IIT-JAM p-Block Chemistry — Anomalies, Oxidation States and Structures
The p-block is the largest inorganic unit in the IIT-JAM syllabus and the one students most often try to memorise rather than understand. Memorising six groups of facts does not work. What does work is holding four organising ideas — the anomalous first member, the inert pair effect, the availability of orbitals beyond the octet, and VSEPR — and deriving the facts from them in the exam hall. This guide sets out those four ideas, then works through the oxidation-number and shape problems that JAM asks most, with the arithmetic done in full.
Idea 1 — the first member of every group is different
Boron, carbon, nitrogen, oxygen and fluorine all behave unlike the rest of their groups, for three linked reasons: they are much smaller, they are much more electronegative, and they have no d orbitals in the valence shell, so they cannot expand beyond an octet. A fourth consequence follows from small size: they form strong pπ–pπ multiple bonds, which the heavier members do not.
| Observation | First member | Heavier member | Reason |
|---|---|---|---|
| Elemental form | N2, a gas with N≡N | P4, a solid with only single bonds | pπ–pπ overlap works only for small atoms |
| Oxide structure | CO2, discrete molecules | SiO2, giant covalent | Si cannot form strong Si=O π bonds |
| Maximum covalency | NF3 only; NF5 does not exist | PF3 and PF5 both exist | N has no d orbitals to expand the octet |
| Catenation | Very strong in carbon | Falls sharply: C ≫ Si > Ge ≈ Sn ≫ Pb | E–E bond enthalpy falls with size |
Idea 2 — the inert pair effect
Down groups 13 to 15, the oxidation state two units below the group state becomes progressively more stable, because the ns² pair becomes increasingly reluctant to take part in bonding (poor shielding by the intervening d and f electrons leaves it more tightly held). So:
- Group 13: Al3+ is normal, but Tl+ is more stable than Tl3+.
- Group 14: Sn2+ is a reducing agent while Pb2+ is the stable state; PbCl4 decomposes to PbCl2 + Cl2, and PbI4 does not exist at all because iodide would simply reduce Pb(IV).
- Group 15: Bi(III) is stable while Bi(V) is a strong oxidant.
The exam use of this is almost always comparative — "which of Tl(I) and Tl(III) is more stable, and why" — so learn the trend as a sentence, not as a list.
Worked example 1 — oxidation numbers with a peroxide linkage
Find the oxidation number of sulphur in H2S2O8 (peroxodisulphuric acid) and in H2SO5 (Caro's acid).
The naive attempt for H2S2O8, treating every oxygen as −2:
2(+1) + 2x + 8(−2) = 0 → 2 + 2x − 16 = 0 → x = +7
But sulphur cannot exceed +6, since it has only six valence electrons. The structure is HO–SO2–O–O–SO2–OH: two of the eight oxygens are in a peroxide linkage and carry −1, not −2. Redo it:
2(+1) + 2x + 6(−2) + 2(−1) = 0 → 2 + 2x − 12 − 2 = 0 → 2x = 12 → x = +6
Now H2SO5, structure HO–O–SO2–OH. Naive: 2 + x − 10 = 0 gives +8, which is impossible. With one peroxide linkage (two O at −1, three at −2):
2(+1) + x + 3(−2) + 2(−1) = 0 → 2 + x − 6 − 2 = 0 → x = +6
The rule to carry into the exam: if an oxidation number comes out higher than the group can support, look for a peroxide (O–O), a superoxide, or an element–element bond. It is a signal, not a slip.
Worked example 2 — shapes by counting electron pairs
Deduce the shapes of XeF2, XeF4, XeO3 and ClF3.
XeF2: Xe has 8 valence electrons; 2 are used in two Xe–F bonds, leaving 6 electrons = 3 lone pairs. Total domains = 2 + 3 = 5 → trigonal bipyramidal arrangement. The three lone pairs occupy the equatorial positions (more room), so the molecule is linear.
XeF4: 4 bonding pairs + (8 − 4)/2 = 2 lone pairs = 6 domains → octahedral arrangement, lone pairs trans to each other, so the shape is square planar.
XeO3: 3 σ bonds + 1 lone pair = 4 domains → tetrahedral arrangement, shape pyramidal.
ClF3: Cl has 7 valence electrons; 3 in bonds leaves 4 = 2 lone pairs. Domains = 3 + 2 = 5 → trigonal bipyramidal arrangement, lone pairs equatorial, shape T-shaped.
Same method gives BrF5 as square pyramidal (5 bp + 1 lp) and XeOF4 as square pyramidal too. Count first, name afterwards — students who try to recall the name directly get square planar and square pyramidal the wrong way round under time pressure.
Worked example 3 — electron counting in diborane
Why is B2H6 called electron deficient, and how is it bonded?
Valence electrons available: 2 B × 3 + 6 H × 1 = 12 electrons.
A conventional structure like ethane would need 8 bonds (6 B–H plus a B–B) = 16 electrons. Only 12 are available, so it cannot be built from ordinary two-centre two-electron bonds.
The real structure uses four terminal B–H bonds (2c–2e, using 8 electrons) and two B–H–B bridges that are three-centre two-electron bonds, using the remaining 4 electrons. The four terminal hydrogens lie in one plane, and the two bridging hydrogens sit above and below it. Bridge B–H bonds are longer and weaker than terminal ones.
Hydrolysis — the classic comparisons
Two questions come back constantly, and both are about orbital availability rather than bond strength.
- CCl4 resists hydrolysis; SiCl4 hydrolyses instantly. Water must first attach to the central atom. Carbon, already at an octet with no low-lying vacant orbitals available, cannot accept the lone pair, so there is no route in. Silicon can expand its coordination number and accept it, so hydrolysis proceeds to hydrated silica and HCl. (The classroom shorthand is "silicon has d orbitals"; the modern description is that silicon has accessible low-lying vacant orbitals. Either wording is accepted, but the physical point is the same.)
- NF3 is inert to hydrolysis; NCl3 hydrolyses to NH3 and HOCl. Chlorine is less electronegative than nitrogen and is the atom attacked by water; fluorine is more electronegative than nitrogen and holds its electrons too tightly.
Noble gas fluorides hydrolyse in ways worth writing out, because the equations themselves are asked:
6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2 (disproportionation)
XeF6 + H2O → XeOF4 + 2HF (partial)
XeF6 + 3H2O → XeO3 + 6HF (complete)
Note that XeF2 and XeF6 hydrolyse without a change in the Xe oxidation state pattern seen for XeF4, whose hydrolysis is a genuine disproportionation into Xe(0) and Xe(VI). Checking A and Z — here, checking every element count — is the only safe way to write these from memory.
Oxoacids — basicity and acid strength
Basicity is the number of ionisable O–H hydrogens, not the number of hydrogens in the formula. The phosphorus oxoacids are the standard test of that distinction:
| Acid | Structure | P–H bonds | Basicity | Note |
|---|---|---|---|---|
| H3PO2 (hypophosphorous) | one OH, two P–H | 2 | Monobasic | Strong reducing agent |
| H3PO3 (phosphorous) | two OH, one P–H | 1 | Dibasic | Also reducing |
| H3PO4 (orthophosphoric) | three OH | 0 | Tribasic | Not reducing |
Hydrogen attached directly to phosphorus is not ionisable, and it is also the source of the reducing power — one fact explaining two properties.
For the chlorine oxoacids, acid strength rises with oxidation state: HOCl < HClO2 < HClO3 < HClO4, because each extra terminal oxygen delocalises the negative charge of the conjugate base more effectively. Boric acid, H3BO3, is the odd one out: it is monobasic and acts as a Lewis acid, taking OH− from water rather than donating a proton — B(OH)3 + 2H2O → [B(OH)4]− + H3O+.
Hydrides, allotropes and interhalogens in brief
- Group 15 hydrides: thermal stability and basicity both fall NH3 > PH3 > AsH3 > SbH3 > BiH3, and the bond angle drops sharply from about 107° in NH3 to roughly 92–94° in the heavier hydrides, where the central atom uses almost pure p orbitals.
- Boiling point anomaly: NH3, H2O and HF boil far above their group trend because of hydrogen bonding. H2O is the most anomalous because each molecule can form up to four hydrogen bonds.
- Allotropes: carbon as diamond, graphite and the fullerenes; phosphorus as white (P4 tetrahedra with strained 60° angles, hence very reactive and stored under water), red and black; sulphur as rhombic and monoclinic S8 rings.
- Interhalogens come in the types AX, AX3, AX5 and AX7; IF7 is the only known AX7, because only the large iodine atom can hold seven fluorines. They are all more reactive than the parent halogens because the A–X bond is weaker than X–X in the corresponding halogen.
- Silicones, (R2SiO)n, come from hydrolysis and condensation of R2SiCl2; their thermal and chemical stability follows from the strong Si–O bond.
Common mistakes that cost marks
- Assigning −2 to every oxygen. In peroxides it is −1, in superoxides −½, and in OF2 it is +2. An impossible oxidation number is telling you which case you are in.
- Counting hydrogens as basicity. H3PO3 is dibasic, not tribasic. Draw the structure before answering.
- Forgetting lone pairs when naming a shape. The arrangement counts all domains; the shape names only the atoms. XeF4 is octahedral in arrangement and square planar in shape.
- Claiming nitrogen forms pentahalides. It cannot expand its octet; PF5 exists, NF5 does not.
- Explaining CCl4's inertness by bond strength. It is about the absence of a route for water to attack, not about how strong the C–Cl bond is.
- Using the inert pair effect in the wrong direction. The lower oxidation state becomes more stable down the group.
- Writing unbalanced xenon fluoride hydrolysis equations. Balance Xe, F, H and O separately every time — the XeF4 equation in particular is easy to get wrong.
How to prepare this unit
| Theme | What you must be able to do without hesitation |
|---|---|
| Anomalous first member | Give the reason and one example for each of groups 13–17 |
| Inert pair effect | Compare two oxidation states of Tl, Sn/Pb or Bi and justify the answer |
| Oxidation numbers | Handle peroxo, superoxo and element–element bonds correctly |
| Shapes | Apply VSEPR to interhalogens and xenon compounds by domain counting |
| Hydrides and halides | Predict stability, basicity, bond angle and hydrolysis behaviour |
| Oxoacids | Deduce basicity from structure; order acid strength by oxidation state |
| Electron-deficient bonding | Count electrons in B2H6 and describe 3c–2e bonds |
Treat that as a revision checklist rather than a forecast of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.
Check your oxidation-number working. The oxidation number calculator applies the standard rules to a formula and shows the assignment, so you can verify ordinary cases in seconds — then handle the peroxo and superoxo exceptions above by hand, since those need the structure and not just the formula.
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