Question

The element vanadium exhibits multiple oxidation states in its compounds, including \(+2\). The compound \(\mathrm{VCl}_{2}\) is known, whereas \(\mathrm{ScCl}_{2}\) is unknown. Use electron configurations and effective nuclear charges to account for this difference in behavior.

Short answer

Vanadium (V) has the electron configuration \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\], and an effective nuclear charge (Z_eff) of 5. This allows it to easily lose two 4s electrons to form VCl₂. Scandium (Sc) has the electron configuration \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\], and a lower Z_eff of 3, making it harder to lose two electrons. Sc forms ScCl₃ with a +3 oxidation state, but ScCl₂ is not formed. The difference in electron configurations and effective nuclear charges account for vanadium forming VCl₂, while scandium does not form ScCl₂.

Step-by-step solution

Determine the electron configurations of vanadium and scandium

First, we need to determine the electron configurations of V and Sc. The atomic numbers of vanadium and scandium are 23 and 21, respectively. Therefore, their electron configurations can be written as: - Vanadium (V): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3 \] - Scandium (Sc): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1 \]

Calculate the effective nuclear charges of vanadium and scandium

Next, we need to calculate the effective nuclear charges (Z_eff) of each element. The effective nuclear charge is a measure of the net positive charge experienced by an electron in a multi-electron atom. It can be calculated using Slater's rules: \[Z_{eff} = Z - S\] where Z is the atomic number, and S is the shielding constant, which accounts for the electron-electron repulsion in the atom. For vanadium: - Z (V) = 23 - S (V) ≈ 2(1s) + 2(2s) + 6(2p) + 2(3s) + 6(3p) = 18 \[Z_{eff} (V) = 23 - 18 = 5\] For scandium: - Z (Sc) = 21 - S (Sc) ≈ 2(1s) + 2(2s) + 6(2p) + 2(3s) + 6(3p) = 18 \[Z_{eff} (Sc) = 21 - 18 = 3\]

Compare the electron configurations and effective nuclear charges to explain the difference in behavior

Now that we have the electron configurations and effective nuclear charges of vanadium and scandium, we can compare them to understand why V forms a compound with a +2 oxidation state (VCl₂), while Sc does not (ScCl₂). Vanadium's electron configuration has three 3d electrons, which makes it easier for V to lose two or more electrons to achieve a stable oxidation state. In the case of VCl₂, vanadium loses two 4s electrons, forming V²⁺. Due to the higher effective nuclear charge (Z_eff = 5) experienced by vanadium, it is more difficult for V to lose additional electrons, and this prevents the formation of VCl₃ or VCl₄ in the +3 or higher oxidation states. On the other hand, scandium's electron configuration only contains one 3d electron. This makes it difficult for Sc to lose two electrons, as it would involve losing one s electron and one d electron. Furthermore, the relatively lower effective nuclear charge (Z_eff = 3) experienced by Sc suggests that it is less likely to form higher oxidation states, as it requires additional energy to remove more loosely bound electrons. As a result, Sc forms ScCl₃ with a +3 oxidation state, but ScCl₂ is not formed. Overall, the difference in electron configurations and effective nuclear charges of vanadium and scandium account for the observed difference in their behavior, allowing vanadium to form VCl₂ while scandium does not form ScCl₂.

Key concepts

Electron Configuration

Electron configuration plays a pivotal role in understanding the chemical behavior of elements, especially transition metals like vanadium and scandium. An electron configuration defines the distribution of electrons among the orbitals of an atom. Each element's position on the periodic table reflects its electron configuration.

For instance, vanadium (V), with an atomic number of 23, has the electron configuration \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\]. This means vanadium has 23 electrons distributed in those energy levels and orbitals. The 3d electrons are significant because they are less tightly bound and can participate in bonding and chemical reactions, including forming various oxidation states.

Scandium (Sc), with an atomic number of 21, has the electron configuration \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\]. Scandium's valence electrons include one 3d electron, making it less versatile in forming multiple oxidation states compared to vanadium.

• Vanadium: More 3d electrons, leading to diverse oxidation states
• Scandium: Only one 3d electron, usually resulting in a +3 oxidation state

The electron configuration is crucial when predicting the atom's chemical behavior, like which compounds it can form.

Effective Nuclear Charge

The effective nuclear charge (\(Z_{eff}\)) is the net positive charge experienced by an electron in an atom. It influences the atom's size and its ability to form ions, which directly affects its chemical properties, including possible oxidation states.

Vanadium, with a higher effective nuclear charge of\( Z_{eff} = 5 \), experiences a stronger pull from the nucleus on its valence electrons. This translates to a greater ability to hold on to these electrons, influencing its ability to form compounds like VCl₂ by losing two electrons. The higher \(Z_{eff}\) makes it energetically more feasible to reach the +2 oxidation state, forming stable compounds.

Meanwhile, scandium has a lower \(Z_{eff} = 3\), meaning its electrons experience less pull from the nucleus. This weaker nuclear attraction results in a preference to lose precisely the desired number of electrons, leading it to predominantly form \(Sc^{3+}\) ions and not ScCl₂. The influence of nuclear charge highlights:

• Stronger effective nuclear charge generally stabilizes lower oxidation states.
• Weaker effective nuclear charge favors removal of more electrons until stable configurations (like d^{0}) are achieved.

Understanding nuclear charge helps explain why transition metals exhibit various oxidation states depending on their electron configurations.

Transition Metals

Transition metals, such as vanadium and scandium, are unique due to their ability to form multiple oxidation states. This is primarily due to their electron configurations and the involvement of d-orbitals. In the periodic table, transition metals occupy groups 3-12 and are characterized by the filling of d-orbitals in their electron configurations.

These metals can lose d and s electrons to form cations of various charges, which allows them to engage in diverse chemical bonding situations. For example, vanadium can be found in several oxidation states, including +2, +3, +4, and +5. This diversity is crucial for catalysis and is utilized in many industrial processes and chemical reactions.

Scandium, while being a transition metal, behaves differently due to its limited d-electron count. Typically losing three electrons to form \(Sc^{3+}\), its chemical versatility is comparatively limited. This variance among transition metals illustrates:

• Ability to form complexes with variable oxidation states.
• Diverse roles in industrial and chemical processes.
• The variability in properties even among metals within the same category.

The versatility and behavior of transition metals are what make them indispensable in both chemistry and industry.