🎯 Curriculum & Exam Overview
The p-block elements chapter encompasses the study of groups 13 to 18, focusing on trends in physical and chemical properties driven by electronic configuration and the inert pair effect. It is a high-yield topic because it bridges inorganic periodic trends with direct application in industrial chemical processes and coordination chemistry.
🚀 NTA JEE Main 2027 Countdown
January Session 1 Sprint
JEE Main 2027 Engineering Focus: High-Yield Problem Solving & Negative Marking Defense
- Session 1 Registration Timeline: Registration for JEE Main 2027 Session 1 opens in November on jeemain.nta.nic.in. January exam performance sets your prime percentile baseline.
- Negative Marking Safeguard (-1): Eliminate blind guesswork. Prioritize question selection, dimensional checks, and options elimination to maximize net marks.
- Multi-Concept Problem Linkages: JEE questions integrate multiple domains (e.g. Calculus with Mechanics, or Thermodynamics with Equilibrium) into single composite problems.
- Timed Shift PYQ Practice: Solve at least 15-20 recent shift questions (2021-2026) under a strict 2.5-minute timer per numerical/MCQ.
⚡ JEE Main Exam Blueprint & Problem-Solving Guidelines
📊 Expected Weightage
Expected 4 to 8 marks in CBSE board exam (typically 1-2 MCQs + 1 Short/Long Answer or Case Study).
✍️ Examiner & Scoring Insight
Examiners award partial step marks for formula definition, intermediate substitution, correct SI units, and neat labeled diagrams.
1. ⚡ High-Speed Formula Sheet & Concept Limits
1
Inert Pair Effect: The reluctance of ns² electrons to participate in bonding as we move down the group, leading to the stability of lower oxidation states in heavier elements.
2
Bonding Trends: The shift from ionic/metallic character to covalent character, and the significant impact of d-orbital contraction (lanthanoid contraction) on atomic size.
3
Lewis Acid-Base Character: The tendency of group 13 elements (e.g., BCl₃) to act as electron-deficient Lewis acids, contrasting with the inertness of N₂ due to strong N≡N triple bond energy.
⚡ High-Yield Formula & Reaction Sheet
- 4NH₃ + 5O₂ → 4NO + 6H₂O: The catalytic oxidation of ammonia (Ostwald’s Process) representing the key industrial synthesis of Nitric Acid.
- B₂H₆ + 3O₂ → B₂O₃ + 3H₂O: The combustion of diborane, showcasing the high enthalpy of formation of boron oxides and the electron-deficient nature of bridging hydrides.
⚠️ Common Board Trap Alert
Confusing the 'Inert Pair Effect' with 'Lanthanoid Contraction'; the former dictates stability of oxidation states (e.g., Pb²⁺ > Pb⁴⁺), while the latter explains why elements of the 4d and 5d series have nearly identical atomic radii.
📺 High-yield video lesson for this chapter will premiere on @qbnexa_official.
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3. ⚡ NTA JEE Shift Practice & Rank-Booster Problems
Question 1 • MCQ
4 Mark(s)
In which of the following compound hybridization of bridging atom is different from hybridisation of central atom:
(A) Al$_{2}$(NH$_{3}$)$_{6}$
(B) I$_{2}$Cl$_{6}$
(C) Solid BeCl$_{2}$
(D) l$_{2}$(OH)$_{6}$
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: b
Explanation: I$_{2}$Cl$_{6}$
Question 2 • MCQ
🔥 JEE Main (2026)
4 Mark(s)
Regarding the hydrides of group 15 elements EH$_{3}$ (E = N, P, As, Sb), select the correct statement from the following:
(a) The stability of hydrides decreases down the group.
(b) The basicity of hydrides decreases down the group.
(c) The reducing character increases down the group.
(d) The boiling point increases down the group.
Choose the **_correct_** answer from the options given below:
**[JEE Main 2026, 28th January, First Shift]**
(A) (a), (b) & (c) only
(B) (a) & (d) only
(C) (a), (b), (c) & (d)
(D) (b) & (c) only
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: a
Explanation: (a), (b) & (c) only Stability : NH$_{3}$ > PH$_{3}$ > AsH$_{3}$ > SbH$_{3}$ > BiH$_{3}$ Basicity : NH$_{3}$ > PH$_{3}$ > AsH$_{3}$ > SbH$_{3}$ > BiH$_{3}$ Reducing character : NH$_{3}$ < PH$_{3}$ < AsH$_{3}$ < SbH$_{3}$ < BiH$_{3}$ Boiling point : PH$_{3}$ < AsH$_{3}$ < NH$_{3}$ < SbH$_{3}$ < BiH$_{3}$
Question 3 • MCQ
4 Mark(s)
Among the following which is the strongest oxidising agent?
(A) r$_{2}$
(B) I$_{2}$
(C) l$_{2}$
(D) F$_{2}$
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: d
Explanation: F$_{2}$ Values of standard reduction potential of halogens are positive and decreases from fluorine to iodine. Therefore halogens act as strong oxidising agent and their oxidising power decreases from fluorine to iodine.
Question 4 • MCQ
4 Mark(s)
When CO$_{2}$ is bubbled through a solution of barium peroxide in water
(A) O$_{2}$ is released
(B) arbonic acid is formed
(C) H$_{2}$O$_{2}$ is formed
(D) No reaction occurs
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: c
Explanation: H$_{2}$O$_{2}$ is formed BaO$_{2}$ (s) + CO$_{2}$ (g) + H$_{2}$O (l) → BaCO$_{3}$ (s) + H$_{2}$O$_{2}$ (aq)
Question 5 • Assertion-Reason
🔥 JEE Main (2024)
1 Mark(s)
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : PH$_{3}$ has lower boiling point than NH$_{3}$.
Reason (R) : In liquid state NH$_{3}$ molecules are associated through van der Waal’s forces, but PH$_{3}$ molecules are associated through hydrogen bonding.
In the light of above statements, choose the most appropriate answer from the options given below.
**[JEE Main 2024, 01st February, First Shift]**
(A) oth (A) and (R) are correct and (R) is the correct explanation of (A)
(B) oth (A) and (R) are correct but (R) is not the correct explanation of (A)
(C) (A) is correct but (R) is not correct
(D) (A) is not correct but (R) is correct
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: c
Explanation: (A) is correct but (R) is not correct $\left(P H\right)_{3}$ molecules. They are unable to form hydrogen bonds among themselves.
4. ❓ Frequently Asked Questions (Exam Strategy)
❓ How many questions appear from The p-Block Elements JEE Main Chemistry: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense in JEE Main Session 1?
Typically 1 to 2 questions (4 to 8 marks) appear from The p-Block Elements JEE Main Chemistry: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense, frequently featuring combined multi-topic concepts.
❓ What is the best approach to avoid negative marks in The p-Block Elements JEE Main Chemistry: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense?
Master standard shortcut formulas, verify boundary states (e.g. t=0, infinity), and skip uncertain numerical value questions.
❓ Why does nitrogen exhibit catenation less than phosphorus?
Nitrogen-nitrogen single bonds are relatively weak due to high inter-electronic repulsion of non-bonding electrons. In contrast, phosphorus forms stronger P-P single bonds, facilitating better catenation.
❓ Is diborane a Lewis acid or base?
Diborane is a Lewis acid because it is an electron-deficient compound with only 12 valence electrons. It readily accepts electron pairs from Lewis bases like ammonia to complete its octet.
❓ Why is Bi(V) a stronger oxidizing agent than Sb(V)?
Due to the inert pair effect, the +3 oxidation state is more stable than +5 for Bismuth. Thus, Bi(V) strongly tends to reduce to Bi(III), making it a powerful oxidizing agent.
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