QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 3.4NCERT Class 11 · Chemistry · Chapter 3

Electronic Configurations & the Periodic Table: Block, Period, Group

The modern periodic table is not a memory chart — it is electronic configuration drawn large. One filling ladder generates the 7 periods and 18 groups; three short rules convert any configuration into block, period and group; and two rebellious atoms (Cr, Cu) teach why symmetry wins.

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 14 min read
01

From Filling Order to the Table — Complete Theory

The Aufbau principle from § 2.6 — fill lowest n + l first, ties broken by lower n — produces one fixed ladder, and the long-form periodic table is simply that ladder laid on its side. Every new period begins when a new shell (new n) starts filling; every group collects elements whose valence configuration matches. Nothing in the table is arbitrary:

Filling order1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s… Period startsnew n begins filling Group rulevalence e⁻ → column
Table 1 — The seven periods: what fills, how long, who books the ends
PeriodShell (n) fillingSubshells filledElementsStart → end
1n = 11s2H → He
2n = 22s, 2p8Li → Ne
3n = 33s, 3p8Na → Ar
4n = 44s, 3d, 4p18K → Kr
5n = 55s, 4d, 5p18Rb → Xe
6n = 66s, 4f, 5d, 6p32Cs → Rn
7n = 77s, 5f, 6d, 7p32Fr → Og (complete)

The period lengths — 2, 8, 8, 18, 18, 32, 32 — are just the summed capacities of the subshells each period fills. This also explains why periods 4 and 5 are suddenly long: the d-subshell (10 electrons) joins the filling, and periods 6–7 stretch further because the f-subshell (14) enters.

Now the three rules that turn any configuration into a table address — the most operational content in this chapter:

Block — named after the subshell receiving the last electron in the Aufbau sequence: ns → s-block, np → p-block, (n−1)d → d-block, (n−2)f → f-block. Period — the highest n occupied. Group — block-dependent: s-block gives group = ns electrons; p-block gives group = 10 + (ns + np) electrons; d-block gives group = (n−1)d + ns electrons; f-block elements all sit in group 3 (as the lanthanoid/actinoid series). Work one: Br = [Ar] 3d¹⁰ 4s² 4p⁵ → last electron in 4p → p-block; highest n = 4 → period 4; group = 10 + 2 + 5 = 17. The lab below runs this machinery on any Z you feed it.

The two rebels. Aufbau predicts Cr (Z = 24) as [Ar] 3d⁴ 4s² and Cu (Z = 29) as [Ar] 3d⁹ 4s². Nature disagrees: Cr = [Ar] 3d⁵ 4s¹, Cu = [Ar] 3d¹⁰ 4s¹. The reason: exactly half-filled and completely filled subshells carry extra stability — their charge distribution is symmetric, and electrons of parallel spin in degenerate orbitals enjoy maximum exchange energy. Shifting one 4s electron into 3d buys that stability at trivial cost. The same logic explains Mo, Ag and Au further down.

02

Visualising the Ladder & Finding Any Address

Ek seedhi, ek finder — the filling ladder first, then type any atomic number and watch block, period and group fall out.

The Aufbau ladder laid out in period runs — the table drawn as a straight line FIG. 1 — THE FILLING LADDER = THE TABLE, UNROLLED 1sP1 · 2e⁻ 2s · 2pP2 · 8 3s · 3pP3 · 8 4s · 3d · 4pP4 · 18 — d joins 5s · 4d · 5pP5 · 18 6s 4f — fourteen seats 5d 6p P6 · 32 — the f-block joins; 57–71 run below the table 7s 5f — fourteen seats 6d 7p P7 · 32 — actinoids run below; Fr → Og now complete period length = Σ capacities of subshells filled 2 · 8 · 8 · 18 · 18 · 32 · 32 same ladder, three readings: block, period, group
FIG. 1 — Unroll the table into the filling ladder and the periods become visible as runs: short runs (s+p only) give 2-8-8, the d-entry stretches periods 4–5 to 18, and the f-entry stretches 6–7 to 32. The exceptions (Cr, Cu) are the two places nature jumps seats on this ladder.
Try it live

Block–Period–Group Finder

BrBromineZ = 35

Blockp
Period4
Group17
Valence e⁻7

Exceptions handled: Cr (3d⁵4s¹), Cu (3d¹⁰4s¹), Mo, Ag, Pd, Pt, Au. A few f-block elements show small 5d/6d occupancy anomalies — grouped f-block by convention (as NCERT does).

03

Solved Examples (Step-by-Step)

Config → block → period → group. Jo chain yahan chalti hai, wahi finder me live chalti hai.

EXAMPLE 01Foundation · Full address

The complete address of bromine (Z = 35)

Write the electronic configuration of bromine (Z = 35) and determine its block, period and group.

  1. ConfigBr: [Ar] 3d¹⁰ 4s² 4p⁵ — Ar core (18) + 17 electrons through 4s, 3d, 4p.
  2. BlockLast electron enters 4p → p-block.
  3. PeriodHighest n occupied = 4 → period 4.
  4. Group10 + (4s² + 4p⁵) = 10 + 7 = 17 — the halogen column ✓ (verify on the finder).

[Ar] 3d¹⁰ 4s² 4p⁵ · p-block · period 4 · group 17

EXAMPLE 02JEE Main · Exception

Why Cr breaks the ladder — and where it sits

The predicted configuration of chromium (Z = 24) is [Ar] 3d⁴ 4s². Write its actual configuration, explain the anomaly, and find its group.

  1. ActualCr: [Ar] 3d⁵ 4s¹ — one 4s electron moves into 3d.
  2. Why3d⁵ is exactly half-filled: symmetric charge distribution + maximum exchange energy among five parallel-spin electrons — stability worth more than the 4s–3d energy gap.
  3. Groupd-block → group = 3d + 4s = 5 + 1 = 6 ✓ — the rule works on the actual config too.
  4. FamilyCu (3d¹⁰ 4s¹, group 11), Mo, Ag, Au — the same half/full-filled driver.

[Ar] 3d⁵ 4s¹ · symmetry wins · group 6

EXAMPLE 03JEE Main · Reverse run

From address back to element

An element belongs to group 14 and period 5. Write its configuration and name the element.

  1. BlockGroup 14 ≥ 13 → p-block.
  2. Valence14 − 10 = 4 valence electrons, period 5 → 5s² 5p².
  3. ConfigCount back: period 5 p-block with 4 valence → inner core [Kr] + 4d¹⁰ (filled in period 5) → [Kr] 4d¹⁰ 5s² 5p²
  4. ElementZ = 36 + 14 = 50 → Sn (tin) ✓ (finder confirms: p, period 5, group 14).

[Kr] 4d¹⁰ 5s² 5p² — tin, group 14

04

Practice Questions (With Solutions)

Attempt first — options lock after one shot, exactly like the real exam. Then read the working, chahe galti ho ya na ho.

Attempted 0/4 · Correct 0

Q1NEET · Group rule

An element has the configuration [Ne] 3s² 3p⁴. Its group number is:

Solution

  1. Last electron enters 3p → p-block → group = 10 + (2 + 4) = 16 — sulfur.
  2. Option D counts only 3p electrons; B uses the d-block-style sum — rule-mixing bait.

(C) 16

Q2JEE Main · Exception config

The ground state electronic configuration of Cu (Z = 29) is:

Solution

  1. [Ar] 3d¹⁰ 4s¹ — the fully-filled 3d¹⁰ plus half-filled 4s¹ beats the aufbau-predicted 3d⁹ 4s².
  2. C (31 electrons) is zinc; A is the naive prediction the exception overrides.

(B) [Ar] 3d¹⁰ 4s¹

Q3JEE Main · He’s address

Which statement about helium (Z = 2) is correct?

Solution

  1. Config 1s² → s-block by the last-electron rule; but its closed shell gives noble-gas inertness → placed in group 18.
  2. The one sanctioned block-vs-position mismatch — statement questions quote it both ways.

(B) s-block, group 18

Q4NEET · Reverse

The element with configuration [Ar] 3d⁶ 4s² lies in:

Solution

  1. Last electron enters 3d → d-block → group = 3d + 4s = 6 + 2 = 8; highest n = 4 → period 4 → Fe.
  2. Option D applies the s-block rule (counting only 4s²) — the classic d-block trap.

(B) Iron, group 8

05

Key Rules & Takeaways

Rule card

Eight lines that solve this topic

Aufbau: n+l ↑, tie → lower nOrder: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s…
Block = last electron’s subshellns → s · np → p · (n−1)d → d · (n−2)f → f.
Period = highest n occupiedEvery period begins when a new shell starts filling.
s-block group = ns e⁻ (1 or 2)Groups 1–2; He is the sanctioned oddity (s-block, group 18).
p-block group = 10 + ns + np e⁻Groups 13–18; Br: 10 + 2 + 5 = 17.
d-block group = (n−1)d + ns e⁻Groups 3–12; Fe: 3d⁶ 4s² → 8.
f-block: all in group 3Lanthanoids (4f) and actinoids (5f) run below the table.
Cr [Ar]3d⁵4s¹ · Cu [Ar]3d¹⁰4s¹Half-filled / fully-filled symmetry + exchange energy beats aufbau.

Period lengths: 2, 8, 8, 18, 18, 32, 32  ·  Subshell capacities: s 2 · p 6 · d 10 · f 14  ·  Exceptions: Cr 24, Cu 29, Mo 42, Ag 47, Pd 46, Pt 78, Au 79

  1. The table is configuration made visible — periods are runs of the Aufbau ladder; group rules are valence-counting per block.
  2. Three rules, three blocks — never mixed — s: ns count · p: 10 + valence · d: (n−1)d + ns; f: pinned at group 3.
  3. The rule chain runs both ways — config → address and address → config (group 14, period 5 → Sn); practise the inverse.
  4. Symmetry beats aufbau twice — Cr and Cu trade a 4s electron for half-filled/full-filled d stability; quote “symmetry + exchange energy”, never just “stability”.
06

FAQs

How do you find the block of an element from its electronic configuration?

The block is named after the subshell that receives the last electron in the Aufbau filling: if the last electron enters an ns subshell, the element is s-block; an np subshell makes it p-block; an (n−1)d subshell makes it d-block; and an (n−2)f subshell makes it f-block. Helium is the special case — its configuration is 1s² (s-block) but it is placed in group 18 with the p-block noble gases.

How do you find the period and group from an electronic configuration?

The period equals the highest principal quantum number n occupied in the configuration. The group depends on the block: s-block, group = number of ns electrons (1 or 2); p-block, group = 10 + the total valence electrons (ns + np); d-block, group = (n−1)d electrons plus ns electrons; f-block, all elements are placed in group 3.

Why is the configuration of chromium [Ar] 3d⁵ 4s¹ instead of 3d⁴ 4s²?

Exactly half-filled and completely filled subshells carry extra stability from their symmetry and from maximum exchange energy among electrons of like spin. Shifting one electron from 4s to 3d gives chromium the symmetrical 3d⁵ 4s¹ arrangement instead of 3d⁴ 4s², so the atom adopts it. Copper shows the same behaviour, settling as [Ar] 3d¹⁰ 4s¹.

How many elements are there in each period of the long form periodic table?

Period 1 holds 2 elements (1s²), periods 2 and 3 hold 8 each (one s and one p subshell), periods 4 and 5 hold 18 each (s, d and p subshells), and periods 6 and 7 hold 32 each (s, f, d and p subshells). The period lengths therefore run 2, 8, 8, 18, 18, 32, 32 — the subshell capacities written across the table.

Your progress

Saved on this device only — no account, no sign-in.

Related in this chapter