Classification of Elements and Periodicity in Properties JEE Main Chemistry: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense
Published: October 27, 2026
Author: NEET & JEE STUDY HUB
🎯 Curriculum & Exam Overview
This chapter establishes the fundamental periodic trends in atomic properties, which serve as the foundation for chemical reactivity and bonding in inorganic chemistry. It carries high weightage as it links microscopic electronic configurations to macroscopic physical and chemical behavior, frequently appearing in competency-based analysis questions.
🚀 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 mentioning the effective nuclear charge (Zeff), citing the electronic configuration, and explaining the shielding effect in terms of 's > p > d > f' orbital penetration.
1. ⚡ High-Speed Formula Sheet & Concept Limits
1
Modern Periodic Law: Physical and chemical properties of elements are periodic functions of their atomic numbers.
2
Effective Nuclear Charge (Zeff): The net positive charge experienced by valence electrons, calculated as Z - σ (where σ is the shielding constant).
3
Periodic Trends: Ionization enthalpy, electron gain enthalpy, and electronegativity exhibit predictable patterns across periods (left to right) and groups (top to bottom) due to changing atomic radii and Zeff.
⚡ High-Yield Formula & Reaction Sheet
- Zeff = Z - S: Z represents atomic number; S (or σ) represents the screening constant calculated via Slater's rules.
- Electronegativity (Pauling Scale): |χA - χB| = 0.1017 √Δ, where Δ is the resonance energy of the A-B bond relative to A-A and B-B bonds.
⚠️ Common Board Trap Alert
Confusing 'Electron Gain Enthalpy' with 'Electronegativity'; students often mistakenly claim Fluorine has the highest electron gain enthalpy, ignoring that Chlorine actually has a more negative value due to inter-electronic repulsion in the small 2p orbital of Fluorine.
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3. ⚡ NTA JEE Shift Practice & Rank-Booster Problems
Question 1 • MCQ
4 Mark(s)
Match List I (Element) with List II (Electronegativity on Pauling Scale) and select the correct answer using the codes given below:
List I (Element) | List II (Electronegativity on Pauling Scale) |
A. Carbon | 1. 0.8 |
B. Nitrogen | 2. 1.6 |
C. Aluminium | 3. 2.5 |
D. Cesium | 4. 3.0 |
| 5. 4.0 |
(A) – 2; B – 4; C – 5; D – 1
(B) – 3; B – 1; C – 2; D – 4
(C) – 2; B – 1; C – 5; D – 4
(D) – 3; B – 4; C – 2; D – 1
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: d
Explanation: A – 3; B – 4; C – 2; D – 1
Element | Electronegativity |
C | 2.5 |
N | 3.0 |
Al | 1.6 |
Cs | 0.8 |
Question 2 • MCQ
4 Mark(s)
For an element X$_{(g)}$ value of negative electron gain
enthalpy is second highest in periodic table elements.
Select CORRECT statement(s) for element X.
(i) Electronegativity of X is higher as compared to Cl$_{(g)}$ on Pauling’s scale.
(ii) Negative electron gain enthalpy of
${\rm{X}}_{\left( {\rm{g}} \right)}^ +$ is more than negative electron gain enthalpy of
${\rm{O}}_{\left( {\rm{g}} \right)}^ + .$
(iii) Negative electron gain enthalpy of
${X_{\left( {\rm{g}} \right)}}$ is more than negative electron gain enthalpy of
(iv) Ionisation energy of
$X_{\left( {\rm{g}} \right)}^ -$ is less than ionisation energy of
(A) (i) ,(iii) and (iv) only
(B) (ii), (iii) and (iv) only
(C) (i),(ii) and (iv) only
(D) (i),(ii) and (iii) only
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: c
Explanation: (i),(ii) and (iv) only
Atom X will be F atom as it has 2<$^{nd}$ highest electron gain enthalpy.
(a) Electronegativity of F > Cl
(b) O<$^{+}$ will have half filled subshell (2p<$^{3}$)
So, it is difficult to add electron in it.
So, its $\Delta H_{e g} < \Delta H_{e g} \text{ } o f \text{ } F^{+}$
$\left(c\right) \text{ } \Delta H e g \text{ } o f \text{ } F < \Delta H e g \text{ } o f \text{ } C l$
(d) It is much carrier to remove electron form F<$^{–}$ due to
electron –electron repulsion and its small size.
Question 3 • MCQ
🔥 JEE Main (2025)
4 Mark(s)
The maximum covalency of a non-metallic group 15 element ' E ' with weakest E–E bond is :
**[JEE Main 2025, 22nd January, Second Shift]**
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: d
Explanation: N – N < P – P : single (σ) bond strength
Due to L.P.-L.P. replusion
and maximum possible covalency of nitrogen is 4.
Question 4 • MCQ
4 Mark(s)
The total number of elements in the group IB is _x_. What will be the value of
$~\rm \frac{x^2}{2} ?~$
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: a
Explanation: 4.50
Group I B contain Cu, Ag and Au
So value of x is 3. Now the value of $\&\text{nbsp}; \frac{x^{2}}{2} \&\text{nbsp};$will be $\&\text{nbsp}; \frac{3^{2}}{2} \&\text{nbsp};$ i.e. 4.50
Question 5 • MCQ
4 Mark(s)
The order of Zeff for Na, K, Rb, Cs is
(A) Na < K < Cs < Rb
(B) s < Rb < K < Na
(C) Na = K = Rb = Cs
(D) K > Na > Rb > Cs
🚀 View Fast Shortcut Solution & Trap Defense
Correct Answer: c
Explanation: Na = K = Rb = Cs
Value of Zeff for Na, K, Rb, Cs = 2.2 (Calculation by slater’s rule)
Zeff = Z – σ
Where Z = Atomic no.
σ = Shielding/screening constant.
4. ❓ Frequently Asked Questions (Exam Strategy)
❓ How many questions appear from Classification of Elements and Periodicity in Properties 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 Classification of Elements and Periodicity in Properties 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 Classification of Elements and Periodicity in Properties 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 is the first ionization enthalpy of Nitrogen higher than that of Oxygen?
Nitrogen has a stable, half-filled 2p³ subshell configuration which offers extra stability. Consequently, removing an electron from Nitrogen requires more energy than from Oxygen, which has a 2p⁴ configuration.
❓ How does atomic radius change moving down a group?
Atomic radius increases down a group because each subsequent element adds a new principal energy shell (n). This increase in the number of shells outweighs the increase in nuclear charge, leading to a larger size.
❓ Why are Noble Gases assigned zero electronegativity?
Electronegativity is defined as the ability of an atom to attract shared pairs of electrons in a covalent bond. Since noble gases generally do not form stable covalent bonds under standard conditions, they are assigned zero or no value.
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