Electronic Distribution of the Transition Elements

Remarks

As shown in the previous example (₁₉K), 4s was filled before the 3d orbitals.
This is because 4s is lower in energy than 3d orbitals.
The following element in the table (₂₀Ca), 4s will be filled with another electron making it:
₂₀Ca: [Ar] 4s²
The 3d orbitals start to be filled with element ₂₁Sc, followed by the rest of the "Transition Elements".
Transition metals are, hence, elements in which the d orbitals are being filled.

Example

Write the summarized & condensed electronic distribution of a ₂₄Cr atom.

Solution:
₂₄Cr: 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d⁴

₂₄Cr: [Ar] 4s², 3d⁴

Remark:
In the previous example, more "Stable" electronic configurations can be written as follows:
₂₄Cr: 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d⁴

₂₄Cr: [Ar] 4s¹, 3d⁵

The reason is that an electron moved from 4s to 3d to make 3d "half filled" with electrons, which is a much more stable configuration (see figure below). 4s.

Another Related Remark:
The more stable electronic configuration of ₂₉Cu is:
₂₉Cu: 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰

₂₉Cu: [Ar] 4s¹, 3d¹⁰

Where The d orbitals are "completely filled".
Hence, achieving "half-filled" or "completelly-filled" d orbitals will give more stable configurations.

Exercise

Exercise on Ground-State Electron Configuration

Check your answers here: Solution to the Exercise on Ground-State Electron Configuration

So, electronic configurations can be written for any element in the periodic table as long as the rules of filling electrons in the orbitals are fulfilled and the rules of "half-and Completely-filled" d orbitals are also fulfilled.

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Date of last modification: 2021