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. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

Short answer

The number of unpaired electrons in the complex ions are: a. 0 unpaired electrons b. 2 unpaired electrons c. 0 unpaired electrons

Step-by-step solution

Identify the oxidation state of the metal ion

a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): Since NH3 is a neutral ligand, the oxidation state of Ru is +2. b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): H2O is a neutral ligand, so the oxidation state of Ni is +2. c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): 'en' refers to ethylenediamine, which is a bidentate neutral ligand. Therefore, the oxidation state of V is +3.

Determine the electron configuration of the metal ion

We will consider the electron configurations of the metal ions in their ground state and then remove the required number of electrons from the outermost orbitals based on their oxidation state: a. Ru has an atomic number of 44. Its electron configuration is [Kr] 4d7 5s1. In an oxidation state of +2, we have Ru(II), with the electron configuration [Kr] 4d6. b. Ni has an atomic number of 28. Its electron configuration is [Ar] 3d8 4s2. In an oxidation state of +2, we have Ni(II), with the electron configuration [Ar] 3d8. c. V has an atomic number of 23. Its electron configuration is [Ar] 3d3 4s2. In an oxidation state of +3, we have V(III), with the electron configuration [Ar] 3d2.

Calculate the number of unpaired electrons

Now we calculate the unpaired electrons considering the electron distribution in d orbitals for each complex ion: a. For \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): It is a low spin complex due to the strong field ligand (NH3) present in the complex ion. Therefore, the d orbitals will pair up before filling, resulting in the electron configuration t2g6 eg0. The number of unpaired electrons is 0. b. For \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): It is a high spin complex due to the weak field ligand (H2O) present in the complex ion. The electron configuration is t2g6 eg2. The number of unpaired electrons is 2. c. For \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): Ethylenediamine is a strong field ligand, making it a low spin complex. The electron configuration is t2g2 eg0. The number of unpaired electrons is 0. Thus, the number of unpaired electrons in the complex ions are as follows: a. 0 unpaired electrons b. 2 unpaired electrons c. 0 unpaired electrons

Key concepts

Electron Configuration

Understanding the electron configuration of an atom is crucial in coordination chemistry. It helps us determine how electrons are distributed around an atom in its ground state. Generally, this distribution follows a specific order: electrons fill the lowest energy levels first. For transition metals, like ruthenium (Ru), nickel (Ni), and vanadium (V), electrons mainly occupy 3d, 4d, and 5s orbitals. Starting with the atomic numbers:

• Ruthenium (Ru) has an atomic number of 44, giving it an electron configuration of \[ \text{[Kr]} 4d^7 5s^1 \] in its neutral state. In the Ru(II) state, it loses two electrons leaving \[ \text{[Kr]} 4d^6 \].
• Nickel (Ni) with atomic number 28 has \[ \text{[Ar]} 3d^8 4s^2 \]. Nickel in Ni(II) loses two electrons resulting in \[ \text{[Ar]} 3d^8 \] configuration.
• Vanadium (V), atomic number 23, thus \[ \text{[Ar]} 3d^3 4s^2 \]. In V(III), losing three electrons results in \[ \text{[Ar]} 3d^2 \].

The electron configurations can change when atoms become ions, especially in transition metals due to the movement of electrons from s to d orbitals. This also affects their magnetic and spectroscopic properties.

Unpaired Electrons

Unpaired electrons are those that occupy an orbital alone, without a partner with opposite spin. The presence or absence of unpaired electrons is a determinant in the magnetic properties of a substance. The electron configurations of complex ions give us a preview of their magnetic behavior.In coordination complexes:

• Low spin complexes like Ru(NH₃)₆²⁺ and V(en)₃³⁺ are more likely to have paired electrons due to strong field ligands, resulting in fewer unpaired electrons.
• For instance, the d orbital in Ru(NH₃)₆²⁺ pairs up as \[ t_{2g}^6 \] resulting in all electrons being paired.
• High spin complexes like Ni(H₂O)₆²⁺ possess weaker field ligands, often leading to more unpaired electrons. Ni(H₂O)₆²⁺ shows up as \[ t_{2g}^6 e_g^2 \], indicating two unpaired electrons.

Understanding the number of unpaired electrons not only affects the expected magnetism but is also crucial in predicting chemical reactivity and stability.

Oxidation State

The oxidation state of a metal in a coordination complex helps us determine how many electrons are lost when forming the complex. It is key to understanding the nature of the complex and its chemical behavior. In the given complexes:

• For Ru(NH₃)₆²⁺, ammonia is neutral, leaving Ru in a +2 oxidation state.
• In Ni(H₂O)₆²⁺, water is neutral as well, resulting in Ni(II) or +2 oxidation state.
• Lastly, for V(en)₃³⁺, ethylenediamine is a neutral ligand, leading to a +3 oxidation state for vanadium.

The oxidation state profoundly influences the electronic configuration and subsequently the properties of the complex like color, magnetism, and reactivity. It's determined by balancing the overall charge on the complex with the known charges of the ligands.

Ligand Field Theory

Ligand field theory (LFT) is an essential concept in coordination chemistry used to predict and understand the behavior of complexes. It evolved from crystal field theory, focusing on how ligands' electron fields split the d-orbitals of the central metal. Key highlights of LFT in complexes:

• Strong field ligands like NH₃ and en cause greater splitting of d-orbitals, leading to low-spin configurations. This results in more electron pairing, as in Ru(NH₃)₆²⁺ and V(en)₃³⁺.
• Weak field ligands such as H₂O, lead to smaller splitting with a high-spin state. This allows more unpaired electrons, as seen in Ni(H₂O)₆²⁺.
• The energy difference caused by ligand field splitting influences many properties of the complex like its spectroscopic behavior and magnetism.

Ligand field theory is a more advanced and nuanced approach than the simpler crystal field theory. It provides an invaluable framework for predicting and explaining the properties of coordination compounds.