Periodic Table — Trends, Properties & Memory Tricks
The periodic table is not just a chart you memorise for the sake of it. Every trend in it has a reason — and once you understand the “why”, the “what” falls into place automatically. This is one of the highest-scoring topics in CBSE Class 11, JEE Main, and NEET, with consistent weightage every year.
JEE Main asks 1–2 direct questions on periodic trends every year. NEET has 2–3 questions from this chapter. CBSE boards have a guaranteed 3–5 mark question. This is free marks if you know the patterns.
The Modern Periodic Table — Structure First
Mendeleev’s periodic table arranged elements by atomic mass. The modern periodic table, proposed by Moseley (1913), arranges them by atomic number — and that small shift fixed almost every anomaly Mendeleev’s table had.
The modern periodic table has 7 periods (horizontal rows), 18 groups (vertical columns), and 118 elements currently.
The Four Blocks
Elements are classified into four blocks based on which subshell receives the last (differentiating) electron.
| Block | Subshell filling | Groups | Example |
|---|---|---|---|
| s-block | 1s, 2s… | 1, 2 (+ He) | Na, Mg, Ca |
| p-block | 2p, 3p… | 13–18 | C, N, O, Cl |
| d-block | 3d, 4d… | 3–12 | Fe, Cu, Zn |
| f-block | 4f, 5f | Lanthanides, Actinides | La, U |
Helium (He) is placed in Group 18 (noble gases) even though it is an s-block element. Its placement is based on chemical properties (inert), not electron configuration. This is a popular NEET trap question.
Memory Tricks for s-Block Groups
Group 1 (Alkali metals): H Li Na K Rb Cs Fr Mnemonic: “Hi, Little Naughty Kids Run Crazy Fast”
Group 2 (Alkaline earth metals): Be Mg Ca Sr Ba Ra Mnemonic: “Beta Mange Car Scooter Baap Raazi” — classic Indian classroom trick that never fails.
Periodic Trends — The Core of This Chapter
1. Atomic Radius
Atomic radius is the measure of the size of an atom. We use covalent radius (half the distance between two bonded same-element atoms) as the standard for non-metals, and metallic radius for metals.
Across a period (left to right): atomic radius decreases.
Nuclear charge increases as we move across a period, but electrons are added to the same shell — so shielding stays roughly constant. The stronger nuclear pull contracts the electron cloud inward.
Down a group (top to bottom): atomic radius increases.
Each new period adds a new electron shell. More shells means more shielding and greater distance from the nucleus — the expanded size wins over the increased nuclear charge.
Across period → nuclear charge ↑, same shell, shielding ≈ constant → radius ↓
Down group → new electron shell added, shielding ↑ → radius ↑
Key sequence for Period 3 (decreasing radius): Na > Mg > Al > Si > P > S > Cl > Ar
Key sequence for Group 1 (increasing radius): Li < Na < K < Rb < Cs
Many students write “atomic radius increases across a period” — this is wrong. It decreases across a period. The confusion happens when students mix up the group trend with the period trend.
2. Ionization Energy (IE)
Ionization energy is the minimum energy required to remove the outermost electron from a gaseous atom in its ground state. Higher IE = more tightly held electron = harder to remove.
Across a period: IE generally increases.
Nuclear charge increases and atomic radius decreases across a period. The outer electron is held more tightly by the nucleus, so more energy is needed to remove it.
Down a group: IE decreases.
Atomic radius increases and inner electrons provide more shielding. The outer electron is farther from the nucleus and more screened — it requires less energy to remove.
Critical IE Anomalies — These Come in Every Exam
IE(Be) > IE(B): Beryllium has configuration [He] 2s² — a fully filled 2s subshell with extra stability. Boron’s differentiating electron is in 2p¹, which is higher in energy and easier to remove. Despite having more nuclear charge, B has lower IE than Be.
IE(N) > IE(O): Nitrogen has [He] 2s² 2p³ — a half-filled 2p subshell with extra stability (Hund’s rule symmetry). Oxygen has [He] 2s² 2p⁴ — one pair of electrons is forced into the same orbital, causing electron–electron repulsion. This repulsion makes one electron easier to remove from oxygen.
The anomalies IE(Be) > IE(B) and IE(N) > IE(O) appear in JEE Main and NEET almost every alternate year. Same logic applies to the P–S pair: IE(P) > IE(S). If you can explain the reason (fully-filled / half-filled subshell stability), you’ll never get these wrong.
Second Ionization Energy (IE₂): Always greater than IE₁ — you’re removing an electron from a positively charged ion. The sharp jump in IE₂ of Na is famous: Na⁺ has the noble gas configuration of Ne (2,8), making the second removal extraordinarily difficult. This pattern helps identify group membership from IE data.
3. Electron Affinity (EA)
Electron affinity is the energy change when a gaseous atom gains one electron to form an anion. A more negative EA means the atom more readily accepts an electron.
Across a period: EA generally becomes more negative (increases in magnitude). Down a group: EA generally becomes less negative (decreases in magnitude).
Key EA Anomalies
EA(F) less negative than EA(Cl): This is the most-tested EA fact in exams. Fluorine should theoretically have the highest EA (most electronegative element), but chlorine’s EA is more negative. Fluorine’s 2p orbitals are extremely compact — adding an electron into the already dense 2p subshell creates significant electron–electron repulsion. Chlorine’s 3p orbitals are larger and can accommodate the extra electron with less repulsion.
EA of noble gases ≈ 0: Fully filled valence shells — no room or desire for another electron.
EA of alkaline earth metals ≈ 0 or positive: Their s-subshells are full; the next electron must go into a higher-energy p subshell — energetically unfavourable.
4. Electronegativity (EN)
Electronegativity is the ability of an atom in a covalent bond to attract shared electrons toward itself. The Pauling scale is standard.
Across a period: EN increases. Down a group: EN decreases.
Fluorine (EN = 4.0) has the highest electronegativity of all elements. Caesium (or Francium) has the lowest.
F = 4.0 | O = 3.5 | Cl = 3.0 | N = 3.0 | Br = 2.8 | C = 2.5 | H = 2.1 | Na = 0.9 | Cs = 0.7
Bond character from EN difference:
- EN difference > 1.7 → Ionic bond (e.g., NaCl: 3.0 − 0.9 = 2.1)
- EN difference 0.4–1.7 → Polar covalent (e.g., HCl: 3.0 − 2.1 = 0.9)
- EN difference < 0.4 → Non-polar covalent (e.g., H₂: 0)
5. Metallic and Non-Metallic Character
Metallic character (tendency to lose electrons) decreases across a period and increases down a group — directly mirroring IE trends.
Non-metallic character (tendency to gain electrons) increases across a period and decreases down a group — mirroring EA and EN trends.
The staircase line on the periodic table separates metals (left) from non-metals (right). Metalloids (Si, Ge, As, Sb, Te) lie along this boundary and have intermediate properties — crucial for NEET.
6. Valency and Oxidation States
For main group elements:
- Group 1: Valency = 1
- Group 2: Valency = 2
- Groups 13–17: Common valency = Group number − 10 (or 18 − Group number, the “alternative” valency)
For transition metals, variable valency is standard because both 4s and 3d electrons can participate in bonding.
The highest oxidation state of a main group element generally equals its group number for Groups 1–7. Mn (Group 7) shows +7 in KMnO₄. Beyond Group 7, very high oxidation states become unstable due to high ionization energies. For halogens, the maximum oxidation state is +7 (in Cl₂O₇, HClO₄).
The d-Block — Transition Metals (Groups 3–12)
Transition metals have incompletely filled d-subshells in their elemental state or in at least one stable oxidation state.
Key properties:
- Variable oxidation states: Fe shows +2 and +3; Mn shows +2, +3, +4, +6, +7
- Coloured ions: Due to d–d electron transitions (e.g., Cu²⁺ is blue, Cr³⁺ is green, MnO₄⁻ is purple)
- Catalytic activity: Fe in Haber process, V₂O₅ in Contact process, MnO₂ in thermal decomposition of KClO₃
- Complex formation: Empty d orbitals accept lone pairs from ligands
Atomic radius across d-block: Does not decrease smoothly. As d electrons are added, they shield each other reasonably well — the radius stays nearly constant from Sc to Cu, then increases slightly at Zn.
Lanthanide Contraction
In lanthanides (Ce to Lu), electrons fill the 4f subshell. The 4f orbitals have complex angular distributions and shield outer electrons very poorly from nuclear charge. So as nuclear charge increases across the lanthanides, the electron cloud contracts progressively.
Consequence: Elements immediately after the lanthanides (Period 6 d-block: Hf, Ta, W, Re…) have almost the same atomic radii as their Period 5 counterparts (Zr, Nb, Mo…). Zr and Hf are virtually identical in size (both ~160 pm) — making them the hardest pair of elements to separate in industry.
Full Trend Comparison Table
| Property | Across Period (→) | Down Group (↓) |
|---|---|---|
| Atomic radius | Decreases | Increases |
| Ionic radius (isoelectronic series) | Decreases with Z | — |
| Ionization energy | Increases (with anomalies) | Decreases |
| Electron affinity | Increases (with anomalies) | Decreases |
| Electronegativity | Increases | Decreases |
| Metallic character | Decreases | Increases |
| Non-metallic character | Increases | Decreases |
5 Common Mistakes Students Make
Mistake 1: Confusing period trend with group trend for atomic radius. Atomic radius decreases going across a period (left to right). It increases going down a group. Students frequently reverse one of these.
Mistake 2: Thinking fluorine has the highest electron affinity. Chlorine has a more negative EA than fluorine. F’s compact 2p orbitals cause electron–electron repulsion when a new electron tries to enter. This is one of the most reliable trick questions in NEET MCQs.
Mistake 3: Ignoring IE anomalies (Be > B, N > O, P > S). These anomalies arise from extra stability of fully-filled or half-filled subshells. Missing these means losing marks on questions that appear almost every year.
Mistake 4: Applying s/p block trends directly to d-block. Transition metal radii don’t decrease smoothly across a period. Their IE and EN trends are also less pronounced. Don’t extrapolate main-group rules blindly to transition metals.
Mistake 5: Treating electronegativity and electron affinity as the same thing. EN is about shared electrons in a bond (relative concept). EA is about an isolated gaseous atom gaining an electron (measurable energy). They often trend the same way, but the F vs Cl anomaly shows they can diverge.
Real-World Examples
Example 1: Why Your Kitchen Catches Fire (But the Lab Doesn’t)
If you’ve ever seen a chemistry teacher drop sodium into a beaker of water, you know the drama — violent fizzing, sometimes a small flame. Magnesium, sitting just to the right in Period 3, does almost nothing in cold water. The reason is atomic size and ionisation energy: sodium’s outermost electron sits far from the nucleus with a low ionisation enthalpy (~496 kJ/mol), so it surrenders that electron to water almost instantly. Magnesium has a higher nuclear charge pulling its two valence electrons much closer — it reacts with steam, not cold water.
Connect to the syllabus: This directly illustrates the periodic trend that ionisation enthalpy increases across a period, making Group 1 metals far more reactive than Group 2 for the same row.
Example 2: The Secret Ingredient in Colgate and Patanjali Dant Kanti
Next time you pick up a toothpaste tube — Colgate, Pepsodent, or Patanjali Dant Kanti — check the back label for “sodium fluoride” or “monofluorophosphate.” Fluorine is chosen deliberately because it is the most electronegative element on the periodic table ( on the Pauling scale). That extreme pull on electrons means the F⁻ ion bonds tightly with calcium in tooth enamel, forming fluorapatite, which is harder and more acid-resistant than natural hydroxyapatite. No other halogen can do this job as well — chlorine is weaker, and anything heavier is toxic or too large.
Connect to the syllabus: Electronegativity reaches its maximum at fluorine because it has the smallest atomic radius in its group combined with the highest effective nuclear charge — a classic exam question on periodic trends.
Example 3: Why Your Nerve Cell Prefers Potassium Over Sodium
Both sodium and potassium are Group 1 elements, yet your neurons pump potassium inside the cell and keep sodium outside — a precise biochemical choice. Potassium’s larger ionic radius ( ≈ 138 pm vs ≈ 102 pm) means it fits the specific protein channel on the nerve membrane while sodium does not, generating the ~−70 mV resting potential that fires every thought, heartbeat, and reflex. ISRO engineers even account for this in designing life-support systems for astronauts, where electrolyte balance is critical in microgravity. Same group, same charge, completely different biological role — purely because of size.
Connect to the syllabus: Ionic radius increases down a group as new shells are added; vs size difference is a favourite NEET application question on periodic trends.
Practice Questions
Q1. Which has the larger atomic radius — Na or Mg? Explain why.
Na has a larger atomic radius than Mg. Both are in Period 3, but Mg has one more proton (Z = 12 vs Z = 11). The increased nuclear charge in Mg pulls the 3s electron cloud closer. Since both have their outermost electrons in the same n=3 shell, the shielding is similar. Net result: the greater nuclear pull in Mg wins, making Mg smaller. Na (186 pm) > Mg (160 pm).
Q2. Why does the ionization energy of oxygen fall below that of nitrogen?
Nitrogen has electron configuration [He] 2s² 2p³ — a half-filled 2p subshell, which carries extra stability due to symmetric electron distribution (all three 2p electrons have parallel spins). Oxygen has [He] 2s² 2p⁴ — one 2p orbital has a pair of electrons. That pair repels each other, making one electron easier to remove than you’d expect. So IE₁(N) = 1402 kJ/mol > IE₁(O) = 1314 kJ/mol despite oxygen having more nuclear charge.
Q3. Arrange in decreasing order of atomic radius: Na, K, Cl, Mg.
K > Na > Mg > Cl
K is in Period 4, so it has one more electron shell than all three Period 3 elements — it is the largest. Among Period 3 elements, atomic radius decreases across the period: Na (Group 1) > Mg (Group 2) > Cl (Group 17). So the final order is K > Na > Mg > Cl.
Q4. The second ionization energy of Na is about 9 times its first. Why?
Na’s first ionization removes the lone 3s¹ electron, giving Na⁺ with the stable noble gas configuration of Ne (1s² 2s² 2p⁶). The second ionization must remove an electron from this stable, compact Ne-like core — an n=2 electron that is much closer to the nucleus and far more strongly held. IE₁(Na) ≈ 496 kJ/mol; IE₂(Na) ≈ 4562 kJ/mol. The massive jump signals that the second electron comes from a complete inner shell.
Q5. Why is the ionic radius of Cl⁻ greater than the atomic radius of Cl?
Neutral Cl has 17 protons and 17 electrons. Cl⁻ has 17 protons but 18 electrons. The extra electron increases electron–electron repulsion in the 3p subshell and reduces the effective nuclear charge per electron (Z_eff per electron drops). The electron cloud expands. Cl⁻ radius ≈ 181 pm vs Cl atomic radius ≈ 99 pm.
General rule: Anions are always larger than their parent atoms. Cations are always smaller than their parent atoms.
Q6. Why is EA(Cl) more negative than EA(F) even though F is more electronegative?
Fluorine’s 2p orbitals are extremely small and already electron-dense with 5 electrons. When an additional electron tries to enter this compact space, strong electron–electron repulsion resists it — reducing the energy released. Chlorine’s 3p orbitals are larger; the incoming electron can settle in with less repulsion, releasing more energy. EA(Cl) = −349 kJ/mol vs EA(F) = −328 kJ/mol.
Electronegativity (EN) measures tendency in a bond — that’s a different context where F wins because of its smaller size and higher Z_eff. EA measures energy for an isolated atom gaining an electron — and here Cl wins due to less repulsion.
Q7. What is lanthanide contraction? Give one important consequence.
Lanthanide contraction is the gradual, steady decrease in atomic and ionic radii across the 14 lanthanide elements (Ce to Lu). As nuclear charge increases across the lanthanide series, electrons are added to the 4f subshell. The 4f orbitals are diffuse and poor at shielding outer electrons from the nucleus — so the effective nuclear charge felt by outer electrons increases steadily, contracting the atom.
Consequence: The Period 6 d-block elements (Hf, Ta, W, Re, Os, Ir, Pt, Au) have almost the same atomic radii as their Period 5 counterparts (Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag). For example, Zr ≈ 160 pm and Hf ≈ 159 pm. Their chemical properties are so similar that Zr and Hf are among the most difficult pairs of elements to separate industrially.
Q8. Which has higher first ionization energy — P or S? Explain.
P has higher first IE than S. Phosphorus has [Ne] 3s² 3p³ — half-filled 3p subshell with extra stability. Sulfur has [Ne] 3s² 3p⁴ — one paired electron in the 3p orbitals introduces repulsion that makes removal easier. IE₁(P) = 1011 kJ/mol > IE₁(S) = 999 kJ/mol.
This is the Period 3 equivalent of the N–O anomaly. Both arise from the same principle: half-filled subshell stability in N and P makes them resist ionization more than the next element in the period.
Frequently Asked Questions
How many elements are in Period 3?
Eight: Na, Mg, Al, Si, P, S, Cl, Ar. Period 3 contains only s-block and p-block elements — the d-block starts in Period 4 with Sc (Z = 21).
Why does Period 6 have 32 elements?
Period 6 fills 6s (2 electrons), 4f (14 electrons), 5d (10 electrons), and 6p (6 electrons) subshells. 2 + 14 + 10 + 6 = 32 elements. The f-block makes its first appearance in Period 6 (lanthanides, Ce–Lu) and Period 7 (actinides, Th–Lr).
What is effective nuclear charge (Z_eff)?
Z_eff = Z − σ, where σ is the shielding constant from inner electrons. It represents the net nuclear charge felt by an outer electron after accounting for repulsion by inner electrons. Slater’s rules give an approximate σ. Higher Z_eff → smaller radius, higher IE, higher EN. This is the single concept that explains almost all periodic trends.
Why are Group 2 elements harder to ionise than Group 1 in the same period?
Group 2 elements have one more proton than Group 1 elements in the same period, with both sets of outermost electrons in the same s-subshell. For example, Mg (Z = 12) vs Na (Z = 11) — both have 3s outer electrons, but Mg’s nucleus pulls them more strongly. IE(Mg) = 738 kJ/mol > IE(Na) = 496 kJ/mol.
Is the periodic table memorisation-heavy for JEE?
Not if you understand trends. JEE doesn’t ask you to recite atomic radii in picometres — it asks you to compare elements or explain anomalies. Know the direction of each trend, know the four key anomalies (Be > B, N > O, P > S for IE; F < Cl for EA), and you can handle any periodic table question without rote learning.
What is the difference between ionization energy and electron affinity?
IE is the energy you must supply to remove an electron from a neutral gaseous atom (always endothermic, always positive). EA is the energy released when a neutral gaseous atom gains an electron (usually exothermic). IE involves losing electrons; EA involves gaining electrons. Both generally increase across a period and decrease down a group — but their anomalies differ. The most important difference in exam context: for IE, N > O; for EA, F < Cl.