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Explain these facts: (a) Copper and iron have several oxidation states, whereas zinc exists in only one. (b) Copper and iron form colored ions, whereas zinc does not.

Short Answer

Expert verified
Copper and iron have multiple oxidation states and form colored ions due to their unfilled d-orbitals. Zinc has a filled d-orbital, thus only one oxidation state and no color in its ions.

Step by step solution

01

Understand Oxidation States

Metals exhibit different oxidation states based on how easily they can lose electrons. Copper (Cu) exists in +1 and +2 oxidation states and iron (Fe) can exist in +2 and +3 states due to their partially filled d-electron configuration that allows variable electron loss. Zinc (Zn), however, has a completely filled 3d subshell (3d鹿鈦) in its ground state, which means it typically displays only one oxidation state, +2, due to the stability of losing two electrons.
02

Colorful Ion Formation in Transition Metals

Colored ions form due to d-d electron transitions in the presence of unpaired electrons in partially filled d-orbitals under the influence of a ligand field. Copper and iron have partially filled 3d orbitals which lead to such electron transitions, creating color. Zinc, with a filled d-shell (3d鹿鈦), does not have unpaired d-electrons available for such transitions, resulting in colorless ions.
03

Conclusion on Properties

Copper and iron can have multiple oxidation states and form colored ions due to partially filled d-orbitals and the presence of unpaired electrons, allowing for electron transitions and variable electron removal. Zinc's full 3d subshell results in a single oxidation state and colorless ions, as it lacks the electronic configuration to undergo d-d transitions.

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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 a fascinating group of elements found in the middle of the periodic table. These metals have a special characteristic: they have partially filled d-electron shells. This feature leads to a variety of interesting properties.
For instance, transition metals like copper (Cu) and iron (Fe) can lose different numbers of electrons, resulting in multiple oxidation states. This variability stems from their complex electron arrangements, specifically the availability of d-electrons that can be easily involved in bonding interactions.
In contrast, zinc (Zn) is somewhat different. Although it is close to transition metals on the periodic table, it typically displays only +2 oxidation state. This is because zinc has a full d鹿鈦 electron configuration, making its electronic structure more stable and less varied than other transition metals.
  • Transition metals have variable oxidation states.
  • Zinc usually shows only a +2 oxidation state.
d-Electron Configuration
The term 'd-electron configuration' refers to how electrons are organized in the d sublevel of an atom. These configurations are crucial in determining the chemical properties of transition metals.
Copper and iron have partially filled d-orbitals. For example, copper can have configurations ending in 3d鈦 or 3d鹿鈦, while iron has 3d鈦, 3d鈦 configurations possible. These arrangements allow for the loss of different numbers of electrons, which is key to their ability to form various oxidation states.
Zinc, however, has a stable 3d鹿鈦 configuration. This filled state means zinc doesn鈥檛 have the unpaired d-electrons needed to exhibit multiple oxidation states, limiting it primarily to the +2 state. Understanding d-electron configuration helps explain why some metals are more reactive or form different types of compounds compared to others.
  • Partially filled d orbitals lead to variable chemical behaviors.
  • A fully filled d鹿鈦 configuration leads to less variability in oxidation states.
Colored Ions
Colored ions in transition metals result from intricate processes involving light and electron movements. When light hits these ions, specific light wavelengths are absorbed, causing electron transitions within the d orbitals.
Both copper and iron have d-orbitals that aren鈥檛 fully filled, allowing for such electron transitions, which results in their characteristic colors. It is the movement of electrons between lower and higher energy levels within the d-orbitals, known as d-d transitions, that typically gives rise to color.
On the other hand, zinc ions are colorless. This is because zinc's d-orbitals are filled (3d鹿鈦), lacking unpaired electrons needed for these d-d transitions, which results in no absorption of visible light necessary to impart color. Thus, zinc-based compounds often appear white or colorless.
  • Color in ions is due to d-d electron transitions.
  • Zinc ions are colorless due to filled d-orbitals.

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Most popular questions from this chapter

What are the systematic names for these ions and compounds: (a) \(\left[\text { cis }-\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{3}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3},(\mathrm{~d})\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2},\) (e) trans- \(\mathrm{Pt}\left(\mathrm{NH}_{1}\right)_{2} \mathrm{Cl}_{2} ?\)

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A concentrated aqueous copper(II) chloride solution is bright green in color. When diluted with water, the solution becomes light blue. Explain.

A student in 1895 prepared three chromium coordination compounds having the same formulas of \(\mathrm{CrCl}_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\) with these properties: \(\mathrm{Cl}^{-}\) Ions in Solution \begin{tabular}{l} Color \\ \hline Violot \end{tabular} $$ \begin{array}{c} \text { per Formula Unit } \\ \hline 3 \\\ 2 \\ 1 \end{array} $$ Light Write modern formulas for these compounds and suggest a method for confirming the number of \(\mathrm{Cl}\) ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates, which are compounds that have a specific number of water molecules attached to them. The Cr has a coordination number of 6 in all three compounds.)

The complex ion \(\left[\mathrm{Ni}(\mathrm{CN})_{2} \mathrm{Br}_{2}\right]^{2-}\) has a square planar geometry. Draw the structures of the geometric isomers of this complex.

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