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Q4.1(i):
Write down the electronic configuration of: (i) \(Cr^{3+}\)

Solution :

Step 1: Establishing the Ground State Electronic Configuration of Chromium ($Cr$)

Chromium ($Cr$) has an atomic number ($Z$) of $24$. According to the Aufbau principle, the electrons fill the orbitals in order of increasing energy. However, due to the extra stability provided by a half-filled $d$-subshell, chromium exhibits an anomalous electronic configuration.

Standard configuration of neutral $Cr$ ($Z=24$):

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$

[Justification: The stability of the $d^5$ arrangement is lower in potential energy compared to the $4s^2 3d^4$ configuration per Hund’s Rule and exchange energy stabilization principles.]

Step 2: Determining the Configuration of the Ion ($Cr^{3+}$)

The formation of a cation involves the removal of electrons from the outermost shell first. For transition metals, electrons are removed from the $4s$ orbital before the $3d$ orbital [Per the principle of orbital energy levels in ionized transition metals].

To reach the $Cr^{3+}$ state, $3$ electrons must be removed from the neutral $Cr$ atom:

  • First electron: Removed from the $4s$ orbital. Configuration becomes $3d^5$.
  • Second electron: Removed from the $3d$ orbital. Configuration becomes $3d^4$.
  • Third electron: Removed from the $3d$ orbital. Configuration becomes $3d^3$.

Step 3: Final Electron Distribution

Subtracting the electrons systematically from the configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$:

Removing $1$ electron from $4s^1$ leaves $0$ in $4s$.

Removing $2$ electrons from $3d^5$ leaves $3$ in $3d$.

Resulting configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$.

Visual Representation of $Cr^{3+}$ Valence Subshells

3d³ ↑ ↑ ↑

Final Solution: The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$ or $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$.


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