Chemistry Class 12 Part I…
🏠 હોમ લૉગિન
ટોપિક 21 / 23

Vbt Cft

🎓 Class 12 Chemistry CBSE Theory Ch 5 – Coordination Compounds ⏱ ~14 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Vbt Cft

આ મૂલ્યાંકન આના પર આધારિત હશે: Vbt Cft

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

Vbt Cft

5.6 Bonding in Coordination Compounds

Werner's theory captured the geometry, but it could not explain why only certain elements form complexes, why coordinate bonds are directional, or why complexes have characteristic colours and magnetic properties. Two complementary theories address these questions at the introductory level: Valence Bond Theory (VBT) and Crystal Field Theory (CFT). Beyond NCERT, ligand field theory and molecular orbital theory go further still.

5.6.1 Valence Bond Theory (VBT)

According to VBT, the metal ion provides a set of empty hybrid orbitals of definite geometry; each ligand donates a lone pair into one of these hybrid orbitals, forming a coordinate (dative) bond. The hybridisation depends on the coordination number and on whether inner-shell (n−1)d orbitals are used.

CNHybridisationGeometryExamples
4sp³Tetrahedral[NiCl4]2−, [Ni(CO)4]
4dsp²Square planar[Ni(CN)4]2−, [Pt(NH3)2Cl2]
5sp³dTrigonal bipyramidal[Fe(CO)5]
6d²sp³ (inner-orbital)Octahedral[Co(NH3)6]3+, [Fe(CN)6]3−
6sp³d² (outer-orbital)Octahedral[CoF6]3−, [FeF6]3−

Inner vs outer orbital octahedral complexes

For Co3+ (3d6), with strong-field ligands like NH3 the six 3d electrons pair up into the lower three 3d orbitals. The two now-empty 3d orbitals combine with 4s and 4p (giving d²sp³ hybridisation) — an inner-orbital (low-spin) complex with no unpaired electrons → diamagnetic, e.g. [Co(NH3)6]3+.

With weak-field ligands such as F−, electrons stay unpaired in 3d. The metal then uses the outer 4d orbitals: hybridisation sp³d² → outer-orbital (high-spin) complex with four unpaired electrons → paramagnetic, e.g. [CoF6]3−.

[Co(NH₃)₆]³⁺ — d²sp³ (inner orbital, diamagnetic) ↑↓↑↓↑↓ 3d (filled paired) 3d empty 4s 4p → 6 d²sp³ hybrids ↑↓↑↓↑↓↑↓↑↓↑↓ six pairs from NH₃ ligands [CoF₆]³⁻ — sp³d² (outer orbital, paramagnetic, 4 unpaired e⁻) ↑↓↑↑↑↑ 3d⁶ (4 unpaired) 4s 4p 4d → 6 sp³d² hybrids
Fig. 5.7: VBT orbital diagrams for the diamagnetic inner-orbital complex [Co(NH₃)₆]³⁺ (d²sp³) and the paramagnetic outer-orbital complex [CoF₆]³⁻ (sp³d²).

Magnetic properties from VBT

Magnetic moment is calculated from the number of unpaired electrons (n) using the spin-only formula:

\[ \mu_{\text{spin-only}} = \sqrt{n(n+2)} \;\; \text{Bohr Magneton (BM)} \]

Examples (first transition series):

  • n = 1 → μ = √3 ≈ 1.73 BM
  • n = 2 → μ = √8 ≈ 2.83 BM
  • n = 3 → μ = √15 ≈ 3.87 BM
  • n = 4 → μ = √24 ≈ 4.90 BM
  • n = 5 → μ = √35 ≈ 5.92 BM

Worked Example 5.7 — Geometry from magnetic moment

The spin-only magnetic moment of [MnBr4]2− is 5.9 BM. Predict the geometry.

For Mn2+, configuration is 3d5. CN = 4, so the geometry is either tetrahedral (sp³) or square planar (dsp²).
μ = 5.9 BM ⇒ √n(n+2) = 5.9 ⇒ n = 5 unpaired electrons.
Square planar dsp² requires pairing of d-electrons (typical n = 1 or 0). Since five electrons remain unpaired, the geometry must be tetrahedral (sp³).

Limitations of VBT

  1. Built on assumptions; quantitative magnetic moment is not predicted exactly.
  2. Cannot explain colour of complexes.
  3. Cannot quantify thermodynamic or kinetic stability.
  4. Does not always predict whether 4-coordinate complexes are tetrahedral or square planar.
  5. Does not distinguish between weak- and strong-field ligands.

5.6.2 Crystal Field Theory (CFT)

CFT is purely electrostatic: ligands are treated as point negative charges (or point dipoles for neutral ligands) that perturb the energy of the metal's five d-orbitals. In an isolated ion all five d-orbitals are degenerate. A spherical field would raise them all equally. But a real ligand field is asymmetrical, and the d-orbitals split.

(a) Octahedral splitting — \(\Delta_o\)

In an octahedral complex six ligands lie along the ±x, ±y, ±z axes. The two d-orbitals that point along the axes (\(d_{x^2-y^2}\) and \(d_{z^2}\)) experience strong electrostatic repulsion and are destabilised — these are the eg set. The three orbitals that point between the axes (\(d_{xy}, d_{yz}, d_{xz}\)) are less repelled and are stabilised — these are the t2g set.

The energy gap is \(\Delta_o\) (the crystal-field splitting energy in an octahedral field). The barycentre rule gives:

\[ E(t_{2g}) = -\tfrac{2}{5}\Delta_o, \qquad E(e_g) = +\tfrac{3}{5}\Delta_o \]
free ion 5 degenerate d spherical field eg d(x²−y²), d(z²) t2g d(xy), d(yz), d(xz) +(3/5)Δₒ −(2/5)Δₒ octahedral field Δₒ
Fig. 5.8: d-orbital splitting in an octahedral crystal field. The energy gap Δₒ separates the lower t₂g set from the higher eₘ set; the t₂g energy is lowered by (2/5)Δₒ and the eₘ raised by (3/5)Δₒ relative to the spherical-field average.

(b) Filling d-orbitals: high spin vs low spin

For d¹, d², d³ ions the electrons occupy t2g singly (Hund's rule). For d4–d7 there are two competing options:

  • If \(\Delta_o < P\) (pairing energy) → 4th electron enters eg, giving t2g3eg1 (high-spin, weak-field).
  • If \(\Delta_o > P\) → 4th electron pairs in t2g, giving t2g4eg0 (low-spin, strong-field).

(c) Spectrochemical series

Ligands ranked by the magnitude of \(\Delta_o\) they produce form the experimentally derived spectrochemical series:

I− < Br− < SCN− < Cl− < S2− < F− < OH− < C2O42− < H2O < NCS− < EDTA4− < NH3 < en < CN− < CO
(weak-field ligands → strong-field ligands)
weak field (small Δₒ, high spin) strong field (large Δₒ, low spin) I⁻ Br⁻ Cl⁻ F⁻ OH⁻ H₂O NH₃ en CN⁻ CO
Fig. 5.9: Spectrochemical series — ligand field strength increases from left (I⁻) to right (CO).

(d) Tetrahedral splitting — \(\Delta_t\)

In a tetrahedral field the four ligands lie between the axes, so now the t2 orbitals (between axes) are destabilised and the e orbitals (along axes) are stabilised — the reverse of the octahedral pattern. The "g" subscript is dropped (no centre of symmetry):

\[ \Delta_t = \tfrac{4}{9}\,\Delta_o \]

Because Δt is much smaller than the pairing energy, tetrahedral complexes are almost always high-spin.

5 degenerate d t2 d(xy), d(yz), d(xz) e d(x²−y²), d(z²) Δₜ tetrahedral field — splitting INVERTED, Δₜ = (4/9) Δₒ
Fig. 5.10: d-orbital splitting in a tetrahedral crystal field. The pattern is inverted relative to octahedral; the gap is much smaller — tetrahedral complexes are usually high-spin.

(e) Colour of coordination compounds

If a complex absorbs light of a particular wavelength to promote an electron from t2g → eg (a d–d transition), the colour we see is the complementary colour. For example [Ti(H2O)6]3+ (d1) absorbs blue-green light (≈498 nm) and so appears violet. Removing the ligand field (e.g. by dehydration) destroys the splitting and the substance turns colourless: [Ti(H2O)6]Cl3 is violet but anhydrous TiCl3 is colourless. Anhydrous CuSO4 is white but CuSO4·5H2O is blue.

Complexλ absorbed (nm)Colour absorbedColour observed
[CoCl(NH3)5]2+535yellowviolet
[Co(NH3)5(H2O)]3+500blue-greenred
[Co(NH3)6]3+475blueyellow-orange
[Co(CN)6]3−310UV (not visible)pale yellow
[Cu(H2O)4]2+600redblue
[Ti(H2O)6]3+498blue-greenviolet
Gem stones: Ruby (red) is Al2O3 with ~0.5–1% Cr3+ in octahedral sites; emerald (green) is the same Cr3+ in beryl Be3Al2Si6O18. Same metal, different ligand field → different colour.

(f) Limitations of CFT

  1. Treats ligands as point charges, so anionic ligands ought to give the largest Δ — but they actually sit at the low end of the spectrochemical series.
  2. Ignores the covalent character of the M–L bond (addressed by ligand field/MO theories).
🧪 Activity 5.3 — Ligands and Colour (Predict → Observe → Explain)

Setup: A solution of NiCl2 in water gives the green [Ni(H2O)6]2+ ion. Ethane-1,2-diamine (en) is added drop-wise in molar ratios 1:1, 2:1 and 3:1.

Predict: What colour will the solution become as more en is added? Why?

Observed: green → pale blue → blue/purple → violet.

Sequence of complexes: [Ni(H2O)6]2+ → [Ni(H2O)4(en)]2+ → [Ni(H2O)2(en)2]2+ → [Ni(en)3]2+. en sits higher than H2O in the spectrochemical series, so each substitution increases Δₒ and shifts the absorbed wavelength to higher energy (shorter λ). The colour we see (the complement) accordingly shifts.

🔧 Interactive: Octahedral Spin-State and Magnetic-Moment Predictor

Pick a metal dn count and a ligand. The simulation predicts the t2g/eg filling, the number of unpaired electrons and the spin-only magnetic moment.

Choose values above.

Worked Example 5.8 — Why is [Ni(CN)₄]²⁻ diamagnetic but [NiCl₄]²⁻ paramagnetic?

Both contain Ni2+ (d8). In [Ni(CN)4]2−, the strong-field CN− ligand forces all eight d-electrons to pair up in four 3d orbitals, leaving one 3d orbital vacant. Hybridisation is dsp² → square planar. No unpaired electrons → diamagnetic.

In [NiCl4]2−, Cl− is a weak-field ligand and cannot pair up the d-electrons. The 3d8 remains as 3d⁶ paired + 2 unpaired. Hybridisation is sp³ → tetrahedral. Two unpaired electrons → paramagnetic (μ ≈ 2.83 BM).

Worked Example 5.9 — Magnetic moment of hexaaqua vs hexacyano Mn²⁺

The hexaqua manganese(II) ion has 5 unpaired electrons but the hexacyano-manganese(III) ion has only one. Explain using CFT.

For Mn2+ (3d5) with the weak-field H2O ligand, Δₒ < P → all five d-electrons go in singly: t2g3eg2; n = 5 unpaired.

For Mn3+ (3d4) with the strong-field CN− ligand the question text refers to [Mn(CN)6]3−: Δₒ > P → low-spin t2g4eg0; n = 2 unpaired (although NCERT data often quote 1, depending on Jahn-Teller distortion). The strong field is responsible for the pairing.

🎯 Competency-Based Questions

Q1. The hybridisation of [Ni(CN)4]2− is: L1 Remember

  • (a) sp³ (b) dsp² (c) sp³d (d) d²sp³
Answer: (b) dsp² — square planar, diamagnetic.

Q2. The number of unpaired electrons in [FeF6]3− is: L3 Apply

  • (a) 1 (b) 3 (c) 4 (d) 5
Answer: (d) 5. Fe3+ = 3d5; F− is weak-field → outer-orbital sp³d², high spin (t2g3eg2) → 5 unpaired electrons.

Q3. Why is [Cr(NH3)6]3+ paramagnetic but [Ni(CN)4]2− diamagnetic? L4 Analyse

Cr3+ is 3d3: regardless of field strength, the three d-electrons sit singly in t2g → 3 unpaired electrons → paramagnetic. Ni2+ is 3d8; with the strong-field CN− the geometry is square planar (dsp²) and all electrons are paired → diamagnetic.

Q4. [Fe(CN)6]4− is yellow but [Fe(H2O)6]2+ is pale green. Account for the difference using CFT. L4 Analyse

CN− is a strong-field ligand; Δₒ is large; the d–d transition needs higher-energy (shorter wavelength) light, absorbing in the violet and reflecting yellow. H2O is weaker; Δₒ smaller; absorption shifts to the red end and the complementary colour is pale green.

Q5. HOT (Create): Sketch a CFT-style splitting diagram for a square-planar complex (derive it conceptually from the octahedral diagram). Where do the four orbitals end up relative to the average? L6 Create

Start with octahedral and remove the two axial (z-axis) ligands. Orbitals with z-character (dz², dxz, dyz) drop in energy because axial repulsion is gone. Orbitals with strong xy character (dx²−y²) climb still higher. Result, in order of increasing energy: dxz, dyz (degenerate, lowest) < dz² < dxy ≪ dx²−y² (highest). The very large gap between dxy and dx²−y² explains why d8 ions like Ni2+ with strong ligands prefer square-planar low-spin geometry.

🧠 Assertion–Reason Questions

Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.

A: [Ni(CO)4] is diamagnetic but [NiCl4]2− is paramagnetic.

R: CO causes pairing of all 3d electrons of Ni; Cl−, being a weak-field ligand, cannot.

Answer: (A). Both true; R explains A. In [Ni(CO)4] Ni is in 0 oxidation state (3d¹⁰) and all electrons are paired. In [NiCl4]2−, Ni is +2 (3d⁸) with two unpaired electrons — Cl− is too weak to pair them.

A: Tetrahedral complexes are almost always high-spin.

R: The tetrahedral splitting Δt is only 4/9 of Δo, so it is usually less than the pairing energy.

Answer: (A). Both true; R correctly explains A.

A: Anhydrous CuSO4 is white but CuSO4·5H2O is blue.

R: Crystal field splitting requires ligands; without ligands no d–d transition is possible.

Answer: (A). Both true; R explains A. With H2O present, Cu2+ sits inside [Cu(H2O)4]2+; the d-orbitals split and a d–d transition absorbs orange light, giving a blue colour. Without ligands, no splitting and no visible colour.

Frequently Asked Questions - Vbt Cft

What is the main concept covered in Vbt Cft?
In NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds), "Vbt Cft" 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 Vbt Cft useful in real-life or applied chemistry?
Real-life applications of "Vbt Cft" from NCERT Class 12 Chemistry Chapter 5 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 Vbt Cft?
Key reactions in "Vbt Cft" (NCERT Class 12 Chemistry Chapter 5 Coordination Compounds) 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 5?
NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds) is structured so each part builds chemical understanding sequentially. "Vbt Cft" 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 Vbt Cft?
CBSE board questions from "Vbt Cft" 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 "Vbt Cft" 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.
AI ટ્યુટર
Chemistry Class 12 Part I – NCERT (2025-26)
તૈયાર
નમસ્તે! 👋 હું ગૌરા છું, Vbt Cft માટે તમારું AI ટ્યુટર. આરામથી પાઠ ભણો — જ્યારે પણ કોઈ શંકા થાય, બસ મને પૂછો! હું મદદ માટે અહીં જ છું.

📝 આ ટોપિકની પ્રેક્ટિસ કરો — મોક પરીક્ષાઓ

🎯 Chemistry ની પ્રેક્ટિસ કરો

તમે જે ભણ્યા તેનું પૂરું પેપર આપો, પ્રશ્ન દીઠ તપાસાયેલું.

મોક પરીક્ષાઓ

આ વિષયની બધી મોક પરીક્ષાઓ →

🎁 Join our community and get free AI credits!
WhatsApp