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Chapter 23: Problem 5

Write the electron configurations of the following ions: \(\mathrm{V}^{5+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Cu}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}\)

Short Answer

Expert verified
The electron configurations of the ions are: V^{5+} is \([Ar]\), Cr^{3+} is \([Ar] 3d^3\), Mn^{2+} is \([Ar] 3d^5\), Fe^{3+} is \([Ar] 3d^5\), Cu^{2+} is \([Ar] 3d^9\), Sc^{3+} is \([Ar]\), and Ti^{4+} is \([Ar]\).

Step by step solution

01

Determine the Atomic Numbers

Firstly locate the elements on the periodic table and identify their atomic numbers: V (Vanadium) is 23, Cr (Chromium) is 24, Mn (Manganese) is 25, Fe (Iron) is 26, Cu (Copper) is 29, Sc (Scandium) is 21, and Ti (Titanium) is 22.
02

Write the Electron Configurations of the Atoms

Write out the electron configurations for the neutral atoms. Recollect the order of filling in the orbitals (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p): V is \([Ar] 4s^2 3d^3\), Cr is \([Ar] 4s^1 3d^5\), Mn is \([Ar] 4s^2 3d^5\), Fe is \([Ar] 4s^2 3d^6\), Cu is \([Ar] 4s^1 3d^{10}\), Sc is \([Ar] 4s^2 3d^1\), and Ti is \([Ar] 4s^2 3d^2\).
03

Adjust for Ion Charge

Now account for the ion charge. Each positive charge means that one electron has been removed. Note that electrons are removed from the outermost shell first. So, V^{5+} would be \([Ar]\), Cr^{3+} would be \([Ar] 3d^3\), Mn^{2+} would be \([Ar] 3d^5\), Fe^{3+} would be \([Ar] 3d^5\), Cu^{2+} would be \([Ar] 3d^9\), Sc^{3+} would be \([Ar]\), and Ti^{4+} would be \([Ar]\).

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Most popular questions from this chapter

Which of the following 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.

Define the following terms: stereoisomers, geometric isomers, optical isomers, and plane-polarized light.

As we read across the first-row transition metals from left to right, the +2 oxidation state becomes more stable in comparison with the +3 state. Why is this so?

What are the differences between geometric isomers and optical isomers?

Explain why atomic radii decrease very gradually from scandium to copper.
See all solutions

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