The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective

Exercises · 2.65

Q. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

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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 ( $Z_{\text{eff}}$ ) increases across a period because the added protons outweigh the shielding from electrons in the same shell. For 3p electrons, Si (atomic number 14) has a higher $Z_{\text{eff}}$ than Al (atomic number 13), so Si’s 3p electrons feel a stronger pull from the nucleus.

The question asks which 3p electrons — those in aluminium (Al) or those in silicon (Si) — experience a greater effective nuclear charge. This is a classic application of periodic trends in $Z_{\text{eff}}$ .

Why effective nuclear charge matters here

Electrons are not just attracted to the nucleus by its full positive charge. Inner electrons (core electrons) “shield” the outer electrons from the full nuclear pull. The net positive charge felt by an outer electron is the effective nuclear charge:

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

where $Z$ is the atomic number (number of protons) and $S$ is the shielding constant (a measure of how much inner electrons block the nuclear charge). For electrons in the same subshell (here, 3p), the shielding from core electrons is nearly the same, so the difference in $Z_{\text{eff}}$ comes mainly from the difference in $Z$ .

Step-by-step reasoning

Locate the elements in the periodic table.

Aluminium (Al) has atomic number 13: electron configuration $1s^2 2s^2 2p^6 3s^2 3p^1$ .

Silicon (Si) has atomic number 14: electron configuration $1s^2 2s^2 2p^6 3s^2 3p^2$ .

Both have their unpaired electrons in the 3p orbital. They are in the same period (Period 3), with Si immediately to the right of Al.
Compare the nuclear charge.

Si has one more proton in its nucleus than Al does ( $Z = 14$ vs $Z = 13$ ). That extra proton adds a positive charge that attracts all electrons, including the 3p electrons.
Consider the shielding effect.

The core electrons (1s, 2s, 2p) are identical in both atoms — 10 core electrons each. The 3s electrons also provide some shielding, but again both have two 3s electrons. The only difference is the extra 3p electron in Si. However, electrons in the same principal shell (n=3) shield each other poorly because they are at similar distances from the nucleus. So the additional 3p electron in Si does not significantly increase the shielding for the other 3p electrons.
Apply the trend.

Across a period, as atomic number increases, $Z_{\text{eff}}$ for outer electrons increases because the added protons are not fully cancelled by the small increase in shielding from same-shell electrons. This is a well-established periodic trend: $Z_{\text{eff}}$ increases from left to right. …

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Q.The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

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