For the electrons of oxygen atom, which of the following statements is correct? (i) Zeff for an electron in a

NCERT Exemplar · Q15

Q. For the electrons of oxygen atom, which of the following statements is correct?

(i) Z_eff for an electron in a 2s orbital is the same as Z_eff for an electron in a 2p orbital.

(ii) An electron in the 2s orbital has the same energy as an electron in the 2p orbital.

(iii) Z_eff for an electron in 1s orbital is the same as Z_eff for an electron in a 2s orbital.

(iv) The two electrons present in the 2s orbital have spin quantum numbers m_s but of opposite sign.

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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 ( $Z_{\text{eff}}$ ) explains why 2s and 2p electrons in oxygen experience different attractions, making option (A) false, (B) false, (C) false, and (D) the correct statement.

The question tests your understanding of effective nuclear charge ( $Z_{\text{eff}}$ ) — the net positive charge experienced by an electron after accounting for shielding by other electrons. In multi-electron atoms, electrons in different orbitals feel different pulls from the nucleus because of differences in penetration and shielding.

For oxygen (atomic number $Z = 8$ ), the electron configuration is $1s^2 2s^2 2p^4$ . Let’s examine each statement.

Statement (A): $Z_{\text{eff}}$ for a 2s electron equals $Z_{\text{eff}}$ for a 2p electron.

This is false. A 2s orbital has a small but real probability of being found near the nucleus (it penetrates the 1s shell more than a 2p orbital does). Greater penetration means less shielding from inner electrons, so a 2s electron experiences a higher $Z_{\text{eff}}$ than a 2p electron in the same atom. For oxygen, $Z_{\text{eff}}(2s) \approx 4.5$ while $Z_{\text{eff}}(2p) \approx 4.0$ (Slater’s rules confirm this difference).
Statement (B): A 2s electron has the same energy as a 2p electron.

False. Because $Z_{\text{eff}}$ is larger for 2s, the 2s orbital is more tightly bound (lower energy) than 2p. In multi-electron atoms, orbitals within the same principal quantum number $n$ split in energy: $E_{2s} < E_{2p}$ .
Statement (C): $Z_{\text{eff}}$ for a 1s electron equals $Z_{\text{eff}}$ for a 2s electron. …

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Q.For the electrons of oxygen atom, which of the following statements is correct?

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