Question

Write the formula for the fluoride corresponding to the highest expected oxidation state for (a) \(\mathrm{Sc}\), (b) \(\mathrm{Co}\), (c) \(\mathrm{Zn}\), (d) Mo.

Short answer

The fluoride compounds corresponding to the highest expected oxidation states for the given elements are: (a) ScF₃, (b) CoF₃, (c) ZnF₂, and (d) MoF₆.

Step-by-step solution

Determine the Electron Configurations of the Given Elements

To determine the highest expected oxidation state for each element, we first need to know their electron configurations. The electron configurations for the given elements are: (a) Scandium (Sc) - \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\) (b) Cobalt (Co) - \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\) (c) Zinc (Zn) - \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1 0\) (d) Molybdenum (Mo) - \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1 4d^5\)

Determine the Highest Expected Oxidation States

Using the electron configurations, we can now find the highest expected oxidation states for each element: (a) Scandium (Sc): Sc loses 2 electrons from the 4s subshell and 1 electron from the 3d subshell, giving an oxidation state of +3. (b) Cobalt (Co): Co can lose up to 2 electrons from the 4s subshell and 7 electrons from the 3d subshell, giving an oxidation state of +9. However, the +9 state is rare, and the more common highest oxidation state for cobalt is +3. (c) Zinc (Zn): Zn loses 2 electrons from the 4s subshell, giving an oxidation state of +2. (d) Molybdenum (Mo): Mo loses 1 electron from the 5s subshell and 5 electrons from the 4d subshell, giving an oxidation state of +6.

Write the Formulas for the Corresponding Fluoride Compounds

With the highest expected oxidation states determined, we can now write the formulas for the fluoride compounds: (a) Scandium fluoride: Sc has an oxidation state of +3, and fluoride (F) has an oxidation state of -1. The formula is: ScF₃. (b) Cobalt fluoride: Although the highest oxidation state for Co is +9, the more common oxidation state is +3. Considering the common oxidation state, the formula for cobalt fluoride is: CoF₃. (c) Zinc fluoride: Zn has an oxidation state of +2, and the formula for zinc fluoride is: ZnF₂. (d) Molybdenum fluoride: Mo has an oxidation state of +6, and the formula for molybdenum fluoride is: MoF₆. In conclusion, the fluoride compounds corresponding to the highest expected oxidation states for the given elements are: (a) ScF₃ (b) CoF₃ (c) ZnF₂ (d) MoF₆

Key concepts

Electron Configuration

Electron configuration is a way to represent the distribution of electrons in an atom. Electrons occupy specific regions called orbitals, and each orbital can hold a certain number of electrons. Understanding electron configuration is crucial for predicting the oxidation states of elements, especially transition metals, due to their complex electron arrangements.
Let's break down a few key points about electron configurations:

• Each electron shell is divided into subshells, named s, p, d, and f.
• s subshells can hold up to 2 electrons, p subshells up to 6, d subshells up to 10, and f subshells up to 14.
• The electron configuration follows the Aufbau principle, filling from lower to higher energy levels.

For example, scandium (Sc) has the electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\), with electrons filling the 3d orbital last, explaining its +3 oxidation state as it loses all 3d and 4s electrons. By understanding the filling order and specific electron configurations, we predict how many electrons an element can lose or gain, thus determining its oxidation potential.

Fluoride Compounds

Fluoride compounds consist of a metal or other element combined with fluorine atoms. Fluorides form when elements react with fluorine, often providing insights into the element's oxidation state. Fluorine is a highly electronegative element, meaning it tends to attract electrons strongly, making it very reactive.
Here are some important points about fluoride compounds:

• Fluoride acts as an oxidizing agent, facilitating the formation of metal fluorides by taking electrons from metals.
• The oxidation state of fluorine is always -1 in its compounds.
• Metallic elements form ionic bonds with fluorine in fluoride compounds.

In the context of the problem, the highest oxidation state of the metal affects the formula of the fluoride.
For instance, scandium can achieve a +3 state, creating the compound ScF₃. Cobalt, in its +3 state, forms CoF₃. Transition metals often have multiple oxidation states, leading to various possible fluoride compounds. Understanding these states allows us to determine the most probable and stable compound for each metal.

Transition Metals

Transition metals are elements found in the d-block of the periodic table. They are known for their ability to form various oxidation states due to their partially filled d orbitals. This characteristic leads to a diverse range of chemical behaviors and compounds.
Key characteristics of transition metals include:

• The ability to form multiple oxidation states. For instance, cobalt can have oxidation states ranging from +2 to +3 and even higher in certain conditions.
• The presence of electrons in d orbitals allows them to form colorful compounds and complex ions.
• They often serve as catalysts in chemical reactions.

Transition metals like cobalt and molybdenum show diverse behaviors due to their electron configurations. In the given exercise, scandium, cobalt, zinc, and molybdenum each have different highest oxidation states used for forming their fluoride compounds. Cobalt typically achieves a +3 state for stability in compounds such as CoF₃. Transition metals continue to be of interest due to their flexibility and role in various chemical processes.