effective nuclear charge and shielding

[905290504]

Could someone please explain the effective nuclear charge trend? I'm just confused on why the trend is that way. And how does it relate to shielding?

[Akash J 1J]

Going down a group increases the radius of the atom. This is because there are more inner orbitals to "shield" the valence electrons from the pull of the nucleus.
Going left to right in a period decreases the radius because electrons are being added to the valence shell (not inner shells so shielding doesn't increase) but there are more protons, increasing ENC pulling the valence electrons closer to the nucleus.

[Sabina House 2A]

Just to add on, when going left to right across a period, the number of shielding electrons does not increase but the number of protons does increase. This means that the valence electrons are being pulled in by a greater force (from more protons) but there are not more shielding electrons to counter this force. This is why the effective nuclear charge increases going left to right across a period.

[Arjun_Anumula_3E]

effective nuclear charge of an element is defined as the atomic number of the element - the number of non-valence electrons.

Going across across a row increases the number of protons but does not increase the number of shielding (non-valence) electrons. Thus there is a higher pull from the nucleus onto electrons, or a higher effective nuclear charge across the row.

The opposite is said from going down a column. While the number of protons do increase, the amount of shielding also vastly increases due to an increased number of electron shells (orbitals) to counteract this. Thus, effective nuclear charge decreases down a group.

[Emily Wan 1l]

Effective nuclear charge refers to the attraction between protons and the outermost valence electrons. Effective nuclear charge increases across a period because of the increase in protons and thus the increase in the attraction between valence electrons and the nucleus. Effective nuclear charge decreases down a group because of the increase in the shielding effect of inner electrons and the increased distance between the nucleus and valence electrons.