Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2

Exercises · 2.64

Q. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) 2s and 3s,

(ii) 4d and 4f,

(iii) 3d and 3p.

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Concept understanding — Effective Nuclear Charge

The Intuition: Why Don't Electrons Just Fly Away?

Imagine you're holding a magnet near a pile of paperclips. The closer the magnet, the stronger the pull. Now imagine you put a sheet of cardboard between the magnet and the paperclips. The pull weakens — the cardboard "shields" the paperclips from the full force of the magnet.

An atom works similarly. The nucleus (positive charge) pulls on the electrons (negative charge). But an electron is not alone — there are other electrons buzzing around between it and the nucleus. Those inner electrons act like the cardboard sheet: they shield or screen the outer electron from feeling the full positive charge of the nucleus.

So an outer electron doesn't "see" the full nuclear charge $Z$ (the atomic number). It sees a smaller, effective charge — the net positive pull after accounting for the repulsion from inner electrons.

That's effective nuclear charge, denoted $Z_{\text{eff}}$ .

The Precise Statement

Z_{\text{eff}} = Z - S

$Z_{\text{eff}}$ is always less than $Z$ (except for hydrogen, which has no other electrons to shield — there $Z_{\text{eff}} = Z$ ).

What Determines the Shielding Constant $S$ ?

Not all electrons shield equally. The key rules:

Inner electrons shield outer electrons very effectively. An electron in the $n=1$ shell completely blocks about 1 unit of charge from an electron in $n=2$ .
Electrons in the same shell shield poorly. They're at roughly the same distance, so they don't block much of the nucleus from each other.
Outer electrons do not shield inner electrons at all. An electron farther out cannot block the nucleus from one closer in.

There are detailed rules (Slater's rules) to calculate $S$ numerically, but the core idea is simple: the more electron shells between an electron and the nucleus, the more shielding, and the lower $Z_{\text{eff}}$ .

Why Does This Matter?

$Z_{\text{eff}}$ explains three fundamental patterns in the periodic table:

Trend	What happens to $Z_{\text{eff}}$	Why
Across a period (left to right)	Increases	Adding protons ( $Z$ up) but electrons go into the same shell (shielding roughly constant). Net pull on outer electrons gets stronger.
Down a group (top to bottom)	Stays roughly constant or decreases slightly	Adding a new shell means much more shielding. The extra protons are almost completely cancelled by the new inner electrons.
Atomic size	Larger $Z_{\text{eff}}$ → smaller atom	Stronger pull pulls electrons closer to nucleus.

Note

This is why fluorine is smaller than lithium, even though fluorine has more protons. The extra protons in fluorine are not fully shielded — the outer electrons feel a much stronger pull.

A Concrete Example: Sodium vs. Chlorine

Sodium ( $Z = 11$ ): Electron configuration $1s^2 2s^2 2p^6 3s^1$

The outermost electron (3s) is shielded by the 10 inner electrons ( $1s^2 2s^2 2p^6$ ). Roughly, $S \approx 10$ , so $Z_{\text{eff}} \approx 11 - 10 = 1$ . The outer electron feels a pull equivalent to just one proton. …

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Effective nuclear charge ( $Z_{\text{eff}}$ ) depends on both distance from the nucleus and penetration ability. The orbital closer to the nucleus or with better penetration experiences the larger $Z_{\text{eff}}$ : (i) 2s, (ii) 4d, (iii) 3p.

The effective nuclear charge is the net positive charge an electron "feels" after accounting for the shielding (screening) by other electrons. An electron in a multi-electron atom doesn't experience the full nuclear charge $Z$ because inner electrons repel it and partially cancel the nuclear attraction.

The key insight: $Z_{\text{eff}}$ increases when an orbital either lies closer to the nucleus or penetrates more deeply through inner electron shells. Penetration means the electron spends more time near the nucleus, experiencing less shielding from inner electrons.

Two principles govern penetration and hence $Z_{\text{eff}}$ :

Distance from nucleus: For orbitals with the same $l$ (angular momentum quantum number), lower $n$ means closer to the nucleus → higher $Z_{\text{eff}}$ .

Penetration for same $n$ : The penetration order is $s > p > d > f$ . An $s$ orbital has no angular nodes and its probability density doesn't vanish at the nucleus, so it penetrates most effectively. As $l$ increases, more angular nodes keep the electron farther from the nucleus on average.

Z_{\text{eff}} = Z - S

Now let's analyze each pair:

(i) 2s vs 3s

Both are $s$ orbitals, so they have identical penetration character. The difference is purely in the principal quantum number.

The 2s orbital ( $n=2$ ) is in the second shell, much closer to the nucleus than 3s ( $n=3$ ).
The 3s electron is shielded by all electrons in $n=1$ and $n=2$ shells, plus it sits farther out where the nuclear attraction is weaker.
The 2s electron experiences less shielding (only the 1s electrons shield it significantly) and is closer to the positive nucleus.

Result: 2s experiences larger $Z_{\text{eff}}$ .

(ii) 4d vs 4f

Both orbitals are in the same shell ( $n=4$ ), so distance from the nucleus is comparable. The deciding factor is penetration.

The 4d orbital has $l=2$ , while 4f has $l=3$ . Lower $l$ means better penetration.
A 4d electron penetrates closer to the nucleus through inner shells, spending more time in regions of high positive charge density.
A 4f electron has three angular nodes and is more diffuse, staying farther from the nucleus on average. It experiences more shielding from inner electrons.
More shielding for 4f means lower $Z_{\text{eff}}$ .

Result: 4d experiences larger $Z_{\text{eff}}$ . …

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Q.Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) 2s vs 3s

(ii) 4d vs 4f

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