spin complex), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (a low-spin complex), (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (a low-spin complex), (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\), (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\).

Short answer

In summary, the crystal-field energy-level diagrams with the placements of d-electrons for the complexes are as follows: (b) Mn(H2O)6(2+) (high-spin): \[ \begin{array}{c} \underline{\uparrow\downarrow} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow} & \underline{\uparrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \] (a) Cr(H2O)6(2+) (high-spin): \[ \begin{array}{c} \underline{\uparrow\downarrow} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow} & \underline{\uparrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \] (c) Ru(NH3)5H2O(2+) (low-spin): \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \] (d) IrCl6(2-) (low-spin): \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \] (e) Cr(en)3(3+) (low-spin): \[ \begin{array}{c} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \] (f) NiF6(4-) (high-spin): \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

Step-by-step solution

(b) Mn(H2O)6(2+)

: First, find the oxidation state of Mn in Mn(H2O)6(2+). Since the complex has an overall charge of +2, Mn is in the +2 oxidation state. Manganese, Mn, has an electronic configuration of [Ar] 4s² 3d⁵. In its +2 oxidation state, Mn loses 2 electrons from the 4s orbital, giving the configuration of [Ar] 3d⁵. This complex is not specified as high-spin or low-spin, but we can assume that Mn(H2O)6(2+) is high-spin since it is an octahedral complex with water ligands. The energy-level diagram for a high-spin, octahedral Mn(II) complex is: \( \) \[ \begin{array}{c} \underline{\uparrow\downarrow} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow} & \underline{\uparrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

(a) Cr(H2O)6(2+) (four unpaired electrons, high-spin complex)

: First, find the oxidation state of Cr in Cr(H2O)6(2+). Since the complex has an overall charge of +2, Cr is in the +2 oxidation state. Chromium, Cr, has an electronic configuration of [Ar] 4s² 3d⁴. In its +2 oxidation state, Cr loses 2 electrons from the 4s orbital, giving the configuration of [Ar] 3d⁴. Since the complex is high-spin, the energy-level diagram for Cr(H2O)6(2+) is: \( \) \[ \begin{array}{c} \underline{\uparrow\downarrow} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow} & \underline{\uparrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

(c) Ru(NH3)5H2O(2+) (low-spin complex)

: First, find the oxidation state of Ru in Ru(NH3)5H2O(2+). Since the complex has an overall charge of +2, Ru is in the +2 oxidation state. Ruthenium, Ru, has an electronic configuration of [Kr] 5s² 4d⁶. In its +2 oxidation state, Ru loses 2 electrons from the 5s orbital, giving the configuration of [Kr] 4d⁶. Since the complex is low-spin, the energy-level diagram is: \( \) \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

(d) IrCl6(2-) (low-spin complex)

: First, find the oxidation state of Ir in IrCl6(2-). Since the complex has an overall charge of -2, Ir is in the +3 oxidation state. Iridium, Ir, has an electronic configuration of [Xe] 6s² 5d⁷. In its +3 oxidation state, Ir loses 2 electrons from the 6s orbital and 1 electron from the 5d orbital, giving the configuration of [Xe] 5d⁶. Since the complex is low-spin, the energy-level diagram is: \( \) \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

(e) Cr(en)3(3+)

: First, find the oxidation state of Cr in Cr(en)3(3+). Since the complex has an overall charge of +3, Cr is in the +3 oxidation state. Chromium, Cr, has an electronic configuration of [Ar] 4s² 3d⁴. In its +3 oxidation state, Cr loses its 2 electrons in the 4s orbital and 1 electron from the 3d orbital, giving the configuration of [Ar] 3d³. The energy-level diagram will be the same as for a low-spin complex since Cr(en)3(3+) is an octahedral complex with bidentate ligands: \( \) \[ \begin{array}{c} & & & \\ e_{g} & & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

(f) NiF6(4-)

: First, find the oxidation state of Ni in NiF6(4-). Since the complex has an overall charge of -4, Ni is in the +2 oxidation state. Nickel, Ni, has an electronic configuration of [Ar] 4s² 3d⁸. In its +2 oxidation state, Ni loses 2 electrons from the 4s orbital, giving the configuration of [Ar] 3d⁸. Since NiF6(4-) is an octahedral complex with fluoride ligands, the energy-level diagram will be the same as for a high-spin complex: \( \) \[ \begin{array}{c} \underline{\uparrow\downarrow} & \underline{\uparrow} & & \\ e_{g} & e_{g} & & \\ \cline{1-1} \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} & \underline{\uparrow\downarrow} \\ t_{2 g} & t_{2 g} & t_{2 g} \\ \end{array} \]

Key concepts

High-spin complex

High-spin complexes are fascinating components of coordination chemistry where the electron configuration in a transition metal's d-orbitals reveals a unique pattern. Here, the energy gap between split d-orbitals is small, allowing more electrons to occupy higher-energy orbitals with unpaired spins. This makes such complexes magnetic since they contain more unpaired electrons.

Consider \([\text{Mn(H}_2\text{O})_6]^{2+}\) as an example of a high-spin complex. Manganese in this complex adopts a +2 oxidation state, leading to a configuration of \[3d^5\]. The arrangement of electrons across the d-orbitals for Mn produces five unpaired electrons.

• This configuration results in a magnetic moment due to the presence of all unpaired spins.
• Ligands with weaker field strength, like water, do not sufficiently split the orbitals to change the spin state, hence maintaining it as high-spin.

Understanding high-spin complexes helps in distinguishing magnetic properties in transition metal complexes.

Low-spin complex

Low-spin complexes, in contrast to their high-spin counterparts, form when the d-orbital electron gap increases, forcing electrons to pair up in the lower energy levels. This scenario arises frequently with strong-field ligands.

Consider the low-spin nature of \(\text{Ru(NH}_3)_5\text{H}_2\text{O})^{2+}\) and \(\text{IrCl}_6^{2-}\). For Ruthenium in \(\text{Ru(NH}_3)_5\text{H}_2\text{O})^{2+}\), the oxidation state is +2, leading to an electron configuration of \[4d^6\].\

• The strong-field ligands like \(\text{NH}_3\) result in greater splitting, forcing electrons to fill lower energy orbitals, minimizing unpaired electrons.
• This prompt for electron pairing leads to complexes exhibiting less magnetism or being diamagnetic.

For a low-spin complex like \(\text{IrCl}_6^{2-}\), you would see only paired electrons forming, resulting in a diamagnetic nature.Low-spin complexes are often used in systems where low magnetism is beneficial, such as in catalysts.

Octahedral complexes

Octahedral complexes are prevalent in coordination chemistry involving six ligands symmetrically arranged around a central metal atom. This geometric arrangement leads to distinct splitting of d-orbitals into two sets: \(e_g\) and \(t_{2g}\).

In an octahedral field, the electron configuration's response determines if a complex will be high-spin or low-spin:

• The presence of weak-field ligands results in small splitting, usually leading to high-spin configurations with several unpaired electrons.
• Conversely, strong-field ligands cause significant splitting, favoring low-spin states with minimal unpaired electrons.

Examples like \(\text{Cr}(\text{H}_2\text{O})_6^{2+}\) and \(\text{NiF}_6^{4-}\), both octahedral, illustrate how ligand type influences electron placement in the split orbitals. Understanding octahedral geometry is crucial for predicting the magnetic and spectroscopic properties of transition metal complexes.

Electronic configuration

Electronic configuration describes the distribution of electrons in an atom's orbitals and plays a key role in chemical properties, especially for transition metals. These metals fill the d-orbitals, which directly influence their chemical behavior.

Taking Chromium in \(\text{Cr}(\text{H}_2\text{O})_6^{2+}\) as an example, in a +2 oxidation state, its configuration is \[3d^4\]. Analyzing this configuration helps predict:

• Magnetic properties through the number of unpaired electrons—here, it results in four unpaired ones.
• The split pattern of electrons in octahedral complexes that can result in either high-spin or low-spin state.

Correctly interpreting electronic configurations informs us about complex formation, reactivity, color, and other vital characteristics.

Transition metals

Transition metals are a group of elements central to many chemical processes due to their unique ability to form variable oxidation states and complex ions. They can efficiently participate in bonding through d-orbitals.

Transition metals often form coordination compounds such as \(\text{Cr}(\text{en})_3^{3+}\) owing to their:

• Ability to exhibit various oxidation states, providing versatility in reactions.
• Formation of colored compounds due to d-d electron transitions influenced by ligand fields.
• Crucial role in biological systems as iron in hemoglobin.

Their widespread application in catalysis, electronics, and materials science underscores the importance of understanding transition metal chemistry.