JEE Main (Engineering) · Chemistry (JEE & NEET)
Classification of Elements and Periodicity
Modern periodic law, periodic table structure and trends in atomic radius, ionisation energy, electron affinity and electronegativity.
Eight concepts on the long-form table: modern periodic law, s/p/d/f blocks from outer configuration, effective nuclear charge, radius, ionisation and electron-gain enthalpy dips, electronegativity/metallic character, and diagonal relationships.
- JEE Main (Engineering)
- Easy level
- 8 concepts
- 5 practice questions
1Modern periodic law and the long-form table
Properties of elements are a periodic function of atomic number Z, not atomic mass. The long-form periodic table arranges elements in 7 periods and 18 groups by increasing Z, so that elements with the same outer-shell configuration sit in the same group.
Blocks mark which subshell is filling: s (groups 1–2), p (13–18), d (3–12) and f (lanthanoids and actinoids). Period lengths follow the subshell capacities: 2, 8, 8, 18, 18, 32.
Figure. Block layout of the long-form table, schematic and not to cell count. s at left, d in the middle (from period 4), p at right, f as a detached strip. Periods run top to bottom; groups share a column and a valence configuration.
Placing an element
- Read ZAtomic number fixes the position; mass does not. Argon before potassium is the classic mass-inversion that Mendeleev's table got wrong and the modern law gets right.
- Find the periodThe period equals the principal quantum number of the valence shell.
- Find the block and groupThe subshell being filled is the block; the group follows from the valence configuration.
Period length from subshells
Why does period 4 contain 18 elements while period 3 contains only 8?
- Period 3 fills 3s then 3p2 + 6 = 8 elements
- Period 4 fills 4s, 3d, then 4p2 + 10 + 6 = 18 elements
- 3d opens only after 4sd-block appears from period 4
- Check against the formula sheetelements per period: 2, 8, 8, 18, 18, 32
Pro tip. Period 6 is 32 long because 4f (14) joins 6s, 5d and 6p. Counting subshell electrons is faster than memorising the sequence of numbers.
The modern periodic law orders elements by
- Atomic mass
- Atomic number
- Mass number
Properties are a periodic function of Z. Atomic mass is close but fails at Ar/K, Co/Ni and Te/I.
2s-, p-, d- and f-block elements
NCERT types the long-form table by which subshell is filling. The s-block is Groups 1 and 2 — outermost configurations ns^1 and ns^2 (alkali and alkaline-earth metals). The p-block is Groups 13–18, where the np subshell fills from np^1 to np^6; it holds metals, metalloids and non-metals, and noble gases close each period with ns^2 np^6.
The d-block (Groups 3–12) is the transition metals: the (n-1)d subshell fills while ns is usually occupied. The f-block sits as the lanthanoids and actinoids (inner transition elements) below the main table, with (n-2)f filling. Locating an element in a block is an outermost-configuration question — the same configuration also predicts the period (highest n) and the group once you know which block you are in.
Figure. Schematic long-form layout only — block widths are not to scale with period lengths. Use the filling-subshell table for exact group numbers.
Assign the block
- Read the outer configEnding in ns¹/ns² → s-block. Ending in np¹–np⁶ → p-block. Filling (n−1)d → d-block. Filling (n−2)f → f-block.
- Period from nHighest principal quantum number in the ground-state configuration is the period.
- Do not confuse with metalsp-Block is not “all non-metals”; it mixes metals, metalloids and non-metals. s-Block metals are a proper subset of metals.
| Block | Groups / rows | Outer filling |
|---|---|---|
| s | 1–2 | ns¹–ns² |
| p | 13–18 | np¹–np⁶ |
| d | 3–12 | (n−1)d with ns |
| f | Lanthanoids / actinoids | (n−2)f |
Place Z = 31
An element has ground-state outer configuration 4s² 4p¹. Name its block, period and group family.
- Filling subshell4p → p-block
- Highest n4 → period 4
- np¹ in p-blockGroup 13 family (like B, Al, Ga)
Pro tip. Count p electrons for the p-block group: np¹ is Group 13, np² Group 14, …, np⁶ Group 18.
An element with outer configuration 3d⁶ 4s² belongs to the
- s-block, because 4s² is written last in some notations
- d-block — the (n−1)d subshell is the one being filled across the series
- f-block, because six d electrons imply inner transition behaviour
Transition (d-block) elements are those for which (n−1)d is filling. The 4s² ending does not move Fe-group species into the s-block.
3Effective nuclear charge and Slater screening
A valence electron does not feel the full nuclear charge Z. Inner electrons screen it, leaving an effective nuclear charge Z_eff = Z − σ, where σ is Slater's screening constant. Rising Z_eff across a period is why atoms shrink and become harder to ionise even though a new proton is added each time.
Down a group a new shell is added; the increase in distance outweighs the modest rise in Z_eff, so atoms grow and ionisation energies fall.
Figure. Z_eff = Z − σ. Both Li and Be use n=2 valence, but Be's higher Z and incomplete screening lift Z_eff from 1.30 to 1.95 — the quantitative reason radius falls and IE rises across the period.
Estimating Z_eff
- Write the configuration in Slater groups(1s), (2s2p), (3s3p), (3d), … — s and p of the same n share a group.
- Apply the contributions to σOther electrons in the same group contribute 0.35 each (0.30 in 1s); n−1 shell contributes 0.85 each; deeper shells contribute 1.00 each.
- SubtractZ_eff = Z − σ. Compare across a period: Z rises by 1 each step while σ rises by less than 1 for a valence electron.
Z_eff for the 2s electron of Li and Be
Estimate Z_eff for a 2s electron in Li (Z = 3) and Be (Z = 4) using Slater's rules.
- Li: 1s² 2s¹; σ = 2 × 0.85 (the 1s pair)σ = 1.70
- Z_eff(Li) = 3 − 1.701.30
- Be: 1s² 2s²; σ = 0.35 (other 2s) + 2 × 0.85σ = 2.05
- Z_eff(Be) = 4 − 2.051.95
Pro tip. Z_eff rises from Li to Be even though screening also rises — that is the across-period story in one pair of numbers.
Across period 2, atomic radius falls mainly because
- Mass increases
- Z_eff on the valence shell rises
- A new shell is added each step
Each added proton raises Z by 1 while the added electron screens by less than 1, so Z_eff rises and the valence shell is pulled in. A new shell is the down-group story, not the across-period one.
4Atomic and ionic radius trends
Atomic radius decreases across a period as Z_eff rises, and increases down a group as a new shell is added. Cations are smaller than their parent atoms (electrons removed, Z_eff on the remainder rises); anions are larger (electrons added, repulsion up, Z_eff down).
For isoelectronic species — same electron count — radius falls as nuclear charge rises. That single rule orders N³⁻, O²⁻, F⁻, Na⁺ and Mg²⁺ without memorising five numbers.
Figure. Covalent radius against Z across period 2. The fall is steep at first (Li → B) and then flattens; Ne sits near F. This plot replaces the undrawable 'atoms as circles of graded size' figure — same trend, checkable coordinates.
Ordering sizes
- Same period?Higher Z → smaller atom.
- Same group?Higher period → larger atom.
- Isoelectronic set?Same electron count: highest Z is smallest.
| Ion | Z | Electrons | Radius trend |
|---|---|---|---|
| N³⁻ | 7 | 10 | largest |
| O²⁻ | 8 | 10 | ↓ |
| F⁻ | 9 | 10 | ↓ |
| Na⁺ | 11 | 10 | ↓ |
| Mg²⁺ | 12 | 10 | smallest |
Ordering isoelectronic ions
Arrange N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ in increasing order of ionic radius.
- Electron countall 10e⁻ (isoelectronic)
- Nuclear charges ZN 7, O 8, F 9, Na 11, Mg 12
- Higher Z → smaller radiusMg²⁺ smallest, N³⁻ largest
- Increasing radiusMg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Pro tip. For isoelectronic species the ion with the highest Z is always the smallest — no radius table required.
Which is largest: Na⁺, Ne, F⁻? (all have 10 electrons)
- Na⁺
- Ne
- F⁻
Same electron count; F has Z = 9, Ne Z = 10, Na Z = 11. Lowest Z wins on size, so F⁻ > Ne > Na⁺.
5Ionisation enthalpy and the Be > B, N > O dips
First ionisation enthalpy generally rises across a period and falls down a group, tracking Z_eff and radius. Successive ionisations always climb: IE₁ < IE₂ < IE₃, because each electron is pulled from an increasingly positive ion.
Two dips break the across-period rise in period 2: Be > B and N > O. Boron starts the 2p subshell (less tightly held than Be's filled 2s²); oxygen has a paired 2p electron whose repulsion makes removal easier than from nitrogen's half-filled 2p³.
Figure. IE₁ against Z for period 2, using standard values (Li 520 through Ne 2081 kJ mol⁻¹). The general rise is broken by two clear dips: Be (899) above B (801), and N (1402) above O (1314). Those four points are marked.
Reading an IE plot
- General trendUp across a period, down a group.
- Spot the dipsGroup 2 → 13 (filled s² vs starting p) and group 15 → 16 (half-filled p³ vs paired p⁴).
- Successive IE jumpsA sharp jump marks removal from a noble-gas core — that is how you read valence from IE data.
Anomalous ionisation energy of nitrogen
Why is the first ionisation energy of nitrogen higher than that of oxygen?
- N configuration[He] 2s² 2p³ (half-filled 2p)
- O configuration[He] 2s² 2p⁴ (one 2p pair)
- Removing one e⁻ from Orelieves pair repulsion → easier
- IE orderIE(N) > IE(O), against the general across-period rise
Pro tip. Half-filled and fully filled subshells give extra stability — the same logic gives Be > B (filled 2s² vs starting 2p).
IE₁ of Be is greater than IE₁ of B mainly because
- Be has a larger atomic radius
- Be has a filled 2s² subshell; B starts 2p
- B has a higher nuclear charge
Be's electron is removed from a stable filled 2s²; B's is the first 2p electron, farther from the nucleus on average and less tightly held despite B's higher Z.
6Electron gain enthalpy: why Cl beats F
Electron gain enthalpy is the enthalpy change when a gaseous atom gains an electron. Halogens have the most negative values — adding an electron completes a noble-gas configuration. The group order is Cl > F > Br > I in magnitude of the negative enthalpy.
Fluorine is less negative than chlorine because the added electron enters a very compact 2p orbital and suffers strong electron–electron repulsion. That size/repulsion effect outweighs fluorine's higher electronegativity for this one quantity.
Figure. Magnitudes of the (negative) electron gain enthalpies of the halogens in kJ mol⁻¹ — taller means more negative Δ_egH. Cl leads; F sits above Br only narrowly. Values are |Δ_egH|, not signed enthalpies.
Comparing halogen EGE
- Default group trendDown the group, the added electron is farther from the nucleus → less negative EGE.
- The F anomalyF is smaller than the down-group trend predicts for attraction, but repulsion in 2p dominates → less negative than Cl.
- Do not confuse with electronegativityF is still the most electronegative element; EGE and χ answer different questions.
The most negative electron gain enthalpy among the halogens belongs to
- F
- Cl
- Br
Cl is more negative than F despite F's higher electronegativity, because the added electron in F meets severe 2p–2p repulsion.
7Electronegativity and metallic character
Electronegativity — the tendency of an atom in a bond to attract shared electrons — rises across a period and falls down a group. Fluorine is the most electronegative element (Pauling 4.0). Metallic character runs the opposite way: strongest at the bottom-left (Cs, Fr) and weakest at the top-right.
Across a period, non-metallic and oxidising character rise with electronegativity; down a group, metallic and reducing character rise. That is the same Z_eff/radius story read from the bonding end rather than the ionisation end.
Figure. Electronegativity climbs toward fluorine at the top-right; metallic character climbs toward caesium at the bottom-left. Same table, opposite arrows — do not read one trend as the other.
Two opposite trends
- ElectronegativityUp across, down a group. F = 4.0 on Pauling's scale.
- Metallic characterDown across (to the left), up a group. Cs is more metallic than Li; Na more than Cl.
- Oxidising vs reducingTop-right elements oxidise; bottom-left elements reduce — the E° story from electrochemistry in periodic clothing.
Comparing two pairs
Which is more electronegative, N or P? Which is more metallic, Na or Mg?
- N and P: same group, N above Pχ(N) > χ(P)
- Na and Mg: same period, Na left of MgNa more metallic
- Check against F = 4.0 extremeboth answers point toward top-right for χ, bottom-left for metallic
- Pauling values (reference)N 3.0 > P 2.1; Na clearly more metallic than Mg
Pro tip. When two trends fight (e.g. diagonal neighbours), charge-to-size ratio usually decides — that is the next concept.
Across period 3, metallic character
- Rises
- Falls
- Stays roughly constant
Na → Ar moves from a strong metal toward a noble gas. Metallic character falls as electronegativity and non-metallic character rise.
8Diagonal relationships
Li–Mg, Be–Al and B–Si form diagonal pairs with similar chemistry. The similarity comes from nearly equal charge-to-size ratios: moving right raises charge density, moving down lowers it, and one step of each roughly cancels.
Consequences show up in compounds more than in the elemental trends: Li and Mg form nitrides, Be and Al oxides are amphoteric, and boron and silicon hydrides are electron-deficient covalent networks rather than saline hydrides.
Figure. Three diagonal links on a fragment of the s/p block. Each arrow runs one group right and one period down — the charge-to-size cancel. Si sits off this fragment to the right of Al; the B–Si label names the pair even though Si is not drawn as a cell.
Why the diagonal works
- Right: higher charge densitySmaller radius and (for cations) higher charge raise polarising power.
- Down: lower charge densityLarger radius lowers polarising power.
- Diagonal cancelLi⁺ is closer to Mg²⁺ in polarising power than to Na⁺; likewise Be²⁺ to Al³⁺.
| Pair | Shared trait | Contrast with vertical neighbour |
|---|---|---|
| Li–Mg | Form nitrides; covalent character in halides | Na does not form a nitride easily |
| Be–Al | Amphoteric oxides; bridge-bonded hydrides/halides | MgO is basic; B₂O₃ is acidic |
| B–Si | Covalent network hydrides/oxides; semiconductor chemistry for Si | Al is metallic; C forms discrete molecules more readily |
Which pair shows a diagonal relationship?
- Na–Mg
- Li–Mg
- B–Al
Li–Mg is the classic diagonal pair. Na–Mg are neighbours in period 3; B–Al are in the same group.
Notes
- Modern periodic law: properties of elements are a periodic function of atomic number; the long-form table has 7 periods and 18 groups arranged by increasing Z.
- Atomic and ionic radius: decreases across a period (rising Z_{eff}) and increases down a group (new shells); cations are smaller and anions larger than the parent atom.
- Ionisation enthalpy: increases across a period and decreases down a group, with dips at group 2→13 and 15→16 due to stable filled and half-filled configurations.
- Electron gain enthalpy: most negative for halogens; Cl is more negative than F because of small size and electron-electron repulsion in the compact 2p of F.
- Electronegativity increases across and decreases down; F is the most electronegative element (Pauling 4.0), and metallic character shows the opposite trend.
Formulas
- Z_{eff}=Z-\sigma (Slater's screening constant \sigma)
- IE_1<IE_2<IE_3 (successive removal is progressively harder)
- Anomalous IE order: Be>B and N>O (stable 2s^2 and 2p^3)
- Electron gain enthalpy order: Cl>F>Br>I
- Elements per period: 2, 8, 8, 18, 18, 32
Exam traps & shortcuts
- Diagonal relationships (Li-Mg, Be-Al, B-Si) arise from similar charge-to-size ratios.
- Across a period, non-metallic and oxidising character rise; down a group, metallic and reducing character rise.
- For isoelectronic species, more nuclear charge means smaller size: O^{2-}>F^->Na^+>Mg^{2+}.
Reference tables
Every 'up' and 'down' here is the general trend; the named anomalies are examined in their own concepts.
| Property | Across a period | Down a group | Named anomaly |
|---|---|---|---|
| Atomic radius | Decreases | Increases | — |
| IE₁ | Increases | Decreases | Be > B, N > O |
| Electron gain enthalpy (magnitude) | Peaks at halogens | Falls (usually) | Cl > F |
| Electronegativity | Increases | Decreases | F is the maximum |
| Metallic character | Decreases | Increases | — |
Short enough to reconstruct, not merely recall.
| Relation | Reads as | Watch for |
|---|---|---|
| Z_eff = Z − σ | Effective nuclear charge | σ from Slater groups, not a free parameter |
| IE₁ < IE₂ < IE₃ | Successive ionisation | A sharp jump marks the noble-gas core |
| Isoelectronic size | Higher Z → smaller | Same electron count only |
| EGE order (halogens) | Cl > F > Br > I | Magnitude of the negative enthalpy |
| Elements per period | 2, 8, 8, 18, 18, 32 | Count subshell electrons |
Recap
Read only this the night before.
- Z, not mass
- Modern law orders by atomic number. Period lengths are subshell capacities: 2, 8, 8, 18, 18, 32.
- s/p/d/f blocks
- s: ns¹–² (Gps 1–2). p: np¹–⁶ (Gps 13–18). d: (n−1)d (Gps 3–12). f: (n−2)f lanthanoids/actinoids.
- Z_eff
- Z_eff = Z − σ. It rises across a period (atoms shrink, IE rises) and loses to shell-adding down a group.
- Isoelectronic
- Same electron count: highest Z is smallest. Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.
- IE dips
- Be > B (filled 2s² vs starting 2p) and N > O (half-filled 2p³ vs paired 2p⁴).
- EGE
- Cl more negative than F because of 2p repulsion in compact fluorine. F is still the electronegativity champion.
- Diagonal
- Li–Mg, Be–Al, B–Si from similar charge-to-size ratios — one step right cancels one step down.
Practise Classification of Elements and Periodicity
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