Anything that influences the valence electrons will affect the chemistry of the element. Which one of the foll

Exercises · 3.35

Q. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

(a) Valence principal quantum number (n)

(b) Nuclear charge (Z)

(d) Number of core electrons.

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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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The effective nuclear charge felt by valence electrons depends on the principal quantum number $n$ , the nuclear charge $Z$ , and the shielding from core electrons — but nuclear mass has no direct effect on the valence shell’s electronic environment.

The question asks: which factor does not affect the valence shell? To answer, we need to think about what actually determines the energy and behaviour of valence electrons. The key idea here is effective nuclear charge ( $Z_{\text{eff}}$ ).

Why effective nuclear charge is the lens

Valence electrons don’t feel the full positive charge of the nucleus. Core electrons “shield” them, reducing the pull. The net pull is:

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

where $Z$ is the nuclear charge (proton number) and $S$ is the shielding constant (roughly the number of core electrons, with some refinements). The principal quantum number $n$ also matters — higher $n$ means the electron is farther out, feeling less attraction.

So anything that changes $Z$ , $S$ , or $n$ will affect the valence shell. Let’s check each option.

Valence principal quantum number ( $n$ )

This directly sets the energy level and average distance of the valence electrons. A higher $n$ means weaker attraction to the nucleus (more shielding, larger orbital). So $n$ absolutely affects the valence shell.
Nuclear charge ( $Z$ )

More protons mean a stronger pull on all electrons, including valence ones. This is the core of periodic trends — across a period, increasing $Z$ (with constant shielding) increases $Z_{\text{eff}}$ , pulling valence electrons tighter. So $Z$ definitely affects the valence shell.
Nuclear mass …

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Q.Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

Why effective nuclear charge is the lens

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