Hybridization: 4-coordinate tetrahedral (sp³, e.g. [NiCl₄]²⁻); square planar (dsp², e.g. [Ni(CN)₄]²⁻); 6-coordinate octahedral inner-orbital (d²sp³, low-spin) or outer-orbital (sp³d², high-spin).
-- NCERT Class 12 Chemistry, Ch. 5, p. 16Valence Bond Coordination
8 MCQs9-step worked example
Source: NCERT Coordination CompoundsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The valence bond theory (VBT) approach to coordination bonding explains how the central metal ion forms bonds with ligands through hybridisation of its atomic orbitals. The key idea: the metal ion provides empty hybrid orbitals, and each ligand donates a lone pair of electrons into one of these orbitals, forming a coordinate (dative) bond.
The core VBT procedure for coordination compounds:
- Write the electronic configuration of the metal ion (not the atom — account for charge).
- Identify the coordination number and geometry from the formula.
- Determine the hybridisation that matches the geometry: sp³ (tetrahedral, CN 4), dsp² (square planar, CN 4), sp³d² or d²sp³ (octahedral, CN 6).
- Check whether inner d-orbitals (d²sp³) or outer d-orbitals (sp³d²) are used — this depends on whether the ligand forces electron pairing.
Inner-orbital vs outer-orbital complexes is where aspirants slip. In an inner-orbital complex (e.g., [Co(NH₃)₆]³⁺), strong-field ligands force (n−1)d electrons to pair, vacating d-orbitals for d²sp³ hybridisation. The complex is diamagnetic or has fewer unpaired electrons. In an outer-orbital complex (e.g., [CoF₆]³⁻), weak-field ligands leave d-electrons unpaired, and the metal uses outer nd orbitals for sp³d² hybridisation — the complex is paramagnetic.
A common confusion: students assume hybridisation type alone determines geometry. In reality, for CN = 4, you must distinguish dsp² (square planar, uses one inner d-orbital) from sp³ (tetrahedral, no inner d used). The ligand field strength and d-electron count together decide which hybridisation the metal adopts.
VBT limitations worth knowing for NEET: VBT explains geometry and magnetic behaviour qualitatively but cannot predict the exact magnitude of magnetic moments, does not explain colour of complexes, and fails to explain why certain complexes are inner-orbital versus outer-orbital without importing spectrochemical series data from crystal field theory.
NCERT Class 12 Chemistry Chapter 5, page 16 presents the VBT framework as the first bonding model before introducing CFT (reference: NCERT Class 12 Physics Chapter 5).
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the valence bond theory of coordination compounds, the bond between a metal ion and a ligand is best described as:
MCQ 2Easy RecallPractice
Which hybridisation corresponds to an octahedral geometry in the valence bond approach?
MCQ 3Easy RecallPractice
In the valence bond treatment, a square planar complex with coordination number 4 involves which hybridisation?
MCQ 4Direct ApplicationPractice
The complex [Ni(CN)₄]²⁻ is diamagnetic and square planar. Using VBT, what is the hybridisation of Ni²⁺ in this complex?
MCQ 5Direct ApplicationPractice
[NiCl₄]²⁻ is paramagnetic and tetrahedral. What hybridisation does VBT predict for Ni²⁺ in this complex?
MCQ 6Direct ApplicationPractice
Co³⁺ (d⁶) forms the complex [Co(NH₃)₆]³⁺ which is diamagnetic. According to VBT, which hybridisation and orbital type does Co³⁺ use?
MCQ 7Concept TrapPractice
Two complexes of the same metal ion have coordination number 4: one is diamagnetic and square planar, the other is paramagnetic and tetrahedral. According to VBT, the key difference is:
MCQ 8CalculationPractice
Fe³⁺ (d⁵) forms an outer-orbital octahedral complex with weak-field ligands. According to VBT, the hybridisation is sp³d² and the number of unpaired electrons is:
Worked Example
- 1
Given
- Metal ion: Co²⁺, configuration [Ar] 3d⁷ - Ligand: Cl⁻ (weak-field) - Coordination number: 4
- 2
Required
Hybridisation of Co²⁺, geometry of the complex, and number of unpaired electrons (magnetic behaviour).
- 3
Concept
In VBT, the hybridisation depends on (a) coordination number, (b) whether inner d-orbitals are available. For CN = 4, the two possibilities are dsp² (square planar, uses one inner d-orbital) or sp³ (tetrahedral, no inner d-orbital). Weak-field ligands cannot force electron pairing.
- 4
Formula / rule
CN = 4 with no inner d-orbital available → sp³ → tetrahedral. CN = 4 with inner d-orbital freed by pairing → dsp² → square planar.
- 5
Substitution / reasoning
Co²⁺ is d⁷: the 3d orbitals hold seven electrons arranged as ↑↓ ↑↓ ↑ ↑ ↑ (three unpaired in high-spin). Cl⁻ is weak-field and cannot force further pairing. All five 3d orbitals are at least singly occupied — none is empty. Therefore, no inner (3d) orbital is available for hybridisation.
- 6
Calculation / determination
Since no 3d orbital is free, Co²⁺ uses 4s + three 4p orbitals → **sp³ hybridisation** → **tetrahedral geometry**. Unpaired electron count: 3d⁷ with no forced pairing retains **3 unpaired electrons**.
- 7
Final answer
[CoCl₄]²⁻: sp³ hybridisation, tetrahedral geometry, paramagnetic with 3 unpaired electrons. *Note on exact values: the coordination number 4 and the electron count 7 are exact integers and do not affect any significant-figure analysis.*
- 8
Common trap
Aspirants sometimes assume CN = 4 always means tetrahedral. It does not — if a strong-field ligand were present and could force pairing to free an inner d-orbital, the geometry would be square planar (dsp²) instead. Always check ligand field strength before assigning hybridisation.
- 9
Similar NEET-style question
"Using VBT, predict the hybridisation and number of unpaired electrons in [Fe(H₂O)₆]³⁺, given that H₂O is a weak-field ligand. Fe³⁺ has a d⁵ configuration." *(Answer: sp³d², octahedral, 5 unpaired electrons.)* ---
Before solving, remember these
Formulas
Crystal field splitting (octahedral vs tetrahedral)
Tetrahedral splitting is smaller than octahedral due to fewer/farther ligands.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_o | octahedral splitting | J or eV |
| Delta_t | tetrahedral splitting | J or eV |
Valid when
- Same metal and same ligand
- Mostly high-spin tetrahedral due to small Δ_t
Magnetic moment of coordination complex
Same spin-only formula but n depends on high-spin/low-spin from CFT.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
Valid when
- High vs low spin determined by Δ_o vs P
- Octahedral (or tetrahedral with Δ_t)
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student defaults to one spin state. Strong-field ligand (CN⁻, CO, NH₃ for some) → low-spin (Δ > P, electrons pair). Weak-field (F⁻, H₂O, Cl⁻) → high-spin.
When it triggers
Coordination compound with given ligand asking for magnetic moment, color, or spin state.
How to avoid
Memorise spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. NH₃, CN⁻, CO usually strong-field. F⁻, H₂O, Cl⁻ usually weak-field.
Root cause: concept gap
Correction
Strong-field ligand (CN⁻, CO, NH₃ when applicable): Δ > P → low-spin. Weak-field (F⁻, Cl⁻, H₂O): Δ < P → high-spin. Use spectrochemical series.
Past Year Questions
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1C < A < D < B
2B < D < A < C
3B < A < D < C
4C < D < A < B
NTA Answer: Option 3(final)
NEET 2025
1Statement I is false but Statement II is true
2Both Statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I is true but Statement II is false
NTA Answer: Option 4(final)
NEET 2024Revised key
1B and C
2A and D
3B and E
4E and D
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
1Both Statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is true but Statement II is false
4Statement I is false but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2023
1A, B, C, D are correct
2A, B, C, E are correct
3A, C, D, E are correct
4B, C, D, E are correct
NTA Answer: Option 2(final)
NEET 2023
Homoleptic complex from the following complexes is
1Diamminechloridonitrito-N-platinum (II)
2Pentaamminecarbonatocobalt (III) chloride
3Triamminetriaquachromium (III) chloride
4Potassium trioxalatoaluminate (III)
NTA Answer: Option 4(final)
NEET 2022
Identify the incorrect statement from the following.
1The shapes of d xy , d yz and d zx orbitals are similar to each other; and d x2−y2 and d z2 are similar to each other.
2All the five 5d orbitals are different in size when compared to the respective 4d orbitals.
3All the five 4d orbitals have shapes similar to the respective 3d orbitals.
4In an atom, all the five 3d orbitals are equal in energy in free state.
NTA Answer: Option 1(final)
NEET 2022
The IUPAC name of the complex- [Ag(H O) ][Ag(CN) ] is: 2 2 2
1diaquasilver(I) dicyanidoargentate(I)
2dicyanidosilver(II) diaquaargentate(II)
3diaquasilver(II) dicyanidoargentate(II)
4dicyanidosilver(I) diaquaargentate(I)
NTA Answer: Option 1(final)
NEET 2022
1(B) > (A) > (C)
2(A) > (B) > (C)
3(C) > (B) > (A)
4(C) > (A) > (B)
NTA Answer: Option 4(final)
NEET 2021
12-methylbutan-2-ol
22-methylpropan-2-ol
3pentan-2-ol
4pentan-3-ol
NTA Answer: Option 1(final)
NEET 2021
Ethylene diaminetetraacetate (EDTA) ion is :
1Tridentate ligand with three "N" donor atoms
2Hexadentate ligand with four "O" and two "N" donor atoms
3Unidentate ligand
4Bidentate ligand with two "N" donor atoms
NTA Answer: Option 2(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Determine high-spin or low-spin from ligand field strength. Strong-field (CN, CO): low-spin. Weak-field (F, H2O): high-spin.
InterpretationMedium
Common distractors
forgets weak vs strong ligand
Default low-spin always
Apply IUPAC rules: ligands alphabetical, prefixes for multiplicity, oxidation state of metal.
RecallMedium
Common distractors
wrong order ligands
Uses non-alphabetical order
Identify primary and secondary valencies, coordination number from formula. Primary = ionizable ions; secondary = ligands.
InterpretationMedium
Common distractors
counts counter ions as ligands
Treats all attached species as ligands
Sources
NCERT refs: Class 12 Chemistry Chapter 5, p.16
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