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Chapter 21: Problem 26

Write electron configurations for each of the following. a. \(\mathrm{Cr}, \mathrm{Cr}^{2+}, \mathrm{Crr}^{3+}\) b. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

Short Answer

Expert verified
The electron configurations for the given elements and ions are as follows: a. Chromium: \(\mathrm{Cr: [Ar]4s^1 3d^5}\), \(\mathrm{Cr^{2+}: [Ar]3d^4}\), \(\mathrm{Cr^{3+}: [Ar]3d^3}\). b. Copper: \(\mathrm{Cu: [Ar]4s^1 3d^{10}}\), \(\mathrm{Cu^{+}: [Ar]3d^9}\), \(\mathrm{Cu^{2+}: [Ar]3d^8}\). c. Vanadium: \(\mathrm{V: [Ar]4s^2 3d^3}\), \(\mathrm{V^{2+}: [Ar]3d^3}\), \(\mathrm{V^{3+}: [Ar]3d^2}\).

Step by step solution

01

Determine the atomic number of Chromium

Chromium (Cr) has an atomic number of 24 which means it has 24 electrons in the neutral form.
02

Write the electron configuration for Chromium

The electron configuration of Chromium is \([Ar]4s^1 3d^5\). This is an exception to the Aufbau's principle. Instead of the expected \([Ar]4s^2 3d^4\), one electron from 4s promotes to the 3d orbital to make a half-filled stable configuration.
03

Write the electron configuration for Chromium ions

For \(\mathrm{Cr^{2+}}\), we lose 2 electrons: \([Ar]3d^4\). For \(\mathrm{Cr^{3+}}\), we lose 3 electrons: \([Ar]3d^3\). b. Copper
04

Determine the atomic number of Copper

Copper (Cu) has an atomic number of 29 which means it has 29 electrons in the neutral form.
05

Write the electron configuration for Copper

The electron configuration of Copper is \([Ar]4s^1 3d^{10}\). This is another exception to the Aufbau's principle. Instead of the expected \([Ar]4s^2 3d^9\), one electron from 4s promotes to the 3d orbital to make a fully filled stable configuration.
06

Write the electron configuration for Copper ions

For \(\mathrm{Cu^{+}}\), we lose 1 electron: \([Ar]3d^9\). For \(\mathrm{Cu^{2+}}\), we lose 2 electrons: \([Ar]3d^8\). c. Vanadium
07

Determine the atomic number of Vanadium

Vanadium (V) has an atomic number of 23 which means it has 23 electrons in the neutral form.
08

Write the electron configuration for Vanadium

The electron configuration of Vanadium follows the Aufbau's principle: \([Ar]4s^2 3d^3\).
09

Write the electron configuration for Vanadium ions

For \(\mathrm{V^{2+}}\), we lose 2 electrons: \([Ar]3d^3\). For \(\mathrm{V^{3+}}\), we lose 3 electrons: \([Ar]3d^2\).

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Most popular questions from this chapter

Consider the complex ions $\mathrm{Co}\left(\mathrm{NH}_{3}\right) 6^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\( and \)\mathrm{CoF}_{6}^{3-} .$ The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm},\) \(440 \mathrm{nm},\) and 290 $\mathrm{nm} .$ Match the complex ion to the wave- length of absorbed electromagnetic radiation.

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