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Position Electronic Config

🎓 Class 12 Chemistry CBSE Theory Ch 4 – The d- and f-Block Elements ⏱ ~14 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Position Electronic Config

આ મૂલ્યાંકન આના પર આધારિત હશે: Position Electronic Config

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Position Electronic Config

Introduction: The Metals That Built Civilisation

Iron, copper, silver and gold — four metals that shaped human history — share something invisible to the eye: each carries a partly filled set of d-orbitals. Throw in titanium for jet engines, vanadium for catalysts, chromium for steel, and you have the d-block elements. Add the radioactive uranium and thorium that power reactors, and you reach the f-block.

This chapter unlocks why a single block in the periodic table — Groups 3 to 12 — produces metals so versatile they appear in every cell of your body (Fe in haemoglobin), in every electronic device (Cu wiring), and in every modern catalyst.

Roadmap of Part 1: position of d-/f-blocks → electronic configurations of the four transition series → reasons for the Cr and Cu anomalies → general physical properties (metallic character, melting points, density).

8.1 Position in the Periodic Table

The d-block sits in the wide central section of the periodic table — flanked on the left by the s-block and on the right by the p-block. Within these elements, it is the d-orbitals of the penultimate (n-1) energy level that progressively fill, giving four horizontal series:

  • 3d series — Sc to Zn (Z = 21 to 30) in Period 4.
  • 4d series — Y to Cd (Z = 39 to 48) in Period 5.
  • 5d series — La and Hf to Hg (Z = 57, 72 to 80) in Period 6.
  • 6d series — Ac and Rf to Cn (Z = 89, 104 to 112) in Period 7.

The two inner-transition (lanthanoid 4f and actinoid 5f) series sit in a separate panel below the main table.

s d-BLOCK (Groups 3–12) p f-BLOCK (Lanthanoids 4f & Actinoids 5f) 3d 4d 5d 6d Penultimate (n-1)d orbitals fill across each row.
Fig 8.1: Position of the d- and f-blocks in the modern periodic table.
IUPAC Definition (Transition Element): A transition metal is a metal whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell. By this rule, Zn, Cd and Hg (d¹⁰ in atom and common +2 ion) are not transition metals — though we still study their chemistry alongside the d-block.
Worked Example 8.1 L2 Understand

Q. On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Step 1: Write the ground-state configurations.

Sc (Z = 21): [Ar] 3d¹ 4s²  →  Sc³⁺: [Ar] 3d⁰ Zn (Z = 30): [Ar] 3d¹⁰ 4s²  →  Zn²⁺: [Ar] 3d¹⁰

Step 2: Sc atom has an incomplete 3d shell (d¹) — it qualifies. Zn atom and Zn²⁺ both carry a complete 3d¹⁰ — it does not qualify, even though it sits in the d-block column.

In-text Q 4.1 — Silver: Ag has 4d¹⁰ in its ground state. Yet Ag can show a +2 oxidation state (4d⁹), in which the d sub-shell is incomplete — therefore Ag still satisfies the IUPAC definition of a transition element.

8.2 Electronic Configurations of the d-Block

The general outer configuration of a d-block element is

(n-1)d1-10 ns0-2

The (n-1) tells us electrons enter the inner d-orbitals of the second-from-last shell, while one or two electrons sit in the outer ns. Because the energy difference between (n-1)d and ns is very small, the simple Aufbau prediction sometimes fails — and we get exceptions driven by the extra stability of half-filled (d⁵) and fully filled (d¹⁰) sub-shells.

8.2.1 The Cr and Cu Anomalies

Naively, Cr (Z = 24) should be 3d⁴4s² and Cu (Z = 29) should be 3d⁹4s². Instead, both promote one 4s electron into 3d, giving:

Cr: [Ar] 3d⁵ 4s¹ ← half-filled d⁵ + half-filled s¹ Cu: [Ar] 3d¹⁰ 4s¹ ← fully filled d¹⁰ + half-filled s¹

The added exchange energy from a half-filled or fully filled d-shell more than pays for the small extra cost of unpairing the 4s electrons.

8.2.2 Outer Configurations Across All Four Series

SeriesElement & Z — outer configuration
3d (Period 4)Sc(21)3d¹4s² · Ti(22)3d²4s² · V(23)3d³4s² · Cr(24)3d⁵4s¹ · Mn(25)3d⁵4s² · Fe(26)3d⁶4s² · Co(27)3d⁷4s² · Ni(28)3d⁸4s² · Cu(29)3d¹⁰4s¹ · Zn(30)3d¹⁰4s²
4d (Period 5)Y(39)4d¹5s² · Zr(40)4d²5s² · Nb(41)4d⁴5s¹ · Mo(42)4d⁵5s¹ · Tc(43)4d⁶5s¹ · Ru(44)4d⁷5s¹ · Rh(45)4d⁸5s¹ · Pd(46)4d¹⁰5s⁰ · Ag(47)4d¹⁰5s¹ · Cd(48)4d¹⁰5s²
5d (Period 6)La(57)5d¹6s² · Hf(72)5d²6s² · Ta(73)5d³6s² · W(74)5d⁴6s² · Re(75)5d⁵6s² · Os(76)5d⁶6s² · Ir(77)5d⁷6s² · Pt(78)5d⁹6s¹ · Au(79)5d¹⁰6s¹ · Hg(80)5d¹⁰6s²
6d (Period 7)Ac(89)6d¹7s² · Rf(104)6d²7s² · Db(105)6d³7s² · Sg(106)6d⁴7s² · Bh(107)6d⁵7s² · Hs(108)6d⁶7s² · Mt(109)6d⁷7s² · Ds(110)6d⁸7s² · Rg(111)6d¹⁰7s¹ · Cn(112)6d¹⁰7s²

Notice the unique Pd exception: it skips its 5s entirely (4d¹⁰ 5s⁰) — a consequence of two paired-up exchange contributions plus relativistic effects.

3d Series — Box-Diagram of Electron Filling Sc ↑↓ 4s 3d¹4s² Ti ↑↓ 4s 3d²4s² V ↑↓ 4s 3d³4s² Cr ★ ↑ 4s 3d⁵4s¹ Mn ↑↓ 4s 3d⁵4s² Fe ↑↓ ↑↓ 4s 3d⁶4s² Co ↑↓ ↑↓ ↑↓ 4s 3d⁷4s² Ni ↑↓ ↑↓ ↑↓ ↑↓ 4s 3d⁸4s² Cu ★ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ 4s 3d¹⁰4s¹ Zn ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s 3d¹⁰4s² ★ Anomalous: one 4s electron promoted to 3d to gain exchange-energy stability of half-filled (Cr) or fully-filled (Cu) d-shell. Orange = 3d sub-shell (5 boxes)  ·  Teal = 4s sub-shell (1 box)
Fig 8.2: Box-and-arrow electron filling for the 3d series Sc → Zn.

Interactive: d-Block Configuration Builder L3 Apply L4 Analyse

Pick any 3d-series element and see (i) its outer configuration, (ii) the number of unpaired electrons, (iii) the spin-only magnetic moment, and (iv) whether it follows or breaks Aufbau.

Pick an element to begin.

8.3 General Properties of the Transition (d-Block) Elements

Almost every transition metal is hard, lustrous, ductile, malleable, with high tensile strength and excellent thermal/electrical conductivity. The exceptions you must remember are the soft, low-melting Zn, Cd, Hg (no unpaired d-electrons available for metallic bonding) and the unusual Mn (complex bcc structure).

8.3.1 Lattice Structures

SeriesScTiVCrMnFeCoNiCuZn
3dhcphcpbccbccXbccccpccpccpX

bcc = body-centred cubic · hcp = hexagonal close-packed · ccp = cubic close-packed · X = atypical metal lattice.

8.3.2 High Melting Points and Enthalpy of Atomisation

Transition metals melt at very high temperatures because both the (n-1)d and ns electrons participate in metallic bonding. In each row the melting point peaks roughly at d⁵ (one electron per d-orbital, all spins parallel) — the textbook example is W (3683 K, the highest of any metal). Mn and Tc dip anomalously, and the curve falls again towards Zn/Cd/Hg.

0 1 2 3 4 M.p. /10³ K Atomic number across 3d row Sc Ti V Cr Mn Fe Co Ni Cu Zn Peak near d⁵ (V/Cr) — strongest metallic bonding Mn dip Zn (3d¹⁰4s²) — no d-electrons in bonding
Fig 8.3: Melting points of the 3d-series transition metals.
In-text Q 4.2 — Why is Zn's enthalpy of atomisation only 126 kJ mol⁻¹? Because Zn has 3d¹⁰4s² — its 3d sub-shell is full, so 3d electrons do not contribute to metallic bonding. Only the two 4s electrons hold the lattice together, giving the lowest ΔₐH of the 3d row.

8.4 Why Are They Called "Transition" Elements?

The historical name comes from their position: they bridge the highly electropositive s-block and the largely non-metallic p-block. Chemically, they are "transitional" in two senses — they share metallic character with the s-block but display the variable oxidation states and complex-forming tendency that some p-block elements show only weakly.

The presence of partly filled d-orbitals gives the d-block elements five hallmark features that we will explore in Part 2:

  1. Variable oxidation states (e.g. Mn from +2 to +7).
  2. Coloured ions (d-d transitions).
  3. Paramagnetism (unpaired d-electrons).
  4. Complex-ion formation (vacant d-orbitals + small size).
  5. Catalytic activity (multiple oxidation states + adsorption sites).
Activity 8.1 — Why Do d⁵ and d¹⁰ Configurations Win?L4 Analyse

Aim: Use the exchange-energy idea to predict why Cr is 3d⁵4s¹ and not 3d⁴4s².

Procedure:

  1. Draw the five 3d boxes and the 4s box for the configuration 3d⁴4s². Assume Hund's rule (all four d-electrons spin-up). Count the number of pairs of parallel-spin electrons in the 3d set.
  2. Now draw the alternative 3d⁵4s¹. Again count parallel-spin pairs (3d set + 4s set together if they happen to be parallel).
  3. The configuration with the larger number of parallel pairs has the larger exchange energy and wins.

Predict: Which configuration should have more parallel-spin pairs and therefore be more stable?

3d⁴4s²: Parallel pairs in 3d = C(4,2) = 6 pairs.

3d⁵4s¹: Parallel pairs in 3d = C(5,2) = 10 pairs. If the 4s¹ electron also has parallel spin to the 3d set, add 5 more cross-set pairs (these contribute, though weighted differently).

Conclusion: 3d⁵4s¹ wins by at least 4 extra parallel-spin pairs — a substantial exchange-energy gain that more than offsets the small promotion energy 4s → 3d. The same accounting predicts 3d¹⁰4s¹ over 3d⁹4s² for Cu.

Worked Example 8.2 L2 Understand

Q. Why do the transition elements exhibit higher enthalpies of atomisation than s-block neighbours?

Transition metals carry a large number of unpaired electrons in their 3d/4s (and equivalent) orbitals. These unpaired electrons form strong, multi-electron metallic bonds in the lattice, requiring large amounts of energy to break — hence high ΔₐH.

Worked Example 8.3 L3 Apply

Q. Write the expected and observed ground-state configurations of the Cu atom and identify the source of stabilisation.

Expected (Aufbau): [Ar] 3d⁹ 4s². Observed: [Ar] 3d¹⁰ 4s¹.

Promoting one 4s electron to fill the last 3d vacancy converts a 3d⁹ shell into a fully-filled 3d¹⁰ shell. The exchange energy stabilisation of d¹⁰ outweighs the small 4s → 3d promotion cost, so the d¹⁰s¹ form is the ground state.

Competency-Based Questions L3 L4

Stimulus: A student tabulates the outer configurations of the 3d series and notices that two elements break the simple "fill 4s then 3d" pattern. He further notes that the Mn–Tc–Re trio dip in melting points within their respective series.

Q1. (MCQ) The two anomalous configurations in the 3d series are:

  • (a) Sc and Ti
  • (b) Cr and Cu
  • (c) Mn and Zn
  • (d) Fe and Ni
(b) Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹) due to extra stability of half-filled and fully-filled d-shells.

Q2. (SA) Predict whether the Pt atom (Z = 78) follows the simple Aufbau order or shows an exception. Justify.

Expected: 5d⁸ 6s². Observed: 5d⁹ 6s¹. Pt promotes one 6s electron to 5d to gain exchange-energy stabilisation in the 5d shell — analogous to the Cr/Cu story.

Q3. (MCQ) Which of these is not classified as a transition element by IUPAC?

  • (a) Mn
  • (b) Cr
  • (c) Zn
  • (d) Fe
(c) Zn. Both Zn and Zn²⁺ have full 3d¹⁰ — no incomplete d sub-shell. Zn is studied with the d-block but is not a "transition element".

Q4. (LA) Write the ground-state outer configurations of (i) V (ii) Cr³⁺ (iii) Cu (iv) Zn²⁺. State which of these has the maximum number of unpaired electrons.

(i) V: 3d³4s² — 3 unpaired. (ii) Cr³⁺: 3d³ — 3 unpaired. (iii) Cu: 3d¹⁰4s¹ — 1 unpaired. (iv) Zn²⁺: 3d¹⁰ — 0 unpaired. Maximum unpaired = V or Cr³⁺ (tie at 3).

Q5. (HOT) Across the 3d row, melting points first rise, peak, then fall. Suggest two reasons why Mn melts at a much lower temperature than its neighbours Cr and Fe.

(i) Mn has the configuration 3d⁵4s² — its half-filled 3d⁵ is so stable that its electrons are reluctant to delocalise into the metallic bond. (ii) Mn adopts a complex (atypical) crystal structure with weaker packing efficiency than the bcc/fcc lattices of Cr and Fe.

Assertion–Reason Questions L4 L5

Choose: A) Both A and R true and R explains A · B) Both true but R does not explain A · C) A true, R false · D) A false, R true.

Assertion (A): Cr has the configuration [Ar]3d⁵4s¹ rather than [Ar]3d⁴4s².

Reason (R): A half-filled d⁵ sub-shell is exceptionally stable due to maximum exchange energy.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. The exchange-energy gain of d⁵ over d⁴ is exactly what drives the 4s → 3d promotion.

Assertion (A): Zinc is not regarded as a transition element.

Reason (R): Zinc has a fully filled d-shell (3d¹⁰) in both its ground state and its common +2 oxidation state.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. The IUPAC test (incomplete d sub-shell in atom or common ion) fails for Zn.

Assertion (A): Pd has the unique outer configuration 4d¹⁰5s⁰.

Reason (R): The energy gap between 4d and 5s in Pd is so large that the 5s lies far above the 4d after both electrons are promoted.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. In Pd, full d¹⁰ stabilisation plus 5s being relatively high in energy makes the d¹⁰s⁰ configuration the most stable choice.

Frequently Asked Questions - Position Electronic Config

What is the main concept covered in Position Electronic Config?
In NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements), "Position Electronic Config" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Position Electronic Config useful in real-life or applied chemistry?
Real-life applications of "Position Electronic Config" from NCERT Class 12 Chemistry Chapter 4 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Position Electronic Config?
Key reactions in "Position Electronic Config" (NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 4?
NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements) is structured so each part builds chemical understanding sequentially. "Position Electronic Config" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Position Electronic Config?
CBSE board questions from "Position Electronic Config" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Position Electronic Config" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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Chemistry Class 12 Part I – NCERT (2025-26)
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