Topic 2 · page 12 of 15

Potassium to krypton, and ions

After argon the 4s sub-shell fills, then 3d, then 4p. Chromium and copper break the pattern. Positive ions lose their outer electrons first, which means 4s electrons before 3d in the d block.

14 min read2.14

Read first: Filling the orbitals

Potassium to krypton 2.14

After argon (3p⁶) the 3p sub-shell is full. The next electron does not go into 3d. The 4s sub-shell is lower in energy, so it fills first. The order then runs:

Element (atomic number) Configuration
K (19) [Ar] 4s¹
Ca (20) [Ar] 4s²
Sc (21) [Ar] 3d¹ 4s²
Ti (22) [Ar] 3d² 4s²
V (23) [Ar] 3d³ 4s²
Cr (24) [Ar] 3d⁵ 4s¹
Mn (25) [Ar] 3d⁵ 4s²
Fe (26) [Ar] 3d⁶ 4s²
Co (27) [Ar] 3d⁷ 4s²
Ni (28) [Ar] 3d⁸ 4s²
Cu (29) [Ar] 3d¹⁰ 4s¹
Zn (30) [Ar] 3d¹⁰ 4s²
Ga (31) [Ar] 3d¹⁰ 4s² 4p¹
Ge (32) [Ar] 3d¹⁰ 4s² 4p²
As (33) [Ar] 3d¹⁰ 4s² 4p³
Se (34) [Ar] 3d¹⁰ 4s² 4p⁴
Br (35) [Ar] 3d¹⁰ 4s² 4p⁵
Kr (36) [Ar] 3d¹⁰ 4s² 4p⁶

The table writes 3d before 4s, which is how the sub-shells are grouped by shell. Writing 4s first, in the order of filling, is also correct, for example Fe: [Ar] 4s² 3d⁶.

Configurations of ions 2.14

An ion has gained or lost electrons, so its configuration is the atom's configuration with electrons added or removed. The number of electrons is the atomic number minus the charge.

Positive ions of s- and p-block elements lose their outer electrons. The result is usually the same arrangement as the noble gas before it:

Negative ions gain electrons into the next free space in the outer sub-shell, again giving a noble gas arrangement:

Positive ions of d-block elements lose their 4s electrons first, then electrons from 3d. The 4s electrons are in the outer shell, so they are the easiest to remove.

Electrons-in-boxes diagrams for the 3d and 4s orbitals of Cr, Cu, Fe, Fe2+, Fe3+ and Cu2+. Cr has five single 3d electrons and one 4s electron; Cu has a full 3d set and one 4s electron; the ions have lost their 4s electrons before any 3d electrons.
The 3d and 4s orbitals of chromium, copper and some ions. The ions have no 4s electrons.

Worked exampleThe configuration of Fe³⁺

  1. Iron has 26 electrons: [Ar] 3d⁶ 4s².
  2. A 3+ ion has 26 − 3 = 23 electrons, so three are removed.
  3. The first two come from 4s, leaving [Ar] 3d⁶.
  4. The third comes from 3d, leaving [Ar] 3d⁵.
  5. Fe³⁺ is [Ar] 3d⁵. In boxes, the five 3d orbitals each hold one electron: [↑] [↑] [↑] [↑] [↑], with five unpaired electrons.

Worked exampleThe configuration of Cu²⁺

  1. Copper has 29 electrons: [Ar] 3d¹⁰ 4s¹.
  2. A 2+ ion has 27 electrons. The 4s electron goes first, then one electron from 3d.
  3. Cu²⁺ is [Ar] 3d⁹, with one unpaired electron.

Check yourself

  1. Write the electronic configurations of chromium and copper in [Ar] form.
  2. Write the configuration of Mn²⁺ in [Ar] form and draw its 3d electrons in boxes.
  3. How many unpaired electrons are there in Cu²⁺? Explain.
  4. Write the full electronic configurations of S²⁻ and Br⁻.
  5. Which atom has the configuration [Ar] 3d¹⁰ 4s² 4p³, and how many unpaired electrons does it have?
  6. A student writes Fe²⁺ as [Ar] 3d⁴ 4s². Explain the error and give the correct configuration.

Answers cover these until you have tried the questions

  1. Cr: [Ar] 3d⁵ 4s¹. Cu: [Ar] 3d¹⁰ 4s¹.
  2. Mn is [Ar] 3d⁵ 4s². Mn²⁺ loses the two 4s electrons: [Ar] 3d⁵, drawn [↑] [↑] [↑] [↑] [↑].
  3. Cu²⁺ is [Ar] 3d⁹. Nine electrons in five orbitals: four orbitals are paired and one has a single electron, so one unpaired electron.
  4. S²⁻ has 18 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶. Br⁻ has 36 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶.
  5. The electrons total 18 + 10 + 2 + 3 = 33, which is arsenic. The three 4p electrons are one in each orbital, so three unpaired electrons.
  6. The student removed two electrons from 3d. The 4s electrons are removed first. Correct: Fe is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶.