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1. Introduction to Multi-Electron Systems

A hydrogen atom contains only one electron. Therefore, its electron is attracted only by the nucleus and the energy of its orbitals depends mainly on the principal quantum number n.

Atoms containing two or more electrons are called multi-electron systems. Examples include helium, lithium, carbon, oxygen, sodium and all heavier atoms.

The situation becomes more complicated because the electrons repel one another. An electron therefore does not experience the full attractive force of the nucleus.

The important ideas used to understand multi-electron atoms are:

  • Electron-electron repulsion
  • Shielding or screening effect
  • Penetration
  • Effective nuclear charge
  • Different orbital energies
  • Electronic configuration
  • Exchange energy
The major difference from hydrogen is that in a multi-electron atom the energy of an orbital depends not only on n but also on the type of orbital, represented by l.

2. Hydrogen Atom vs Multi-Electron Atoms

Hydrogen-like system Multi-electron system
Only one electron. Two or more electrons.
No electron-electron repulsion. Electron-electron repulsion is present.
Orbitals having same n have the same energy. Orbitals with same n but different l can have different energies.
Energy depends mainly on n. Energy depends on n and l and on electron interactions.

For hydrogen, the 2s and 2p orbitals have the same energy. They are called degenerate orbitals.

In a multi-electron atom, 2s is generally lower in energy than 2p because the 2s electron penetrates closer to the nucleus and experiences a greater effective nuclear attraction.

3. Electron-Electron Repulsion

Electrons have the same negative charge and therefore repel one another.

In a multi-electron atom, an electron is attracted by the positively charged nucleus but repelled by other electrons.

Therefore, the actual attraction experienced by an electron is smaller than the attraction expected from the full nuclear charge.

This is why the concept of effective nuclear charge is important.

Zeff = Z − S

Here Z is the actual nuclear charge and S represents the shielding constant.

4. Shielding or Screening Effect

The reduction in the attractive force of the nucleus experienced by an electron due to the presence of other electrons is called shielding or screening effect.

Inner-shell electrons shield outer-shell electrons from the full nuclear charge.

Example: Sodium

Sodium has atomic number 11 and electronic configuration:

1s² 2s² 2p⁶ 3s¹

The 3s electron is outside the first two shells. Therefore, the ten inner electrons shield it from much of the attraction of the nucleus.

Factors affecting shielding

  • Number of inner electrons
  • Orbital type
  • Distance from nucleus
  • Penetration of the orbital

Greater shielding generally means smaller effective nuclear charge.

5. Penetration Effect

Penetration refers to the ability of an electron in an orbital to approach close to the nucleus.

For orbitals belonging to the same principal shell, penetration follows approximately:

s > p > d > f

Thus, an s electron has the greatest probability of being found close to the nucleus.

Why is 2s lower than 2p?

Both 2s and 2p belong to the second shell, but the 2s orbital penetrates closer to the nucleus than 2p.

Therefore, a 2s electron experiences greater nuclear attraction and is lower in energy.

Penetration: 2s > 2p

Penetration order

ns > np > nd > nf

6. Effective Nuclear Charge

Effective nuclear charge is the net positive charge experienced by an electron after considering the shielding produced by other electrons.

Zeff = Z − S

where:

  • Zeff = effective nuclear charge
  • Z = atomic number
  • S = shielding constant

Example

Suppose an electron in an atom with Z = 11 experiences a shielding constant of approximately 8.8.

Zeff = 11 − 8.8
Zeff = 2.2

This does not mean that the nucleus has changed from +11 to +2.2. It means that the particular electron experiences a net attraction approximately corresponding to a nuclear charge of +2.2.

7. Slater's Rules

Slater's rules provide an approximate method for calculating the shielding constant S and therefore the effective nuclear charge.

Step 1: Write the configuration in Slater groups

The configuration is grouped as:

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...

For an ns or np electron

The following approximate contributions are used:

  • Other electrons in the same group: 0.35 each.
  • Electrons in the n−1 shell: 0.85 each.
  • Electrons in n−2 or lower shells: 1.00 each.

For the 1s group, the other 1s electron contributes approximately 0.30.

For an nd or nf electron

  • Other electrons in the same nd or nf group: 0.35 each.
  • All electrons in groups to the left contribute approximately 1.00 each.
  • Electrons to the right do not contribute significantly to shielding.
Slater's rules give an approximate value. They are useful for comparing effective nuclear charge and explaining periodic trends, not for treating the calculated value as an exact experimental quantity.

8. Slater Rule Examples

Example 1: Effective nuclear charge for a 3s electron of sodium

Sodium:

1s² 2s² 2p⁶ 3s¹

For the 3s electron, the other electrons in the same 3s group = 0.

Electrons in the n−1 shell:

2s² 2p⁶ = 8 electrons

Their contribution:

8 × 0.85 = 6.80

Electrons in n−2 or lower:

1s² = 2 electrons

Their contribution:

2 × 1.00 = 2.00

Therefore:

S = 6.80 + 2.00 = 8.80

Atomic number:

Z = 11

Hence:

Zeff = Z − S
Zeff = 11 − 8.80
Zeff = 2.20

Answer: The approximate effective nuclear charge experienced by the 3s electron is 2.20.

Example 2: Effective nuclear charge for a 2p electron of fluorine

Fluorine:

1s² 2s² 2p⁵

Consider one 2p electron.

Other electrons in the same 2s,2p group:

6 × 0.35 = 2.10

The 1s² electrons are in the n−1 shell:

2 × 0.85 = 1.70

Therefore:

S = 2.10 + 1.70 = 3.80

Fluorine has Z = 9.

Zeff = 9 − 3.80
Zeff = 5.20

Answer: Approximate effective nuclear charge = 5.20.

9. Energy of Orbitals in Multi-Electron Atoms

In hydrogen-like atoms, orbitals with the same principal quantum number have the same energy.

In multi-electron atoms, electron-electron repulsion, shielding and penetration remove this degeneracy.

Therefore, orbital energy generally depends on both n and l.

Important example

2s < 2p

The 2s orbital is lower in energy than 2p.

Similarly, in many atoms:

3s < 3p < 3d

because s orbitals penetrate more strongly than p orbitals, and p more strongly than d orbitals.

10. Energy Order of Orbitals

A useful general ordering for electron filling is given by the Aufbau sequence:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p

n + l rule

The orbital with the lower value of n+l generally has lower energy.

If two orbitals have the same value of n+l, the orbital with the lower n generally has lower energy.

Example: 4s and 3d

For 4s:

n + l = 4 + 0 = 4

For 3d:

n + l = 3 + 2 = 5

Therefore, 4s fills before 3d in the usual Aufbau sequence.

11. Quantum Numbers

The state of an electron in an atom is described by four quantum numbers.

1. Principal quantum number (n)

It represents the main energy level or shell.

n = 1, 2, 3, 4, ...

Maximum number of electrons in a shell:

Maximum electrons = 2n²

2. Azimuthal quantum number (l)

It determines the subshell and orbital shape.

l = 0, 1, 2, ... (n−1)
l Subshell Maximum electrons
0 s 2
1 p 6
2 d 10
3 f 14

3. Magnetic quantum number (ml)

It describes the orientation of an orbital.

ml = −l, ... , 0, ... , +l

Number of orbitals in a subshell:

Number of orbitals = 2l + 1

4. Spin quantum number (ms)

It describes the spin of the electron.

ms = +1/2 or −1/2

12. Orbitals and Their Shapes

s orbital

An s orbital is approximately spherical.

Each s subshell contains one orbital and can hold a maximum of two electrons.

p orbitals

A p subshell contains three orbitals:

px, py, pz

They are oriented along three mutually perpendicular axes.

A p subshell can contain six electrons.

d orbitals

A d subshell contains five orbitals and can hold ten electrons.

f orbitals

An f subshell contains seven orbitals and can hold fourteen electrons.

13. Nodes

A node is a region where the probability of finding an electron is zero.

Total number of nodes

Total nodes = n − 1

Angular nodes

Angular nodes = l

Radial nodes

Radial nodes = n − l − 1

Example: 3p orbital

For 3p:

n = 3
l = 1

Total nodes:

n − 1 = 3 − 1 = 2

Angular nodes:

l = 1

Radial nodes:

n − l − 1 = 3 − 1 − 1 = 1

Therefore, a 3p orbital has one radial node and one angular node.

14. Aufbau Principle

The Aufbau principle states that electrons occupy available orbitals in order of increasing energy.

The word Aufbau means "building up".

Therefore, lower-energy orbitals are filled before higher-energy orbitals.

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p ...

This principle is useful for writing the ground-state electronic configuration of atoms.

15. Pauli Exclusion Principle

According to the Pauli exclusion principle, no two electrons in an atom can have the same set of all four quantum numbers.

Since an orbital has only one set of n, l and m values, it can contain a maximum of two electrons.

The two electrons must have opposite spins.

↑↓

An orbital cannot contain:

↑↑

because two electrons in the same orbital must have opposite spin.

16. Hund's Rule of Maximum Multiplicity

Hund's rule states that electrons occupy degenerate orbitals singly with parallel spins before pairing occurs.

Consider three p orbitals.

Correct arrangement for p³:

↑ ↑ ↑

Incorrect arrangement:

↑↓ ↑ _

The singly occupied arrangement provides greater stability because of exchange effects and reduced electron-electron repulsion.

17. Electronic Configuration

Electronic configuration describes the distribution of electrons among the available orbitals.

Example: Carbon

Atomic number of carbon = 6.

1s² 2s² 2p²

Orbital diagram:

1s: ↑↓    2s: ↑↓    2p: ↑ ↑ _

Example: Oxygen

Oxygen has Z = 8.

1s² 2s² 2p⁴

According to Hund's rule:

2p: ↑↓ ↑ ↑

Example: Chlorine

1s² 2s² 2p⁶ 3s² 3p⁵

Example: Calcium

1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Noble-gas notation:

[Ar] 4s²

18. Exceptions in Electronic Configuration

Some atoms do not follow the simple expected Aufbau arrangement exactly.

The most familiar examples are chromium and copper.

Chromium

Expected arrangement:

[Ar] 3d⁴ 4s²

Actual ground-state arrangement:

[Ar] 3d⁵ 4s¹

The half-filled d subshell provides additional stability.

Copper

Expected arrangement:

[Ar] 3d⁹ 4s²

Actual arrangement:

[Ar] 3d¹⁰ 4s¹

The completely filled d subshell provides additional stability.

The common shortcut is that half-filled and completely filled subshells can have extra stability. However, electronic configuration should ultimately be based on the actual energy difference between the competing arrangements.

19. Exchange Energy

Exchange energy is associated with the exchange of electrons having parallel spins between degenerate orbitals.

Greater possible exchange interactions can contribute to greater stability.

This helps explain why half-filled and fully filled subshells often show extra stability.

Example: d⁵

A d subshell contains five orbitals. In d⁵, each orbital can contain one electron with parallel spin:

↑ ↑ ↑ ↑ ↑

There are many possible exchanges among these parallel electrons.

Example: d¹⁰

A completely filled d subshell is:

↑↓ ↑↓ ↑↓ ↑↓ ↑↓

Filled-shell symmetry and exchange-related stabilization contribute to the stability of the configuration.

20. Stability of Electronic Configurations

Several factors determine the stability of an electronic configuration.

  • Lower total energy
  • Effective nuclear attraction
  • Reduced electron-electron repulsion
  • Exchange energy
  • Symmetrical distribution
  • Half-filled subshell stability
  • Completely filled subshell stability

Important stable arrangements

p⁰, p³, p⁶
d⁰, d⁵, d¹⁰
f⁰, f⁷, f¹⁴

These configurations are especially useful when explaining irregular electronic configurations.

21. Ionization and Electronic Configuration

When an atom forms a positive ion, electrons are removed.

For transition elements, electrons are generally removed from the ns orbital before the (n−1)d orbital even though the ns orbital was filled first.

Example: Iron

Neutral iron:

Fe = [Ar] 3d⁶ 4s²

Fe²⁺ is formed by removing the two 4s electrons:

Fe²⁺ = [Ar] 3d⁶

Fe³⁺:

Fe³⁺ = [Ar] 3d⁵

The d⁵ arrangement is relatively stable.

22. Magnetic Behaviour of Multi-Electron Atoms

Magnetic behaviour is closely related to the number of unpaired electrons.

Diamagnetic

A species containing only paired electrons is diamagnetic.

Paramagnetic

A species containing one or more unpaired electrons is paramagnetic.

Spin-only magnetic moment

The approximate magnetic moment can be calculated using:

μ = √[n(n + 2)] BM

Here n is the number of unpaired electrons and BM means Bohr magneton.

23. Solved Numericals

Numerical 1: Maximum number of electrons in a shell

Calculate the maximum number of electrons that can be accommodated in the fourth shell.

Given:

n = 4

Formula:

Maximum electrons = 2n²

Substitution:

= 2(4)²
= 2 × 16
= 32

Answer: The fourth shell can contain a maximum of 32 electrons.

Numerical 2: Number of orbitals in a d subshell

Calculate the number of orbitals in a d subshell.

For d:

l = 2

Formula:

Number of orbitals = 2l + 1

Substitution:

= 2(2) + 1
= 5

Answer: A d subshell contains 5 orbitals.

Numerical 3: Maximum electrons in a d subshell

Calculate the maximum number of electrons that can occupy a d subshell.

A d subshell contains 5 orbitals.

Each orbital can accommodate 2 electrons.

Maximum electrons = 5 × 2
= 10

Answer: Maximum electrons = 10.

Numerical 4: Number of possible ml values

Find the possible magnetic quantum numbers for a d electron.

For d:

l = 2

The magnetic quantum number ranges from −l to +l.

ml = −2, −1, 0, +1, +2

Answer: There are five possible values.

Numerical 5: Total, radial and angular nodes

Calculate the total number of nodes, radial nodes and angular nodes in a 4d orbital.

Given:

n = 4

For d:

l = 2

Total nodes:

n − 1 = 4 − 1 = 3

Angular nodes:

l = 2

Radial nodes:

n − l − 1
= 4 − 2 − 1 = 1

Answer:

  • Total nodes = 3
  • Angular nodes = 2
  • Radial nodes = 1

Numerical 6: Effective nuclear charge using a simple model

An electron experiences a nuclear charge of +15 and a shielding constant of 10.5. Calculate its effective nuclear charge.

Given:

Z = 15
S = 10.5

Formula:

Zeff = Z − S

Substitution:

Zeff = 15 − 10.5
Zeff = 4.5

Answer: Effective nuclear charge = 4.5.

Numerical 7: Slater calculation for sodium 3s electron

Calculate the approximate effective nuclear charge experienced by the 3s electron of sodium using Slater's rules.

Step 1: Electronic configuration

Na = 1s² 2s² 2p⁶ 3s¹

For the 3s electron:

Same group electrons = 0

n−1 electrons = 8

n−2 or lower electrons = 2

Step 2: Shielding contribution

S = (8 × 0.85) + (2 × 1.00)
S = 6.80 + 2.00
S = 8.80

Step 3: Effective nuclear charge

Zeff = 11 − 8.80
Zeff = 2.20

Answer: Zeff ≈ 2.20.

Numerical 8: Magnetic moment

Calculate the spin-only magnetic moment of a species containing three unpaired electrons.

Given:

n = 3

Formula:

μ = √[n(n + 2)] BM

Substitution:

μ = √[3(3 + 2)]
μ = √15
μ ≈ 3.87 BM

Answer: Magnetic moment ≈ 3.87 BM.

Numerical 9: Magnetic moment of two unpaired electrons

Calculate the spin-only magnetic moment for an atom or ion having two unpaired electrons.

Given:

n = 2

Formula:

μ = √[n(n + 2)]

Substitution:

μ = √[2(2 + 2)]
μ = √8
μ ≈ 2.83 BM

Answer: Magnetic moment ≈ 2.83 BM.

Numerical 10: Comparing 4s and 3d using n+l rule

Which orbital is expected to fill first, 4s or 3d?

For 4s:

n = 4, l = 0
n + l = 4

For 3d:

n = 3, l = 2
n + l = 5

The orbital with lower n+l generally has lower energy.

Therefore:

4s < 3d

Answer: 4s fills before 3d in the usual Aufbau sequence.

Numerical 11: Identifying an invalid set of quantum numbers

Determine whether the following set is possible:

n = 2, l = 2, ml = 0, ms = +1/2

For a given n:

l = 0 to n−1

If n = 2:

l = 0 or 1

Therefore l = 2 is not allowed.

Answer: The set is invalid.

Numerical 12: Number of orbitals in the fourth shell

Calculate the total number of orbitals in the fourth shell.

For a shell with principal quantum number n:

Number of orbitals = n²

For n = 4:

Number of orbitals = 4² = 16

Answer: The fourth shell contains 16 orbitals.

24. Practice Problems

1. Calculate the maximum number of electrons in the fifth shell.
Answer: 50 electrons.
2. Calculate the number of orbitals in an f subshell.
Answer: 7 orbitals.
3. Calculate the number of angular and radial nodes in a 5d orbital.
Answer: Angular nodes = 2; radial nodes = 2.
4. Calculate the magnetic moment of a species containing four unpaired electrons.
Answer: √24 ≈ 4.90 BM.
5. Write the electronic configuration of Fe and Fe²⁺.
Fe = [Ar] 3d⁶ 4s²; Fe²⁺ = [Ar] 3d⁶.
6. Calculate the effective nuclear charge if Z = 17 and S = 10.2.
Answer: Zeff = 6.8.

25. Common Mistakes

  • Do not say that all orbitals having the same n have equal energy in multi-electron atoms.
  • Remember that 2s is generally lower in energy than 2p in multi-electron atoms.
  • Do not confuse shielding effect with penetration.
  • Shielding reduces the nuclear attraction experienced by an electron.
  • Greater penetration means greater probability of finding the electron close to the nucleus.
  • Do not forget that l ranges from 0 to n−1.
  • For a d subshell, l = 2 and there are five orbitals.
  • Do not pair electrons in degenerate orbitals before applying Hund's rule.
  • Remember that chromium is [Ar] 3d⁵ 4s¹, not simply [Ar] 3d⁴ 4s².
  • Copper is [Ar] 3d¹⁰ 4s¹.
  • When transition-metal ions form, ns electrons are generally removed before (n−1)d electrons.
  • The magnetic-moment equation uses the number of unpaired electrons, not the total number of electrons.

26. TU Examination Preparation

Important 2-mark questions

  1. Define shielding effect.
  2. What is penetration effect?
  3. Define effective nuclear charge.
  4. State Aufbau principle.
  5. State Pauli exclusion principle.
  6. State Hund's rule.
  7. What is exchange energy?
  8. What is a multi-electron atom?
  9. Why is 2s lower in energy than 2p?
  10. What is a degenerate orbital?

Important 5-mark questions

  1. Explain shielding and penetration effects.
  2. Explain effective nuclear charge.
  3. Explain the energy order of orbitals in multi-electron atoms.
  4. Explain Aufbau principle, Pauli exclusion principle and Hund's rule.
  5. Explain Slater's rules with an example.
  6. Explain exchange energy and its role in electronic configuration.
  7. Explain the anomalous electronic configurations of chromium and copper.

Important 10-mark questions

  1. Discuss the structure and electronic arrangement of multi-electron atoms with reference to shielding, penetration and effective nuclear charge.
  2. Explain Slater's rules and calculate the effective nuclear charge for selected electrons.
  3. Explain the rules governing electronic configuration and discuss the stability of half-filled and completely filled subshells.
  4. Explain quantum numbers, orbital energies and electronic configuration in multi-electron atoms.

Important numerical areas

  • Maximum electrons in a shell
  • Number of orbitals
  • Possible quantum numbers
  • Number of nodes
  • Effective nuclear charge
  • Slater-rule calculations
  • Magnetic moment
  • n+l comparison
  • Electronic configuration of atoms and ions

27. Quick Revision

Important relations

Zeff = Z − S
Maximum electrons in shell = 2n²
Number of orbitals in shell = n²
Number of orbitals in subshell = 2l + 1
Maximum electrons in subshell = 2(2l + 1)
Total nodes = n − 1
Angular nodes = l
Radial nodes = n − l − 1
μ = √[n(n + 2)] BM

Penetration order

s > p > d > f

Common filling order

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s

Important configurations

Element Configuration
Cr [Ar] 3d⁵ 4s¹
Cu [Ar] 3d¹⁰ 4s¹
Fe [Ar] 3d⁶ 4s²
Fe²⁺ [Ar] 3d⁶
Fe³⁺ [Ar] 3d⁵

Most important concepts to remember

  1. Multi-electron atoms contain electron-electron repulsion.
  2. Inner electrons shield outer electrons.
  3. s orbitals penetrate more strongly than p, d and f.
  4. Effective nuclear charge is approximately Z − S.
  5. Orbital energy in multi-electron atoms depends on both n and l.
  6. Aufbau tells us the filling order.
  7. Pauli limits an orbital to two electrons with opposite spins.
  8. Hund's rule gives singly occupied degenerate orbitals before pairing.
  9. Half-filled and fully filled subshells can have additional stability.

28. Quick Check

Choose the correct answer.

1. Which orbital has the greatest penetration?

2. What is the maximum number of electrons in the third shell?

3. How many orbitals are present in a d subshell?

4. Which principle states that electrons fill lower-energy orbitals first?

5. Which configuration represents chromium?

6. A species has three unpaired electrons. Its spin-only magnetic moment is approximately: