CBSE Class 12 p-Block — The Oxidation-State Pattern Explained
Most students try to memorise the p-block as a list of compounds. That is why it feels endless. In reality the whole of groups 15 to 18 runs on one pattern: every element has a maximum positive oxidation state fixed by its group, a set of lower states two units apart, and a steady drift down the group towards the lower state. Once you see that pattern, the reactions stop being random facts and start being predictions.
This guide gives you the rule, explains the inert pair effect that causes the drift, and works through the oxidation-number calculations that exams actually ask.
The rule that fixes the maximum oxidation state
A p-block atom can, at most, use all of its valence electrons — the ns and np electrons together. So:
Group 15 → +5 · Group 16 → +6 · Group 17 → +7 · Group 18 → +8
Minimum (most negative) state = (group number) − 18
Group 15 → −3 · Group 16 → −2 · Group 17 → −1
The negative state is what the atom shows when it gains enough electrons to complete its octet. Between the two extremes the common states usually differ by two units (+5, +3, +1 for group 15 and 17; +6, +4, +2 for group 16), because the two ns electrons behave as a pair — they are either both used in bonding or both left alone.
The inert pair effect — why the pattern shifts down a group
Going down a group, the higher oxidation state becomes less stable and the lower one becomes more stable. This reluctance of the outermost ns² pair to take part in bonding is called the inert pair effect.
It has two honest reasons, and NCERT expects both:
- Poor shielding. Heavier elements have filled d (and, lower down, f) subshells in between. d and f electrons shield the nucleus badly, so the ns pair feels a higher effective nuclear charge, is held more tightly and is harder to unpair and promote.
- Weaker bonds don't pay the bill. Using the ns pair costs energy (promotion energy). Down the group the atoms are bigger, so the bonds they form are weaker and release less energy — not enough to repay that cost.
You meet the effect first in groups 13 and 14 (Tl⁺ is more stable than Tl³⁺; Pb²⁺ more stable than Pb⁴⁺), but it runs right across the p-block.
Group by group — what the pattern actually looks like
| Group | States shown | What happens down the group |
|---|---|---|
| 15 (N, P, As, Sb, Bi) | −3 to +5; commonly +3 and +5 | +5 becomes unstable; Bi is stable as +3. Bi(V), e.g. sodium bismuthate NaBiO₃, is a very strong oxidising agent because it "wants" to fall back to +3. |
| 16 (O, S, Se, Te, Po) | −2, +2, +4, +6 | +6 becomes less stable. H₂SO₄ is a stable acid, while H₂SeO₄ is a strong oxidising agent. SO₂ acts mainly as a reducing agent, but TeO₂ acts as an oxidising agent. |
| 17 (F, Cl, Br, I) | −1; and +1, +3, +5, +7 for Cl, Br, I | Fluorine shows only −1. The higher states appear in oxoacids and oxoanions (ClO⁻, ClO₂⁻, ClO₃⁻, ClO₄⁻). |
| 18 (He…Rn) | 0 mainly; Xe shows +2, +4, +6, +8 | Only the heavier, more easily ionised members react: XeF₂ (+2), XeF₄ (+4), XeF₆ and XeO₃ (+6), XeO₄ (+8). Krypton gives KrF₂ (+2). |
Two special cases that are examined every year
Nitrogen cannot expand its octet. Its valence shell is n = 2, which has no d orbitals, so nitrogen's maximum covalency is 4. That is why NCl₅ and NF₅ do not exist while PCl₅ does, and why nitrogen forms the pπ–pπ multiple bonds (N≡N) that phosphorus cannot.
Fluorine is locked at −1. It is the most electronegative element, so nothing can pull electrons away from it, and like nitrogen it has no d orbitals. So there is no HFO₄ to match HClO₄.
A note on textbook editions: NCERT explains PCl₅ and SF₆ using sp³d and sp³d² hybridisation, i.e. d-orbital participation. Modern university-level bonding theory explains the same molecules with three-centre four-electron bonding and says d-orbital involvement is small. Both descriptions predict the same shapes. In a CBSE answer, write the NCERT hybridisation explanation.
Worked example 1 — phosphorus in H₃PO₃
Take H = +1 and O = −2, and let the oxidation number of P be x.
3(+1) + x + 3(−2) = 0
3 + x − 6 = 0
x − 3 = 0 → x = +3
The trap that follows: H₃PO₃ has three hydrogens but is dibasic, not tribasic. One H is attached directly to phosphorus (a P–H bond) and is not released as H⁺. Only the two O–H hydrogens ionise. Similarly H₃PO₂ is monobasic and H₃PO₄ is tribasic.
Worked example 2 — nitrogen in NH₄NO₃
Treating the whole formula at once gives an average. Do it properly, ion by ion.
In NH₄⁺: x + 4(+1) = +1 → x + 4 = +1 → x = −3
In NO₃⁻: x + 3(−2) = −1 → x − 6 = −1 → x = +5
Average = (−3 + 5) ÷ 2 = +1. Both answers are correct, but they answer different questions. If the question says "the oxidation states of nitrogen in ammonium nitrate", give −3 and +5, not +1.
Worked example 3 — chlorine in HClO₄ and sulfur in Na₂S₂O₃
HClO₄: (+1) + x + 4(−2) = 0 → 1 + x − 8 = 0 → x = +7, the group maximum, exactly as the rule predicts.
Sodium thiosulphate, Na₂S₂O₃: 2(+1) + 2x + 3(−2) = 0 → 2 + 2x − 6 = 0 → 2x = 4 → x = +2.
Say honestly what that +2 is: an average. The two sulfur atoms are not equivalent in the structure — one is the central sulfur and one is the terminal ("thio") sulfur, and they carry different oxidation numbers that average to +2.
Worked example 4 — xenon in XeO₃ and XeOF₄
XeO₃: x + 3(−2) = 0 → x − 6 = 0 → x = +6
XeOF₄: x + (−2) + 4(−1) = 0 → x − 2 − 4 = 0 → x = +6
Both are +6, which is why XeOF₄ can be thought of as XeF₆ with two F replaced by one O. Checking that two related compounds give the same oxidation state is a fast way to catch an arithmetic slip.
Mistakes that cost marks
- Assuming oxygen is always −2. It is −1 in peroxides (H₂O₂, Na₂O₂), −½ in superoxides (KO₂), and +2 in OF₂, because fluorine is more electronegative than oxygen.
- Giving fluorine a positive state. There is no such compound. If your working produces F as +1, the error is elsewhere.
- Writing NCl₅ or NF₅. Nitrogen has no d orbitals and a maximum covalency of 4.
- Getting the inert pair direction backwards. Down the group the lower state gets more stable. So Bi(V) and Tl(III) are oxidising agents, not stable resting states.
- Thinking "inert pair" means the electrons vanish. They are still there; they are simply held too tightly to be worth using in a bond.
- Quoting an average as a real state. In NH₄NO₃, Na₂S₂O₃ and Fe₃O₄, the single number you calculate is an average over non-equivalent atoms.
Where this appears in your exam
| Question type | What is really being tested |
|---|---|
| "Calculate the oxidation number of X in …" | The −2 for O / +1 for H convention and careful algebra |
| "Why is Bi(V) a strong oxidant?" / "Why is +3 stable for Bi?" | The inert pair effect, stated with both reasons |
| "Why does nitrogen not form NCl₅?" | Absence of d orbitals in the second shell |
| "Why is H₃PO₃ dibasic?" | Structure, not formula — the P–H bond |
| "Arrange the oxidising power of …" | Position in the group plus the inert pair trend |
The p-block chapter carries real weight in the Class 12 paper, and reasoning questions of the "why" type are common — but always confirm the current pattern and syllabus from the official CBSE notification and the latest NCERT textbook rather than from any guide, including this one.
Check every oxidation number in seconds. The free Oxidation Number calculator takes a formula or an ion and works out the oxidation state of each element, so you can verify a whole exercise after solving it by hand — which is exactly the right order.
Open the Oxidation Number Calculator →Want the p-block taught as one pattern instead of a list to cram? ABC Chemistry runs Class 11–12 chemistry coaching from its Gurugram centre and online classes across India — details at abcchemistry.in. Families in Delhi, Noida or Gurgaon who prefer one-to-one teaching at home can also arrange home tuition.