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Give the highest oxidation states for scandium to copper.

Short Answer

Expert verified
The highest oxidation states for Sc to Cu from +3 to +7.

Step by step solution

01

Understand the Periodic Table Positioning

Scandium to copper are transition metals located in the fourth period of the periodic table with atomic numbers 21 to 29. These elements have partially filled d sub-shells which allow them to exhibit multiple oxidation states.
02

Identify Electron Configurations

Determine the electron configurations for each element from Scandium (Sc) to Copper (Cu):\[ \begin{align*} & \text{Scandium (Sc): } [\text{Ar}] 3d^1 4s^2 \ & \text{Titanium (Ti): } [\text{Ar}] 3d^2 4s^2 \ & \text{Vanadium (V): } [\text{Ar}] 3d^3 4s^2 \ & \text{Chromium (Cr): } [\text{Ar}] 3d^5 4s^1 \ & \text{Manganese (Mn): } [\text{Ar}] 3d^5 4s^2 \ & \text{Iron (Fe): } [\text{Ar}] 3d^6 4s^2 \ & \text{Cobalt (Co): } [\text{Ar}] 3d^7 4s^2 \ & \text{Nickel (Ni): } [\text{Ar}] 3d^8 4s^2 \ & \text{Copper (Cu): } [\text{Ar}] 3d^{10} 4s^1 \end{align*} \] The varying d and s electrons provide flexibility in oxidation states.
03

Determine the Maximum Oxidation State

The highest oxidation state generally corresponds to the total number of electrons available from the outermost s and d sub-shells. Count these electrons for each element: \[ \begin{align*} & \text{Scandium (Sc): } +3 \ & \text{Titanium (Ti): } +4 \ & \text{Vanadium (V): } +5 \ & \text{Chromium (Cr): } +6 \ & \text{Manganese (Mn): } +7 \ & \text{Iron (Fe): } +6 \ & \text{Cobalt (Co): } +5 \ & \text{Nickel (Ni): } +4 \ & \text{Copper (Cu): } +2 \end{align*} \] Note that while some elements can theoretically have higher oxidation states, these are the most commonly observed highest states in stable compounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transition Metals
Transition metals are an intriguing group of elements found in the central section of the periodic table. These elements, ranging from scandium to copper for instance, belong to the d-block, which means they have d-electrons involved in their electron configuration. Transition metals are unique because they often have partially filled d orbitals, allowing them to exhibit a variety of oxidation states. This is why they can form complex ions, conduct electricity, and serve as catalysts in many chemical reactions.

Here are some standout features of transition metals:
  • They can adopt more than one oxidation state.
  • They typically form colored compounds.
  • They can conduct electricity.
  • They often have high melting and boiling points.
Understanding these features helps in comprehending the chemical and physical properties of transition metals, as well as their wide applications in various industries and technologies.

When learning about transition metals, remember they are not just limited to the number of electrons they can lose or gain, but can also include their ability to form various coordination complexes.
Electron Configurations
Electron configurations are vital in determining the chemical behavior of an element. Transition metals from scandium to copper display interesting electron arrangements, mainly due to their d orbitals. Their electrons fill up the 3d sub-level after the 4s, making their configurations unique.

For instance:
  • Scandium: \( [\text{Ar}] 3d^1 4s^2 \)
  • Chromium: \( [\text{Ar}] 3d^5 4s^1 \)
  • Copper: \( [\text{Ar}] 3d^{10} 4s^1 \)
These configurations illustrate why these metals can exhibit multiple oxidation states. The presence of d electrons adds flexibility, allowing the elements to lose various numbers of electrons. This characteristic is key in forming different compounds and influencing reactivity.

Transition metals often have their s and d electrons participate in bonding. This contributes to the rich chemistry associated with them, such as variable oxidation states and complex ion formation.
Periodic Table Positioning
Understanding the periodic table positioning gives insight into the trends and behaviors of elements. Transition metals like scandium through copper are located in the middle of the periodic table, specifically in groups 3 through 12 and periods such as the fourth period. This particular placement informs us about their similarities in physical and chemical properties stemming from partially filled d orbitals.

Positioning in the periodic table can tell us about:
  • The types of chemical bonds they can form.
  • The potential oxidation states they might adopt.
  • Their overall reactivity compared to other sections, like the alkali metals.
Moreover, the periodic table layout highlights the gradual filling of electronic sub-levels, illustrating why the 3d sub-level appears after filling the 4s in transition metals. This positioning aspect further explains specific properties such as color, magnetism, and the ability to act as catalysts.

These elements' properties and behaviors are interconnected with their place on the periodic table, providing valuable insights into their role in both simple and complex chemical reactions.

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Most popular questions from this chapter

Complete the following statements for the complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{CN}\right]^{2+}\). (a) en is the abbreviation for______ (b) The oxidation number of \(\mathrm{Co}\) is _________ (c) The coordination number of Co is ________ (d) _________ is a bidentate ligand.

Write the formulas for each of the following ions and compounds: (a) dichlorobis(ethylenediamine) chromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammineaquachlorocobalt(III) chloride.

Plastocyanin, a copper-containing protein found in photosynthetic systems, is involved in electron transport, with the copper ion switching between the +1 and +2 oxidation states. The copper ion is coordinated with two histidine residues, a cysteine residue, and a methionine residue in a tetrahedral configuration. How does the crystal field splitting \((\Delta)\) change between these two oxidation states?

Consider the following two ligand exchange reactions: $$ \begin{aligned} \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+6 \mathrm{NH}_{3} & \rightleftarrows\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} \\ \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+3 \mathrm{en} & \rightleftarrows\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} \end{aligned} $$ (a) Which of the reactions should have a larger \(\Delta S^{\circ}\) ? (b) Given that the \(\mathrm{Co}-\mathrm{N}\) bond strength is approximately the same in both complexes, which reaction will have a larger equilibrium constant? Explain your choices.

Explain why a thermodynamically stable species may be chemically reactive and a thermodynamically unstable species may be unreactive.

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