## Shielding Electrons and Slater’s Rule

**Shielding electrons** or **screening electrons** decreases the attractive force between the valence shell electron and the nucleus of the atom. **Slater’s rule** uses for calculating the shielding or screening constant and effective nuclear charge for the outer orbital electron of an atom or ion in chemistry.

The semi-empirical rules proposed by John C. Slater in 1930 provide numerical values for the effective nuclear charge in a many-electron atom. A hydrogen atom contains one electron, hence the hydrogen atom has no **shielding electron** or effect.

The valence electrons for multi-electron atoms are attracted by the nucleus of the atom and repelled by the electrons from inner shells. These attractive and repulsive forces acting on the valence electrons experience less attraction from the nucleus of an atom.

## What is Shielding in Chemistry?

The inner electrons that shield the higher energy electron are called shielding electrons which are calculated by Slater’s rule, and the effect is called the shielding effect or screening effect. In learning chemistry, the larger the number of inner or shielding electrons, the lesser will be the attraction between the nucleus and outer orbitals electrons.

When the number of inner electrons increases, the shielding or screening constant also increases. Therefore, effective nuclear charge decreases which affects the chemical properties of the atom or molecule of the matter and is calculated by Slater’s rule.

## What is Effective Nuclear Charge?

Effective nuclear change means the net positive charge which affects the attraction of outer electron particles from the nucleus of a polyelectronic atom. This term is used because the shielding electrons prevent the attraction of the outer orbital electron of an atom.

The effective nuclear charge (Z_{eff} ) is calculated from Slater’s rule = Z âˆ’ Ïƒ, where Ïƒ = shielding or screening constant.

## Slater’s Rule for Calculating Shielding Constant

Slater proposed some empirical set of rules to calculate the screening constant (Ïƒ) of various electrons present in different orbitals of an atom or ion. Slater’s rule for calculating shielding constant for s and p-orbitals is slightly different from d and f-orbitals.

The electrons are arranged in a sequence of groups in order of increasing principal quantum number (n). In such electron arrangement s and p-orbitals are kept together. Therefore, the electron arrangement order for calculating shielding and screening constant is:

[1s] [2s, 2p] [3s, 3p] [3d] [4s, 4p] [4d] [4f] [5s, 5p], etc.

Such arrangement contains three groups such as [1s], [ns, np], and [nd] or [nf]. Each group has a definite shielding constant which is given below in the table,

Group |
Other electrons in the same group with principal quantum number n |
Electrons in the group with the principal quantum number (nâ€“1) |
Electrons in all group(s) with principal quantum number â‰¤ nâ€“2 |

[1s] | 0.30 | ||

[ns, np] | 0.35 | 0.85 | 1 |

[nd] or [nf] | 0.35 | 1 | 1 |

### s or p-Orbital Electrons of an Atom or Ion

Rules for calculating screening constant or shielding constant for s or p-orbital electron of an atom or ion are given below:

- First, we write the electronic configuration of the atom or ion by following order and grouping, [1s] [2s, 2p] [3s, 3p] [3d] [4s, 4p] [4d] [4f] [5s, 5p], [5d], etc.
- Electrons in a certain ns, np-level are screened only by electrons of the same energy level and by the electrons of lower energy levels.
- Electrons lying above the energy level do not screen any electron to any extent. Therefore, the higher energy electrons have no screening effect on the lower energy electrons
- Electrons of an (ns np) level shield the valence electron in the same group by 0.35 each. This rule is also true for the electrons of the nd or nf level for electrons in the same group.
- Electrons belonging to one lower quantum shell or (nâˆ’1) shell shield the valence electron by 0.85 each.
- Electrons belonging to (nâˆ’2) or still lower quantum energy levels shield the valence electron by 1.0 each.

### Shielding or Screening Constant of Sodium Atom

For calculating the value shielding constant of inner electrons of the sodium atom, the electron configuration according to Slater’s rule, [1s]^{2} [2s, 2p]^{8} [3s]^{1}.

Therefore, by using Slater’s rule shielding constant and effective nuclear charge for 3s-electron of sodium atom,

Ïƒ = (2 Ã— 1) + (8 Ã— 0.85) + (0 Ã— 0.35)

= 8.8

Hence the effective nuclear charge of sodium,

Z_{eff} = (11 âˆ’ 8.8)

= 2.2

### Shielding or Screening Constant of Sodium and Magnesium Ions

Electron configuration of sodium and magnesium ions according to Slater’s rule:

Na^{+} ion: [1s]^{2} [2s, 2p]^{8}

Mg^{+2} ion: [1s]^{2} [2s, 2p]^{8}

Screening constant for Na^{+} ion,

Ïƒ (Na^{+}) = (2 Ã— 0.85) + (8 Ã— 0.35)

= 4.5

Similarly, the screening constant for Mg^{+2} ion,

Ïƒ (Mg^{+}) = (2 Ã— 0.85) + (8 Ã— 0.35)

= 4.5

Therefore, the screening and shielding constant for sodium and magnesium ions are similar but the effective nuclear charge of these two ions are different.

Effective nuclear charge for Na^{+} ion = (11 âˆ’ 4.5) = 6.5

Effective nuclear charge for Mg^{+2} ion = (11 âˆ’ 4.5) = 6.5

**Problem:** Calculate the shielding or screening constant for the 2p-electron of carbon and oxygen atoms.

**Solution:** Electron configuration of carbon and oxygen according to Slater’s rule for shielding electrons are:

Carbon (atomic number 6): [1s]^{2} [2s, 2p]^{4}

Oxygen (atomic number 8): [1s]^{2} [2s, 2p]^{6}

The shielding constant for carbon atom,

= (2 Ã— 0.85) + (4 Ã— 0.35)

= 3.10

Similarly, shielding constant of oxygen atom,

= (2 Ã— 0.85) + (6 Ã— 0.35)

= 3.80

**Problem:** How to calculate Ïƒ and Z_{eff} for the fluorine and fluoride ion?

**Solution:** Electron configuration of fluorine and fluoride ion according to Slater’s rule for shielding are:

Fluorine atom (number of electrons 9): [1s]^{2} [2s, 2p]^{7}

Fluoride ion (number of electrons 10): [1s]^{2} [2s, 2p]^{8}

The shielding constant (Ïƒ) for the Fluorine atom,

= (2 Ã— 0.85) + (7 Ã— 0.35)

= 4.15

Similarly, the shielding constant (Ïƒ) of fluoride ion,

= (2 Ã— 0.85) + (8 Ã— 0.35)

= 4.5

Therefore, the effective nuclear charge for fluorine atom = (9 âˆ’ 4.15) = 4.85 and fluoride ion = (9 âˆ’ 4.5) = 4.5.

## Shielding Constant for d or f-Orbital Electron

Slater’s rule for s or p-electron is quite good for estimating the screening constant of s and p-orbital. However, Slater’s rule for d or f-orbital electrons the five and six rules are replaced by new rules for the estimation of screening or shielding effect and effective nuclear charge.

The new rule is all electrons below the nd subshell or nf-subshell contribute 1.0 each towards the screening constant.

### Screening Constant and Z_{eff} for Vanadium

Vanadium has atomic number 23 and the electron configuration according to Slater’s rules for shielding electrons,

[1s]^{2} [2s 2p]^{8} [3s 3p]^{8} [3d]^{3} [4s]^{2}

#### For 4s Electron

The screening constant for 4s electron of vanadium atom,

= (2 Ã— 1.0) + (8 Ã— 1.0) + (8 Ã— 0.85) + (3 Ã— 0.85) + (1 Ã— 0.35)

= 19.7

Hence the effective nuclear charge (Z_{eff}) for the 4s-electron of the vanadium atom,

= 23 âˆ’ 19.7

= 3.3

#### For 3d Electron

Screening constant for 3d electron of the vanadium atom,

= (2 Ã— 1.0) + (8 Ã— 1.0) + (8 Ã— 1.0) + (2 Ã— 0.35)

= 18.70

Hence the effective nuclear charge (Z_{eff}) for the 3d electron of the vanadium atom,

= (23 âˆ’ 18.70)

= 4.30

From the above calculation, the screening constant for the 4s electron of vanadium is higher than the 3d electron. Therefore, the 3d electron is more tightly bound than the 4s electron. Hence during ionization, the 4s electron will be lost in preference to the 3d electron.

### Screening Constant and Z_{eff} for Chromium

Chromium has atomic number 24 and the electron configuration according to Slater’s rule for shielding electrons,

[1s]^{2} [2s, 2p]^{8} [3s, 3p]^{8} [3d]^{5} [4s]^{1}

The screening constant for 4s electron of chromium,

= (2 Ã— 1.0) + (8 Ã— 1.0) + (8 Ã— 0.85) + (5 Ã— 0.85) + (0 Ã— 0.35)

= 21.05

Hence Z_{eff} for 4s-electron of chromium,

= (24 âˆ’ 21.05)

= 2.95

The screening constant for 3d-electron of chromium,

= (2Ã—1.0) + (8 Ã— 1.0) + (8 Ã— 1.0) + (3 Ã— 0.85) + (4 Ã— 0.35)

= 19.40

Therefore, Z_{eff} for 3d-electron of chromium,

= (24 – 19.40)

= 4.60

## Shielding Effect and Ionization Energy

The larger the number of electrons in the inner shell, the lesser the attractive force holding the valence electron to the nucleus, and the lower will be the value of ionization energy.

When we move down in the group of the periodic table, the number of shielding electrons increases, and the effective nuclear charge for valence electrons calculated from Slater’s rule decreases, and hence the ionization trends also decrease.

The ionization trends and shielding or screening effect for the group-2 chemical elements, beryllium > magnesium > calcium > strontium > barium. The formation of ionic bonding or polarity trends also increases from Be to Ba.

## Frequently Asked Questions (FAQs)

### Which electrons are shielding?

A valence electron in a multi-electron atom is attracted by the nucleus but repelled by the electrons of inner shells.

The combined effect of such attractive and repulsive force acting on the valence electron. Therefore, the valence electron of an atom is shielded by inner-shell electrons.

### Which electron has the greatest shielding?

The s-orbital has the greatest shielding effect followed by p, d, and f orbitals because the s-orbital has a higher density of electrons.

### Which element in a group has a higher shielding effect?

As we move down a group, the number of inner shells increases and the shielding effect also increses. Therefore, in the alkali metals group, rubidium has the greatest shielding effect and lithium has the least shielding effect.

### How to calculate the screening constant of potassium ion?

The electron arrangement of potassium ion is

[1s]^{2} [2s, 2p]^{8} [3s, 3p]^{8}

From Slater’s rules, the screening or shielding constant for potassium ion,

= (2 Ã— 1) + (8 Ã— 0.85) + (8 Ã— 0.35)

= 11.6