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Chapter 20: Problem 8

Chromium exhibits several oxidation states in its compounds, whereas aluminum exhibits only the +3 oxidation state. Explain.

Short Answer

Expert verified
Aluminum, with its 3 valence electrons, tends to lose them when bonding to have a stable and full outer shell, achieving a consistent +3 oxidation state. However, Chromium's complex electron configuration with 's' and 'd' electrons available for bonding enables it to have variable oxidation states in its compounds.

Step by step solution

01

Understand the Electronic Configuration

The oxidation state is determined by the valence electrons, which are the electrons in the outermost energy level of an atom. Aluminum, being in Group 13 of the periodic table, has 3 electrons in its outermost shell. These are available for bonding. Chromium, on the other hand, is a transition metal in Group 6. Its electron configuration is more complex. Its outer configuration consists of 4s and 3d electrons.
02

Explain Aluminum's Oxidation State

As Aluminum has 3 electrons in its outermost shell (shell 3, the 'p' subshell), when it forms compounds it tends to lose these three electrons to have a stable electron configuration. After losing these 3 electrons, it achieves the electron configuration of the noble gas before it (Neon), which is a stable configuration with filled energy levels. This results in an oxidation state of +3. Aluminum can't exhibit any other oxidation state because it can only lose these three valence electrons, it does not have readily accessible electrons in lower energy levels (the 'd' or 'f' subshells are empty).
03

Elaborate on Chromium's Multiple Oxidation States

Chromium has a more complex electron configuration, with both 4s and 3d electrons available for bonding. Transition metals like Chromium often utilize their 'd' electrons along with their 's' electrons for bonding in compounds. It is common for transition metals to exhibit variable oxidation states within their compounds, and this is a direct result of the involvement of the inner 'd' shell in bonding. Chromium commonly exhibits several oxidation states (like +2, +3, and +6) based on how it can lose both 's' and varying amounts of 'd' electrons for bonding.

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Most popular questions from this chapter

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

Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

The absorption maximum for the complex ion \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) occurs at \(470 \mathrm{nm}\). (a) Predict the color of the complex and (b) calculate the crystal field splitting in kilojoules per mole.

Write the formulas for each of these ions and compounds: (a) tetrahydroxozincate(II), (b) pentaaquochlorochromium(III) chloride, (c) tetrabromocuprate(II), (d) ethylenediaminetetraacetatoferrate(II).

Which of these hydrated cations are colorless: \(\mathrm{Fe}^{2+}(a q), \mathrm{Zn}^{2+}(a q), \mathrm{Cu}^{+}(a q), \mathrm{Cu}^{2+}(a q), \mathrm{V}^{5+}(a q)\) \(\mathrm{Ca}^{2+}(a q), \mathrm{Co}^{2+}(a q), \mathrm{Sc}^{3+}(a q), \mathrm{Pb}^{2+}(a q) ?\) Explain your choice.
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