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. 133Valence Bond Coordination
8 MCQs9-step worked example
Source: NCERT Coordination CompoundsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 24 Sep 2026
Valence Bond Coordination, explained for NEET
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 133 presents the VBT framework as the first bonding model before introducing CFT (reference: NCERT Class 12 Chemistry Chapter 5).
Can you answer these Valence Bond Coordination 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:
Show answer and why every option is right or wrong
Answer: B. B is correct. VBT describes coordination bonding as the ligand donating a lone pair into an empty hybridised orbital of the metal ion — a coordinate or dative bond (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because the electrons are not equally shared — both electrons in the bond come from the ligand's lone pair, not one from each atom.
Why C is wrong: C is wrong because VBT does not describe coordination bonds as ionic; no electron transfer from metal to ligand occurs. The ligand donates electrons to the metal, not the reverse.
Why D is wrong: D is wrong because metallic bonding involves a sea of delocalised electrons in bulk metals, which has no relevance to discrete coordination compound bonding.
MCQ 2Easy RecallPractice
Which hybridisation corresponds to an octahedral geometry in the valence bond approach?
Show answer and why every option is right or wrong
Answer: A. A is correct. Octahedral geometry (coordination number 6) requires six equivalent hybrid orbitals, achieved by sp³d² (outer-orbital) or d²sp³ (inner-orbital) hybridisation. sp³d² is the standard designation when outer d-orbitals are used (NCERT Class 12 Chemistry Chapter 5, page 133).
Why B is wrong: B is wrong because dsp² hybridisation produces four hybrid orbitals in a square planar geometry (CN = 4), not six.
Why C is wrong: C is wrong because sp³ hybridisation produces four hybrid orbitals arranged tetrahedrally (CN = 4), not six.
Why D is wrong: D is wrong because sp² hybridisation produces three hybrid orbitals in a trigonal planar geometry (CN = 3), not six.
MCQ 3Easy RecallPractice
In the valence bond treatment, a square planar complex with coordination number 4 involves which hybridisation?
Show answer and why every option is right or wrong
Answer: C. C is correct. Square planar geometry (CN = 4) arises from dsp² hybridisation, which uses one (n−1)d orbital, one s orbital, and two p orbitals. This is distinct from sp³ (tetrahedral) even though both have CN = 4 (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because sp³ produces a tetrahedral geometry, not square planar. Both are CN = 4 but differ in hybridisation and geometry.
Why B is wrong: B is wrong because sp³d hybridisation gives five hybrid orbitals (trigonal bipyramidal geometry, CN = 5), not four.
Why D is wrong: D is wrong because d²sp³ hybridisation produces six hybrid orbitals (octahedral geometry, CN = 6), not four.
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?
Show answer and why every option is right or wrong
Answer: D. D is correct. Ni²⁺ has d⁸ configuration. CN⁻ is a strong-field ligand that forces the two unpaired 3d electrons to pair, vacating one 3d orbital. This 3d orbital combines with 4s and two 4p orbitals → dsp² hybridisation → square planar, diamagnetic (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because sp³ tetrahedral hybridisation would leave two unpaired electrons in d⁸ Ni²⁺ (paramagnetic), contradicting the given diamagnetic observation. Strong-field CN⁻ forces pairing and inner d-orbital use.
Why B is wrong: B is wrong because d²sp³ is also an octahedral hybridisation (CN = 6), not applicable to a four-coordinate complex.
Why C is wrong: C is wrong because sp³d² is for octahedral geometry (CN = 6). This complex has CN = 4, ruling out any six-coordinate hybridisation.
MCQ 5Direct ApplicationPractice
[NiCl₄]²⁻ is paramagnetic and tetrahedral. What hybridisation does VBT predict for Ni²⁺ in this complex?
Show answer and why every option is right or wrong
Answer: A. A is correct. Ni²⁺ (d⁸) with weak-field Cl⁻ does not force electron pairing. Two unpaired electrons remain in the 3d orbitals. No inner d-orbital is vacated, so the hybridisation uses 4s and three 4p orbitals → sp³ → tetrahedral, paramagnetic (NCERT Class 12 Chemistry Chapter 5, page 133).
Why B is wrong: B is wrong because dsp² requires one inner d-orbital to be empty, which means electrons must pair. Cl⁻ is a weak-field ligand that cannot force pairing in d⁸ Ni²⁺, so dsp² (square planar) is not achievable.
Why C is wrong: C is wrong because d²sp³ is an octahedral hybridisation (CN = 6). This complex has only four ligands (CN = 4).
Why D is wrong: D is wrong because sp³d² is also an octahedral hybridisation (CN = 6), inconsistent with CN = 4.
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?
Show answer and why every option is right or wrong
Answer: C. C is correct. Co³⁺ has d⁶ configuration. NH₃ is a strong-field ligand that forces all six 3d electrons to pair into three orbitals, leaving two 3d orbitals empty. These two 3d + one 4s + three 4p = d²sp³ → octahedral, diamagnetic (inner-orbital complex). NCERT Class 12 Chemistry Chapter 5, page 133.
Why A is wrong: A is wrong because sp³d² (outer-orbital) would use higher-energy 4d orbitals without forcing 3d electron pairing. This would leave unpaired electrons in d⁶, making the complex paramagnetic — contradicting the diamagnetic observation.
Why B is wrong: B is wrong because sp³ produces only four hybrid orbitals (tetrahedral geometry), but this complex has six NH₃ ligands (CN = 6) requiring six hybrid orbitals.
Why D is wrong: D is wrong because dsp² produces four hybrid orbitals (square planar, CN = 4), not six. [Co(NH₃)₆]³⁺ is an octahedral complex with CN = 6.
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:
Show answer and why every option is right or wrong
Answer: B. B is correct. In VBT, square planar (diamagnetic) arises when a strong-field ligand forces d-electron pairing, freeing an inner (n−1)d orbital for dsp² hybridisation. Tetrahedral (paramagnetic) arises when a weak-field ligand cannot force pairing, so no inner d-orbital is available and sp³ hybridisation occurs instead (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because the relationship is reversed. The diamagnetic complex uses inner d-orbitals (dsp², with one (n−1)d orbital participating), while the paramagnetic complex avoids inner d-orbitals (sp³).
Why C is wrong: C is wrong because the two geometries require different hybridisations: dsp² for square planar and sp³ for tetrahedral. Claiming both use sp³ ignores the fundamental VBT distinction between these CN = 4 geometries.
Why D is wrong: D is wrong because both complexes have the same coordination number (4, as stated in the stem). They differ in hybridisation type, not coordination number.
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:
Show answer and why every option is right or wrong
Answer: D. D is correct. Step 1: Fe³⁺ has configuration [Ar] 3d⁵ — five electrons each in a separate 3d orbital (all unpaired). Step 2: Weak-field ligands cannot force pairing, so all five 3d electrons remain unpaired. Step 3: Since no 3d orbitals are vacated, outer 4d orbitals are used → sp³d² hybridisation. The five unpaired electrons persist (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because zero unpaired electrons would mean all d-electrons are paired, which requires strong-field ligands forcing d²sp³ (inner-orbital) hybridisation. Weak-field ligands explicitly cannot achieve this for d⁵.
Why B is wrong: B is wrong because three unpaired electrons would imply partial pairing of the five 3d electrons. With weak-field ligands, no pairing is forced — all five remain unpaired. Three unpaired electrons is not an achievable configuration for d⁵ under VBT with weak-field ligands.
Why C is wrong: C is wrong because one unpaired electron would require pairing four of the five d-electrons, which demands a very strong-field ligand. In an outer-orbital complex with weak-field ligands, no pairing occurs.
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How do you solve a Valence Bond Coordination question? A 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.)*
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What to remember before solving Valence Bond Coordination questions
More in Coordination Compounds: 2 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Valence Bond Coordination questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
1A, D and E only
2A and C only
3B and E only
4A and D only
NTA Answer: Option 4(final)
Sources
NCERT refs: Class 12 Chemistry Chapter 5, p.133
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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