Questions: Write the condensed (noble-gas) electron configuration of Cr^3+.

Write the condensed (noble-gas) electron configuration of Cr^3+.
Transcript text: Write the condensed (noble-gas) electron configuration of $\mathrm{Cr}^{3+}$.
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Solution

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Solution Steps

Step 1: Determine the Electron Configuration of Neutral Chromium (Cr)

Chromium (Cr) has an atomic number of 24, which means it has 24 electrons in its neutral state. The electron configuration of neutral chromium is:

\[ [\mathrm{Ar}] \, 3d^5 \, 4s^1 \]

This configuration is due to the stability provided by a half-filled \(3d\) subshell.

Step 2: Determine the Electron Configuration of $\mathrm{Cr}^{3+}$

The \(\mathrm{Cr}^{3+}\) ion is formed by removing three electrons from the neutral chromium atom. Electrons are removed first from the outermost shell, which is the \(4s\) orbital, followed by the \(3d\) orbital. Therefore, the electron removal process is as follows:

  1. Remove 1 electron from \(4s^1\).
  2. Remove 2 electrons from \(3d^5\).

This results in the electron configuration:

\[ [\mathrm{Ar}] \, 3d^3 \]

Final Answer

The condensed (noble-gas) electron configuration of \(\mathrm{Cr}^{3+}\) is:

\[ \boxed{[\mathrm{Ar}] \, 3d^3} \]

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