Problem 55 - Chemistry An Atoms First Approach

Question

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\)

Step-by-Step Solution

Answer

The number of unpaired electrons for each complex ion are: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 unpaired electrons b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 unpaired electrons c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\): 0 unpaired electrons

Step 1: Identify the central atom and its electron configuration

For each complex ion: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) - Central atom: Ru (Ruthenium) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) - Central atom: Ni (Nickel) c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\) - Central atom: V (Vanadium) Now we can find the electron configuration of these atoms: a. Ru: [Kr]4d^75s^1 b. Ni: [Ar]3d^84s^2 c. V: [Ar]3d^34s^2

Step 2: Determine the oxidation states of the central atoms

Now we need to find the oxidation states of the central atoms: a. Ru: \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) - Since ammonia (NH3) is a neutral ligand, the oxidation state of Ru is +2. b. Ni: \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) - Since water (H2O) is a neutral ligand, the oxidation state of Ni is +2. c. V: \(\mathrm{V}(\mathrm{en})_{3}^{3+}\) - Since ethylenediamine (en) is a bidentate neutral ligand, the oxidation state of V is +3.

Step 3: Account for the oxidation state and find the electron configuration of the central ion

Now we can find the electron configuration of the central ion with the corresponding oxidation state: a. Ruthenium(II) ion (Ru2+): [Kr]4d^6 b. Nickel(II) ion (Ni2+): [Ar]3d^8 c. Vanadium(III) ion (V3+): [Ar]3d^2

Step 4: Determine the effect of ligands on d-orbital splitting and find the unpaired electrons

To find the number of unpaired electrons, we must consider the effect of ligands on the d-orbital splitting. a. Since we are considering the low-spin case for \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\), the NH3 ligands, being a strong-field ligand, cause the d-electrons to pair. The electron configuration in the low-spin case will be t2g^6 eg^0. Thus: - Unpaired electrons in \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 b. In \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), the water is a weak-field ligand. The electron configuration will be t2g^6 eg^2. Thus: - Unpaired electrons in \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 c. In \(\mathrm{V}(\mathrm{en})_{3}^{3+}\), the ethylenediamine (en) is a strong-field ligand. The electron configuration will be t2g^2 eg^0. Thus: - Unpaired electrons in \(\mathrm{V}(\mathrm{en})_{3}^{3+}\): 0

Key Concepts

Complex Ions

Complex ions are a central concept in coordination chemistry. These are entities made up of a central metal atom or ion connected to surrounding molecules or ions, called ligands. The resulting combination forms a coordination complex that possesses its own unique set of properties.
A metal in a complex ion is usually a transition metal, which has a partially filled d subshell, allowing it to form various coordination bonds with ligands.
A common feature of complex ions is their charge, which results from the sum of the formal charges of the central ion and its ligands.

• The number of ligands attached can vary, and they are referred to as the coordination number of the metal ion.
• Ligands donate pairs of electrons to the metal, forming coordinated covalent bonds.

These characteristics greatly influence the chemical behavior of the complex, including its reactivity, color, and magnetic properties.

Oxidation States

Understanding oxidation states is crucial for chemists as they describe the degree of oxidation of a metal ion within a complex. The oxidation state is a theoretical charge on an atom if all bonds were purely ionic. This helps predict the formula and structure of the compound.
The oxidation state of a central atom influences the overall charge of the complex ion. For instance:

• In \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\), since ammonia (NH\(_3\)) is neutral, the oxidation state of Ru is +2.
• For \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), again, water (H\(_2\)O) is neutral, leaving Ni with a +2 oxidation state.
• In \(\mathrm{V}(\mathrm{en})_{3}^{3+}\), the bidentate ligand ethylenediamine (en) is neutral, resulting in a +3 oxidation state for V.

These oxidation states are determined by considering the known charges of common ligands and the desired charge of the complex.

d-orbital Splitting

When ligands bond to the metal ion in a complex, they affect the energy levels of the d orbitals, a concept known as crystal field splitting. The electronic structure of metal ions plays a crucial role in determining the properties of the complex:
d-orbital splitting occurs because when ligands are present, the degeneracy of the d orbitals (their being at the same energy levels) breaks. This results in different energies for the orbitals.

• For example, in octahedral complexes, such as \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\), the d-orbitals split into two levels: t\(_{2g}\) and e\(_g\).
• The degree of splitting is influenced by the field strength of the ligands, which can be strong-field or weak-field.

Strong-field ligands, like NH\(_3\), cause a large splitting, resulting in more paired electrons and fewer unpaired electrons, defining it as a low-spin complex. In contrast, weak-field ligands like water result in less splitting, often leading to more unpaired electrons in a high-spin complex.

Ligands

Ligands are ions or molecules that donate pairs of electrons to the central metal ion in a complex, forming a coordinate covalent bond. The nature of ligands and their arrangement directly impact the behavior and characteristics of the complex.
Ligands can broadly be categorized into neutral or charged, and by their denticity, which refers to the number of coordinating atoms they contain.

• Monodentate ligands, like ammonia (NH\(_3\)) and water (H\(_2\)O), have one donor atom.
• Bidentate ligands, such as ethylenediamine (en), involve two donor atoms that can form a stable two-point connection with the metal ion.

The type of ligand affects properties like solubility, geometric structure, and magnetic properties of the complex, and is critical in determining the strength and color of metal complexes due to their field strength. Strong-field ligands often lead to low-spin configurations, while weak-field leads to high-spin configurations.