Periodic Table and Periodicity JEE Notes: Download Now

Periodic Table and Periodicity JEE Notes give you the quickest way to revise the entire chapter, brush up essential trends and master the exceptions that JEE loves to test. Everything that really matters for JEE Main and Advanced is collected here in one scroll-friendly sheet.

Periodic Table and Periodicity JEE Notes: Important Concepts

Modern Periodic Law: “The physical and chemical properties of the elements are periodic functions of their atomic numbers $$Z$$.” This places hydrogen ($$Z=1$$) at the start and oganesson ($$Z=118$$) at the end.

• Periods: Horizontal rows, numbered 1 to 7. A period ends when the next electron must enter a new principal shell.
• Groups: Vertical columns, 1 to 18. Elements in the same group show similar valence-shell configurations and hence similar chemistry.
• Blocks: s-block (groups 1,2), p-block (13-18), d-block (3-12, transition metals), f-block (lanthanides and actinides).
• Long form of the periodic table obeys the sequence in which subshells are filled according to the n + ℓ rule.

Historical Milestones (Short Recall)

1. Dobereiner’s triads → Law of octaves → Mendeleev’s periodic law (properties periodic with atomic mass) → Moseley’s X-ray work that set today’s atomic number basis.
2. Mendeleev left gaps and predicted undiscovered elements (gallium, germanium). Moseley resolved mismatches such as Ar–K, Co–Ni.

The n + ℓ Rule in One Line

The subshell with the lower $$n+\ell$$ value fills first; if equal, the subshell with lower $$n$$ fills first. Eg: 4s ($$n+\ell=4+0=4$$) fills before 3d ($$n+\ell=3+2=5$$).

Position from Configuration (Fast Check)

• Last electron in $$ns^1$$ ⇒ Group 1, Period n.
• Last electron in $$np^5$$ ⇒ Group 17, Period n.
• Last electron in $$nd^{1-10}$$ ⇒ Transition series, Group (3 to 12), Period n+1.
• Lanthanides: $$4f^{1-14}5d^{0-1}6s^2$$, Period 6, placed separately.

Electronic Configuration and Table Architecture

Every periodic trend is traceable to electronic configuration. Practice identifying period, group and block directly from the outer configuration before touching trends-based questions. Solved examples in this section reflect the typical JEE Questions you will face.

Worked Example 1 – Spot the Position

Element $$X$$ has the configuration $$[Ar]\,3d^{6}\,4s^{2}$$. Find its period, group and block.

Solution: Principal quantum number of the outermost shell is 4 ⇒ Period 4. The electrons after argon are in 3d and 4s, so it is a d-block element. For transition metals, group number = d electrons + s electrons = 6 + 2 = 8 ⇒ Group 8. Answer: Period 4, Group 8, d-block.

Worked Example 2 – Configuration from Position

An element lies in Period 5 and Group 17. Write the outer electronic configuration and name it.

Solution: Group 17 ⇒ $$ns^2 np^5$$. Period 5 ⇒ principal quantum number $$n=5$$. Hence outer configuration $$5s^2 5p^5$$. The element is iodine, $$I$$. Answer: $$[Kr]\,4d^{10}5s^{2}5p^{5}$$ (iodine).

Periodic Trends, Exceptions and Diagonal Relationships

The properties below move in predictable ways across periods and down groups, but JEE frequently checks whether you also know the exceptions. After revising each trend, solve at least ten mixed numerical or assertion-reason problems pulled from JEE Mains Previous Papers.

Atomic and Ionic Radii

• Across a period: radius decreases due to increasing nuclear charge without added shielding.
• Down a group: radius increases because a new shell is introduced.
• Key exceptions: Lanthanide contraction causes radii of 5d metals to be nearly equal to 4d. d-block contraction makes Ga < Al.

Numerical Angle – Radius Ratio

If $$r_{Na}=186\,\text{pm}$$ and $$r_{Mg^{2+}}=72\,\text{pm}$$, estimate the ratio $$r_{Na^{+}}/r_{Mg^{2+}}$$ given that both ions are isoelectronic (10 e⁻) and the charge difference dominates.

Approx answer: $$r_{Na^{+}}\approx 95\,\text{pm}$$, so $$95/72 \approx 1.32$$.

Ionisation Enthalpy ($$IE$$)

• Increases left → right; decreases top → bottom.
• Exceptions: Be > B and N > O because half-filled/full-filled subshells add extra stability.
• $$IE_2$$ > $$IE_1$$ for any element; the jump is dramatic once an inner noble-gas core is exposed.

Electron Gain Enthalpy ($$\Delta H_{eg}$$) and Electronegativity

• Most negative for halogens, least negative for noble gases (positive).
• O and F are less negative than S and Cl due to small size and e⁻–e⁻ repulsion.
• Various scales: Pauling ($$χ$$), Mulliken ($$(IE+EA)/2$$), Allred-Rochow (electrostatic model).

Reactivity in s- and p-Block

Metals (Group 1, 2) become more reactive down the group as $$IE$$ drops. Non-metals (Group 17) show the opposite.

Oxidation States and Valency

Highest oxidation state of p-block equals group number − 10. Example: $$Cl$$ exhibits +7. Variable valency in d-block arises from similar energies of (n−1)d and ns electrons.

Diagonal Relationship

Inert Pair Effect & Relativistic Effects (Advanced Corner)

• Stability of +1 for Tl, +2 for Pb, +3 for Bi increases down the group as the ns² pair becomes inert.
• Relativistic contraction of 6s holds Au in +1 and Hg in +2/+1 unusual states.

Worked Example 3 – Trend Exception MCQ

Arrange the following in increasing $$IE_1$$: O, S, Se. Give reason.

Answer: S < Se < O. Across the group, $$IE$$ decreases, but O is anomalously high relative to S because of stronger Zeff with no additional shielding.

Important Formulas and Results at a Glance

Print this table and keep it stuck next to your study desk. A parallel detailed PDF is available inside our JEE Formula Sheets library.

$$χ_P = 0.336\,(IE)^{1/2} + 0.744$$ (Pauling empirical relation)

Energy gap between $$IE_{n}$$ and $$IE_{n+1}$$ pinpoints valence: a huge jump means the core noble gas has been uncovered.

Lanthanide contraction ⇒ $$r_{Hf} \approx r_{Zr}$$, influencing stability of complexes in coordination chemistry.

Solved Numerical – Effective Nuclear Charge

Calculate $$Z_{eff}$$ for the valence electron of the phosphorus atom ($$Z=15$$) using Slater’s rules.

1. Configuration: $$1s^2\,2s^2\,2p^6\,3s^2\,3p^3$$.
2. Same shell contribution: $$(3p^3)$$ other electrons: $$3 \times 0.35 = 1.05$$.
3. n = 2 shell: 8 electrons each with 0.85 ⇒ $$8 \times 0.85 = 6.80$$.
4. n = 1 shell: 2 electrons each with 1.00 ⇒ 2.00.
5. $$σ = 1.05 + 6.80 + 2.00 = 9.85$$.
6. $$Z_{eff}=15 - 9.85 = 5.15$$.

Answer: $$Z_{eff}\approx 5.15$$.

JEE Important Points, Common Mistakes and Quick Revision

• Don’t confuse period number with valence shell electron count. Period is highest principal quantum number present.
• Memorise the three exception pairs: Be–B, Mg–Al, N–O for $$IE$$; O, F vs S, Cl for $$EA$$; Ga radius < Al due to d-block contraction.
• While writing configurations for transition elements, remember 4s empties before 3d removes in oxidation. Fe: [Ar]3d⁶4s²; Fe³⁺: [Ar]3d⁵.
• Questions often ask the reason for the stability of +2 state of Pb. Anchor it to the inert pair effect plus relativistic contraction.
• Revision tip: After the formula table, close the book and write down all trends from memory in under five minutes. This active recall halves silly mistakes.
• Upgrade practice from topic-wise sheets to full-length mocks available inside our JEE Mains Online Coaching dashboard once you score > 80 % accuracy in chapter drills.

30-Second Last-Day Drill

1. Recite the period numbers: 2 has 8 elements, 6 has 32.
2. Whisper “radius D-D, IE I-D, EA N-D” (Decrease-Down etc.).
3. Write Li–Mg and Be–Al on a scrap; recollect why diagonal.
4. Close with noble-gas core jumps for $$IE$$.

Flash-Revision Table – What JEE Asks Most