Why do electrons fill 4s before 3d

The order in which electrons fill subshells is governed by their relative energies, not by the principal quantum number alone. For most neutral atoms, the 4s orbital lies lower in energy than the 3d, so it is filled first. This ordering explains why potassium (Z=19) and calcium (Z=20) have configurations ending in 4s¹ and 4s² respectively, even though the 3d shell is empty.

n + l rule and its consequences

The n + l rule assigns each subshell a value equal to the sum of its principal quantum number n and azimuthal quantum number l. Subshells with lower n + l are filled first; if two subshells share the same n + l, the one with lower n fills first. For 4s, n + l = 4 + 0 = 4; for 3d, n + l = 3 + 2 = 5, so 4s is energetically favored until the 3d electrons begin to shield the nucleus.

Key reasons the 4s orbital is lower than 3d in early transition metals:

• Lower n + l value (4 < 5)
• Greater radial extension reduces electron‑electron repulsion
• Effective nuclear charge felt by 4s electrons is slightly higher because inner 3d electrons are not yet present

To predict the filling order for any element:

1. 1Write down all subshells up to the element’s atomic number
2. 2Calculate n + l for each subshell
3. 3Sort by increasing n + l, breaking ties with lower n
4. 4Fill electrons sequentially according to this list

Typical orbital energies (in electron‑volts) for a neutral calcium atom:

Experimental spectroscopy confirms that the 4s orbital lies about 0.7 eV lower than the 3d in calcium, matching the theoretical ordering. When the 3d begins to fill (starting with scandium, Z=21), the added electrons increase shielding of the nucleus, raising the energy of the already‑occupied 4s electrons. Consequently, in ions such as Fe²⁺ the 4s electrons are lost before any 3d electrons, illustrating the reversal of relative energies after the d‑block is entered.

Worked example: Determine the ground‑state configuration of potassium (Z=19). List subshells up to 4s: 1s, 2s, 2p, 3s, 3p, 4s, 3d. Compute n + l values: 1s(1), 2s(2), 2p(3), 3s(3), 3p(4), 4s(4), 3d(5). Fill electrons following the sorted list: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The 3d subshell remains empty because its n + l value is higher, confirming why the electron enters 4s first.