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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.

Why this block is a rank-decider. The p-Block gives 3 to 5 NEET questions a year, 12 to 20 marks, the single biggest cluster in Inorganic Chemistry. There are no numericals. It is pure trend-ordering, structure, and NCERT one-liners. Syllabus note: the NMC kept Unit 10 but cut the industrial processes, silicones, silicates, and zeolites. The trends, bonding, oxoacids, and salt analysis are all still tested. Confirm against your current NCERT edition.

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.

RankSub-topic clusterCountShareMain groups
1Oxoacids: basicity, linkages, acid strength1218.8%15, 16, 17
2VSEPR shapes and lone pairs (Xe, interhalogens)1117.2%17, 18, 15
3Oxidation states and the inert-pair effect1015.6%13, 14, 15, 16
4Hydride trends (stability, BP, pKa, B₂H₆)1015.6%13, 15, 16, 17
5Anomalous first element and carbon allotropes812.5%13, 14, 15, 17
6Periodic property exceptions (radius, IE, bond energy)710.9%13, 16, 17, 18
7Oxides, halides, salt analysis69.4%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 ionsIndustrial 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.

P O H H O H H₃PO₂ : 1 O-H, 2 P-H monobasic, strong reducer P O H O H O H H₃PO₃ : 2 O-H, 1 P-H dibasic, moderate reducer P O O H O H O H H₃PO₄ : 3 O-H, 0 P-H tribasic, no reducing power
The green H atoms are bonded to P, so they never ionise. They set the reducing power, not the basicity.

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
BasicityCount O-H bonds only. P-H and S-H never leave.
Reducing powerComes 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.
FluorineOnly 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.

F F F F Xe lone pair XeF₄ square planar 2 lone pairs, sp³d² F F F F F F Xe 1 lp XeF₆ distorted octahedral 1 lone pair, sp³d³ F F F Cl 2 lp ClF₃ bent T-shape 2 lone pairs, sp³d
The green lobes are lone pairs. They push the atoms into place. XeF₄ is not tetrahedral, and XeF₆ is not a regular octahedron.

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 pairsLinear
XeF₄2 lone pairsSquare planar
XeF₆1 lone pairDistorted octahedral
XeO₃1 lone pairTrigonal pyramidal
XeO₄0 lone pairsTetrahedral
ClF₃ (XX₃′)2 lone pairsBent T-shape
BrF₅ (XX₅′)1 lone pairSquare pyramidal
IF₇ (XX₇′)0 lone pairsPentagonal 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:

SpeciesWhat NEET asksThe answer
Sn²⁺ vs Pb⁴⁺Which is reducing, which is oxidising?Sn²⁺ is reducing (prefers +4). Pb⁴⁺ is oxidising (prefers +2).
TlI vs TlI₃Which is more stable?TlI. Tl⁺ beats Tl³⁺. TlI₃ is really Tl⁺[I₃]⁻.
PbI₄Does it exist?No. Pb⁴⁺ would oxidise I⁻ to I₂.
Polonium, -2 stateDoes Po show -2 like S and O?No. Po is a metal. It shows +2 and +4 only.
Sb₂O₅ vs BiCl₅Which 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 stabilityFalls down the group. NH₃ > PH₃ > ... ; H₂O > H₂S > ...
Acidity (Kₐ)Rises down the group. HF ≪ HCl < HBr < HI.
pKₐFalls down the group. Exact reverse of Kₐ.
Boiling point, Group 15PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃. NH₃ in the middle.
Bond angleDrops near 90° after Period 2 (Drago's rule).
Basicity, Group 15NH₃ > 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.

ElementThe signature anomalies NEET tests
BoronOnly non-metal in Group 13. B₂O₃ is purely acidic. Cannot form [BF₆]³⁻, caps at [BF₄]⁻.
CarbonHighest catenation. CO₂ is a linear gas, SiO₂ is a 3D solid. CCl₄ does not hydrolyse, no d orbitals.
NitrogenN₂ is a gas with a triple bond, P₄ is a solid. Cannot form NCl₅, no d orbitals. No dπ-pπ bonds.
OxygenO₂ is a diatomic gas, S₈ is a ring solid. Max covalency 2, never SF₆-like 6.
FluorineOnly -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.