p-Block Elements
3 lessons
Topic index in NCERT order
3 of 3 lessons by the NCERT chapter they teach from, in book order. The page is the first printed page of your NCERT book the lesson cites; PYQs are the past NEET questions on that topic.
Class 12 Chemistry (pre-2023 edition), Chapter 7
- p-Block Groups 13 to 18p. 1709 PYQs
- First Element Anomalyp. 1721 PYQ
- p-Block Trendsp. 1768 PYQs
First Element Anomaly
p-Block Groups 13 to 18
p-Block Trends
Past-paper questions from this unit
18 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025, 2026. Answers verified against NTA official keys.
By year in our set: 2020 (3) · 2021 (3) · 2022 (4) · 2023 (3) · 2024 (2) · 2025 (1) · 2026 (2)
Lesson: p-Block Groups 13 to 18
Lesson: First Element Anomaly
Lesson: p-Block Groups 13 to 18
Lesson: p-Block Trends
Lesson: p-Block Groups 13 to 18
Among Group 16 elements, which one does NOT show –2 oxidation state?
Lesson: p-Block Trends
Taking stability as the factor, which one of the following represents correct relationship?
Lesson: p-Block Trends
Lesson: p-Block Groups 13 to 18
Lesson: p-Block Groups 13 to 18
Which of the following statement is not correct about diborane?
Lesson: p-Block Trends
Lesson: p-Block Trends
Lesson: p-Block Trends
Lesson: p-Block Trends
Lesson: p-Block Groups 13 to 18
Lesson: p-Block Trends
Lesson: p-Block Groups 13 to 18
Which of the following oxoacid of sulphur has −O−O− linkage ?
Lesson: p-Block Groups 13 to 18
Lesson: p-Block Groups 13 to 18
Which of the following is not correct about carbon monoxide ?
Exam traps and common mistakes in this unit
Lesson: p-Block Trends
Category: Inorganic Exception
Students assume boiling point increases monotonically with molar mass across a group's hydrides. The first-row hydrides H2O, NH3 and HF boil far higher than that trend predicts because of strong intermolecular H-bonding. H2O is the highest-boiling Group 16 hydride (NCERT Table 7.7: 373 K) and HF the highest-boiling hydrogen halide (Table 7.9: 293 K). NH3 is NOT the highest in Group 15: it boils above PH3 and AsH3 but below SbH3 and BiH3 (Table 7.2: 238.5, 185.5, 210.6, 254.6, 290 K). Any statement that places H2O or HF at the low end of their series is incorrect.
When it triggers
Question presents a boiling-point order (or a statement about order) for a set of Group 15, 16, or 17 hydrides and includes H2O, NH3, or HF. The trap fires when an option treats the first-row hydride as the lowest-boiling member.
How to avoid
Identify whether H2O, NH3 or HF is in the series. For H2O and HF, put the first-row hydride at the TOP of the boiling-point ranking; the heavier hydrides then follow molar mass (H2S < H2Se < H2Te; HCl < HBr < HI). For Group 15, NH3 only beats PH3 and AsH3: the order is PH3 < AsH3 < NH3 < SbH3 < BiH3.
Lesson: p-Block Trends
Category: Similar Terms
Noble gases are chemically inert (very high ionization enthalpy, negligible tendency to react), but this chemical stability does not carry over into thermal stability. Because the only intermolecular forces present are weak London dispersion forces, noble gases have very LOW melting and boiling points. Students who associate 'inert' or 'stable' with high thermal resistance will select a statement claiming high m.p./b.p. as correct, when in fact it is the canonical incorrect statement about noble gases.
When it triggers
Question asks about physical properties (melting point, boiling point) of noble gases, or asks to identify an incorrect statement about them. The distractor option asserts that noble gases have high melting or boiling points.
How to avoid
Separate chemical reactivity from physical state properties. Noble gases: chemically unreactive (true), large positive electron-gain enthalpy (true), sparingly soluble in water (true), weak dispersion forces (true), very LOW m.p. and b.p. (true). High m.p./b.p. is the false claim.
Lesson: p-Block Trends
Category: Inorganic Exception
Across period 3, first ionization enthalpy generally increases from Na to Ar, but there is a reversal at Mg/Al: IE1(Mg) > IE1(Al). Removing a 3p electron (Al) requires less energy than removing a 3s electron (Mg) because 3p orbitals are higher in energy and experience more shielding from the inner 3s electrons. Students who apply the blanket 'IE increases uniformly across a period' rule expect Al's IE to exceed Mg's, leading them to accept an incorrect statement or order.
When it triggers
Question presents the first ionization enthalpy order for Na, Mg, Al, Si (or a subset). Any option giving a strictly monotonic order Na < Mg < Al < Si is wrong; the correct order has Mg > Al.
How to avoid
Memorise the two period-3 IE1 anomalies: (1) Mg > Al, because a 3p electron is more shielded and easier to remove than a 3s electron (NCERT Class XI Unit 3 explains the same s-versus-p effect for Be > B); (2) P > S, because of the extra-stable half-filled 3p3 configuration (Class XII Unit 7 page 18). Correct period-3 order: Na < Al < Mg < Si < S < P < Cl < Ar.
Lesson: p-Block Trends
Category: Negative Marking
NEET Assertion-Reason questions require evaluating A and R as independent statements and then determining whether R actually explains A. Students who read a plausible-sounding R alongside a correct A default to option 2 (both correct, R explains A) without separately verifying that R's specific chemical claim is factually accurate. When R is false or irrelevant, selecting option 2 instead of option 3 or 4 costs 1 negative mark.
When it triggers
Any Assertion-Reason format question in chemistry where R invokes a chemical mechanism or property that sounds consistent with A. The trap fires when students accept R without checking whether its chemistry is correct.
How to avoid
Evaluate A and R as completely separate statements before considering their relationship. Protocol: (1) Is A true on its own? (2) Is R's specific chemical claim factually correct on its own? (3) If both true, does R mechanistically explain A? Skipping step 2 is the trap.
Lesson: p-Block Trends
Category: Similar Terms
A weaker bond is easier to break (lower bond dissociation energy = lower activation barrier), so a compound with a weaker bond is MORE reactive under comparable conditions, not less. Students sometimes invert this, concluding that a compound with a weaker bond is less reactive because they associate lower energy content with lower driving force. The correct relationship is: bond strength down = reactivity up.
When it triggers
Question states or asks about the relative reactivity of two compounds where the reason involves comparing bond strengths (e.g. interhalogen vs parent dihalogen, or a bond in acid vs its conjugate). Distractor options invert the bond strength to reactivity mapping.
How to avoid
Bond strength up means harder to break, so reactivity down. Bond strength down means easier to break, so reactivity up. Apply to interhalogens: I-Cl is weaker than I-I, therefore ICl is more reactive than I2 (the R correctly explains the A).
Lesson: p-Block Trends
Category: Inorganic Exception
In p-block groups, the element that deviates from the group standard oxidation state is almost always the HEAVIEST congener (due to metallic character, relativistic effects, and inert-pair tendency at high Z). Students make two opposite errors: (1) they pick the LIGHTEST member (e.g. O in Group 16) as the anomaly because oxygen is already known to have many unique properties; (2) they pick the second-heaviest (e.g. Te instead of Po) because the bottom congener is less familiar from standard syllabus coverage.
When it triggers
Question asks which element in a p-block group does NOT exhibit an oxidation state that all other members readily show. The group extends to a 5th- or 6th-period element (Po, At, Tl, Pb, Bi). Distractors offer the lightest or second-heaviest congener.
How to avoid
The exception is the HEAVIEST (most metallic) congener. Learn by group: Group 16 exception = Po; Group 15 exception = Bi (reduced -3 tendency); Group 17 exception = At. The lightest members (O, N, F) have different anomalies unrelated to this oxidation-state trend.
Lesson: p-Block Trends
Category: Inorganic Exception
The inert pair effect is NCERT's name for the growing stability, down a group, of the oxidation state two units below the group oxidation state (Class XI Unit 11 page 1): Tl +1 over +3 (group 13), Pb +2 over +4 (group 14), Bi +3 over +5 (group 15). Students misapply it to negative oxidation states, e.g. explaining why polonium hardly shows -2. That is a different trend: NCERT states only that the stability of the -2 state decreases down Group 16 (Class XII Unit 7 page 18), alongside rising metallic character.
When it triggers
Question asks which element in a p-block group does not show an oxidation state (especially -2 in Group 16), or asks for the reason, and an option or explanation invokes the inert pair effect for a negative oxidation state or for a mid-group element.
How to avoid
Use the inert pair effect only for positive states two below the group state in the heavier members (Tl+, Sn2+/Pb2+, Bi3+). For the loss of the -2 state in Group 16, the answer is the bottom member, Po, and the reason is the decreasing stability of -2 down the group (increasing metallic character), not the inert pair.
Lesson: p-Block Trends
Hydride electron-counting: Group 13 (electron-deficient) vs Group 14 (electron-precise) confused
Category: Similar Terms
B2H6 (Group 13) is electron-deficient: it 'has too few electrons for writing its conventional Lewis structure', and its two bridge B-H-B bonds are 3-centre-2-electron bonds. GeH4 (Group 14) is electron-precise: group 14 compounds 'have the required number of electrons to write their conventional Lewis structures'. The two hydrides sit side by side in a match-the-column list and look alike, so students carry the electron-deficient label from B2H6 over to GeH4.
When it triggers
Classification or match-the-column question presents a Group 13 hydride (B2H6, AlH3) and a Group 14 hydride (CH4, GeH4) alongside the labels electron-deficient and electron-precise. The trap fires when the electron-deficient label is placed on the Group 14 hydride.
How to avoid
The group number tells the category directly (NCERT Hydrogen unit 9.5.2): all group 13 elements form electron-deficient compounds (Lewis acids); all group 14 elements form electron-precise compounds, tetrahedral (e.g. CH4, GeH4); groups 15-17 form electron-rich hydrides with lone pairs (NH3 1, H2O 2, HF 3), which act as Lewis bases. Use the group number, not the element's metallic character.
Lesson: p-Block Trends
Category: Similar Terms
Both graphite and fullerene are non-diamond carbon allotropes with sp2-hybridised carbon and delocalised electrons, so students who remember only 'non-diamond carbon allotrope' as the distinguishing property may swap their structural labels. Graphite has a 2D layered hexagonal structure; layers slide over each other making it a dry lubricant. Fullerene (C60) is a closed cage molecule (soccer-ball shaped); its defining NEET descriptor is cage-like structure, not lubricant. Assigning cage-structure to graphite, or lubricant to fullerene, is wrong.
When it triggers
Match-the-column or property-identification question lists graphite and fullerene alongside descriptors including cage-like molecule, dry lubricant, sp3 hybridised, or used as reducing agent.
How to avoid
Four-word anchors per allotrope: Diamond = sp3, hard; Graphite = sp2, layers, lubricant; Fullerene = sp2, cage (C60); Coke = amorphous, reducing agent. If asked about cage structure, the answer is always fullerene. If asked about lubricant, the answer is always graphite.
Lesson: p-Block Trends
Category: Similar Terms
MO questions often bundle two distinct properties (bond order AND magnetic character) within a single statement about the same species. Students who have memorised the bond-order series for oxygen species (O2+: 2.5, O2: 2, O2-: 1.5, O22-: 1) confirm the bond-order part and then accept the entire multi-claim statement as correct, missing a separate magnetic-character error in the same statement. Bond order and magnetic character are derived from the same electron configuration but require independent checks: bond order from the difference of bonding vs antibonding occupancies; magnetic character from counting any residual unpaired electrons.
When it triggers
An incorrect-statement MO question includes an option that pairs a correct bond-order value with a wrong magnetic-character claim (or vice versa). Students who verify only the familiar bond-order fact accept the full option.
How to avoid
For every MO option: (1) write out the full electron configuration from scratch; (2) compute bond order = (bonding - antibonding)/2; (3) count unpaired electrons for magnetism. Treat bond order and magnetic character as two separate sub-checks on the same configuration, not as a single fact.
Lesson: p-Block Trends
Category: Similar Terms
O2 having two unpaired electrons (paramagnetic) is a landmark MO result taught explicitly in NEET preparation. When O2+ appears in a question, students recall the O2 lesson and unconsciously carry over its configuration, missing that O2+ has one fewer electron. O2+ (15 electrons) has 1 unpaired electron in pi*2p, making it paramagnetic but with a different unpaired-electron count than O2. The canonical NEET trap is a statement claiming O2+ is diamagnetic, which students fail to challenge because they confuse it with O2.
When it triggers
Statement or incorrect-statement question about the magnetic character of oxygen species (O2, O2+, O2-, O22-). The option claiming O2+ is diamagnetic is the wrong one.
How to avoid
Do NOT reuse O2's configuration for O2+. Re-derive from scratch: O2 has 16 electrons (2 unpaired in pi*2p). O2+ has 15 electrons: remove 1 from pi*2p, leaving 1 unpaired. Still paramagnetic, NOT diamagnetic. O22- has 18 electrons: pi*2p fully filled, 0 unpaired, diamagnetic.
Lesson: p-Block Trends
Category: Inorganic Exception
For diatomics from B2 to N2 (low-Z second-period elements), the pi2p MOs lie LOWER in energy than the sigma2p MO. This is the inversion relative to the O2-type diagram where sigma2p is below pi2p. Students who learn only the O2-type ordering misplace the sigma2p electrons in C2, arriving at wrong electron occupancy for pi2p and a wrong conclusion about C2's magnetic character or number of pi bonds. C2 has 4 electrons in the two degenerate pi2p MOs (all paired, diamagnetic) precisely because pi2p is filled before sigma2p.
When it triggers
Question about C2, B2, or N2 molecular properties (magnetic character, bond order, number of pi bonds, number of electrons in pi orbitals). Any derivation that uses sigma2p < pi2p ordering for these molecules gives a wrong result.
How to avoid
Memorise the crossover: for Z <= 7 (up to N2), energy order is: sigma1s < sigma*1s < sigma2s < sigma*2s < pi2p (x,y degenerate) < sigma2p < pi*2p < sigma*2p. The order flips at O2 (Z = 8): sigma2p moves below pi2p. When a question involves C2, always use the pi-first ordering.
Questions about this unit
- What does p-Block Elements cover for NEET Chemistry?
- 3 lessons: First Element Anomaly, p-Block Groups 13 to 18 and p-Block Trends.
- How often has p-Block Elements come up in NEET past papers?
- Our set of verified past papers has 18 questions from this unit, from NEET 2020, 2021, 2022, 2023, 2024, 2025 and 2026. Each is answered against the official NTA key.
- Is the p-Block Elements material free?
- Yes. All 3 lessons and 24 practice questions are free, with no login needed.