p-Block Elements NEET PYQ — The Oxoacid, VSEPR and Inert-Pair Traps That Decide This Block (2015-2026)
p-Block Elements NEET PYQ analysis (2015-2026). Oxoacid basicity, Xe and interhalogen VSEPR shapes, the inert-pair effect, hydride trends, the deleted syllabus, reference tables, and 12 exam-level PYQs with traps.
Six groups, one scoring block, and traps that repeat almost word for word.
If you want safe Inorganic marks in NEET, learn the p-Block cold. It is the biggest single source of Chemistry questions, and the patterns barely change year to year. The same handful of ideas decide it: oxoacid structure, the shapes of xenon and interhalogen compounds, the inert-pair effect, and the hydride trends.
The data is blunt. Just five clusters carry 83 percent of all p-Block questions. Oxoacids lead, then VSEPR shapes, then inert-pair oxidation states, then hydride trends, then the anomalous first element. Master those five and you cover almost the whole block.
The format has shifted hard. NEET now leans on Assertion-Reason, two-statement, "how many are correct", and match-the-following. These punish a student who half-remembers an NCERT line. Since 2025 the paper is 180 compulsory questions, so you cannot skip the ones you dislike. Sharp, exact recall is the only thing that survives.
We analysed every p-Block question NEET asked from 2015 to 2026, including NEET 2026 and Re-NEET 2026. That is 64 questions across 20 papers. This is part of Logic Bloom's NEET PYQ analysis series.
| 🎯 We mapped every NEET p-Block question of the decade. The app has them all, ready to play and practice. | |
|---|---|
| This block is won by drilling trends and structures until they are reflex, not by re-reading NCERT. Logic Bloom's Playground turns it into practice: rotate XeF4 and XeF6 to see square planar vs distorted octahedral, count P-H and P-OH bonds in each phosphorus oxoacid, and order the hydrides by boiling point with the NH3 dip in place. Then drill every PYQ mapped by year and trap. When the inert-pair reversal or the pKa inversion catches you, TarQ teaches the fix, and your Mistake Book logs it. | Get the app → Free to start. |
The decade at a glance: where the marks sit
Across 2015 to 2026 the block held 3 to 5 questions per paper. Five clusters do almost all the work. Here is the split, by how often each appeared.
| Rank | Sub-topic cluster | Count | Share | Main groups |
|---|---|---|---|---|
| 1 | Oxoacids: basicity, linkages, acid strength | 12 | 18.8% | 15, 16, 17 |
| 2 | VSEPR shapes and lone pairs (Xe, interhalogens) | 11 | 17.2% | 17, 18, 15 |
| 3 | Oxidation states and the inert-pair effect | 10 | 15.6% | 13, 14, 15, 16 |
| 4 | Hydride trends (stability, BP, pKa, B₂H₆) | 10 | 15.6% | 13, 15, 16, 17 |
| 5 | Anomalous first element and carbon allotropes | 8 | 12.5% | 13, 14, 15, 17 |
| 6 | Periodic property exceptions (radius, IE, bond energy) | 7 | 10.9% | 13, 16, 17, 18 |
| 7 | Oxides, halides, salt analysis | 6 | 9.4% | 13, 14, 15, 16 |
The plan follows the table. Spend the most time on oxoacids and VSEPR shapes, because together they are a third of the block and the most trap-heavy. Lock the inert-pair effect and the hydride trends next, since they repeat almost every year. The anomalous first element is short and high-value, so nail it and move on.
What is still tested, and what is dead
This matters before you revise a single line. The NMC kept the p-Block, but trimmed it. Studying the dead parts wastes time you do not have.
| Still tested (revise this) | Deleted (do not study) |
|---|---|
| General trends across periods and down groups; anomalous first element; inert-pair effect; oxide and hydroxide nature; hydride trends; oxoacid structures and basicity; VSEPR shapes; interhalogens and xenon compounds; salt analysis of p-block ions | Industrial flowcharts (Haber, Ostwald, Contact, Deacon); silicones, silicates, zeolites; borax dehydration chain; bleaching powder synthesis; phosphine lab prep; sulphur allotrope transition temperatures |
The rule of thumb: if it is a trend, a structure, or an NCERT exception, it is live. If it is a factory process or a named preparation, it is probably gone. Confirm against your current NCERT edition.
Cluster 1: Oxoacids, the most-tested idea in the block
Oxoacids are the number-one source of p-Block marks. NEET tests three things: how many protons an acid gives, how strong it is, and what bonds it hides. The rule that unlocks all three is simple.
The master rule. In any oxoacid, only hydrogen on an oxygen (an O-H bond) can leave as H⁺. That O-H count is the basicity. Hydrogen bonded straight to the central atom (a P-H or S-H bond) never leaves. Instead it makes the acid a reducing agent. Count the bonds, and both answers fall out.
The phosphorus oxoacid trap
This is the most punished error in Group 15. Students see "H₃" in H₃PO₂ and H₃PO₃ and call them tribasic. Wrong. The structures tell the real story.
So the basicity order is H₃PO₂ (1) < H₃PO₃ (2) < H₃PO₄ (3), and the reducing order is the reverse: H₃PO₂ > H₃PO₃ > H₃PO₄. A second trap asks which oxoacid has both a P-H and a P-P bond. The answer is none. H₄P₂O₅ has two P-H bonds with a P-O-P bridge, and H₄P₂O₆ has a P-P bond with zero P-H. No NCERT acid has both.
The halogen oxoacid trap
Here NEET tests an inverse. As you go HOCl, HClO₂, HClO₃, HClO₄, the number of Cl=O bonds rises, so the conjugate base is more stabilised. That makes acid strength rise: HOCl < HClO₂ < HClO₃ < HClO₄. But oxidising power falls the opposite way: HOCl > HClO₂ > HClO₃ > HClO₄. Students pick HClO₄ as the strongest oxidiser because Cl is +7. That is the trap. It is the strongest acid, the weakest oxidiser.
| ⚡ Oxoacid memory hooks | |
|---|---|
| Basicity | Count O-H bonds only. P-H and S-H never leave. |
| Reducing power | Comes from P-H bonds. H₃PO₂ has two, so it is strongest. |
| Acid strength (Cl) | HOCl < HClO₂ < HClO₃ < HClO₄. More O, stronger acid. |
| Oxidising power (Cl) | Exact reverse. HOCl is the strongest oxidiser. |
| Fluorine | Only one oxoacid, HOF. No HFO₄, since F has no d orbitals. |
Cluster 2: VSEPR shapes of xenon and interhalogen compounds
This is the second biggest cluster, and it is pure structure. The trap is always the same: students match the number of atoms to a shape and forget the lone pairs. Xenon has 8 valence electrons, so lone pairs are everywhere. Draw them, and the shape is forced.
Two classic confusions to kill now. XeF₄ is square planar, not tetrahedral. The tetrahedral one is XeO₄, which has no lone pairs. And XeF₆ is a distorted octahedron, not a regular one. The regular octahedron is SF₆, which also has no lone pairs. The lone pair is the whole difference.
| 🧠 Shape table, by lone-pair count on the central atom | ||
|---|---|---|
| XeF₂ / I₃⁻ | 3 lone pairs | Linear |
| XeF₄ | 2 lone pairs | Square planar |
| XeF₆ | 1 lone pair | Distorted octahedral |
| XeO₃ | 1 lone pair | Trigonal pyramidal |
| XeO₄ | 0 lone pairs | Tetrahedral |
| ClF₃ (XX₃′) | 2 lone pairs | Bent T-shape |
| BrF₅ (XX₅′) | 1 lone pair | Square pyramidal |
| IF₇ (XX₇′) | 0 lone pairs | Pentagonal bipyramidal |
Cluster 3: The inert-pair effect and redox reversals
Going down Groups 13 to 16, the outer ns² pair gets held tighter and resists bonding. So the lower oxidation state gets more stable down the group, and the higher one gets less stable. That one idea drives a stack of NEET traps.
The reversals you must know:
| Species | What NEET asks | The answer |
|---|---|---|
| Sn²⁺ vs Pb⁴⁺ | Which is reducing, which is oxidising? | Sn²⁺ is reducing (prefers +4). Pb⁴⁺ is oxidising (prefers +2). |
| TlI vs TlI₃ | Which is more stable? | TlI. Tl⁺ beats Tl³⁺. TlI₃ is really Tl⁺[I₃]⁻. |
| PbI₄ | Does it exist? | No. Pb⁴⁺ would oxidise I⁻ to I₂. |
| Polonium, -2 state | Does Po show -2 like S and O? | No. Po is a metal. It shows +2 and +4 only. |
| Sb₂O₅ vs BiCl₅ | Which pentavalent species exists? | Sb₂O₅ exists. BiCl₅ does not. Only BiF₅ does. |
The pattern is one sentence: the heavier the element, the more it clings to the lower oxidation state. Lead, thallium, bismuth, and polonium all sit in their lower state, and their higher state turns oxidising.
Cluster 4: Hydride trends, three traps in one
Hydrides of Groups 15, 16, and 17 pack three ordering traps that NEET sets again and again.
Trap A, the pKa inversion. Down a group the E-H bond weakens, so acid strength (Kₐ) rises: H₂O < H₂S < H₂Se < H₂Te. But pKₐ is the reverse of Kₐ, so pKₐ falls: H₂O > H₂S > H₂Se > H₂Te. In 2021 NEET wrote the acidity order and labelled it "increasing pKₐ". Students who equate acidity with pKₐ got it wrong.
Trap B, the boiling-point dip. Students put NH₃ at the top of Group 15 because of hydrogen bonding. But NH₃'s H-bonding is weaker, and the heavy hydrides have strong dispersion forces. The real order has NH₃ in the middle: PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃. Note H₂O and HF still top their own groups.
Trap C, the bond-angle drop. NH₃ is 107.8° and H₂O is 104.5°, both sp³. But PH₃ drops to about 93.6° and H₂S to 92.1°. After Period 2 the central atom bonds through nearly pure p orbitals, so the angle falls near 90°. This is Drago's rule.
| ⚡ Hydride trends, locked down | |
|---|---|
| Thermal stability | Falls down the group. NH₃ > PH₃ > ... ; H₂O > H₂S > ... |
| Acidity (Kₐ) | Rises down the group. HF ≪ HCl < HBr < HI. |
| pKₐ | Falls down the group. Exact reverse of Kₐ. |
| Boiling point, Group 15 | PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃. NH₃ in the middle. |
| Bond angle | Drops near 90° after Period 2 (Drago's rule). |
| Basicity, Group 15 | NH₃ > PH₃ > AsH₃ > SbH₃. |
Cluster 5: The anomalous first element
The first element of each group breaks its own group's pattern. There are three reasons. It is the smallest, and it has the highest electronegativity. It also has no d orbitals, so its covalency caps at 4. NEET lifts these facts almost word for word from NCERT.
| Element | The signature anomalies NEET tests |
|---|---|
| Boron | Only non-metal in Group 13. B₂O₃ is purely acidic. Cannot form [BF₆]³⁻, caps at [BF₄]⁻. |
| Carbon | Highest catenation. CO₂ is a linear gas, SiO₂ is a 3D solid. CCl₄ does not hydrolyse, no d orbitals. |
| Nitrogen | N₂ is a gas with a triple bond, P₄ is a solid. Cannot form NCl₅, no d orbitals. No dπ-pπ bonds. |
| Oxygen | O₂ is a diatomic gas, S₈ is a ring solid. Max covalency 2, never SF₆-like 6. |
| Fluorine | Only -1 state (except +1 in HOF). Only one oxoacid, HOF. Low F-F bond energy. |
The Group 13 radius and ionisation traps
Two famous exceptions here. Atomic radius is B < Ga < Al < In < Tl, not the smooth order. Gallium is smaller than aluminium because the 3d¹⁰ electrons shield poorly. First ionisation enthalpy makes a W-curve: B > Tl > Ga > Al > In. And BX₃ Lewis acidity is BF₃ < BCl₃ < BBr₃ < BI₃, because strong back-bonding in BF₃ makes it the weakest acid, not the strongest.
| 🎯 Drill the whole decade of p-Block PYQs, sorted by group and trap. | |
|---|---|
| Reading a trend once is not the same as recalling it in the hall under the clock. Logic Bloom's Playground has every NEET p-Block PYQ from 2015 to 2026, tagged by group, cluster, and the exact trap it hides. Miss the pKa inversion or the XeF4-vs-XeO4 shape and TarQ walks you through the fix, then your Mistake Book brings it back until it sticks. Deleted topics are stripped out, so you never waste a rep. | Get the app → Free to start. |
12 must-attempt NEET p-Block PYQs, with the traps named
These twelve cover the real spread, single-correct, assertion-reason, two-statement, and match-the-following. Each is written at NEET level and turns on a real decision, not plain recall. Try each before reading the answer.
1. Which oxoacid of phosphorus is dibasic and still acts as a reducing agent?
(A) H₃PO₄ (B) H₃PO₃ (C) H₄P₂O₆ (D) (HPO₃)₃
Work. Dibasic means two O-H bonds. Reducing means at least one P-H bond. H₃PO₃ has two O-H and one P-H, so it fits both. H₃PO₄ has no P-H. H₄P₂O₆ has a P-P bond but no P-H, so it does not reduce.
Answer: (B). Trap: (C) tempts students who think the P-P bond gives reducing power; only P-H does.
2. Arrange the chlorine oxoacids by increasing acid strength and state which is the strongest oxidiser.
(A) HClO₄ < HClO₃ < HClO₂ < HOCl; HClO₄ strongest oxidiser
(B) HOCl < HClO₂ < HClO₃ < HClO₄; HOCl strongest oxidiser
(C) HOCl < HClO₂ < HClO₃ < HClO₄; HClO₄ strongest oxidiser
(D) HClO₄ < HClO₃ < HClO₂ < HOCl; HOCl strongest oxidiser
Work. More terminal Cl=O bonds stabilise the conjugate base, so acid strength rises HOCl to HClO₄. Oxidising power is the reverse, so HOCl is the strongest oxidiser. Only (B) has both directions right.
Answer: (B). Trap: (C) gets the acid order right but keeps HClO₄ as the oxidiser, the most common slip.
3. Which pair has the same shape?
(A) XeF₄ and XeO₄ (B) XeF₄ and ICl₄⁻ (C) XeF₆ and SF₆ (D) XeO₃ and BF₃
Work. XeF₄ is square planar with 2 lone pairs. ICl₄⁻ also has 2 lone pairs and is square planar, so (B) matches. XeO₄ is tetrahedral. SF₆ is regular octahedral but XeF₆ is distorted. XeO₃ is pyramidal but BF₃ is trigonal planar.
Answer: (B). Trap: (A) is the classic XeF₄-equals-XeO₄ error.
4. Two statements are given. Statement I: Sn²⁺ acts as a reducing agent. Statement II: For tin, the +4 state is more stable than the +2 state. Choose the correct option.
(A) Both true, II explains I (B) Both true, II does not explain I (C) I true, II false (D) I false, II true
Work. Tin sits high enough in Group 14 that +4 is still the stable state. So Sn²⁺ readily loses two electrons to reach +4, which makes it a reducing agent. Statement I is true, Statement II is true, and II is exactly why I holds.
Answer: (A). Trap: (B) accepts both facts but misses that II is the cause of I.
5. Which arrangement is NOT correct for the property stated?
(A) Thermal stability: NH₃ > PH₃ > AsH₃ > SbH₃
(B) Acidic strength: HF < HCl < HBr < HI
(C) pKₐ: H₂O < H₂S < H₂Se < H₂Te
(D) Bond angle: NH₃ > PH₃ > AsH₃
Work. Acidity rises down Group 16, so Kₐ rises. pKₐ is the reverse, so it must fall: H₂O > H₂S > H₂Se > H₂Te. Option (C) writes it increasing, which is wrong. The other three are correct.
Answer: (C). Trap: equating higher acidity with higher pKₐ. It is the exact opposite.
6. The correct order of atomic radius for Group 13 is:
(A) B < Al < Ga < In < Tl (B) B < Ga < Al < In < Tl (C) B < Al < Ga < Tl < In (D) Tl < In < Ga < Al < B
Work. Gallium follows the 3d¹⁰ series. The poorly shielding d electrons let the nucleus pull the shell in, so Ga is smaller than Al. The order becomes B < Ga < Al < In < Tl.
Answer: (B). Trap: (A) is the naive smooth order that ignores the d-block contraction.
7. Which statement about diborane (B₂H₆) is correct?
(A) Both B atoms are sp² hybridised
(B) It has four 3c-2e bonds
(C) Both B atoms are sp³ hybridised, with two 3c-2e bridge bonds
(D) All six B-H bonds are identical
Work. In B₂H₆ each boron is sp³, not sp². The four terminal B-H bonds are normal 2c-2e bonds. The two bridging B-H-B bonds are 3c-2e banana bonds. So (C) is right. The terminal and bridge bonds differ, so (D) is false.
Answer: (C). Trap: (A) copies the sp² of monomeric BH₃, which is not how the dimer bonds.
8. Assertion: HClO₄ is a stronger acid than HOCl but a weaker oxidising agent. Reason: The ClO₄⁻ ion is stabilised by charge delocalisation over four oxygen atoms. Choose the correct option.
(A) Both true, R explains A (B) Both true, R does not explain A (C) A true, R false (D) A false, R true
Work. ClO₄⁻ spreads its charge over four oxygens, so it is very stable. A stable conjugate base means a stronger acid and a base that resists being reduced, so a weaker oxidiser. The assertion is true and the reason is its exact cause.
Answer: (A). Trap: (B) accepts both but misses that the stability is the reason for both halves.
9. Which does NOT exist?
(A) [BF₄]⁻ (B) [SiF₆]²⁻ (C) [SiCl₆]²⁻ (D) [AlF₆]³⁻
Work. Silicon has 3d orbitals, so [SiF₆]²⁻ forms. But six large Cl⁻ ligands cannot fit around the small Si⁴⁺, so [SiCl₆]²⁻ does not exist. Boron caps at four, so [BF₄]⁻ is fine and [BF₆]³⁻ is not.
Answer: (C). Trap: assuming Cl behaves like F around Si. Size blocks it.
10. The boiling point order of Group 15 hydrides is:
(A) PH₃ < AsH₃ < SbH₃ < BiH₃ < NH₃
(B) NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃
(C) PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃
(D) BiH₃ < SbH₃ < NH₃ < AsH₃ < PH₃
Work. NH₃ has hydrogen bonding, so it is not the lowest. But its H-bonding is weaker than the dispersion forces of heavy SbH₃ and BiH₃. So NH₃ lands in the middle: PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃.
Answer: (C). Trap: (A) puts NH₃ on top, copying the H₂O and HF pattern.
11. Which element does NOT show the -2 oxidation state?
(A) O (B) S (C) Se (D) Po
Work. Oxygen, sulphur, and selenium are non-metals that reach -2 easily. Polonium is a heavy metal, and the inert-pair effect restricts it to +2 and +4. It does not form the -2 state.
Answer: (D). Trap: assuming every Group 16 element behaves like oxygen.
12. Match the compound with its sulphur linkage, then pick the correct option.
(i) H₂S₂O₇ (ii) H₂S₂O₈ (iii) H₂S₂O₆ (iv) H₂SO₅
(P) peroxy, -O-O- (Q) S-O-S bridge (R) direct S-S bond (S) one peroxy, -O-OH
(A) i-Q, ii-P, iii-R, iv-S (B) i-P, ii-Q, iii-S, iv-R (C) i-R, ii-S, iii-Q, iv-P (D) i-Q, ii-R, iii-P, iv-S
Work. H₂S₂O₇ (oleum) has an S-O-S oxo bridge. H₂S₂O₈ (Marshall's acid) has a peroxy -O-O- bridge. H₂S₂O₆ (dithionic) has a direct S-S bond. H₂SO₅ (Caro's acid) has one peroxy -O-OH group. That gives i-Q, ii-P, iii-R, iv-S.
Answer: (A). Trap: swapping the two peroxy acids, H₂S₂O₈ and H₂SO₅.
What to expect in NEET 2027 and 2028
The 2024 to 2026 papers set a clear pattern. NTA respects the deleted industrial content, but tests the trends, bonding, and structures at full depth. Five archetypes are most likely next.
1. Inert-pair redox, as Assertion-Reason (over 90% likely). Expect statements on why PbI₄, BiCl₅, [BF₆]³⁻, and NCl₅ do not exist, and why Sn²⁺ reduces while Pb⁴⁺ oxidises. This line ran through 2023, 2024, and 2025.
2. VSEPR match-the-following (over 85% likely). Xenon compounds and interhalogens paired with lone-pair counts, hybridisation, or shape. Learn the shape table cold.
3. A "which order is wrong" sequence check (about 80% likely). Prime targets: the Group 15 boiling-point dip, the Group 16 electron-gain anomaly (S > Se > Te > Po > O), the halogen bond energy (Cl₂ > Br₂ > F₂ > I₂), and the Group 13 radius and IE exceptions.
4. Oxoacid linkage counting (about 75% likely). P-H vs P-OH vs P-P across the phosphorus acids, and S-S vs oxo vs peroxy across the sulphur acids.
5. Salt analysis crossover (about 75% likely). p-Block ions like Pb²⁺, As³⁺, and Al³⁺ placed into their qualitative analysis groups, plus the brown ring test.
Do not revise: the Haber, Ostwald, Contact, and Deacon flowcharts, silicones, silicates, zeolites, and named lab preparations. All deleted. Confirm against your current NCERT edition.
| 🚀 Turn this analysis into a real score. | |
|---|---|
| You now know the five clusters and the traps in each. Close the gap between knowing and scoring. Logic Bloom's Playground drills every NEET p-Block PYQ of the decade, and Battleground lets you test the block in 1v1 duels against other aspirants. Understand through games. Score through practice. | Get the app → Free to start. |