ChemistryISC Class 12 · Coordination Compounds

Valence Bond Theory of Coordination Compounds: ISC Class 12

Updated 15 October 2026 · Fact-checked

Valence bond theory says the metal ion uses hybrid orbitals to accept lone pairs from ligands. Hybridisation fixes the shape: sp³ tetrahedral, dsp² square planar, d²sp³ or sp³d² octahedral. Using inner 3d orbitals gives inner-orbital (low-spin) complexes; using outer 4d gives outer-orbital (high-spin) ones. Unpaired electrons give magnetism.

Understand Valence Bond Theory of Coordination Compounds

The idea. In valence bond theory (VBT) the central metal atom or ion makes available a number of vacant orbitals equal to its coordination number. These orbitals mix (hybridise) to give a set of equivalent orbitals pointing in definite directions, and each receives a lone pair from a ligand (a coordinate bond). The type of hybridisation fixes the geometry:

CN 4: $s p^{3}$ , tetrahedral; $d s p^{2}$ (using one inner 3d, 4s and two 4p), square planar.
CN 5: $s p^{3} d$ (or $d s p^{3}$ ), trigonal bipyramidal.
CN 6: $d^{2} s p^{3}$ (two inner 3d, 4s, three 4p), octahedral; $s p^{3} d^{2}$ (4s, three 4p, two outer 4d), octahedral.

Inner and outer orbital complexes. If the metal uses its inner $(n - 1) d$ orbitals ( $d^{2} s p^{3}$ ), the complex is an inner orbital (low spin, spin paired) complex. To free two 3d orbitals the electrons may have to pair up, so fewer unpaired electrons remain. If it uses outer $n d$ orbitals ( $s p^{3} d^{2}$ ), the 3d electrons are undisturbed, giving an outer orbital (high spin, spin free) complex with the same number of unpaired electrons as the free ion. Strong ligands such as CN⁻, CO, NH₃ (with Co³⁺), en and oxalate tend to cause pairing; weak ligands such as F⁻, Cl⁻ and H₂O usually do not.

Worked patterns (3d series).
• $d^{1}$ – $d^{3}$ ions (Ti³⁺, V³⁺, Cr³⁺) always have two empty 3d orbitals, so they form inner orbital $d^{2} s p^{3}$ octahedral complexes with 1, 2 or 3 unpaired electrons: $[C r (N H_{3})_{6}]^{3 +}$ has 3 unpaired electrons, μ = 3.87 BM.
• $d^{6}$ Co³⁺: with NH₃ or oxalate, the six 3d electrons pair into three orbitals: $d^{2} s p^{3}$ , diamagnetic ( $[C o (N H_{3})_{6}]^{3 +}$ , $[C o (C_{2} O_{4})_{3}]^{3 -}$ ). With F⁻: $s p^{3} d^{2}$ , 4 unpaired ( $[C o F_{6}]^{3 -}$ ).
• $d^{5}$ Fe³⁺: $[F e (C N)_{6}]^{3 -}$ is $d^{2} s p^{3}$ with 1 unpaired electron; $[F e F_{6}]^{3 -}$ is $s p^{3} d^{2}$ with 5. $d^{4}$ Mn³⁺: $[M n (C N)_{6}]^{3 -}$ inner (2 unpaired), $[M n C l_{6}]^{3 -}$ outer (4 unpaired).
• $d^{8}$ Ni²⁺: in octahedral complexes ( $[N i (N H_{3})_{6}]^{2 +}$ ) only one 3d orbital could ever be emptied, so it must use $s p^{3} d^{2}$ (outer orbital, 2 unpaired). With CN⁻ the electrons pair into four 3d orbitals and $d s p^{2}$ gives square planar, diamagnetic $[N i (C N)_{4}]^{2 -}$ ; with Cl⁻, $s p^{3}$ gives tetrahedral, paramagnetic $[N i C l_{4}]^{2 -}$ (2 unpaired).
• Ni(0) in $[N i (C O)_{4}]$ : the $4 s^{2}$ electrons move into 3d under the influence of CO, giving $3 d^{10}$ ; $s p^{3}$ , tetrahedral, diamagnetic. Similarly Fe(0) in $[F e (C O)_{5}]$ : $3 d^{8}$ after pairing, $d s p^{3}$ , trigonal bipyramidal, diamagnetic.
• $d^{10}$ Zn²⁺, Cu⁺, Ag⁺: $[Z n (N H_{3})_{4}]^{2 +}$ is $s p^{3}$ tetrahedral and $[A g (N H_{3})_{2}]^{+}$ is $s p$ linear, both diamagnetic.

Magnetic evidence. VBT predictions are checked with magnetic moments, $μ = n (n + 2)$ BM. A measured moment tells you the number of unpaired electrons and hence the hybridisation: for example, $[M n B r_{4}]^{2 -}$ (5.9 BM, 5 unpaired) must be tetrahedral $s p^{3}$ , because $d s p^{2}$ would need a vacant 3d orbital that a $d^{5}$ ion cannot provide without pairing.

Limitations of VBT. (i) It involves a number of assumptions. (ii) It gives no quantitative interpretation of magnetic data. (iii) It does not explain the colour of complexes. (iv) It gives no quantitative account of thermodynamic or kinetic stability. (v) It cannot predict reliably whether a 4-coordinate complex is tetrahedral or square planar. (vi) It does not distinguish between weak and strong ligands. Crystal field theory removes several of these weaknesses.

Valence Bond Theory of Coordination Compounds: formulas and key facts

Hybridisation and shape: $s p^{3} : tetrahedral; d s p^{2} : square planar; d s p^{3} : trigonal bipyramidal; d^{2} s p^{3}, s p^{3} d^{2} : octahedral$; Number of hybrid orbitals = coordination number.
Inner versus outer orbital: $d^{2} s p^{3} = (n - 1) d^{2} n s n p^{3} (inner, low spin); s p^{3} d^{2} = n s n p^{3} n d^{2} (outer, high spin)$; For the 3d series: inner uses 3d, outer uses 4d.
Spin-only magnetic moment: $μ = n (n + 2) BM$; n = unpaired electrons left after hybridisation.
Oxidation state in a complex: $x + Σ (ligand charges) = charge on the complex ion$; Neutral ligands (NH₃, H₂O, CO, en) count as 0.
Typical pairing ligands: $C N^{-}, C O, e n, N H_{3} (with C o^{3 +}), C_{2} O_{4}^{2 -} (with C o^{3 +})$; Weak, non-pairing: F⁻, Cl⁻, Br⁻, H₂O (with most ions).

How to solve Valence Bond Theory of Coordination Compounds questions

ISC VBT questions ask for hybridisation, shape, inner or outer orbital type, magnetic behaviour and moment. Draw the orbital boxes; every answer can be read from the diagram.

1Work out the oxidation state and the d electron count of the metal.
2Draw boxes: five 3d, one 4s, three 4p (and five 4d if needed). Fill the 3d electrons by Hund's rule for the free ion.
3Decide whether the ligand forces pairing (strong ligand) and, if so, pair electrons to empty the 3d orbitals needed.
4Count the empty orbitals of lowest energy to match the coordination number: two 3d + 4s + three 4p = $d^{2} s p^{3}$ ; 4s + 4p + two 4d = $s p^{3} d^{2}$ ; one 3d + 4s + two 4p = $d s p^{2}$ ; 4s + three 4p = $s p^{3}$ .
5Write the shape, inner or outer orbital label, and the number of unpaired electrons.
6Give μ = √(n(n+2)) BM and say paramagnetic (n > 0) or diamagnetic (n = 0).

Valence Bond Theory of Coordination Compounds in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A magnet shows unpaired electrons in a complex

Suppose a teacher hangs two small tubes from a sensitive balance between the poles of a strong magnet. One holds potassium ferricyanide, K₃[Fe(CN)₆] (red crystals); the other holds potassium ferrocyanide, K₄[Fe(CN)₆] (yellow crystals). The first appears slightly heavier when the magnet is on; the second appears slightly lighter. (Fe Z = 26)

Explain the observations using valence bond theory and calculate the spin-only magnetic moment of the ferricyanide ion.

Try it first, then show the solutionSolution

In $[F e (C N)_{6}]^{3 -}$ , Fe is +3 ( $3 d^{5}$ ). CN⁻ pairs the electrons into three 3d orbitals (2 pairs + 1 single), leaving two 3d orbitals for $d^{2} s p^{3}$ : inner orbital, octahedral, 1 unpaired electron.
In $[F e (C N)_{6}]^{4 -}$ , Fe is +2 ( $3 d^{6}$ ). All six electrons pair in three 3d orbitals: $d^{2} s p^{3}$ , no unpaired electrons.
Unpaired electrons make ferricyanide paramagnetic, so it is pulled into the field (appears heavier); ferrocyanide is diamagnetic and is weakly pushed out (appears lighter).
For ferricyanide, $n = 1$ : $μ = 3 = 1.73$ BM.

Answer: Ferricyanide ( $d^{2} s p^{3}$ , 1 unpaired electron) is paramagnetic, μ = 1.73 BM; ferrocyanide ( $d^{2} s p^{3}$ , no unpaired electrons) is diamagnetic.

Memory hook: Pulled in means unpaired electrons; pushed out means all paired.

Tricks to remember Valence Bond Theory of Coordination Compounds

Short memory aids. Say each one aloud once, then use it on a question the same day.

d first means inside
In $d^{2} s p^{3}$ the d is written FIRST because inner (n−1)d orbitals are used: inner orbital. In $s p^{3} d^{2}$ the d comes LAST: outer nd orbitals, outer orbital.
Use it when: Labelling complexes inner or outer.
Three or fewer, always inner
$d^{1}$ , $d^{2}$ and $d^{3}$ ions always leave two 3d orbitals empty, so their octahedral complexes are always $d^{2} s p^{3}$ .
Use it when: Ti³⁺, V³⁺, Cr³⁺ complexes.
Eight is too many for inside
A $d^{8}$ ion can never free two 3d orbitals, so its octahedral complexes are always outer ( $s p^{3} d^{2}$ ); for CN 4 it chooses $d s p^{2}$ (strong ligand) or $s p^{3}$ (weak ligand).
Use it when: Nickel(II) complexes.

Valence Bond Theory of Coordination Compounds: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A cobalt(II) chloro complex · 1 mark · MCQBoard level

The hybridisation and magnetic behaviour of the tetrahedral ion $[C o C l_{4}]^{2 -}$ (Co Z = 27) are:
(a) $d s p^{2}$ , diamagnetic
(b) $s p^{3}$ , paramagnetic with 3 unpaired electrons
(c) $s p^{3}$ , diamagnetic
(d) $d^{2} s p^{3}$ , paramagnetic

Try it first, then show the solutionSolution

Co²⁺ is $3 d^{7}$ : 3 unpaired electrons; Cl⁻ does not cause pairing.
Tetrahedral uses 4s and three 4p: $s p^{3}$ .

Answer: (b).

Example 2: Hexaamminechromium(III) · 2 marks · Short answerBoard level

Using valence bond theory, give the hybridisation, shape, type (inner or outer orbital) and magnetic moment of $[C r (N H_{3})_{6}]^{3 +}$ (Cr Z = 24).

Try it first, then show the solutionSolution

Cr³⁺ is $3 d^{3}$ : three unpaired electrons in three 3d orbitals; two 3d orbitals remain empty (½ mark).
Two 3d, 4s and three 4p accept six NH₃ lone pairs: $d^{2} s p^{3}$ , octahedral, inner orbital complex (1 mark).
$n = 3$ , $μ = 15 = 3.87$ BM, paramagnetic (½ mark).

Answer: $d^{2} s p^{3}$ , octahedral, inner orbital; 3.87 BM.

Example 3: Why Ni(II) octahedral complexes are outer orbital · 3 marks · Short answerHarder board level

Explain, using orbital diagrams, why $[N i (N H_{3})_{6}]^{2 +}$ is an outer orbital complex. Give its hybridisation and number of unpaired electrons, and calculate its spin-only magnetic moment. (Ni Z = 28)

Try it first, then show the solutionSolution

Ni²⁺ is $3 d^{8}$ : three 3d orbitals are full and two hold one electron each (½ mark).
Even if pairing occurred, eight electrons would fill four 3d orbitals, leaving only one empty: two 3d orbitals can never be freed, so $d^{2} s p^{3}$ is impossible (1 mark).
The ion uses 4s, three 4p and two 4d orbitals: $s p^{3} d^{2}$ , octahedral, outer orbital complex (1 mark).
Two unpaired electrons remain: $μ = 8 = 2.83$ BM (½ mark).

Answer: $s p^{3} d^{2}$ , outer orbital, octahedral; 2 unpaired; 2.83 BM.

Example 4: Iron pentacarbonyl · 3 marks · Short answerHarder board level

$[F e (C O)_{5}]$ is diamagnetic. Using valence bond theory, explain its hybridisation and shape. (Fe Z = 26)

Try it first, then show the solutionSolution

Fe is in the zero oxidation state: $3 d^{6} 4 s^{2}$ (½ mark).
CO is a strong ligand; the 4s electrons move into 3d and all eight 3d electrons pair in four 3d orbitals, leaving one 3d empty (1 mark).
One 3d, 4s and three 4p orbitals hybridise: $d s p^{3}$ , trigonal bipyramidal, accepting five CO lone pairs (1 mark).
No unpaired electrons: diamagnetic (½ mark).

Answer: $d s p^{3}$ , trigonal bipyramidal, diamagnetic (all 3d electrons paired).

Example 5: Two cyanide complexes and the limits of VBT · 5 marks · Long answerHOTS / competency

(a) For $[C r (C N)_{6}]^{3 -}$ and $[F e (C N)_{6}]^{4 -}$ (Z: Cr 24, Fe 26), give the oxidation state, d count, hybridisation, number of unpaired electrons and magnetic moment. [3]
(b) Both complexes are coloured. Can valence bond theory explain this? State two other limitations of VBT. [2]

Try it first, then show the solutionSolution

(a) $[C r (C N)_{6}]^{3 -}$ : $x - 6 = - 3$ , Cr +3, $3 d^{3}$ ; $d^{2} s p^{3}$ (inner orbital), 3 unpaired, μ = 3.87 BM (1½ marks).
$[F e (C N)_{6}]^{4 -}$ : $x - 6 = - 4$ , Fe +2, $3 d^{6}$ ; CN⁻ pairs all six electrons, $d^{2} s p^{3}$ (inner orbital), 0 unpaired, diamagnetic (1½ marks).
(b) No. VBT does not explain the colour of complexes (1 mark).
It gives no quantitative interpretation of magnetic data and does not distinguish between weak and strong field ligands (any two limitations) (1 mark).

Answer: (a) Cr +3, $d^{3}$ , $d^{2} s p^{3}$ , 3 unpaired, 3.87 BM; Fe +2, $d^{6}$ , $d^{2} s p^{3}$ , 0 unpaired, diamagnetic. (b) No; e.g. no quantitative magnetic interpretation, no distinction between weak and strong ligands.

Where students lose marks in Valence Bond Theory of Coordination Compounds

The slips examiners see most often, and what to write instead.

Mistake: Calling [Ni(CO)₄] square planar
ISC examiners noted 'square planar' written instead of 'tetrahedral' (2025).
Do this: Ni(0) becomes $3 d^{10}$ ; with no 3d orbital free, it must use $s p^{3}$ : tetrahedral.
Usually costs: 1 mark.
Mistake: Choosing a diamagnetic complex as paramagnetic
In a 2024 MCQ many candidates chose diamagnetic complexes; examiners asked for more practice with VBT and magnetic behaviour.
Do this: Always find the d count and decide whether the ligand pairs electrons before judging magnetism.
Usually costs: 1 mark.
Mistake: Wrong oxidation state or coordination number with bidentate ligands
Examiners advised practice in finding the oxidation state of the metal in complexes such as $[D (e n)_{2} (H_{2} O)_{2}] C l_{3}$ (2025).
Wrong working
$[M (e n)_{2} (H_{2} O)_{2}]^{3 +}$ : CN = 4
Right working
Each en gives 2 donor atoms: CN = 2 × 2 + 2 = 6; oxidation state +3.
Do this: Count donor atoms, not ligands, for coordination number.
Usually costs: 1 mark.
Mistake: Writing $s p^{3} d^{2}$ for a d³ ion
Students assume all H₂O complexes are outer orbital.
Do this: $d^{1}$ – $d^{3}$ ions have two empty 3d orbitals and always form $d^{2} s p^{3}$ complexes.
Usually costs: 1 mark.

Exam tips

Draw the orbital box diagram for the free ion and for the complex; examiners give marks for it.
Label which electron pairs come from the ligands, for example with crosses.
Use 'inner orbital/low spin' and 'outer orbital/high spin' together in answers.
Quote NCERT pairs: $[C o (N H_{3})_{6}]^{3 +}$ / $[C o F_{6}]^{3 -}$ , $[N i (C N)_{4}]^{2 -}$ / $[N i C l_{4}]^{2 -}$ , $[F e (C N)_{6}]^{3 -}$ / $[F e F_{6}]^{3 -}$ .
When a question gives a magnetic moment, start by finding n; it decides the hybridisation.

Practice questions on Valence Bond Theory of Coordination Compounds

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following is an inner orbital complex? (Z: Cr 24, Fe 26, Ni 28, Zn 30)
(a) $[C r (H_{2} O)_{6}]^{3 +}$
(b) $[F e F_{6}]^{3 -}$
(c) $[N i (N H_{3})_{6}]^{2 +}$
(d) $[Z n (N H_{3})_{4}]^{2 +}$
Hint
A $d^{3}$ ion always has two empty 3d orbitals.
Answer
(a). Cr³⁺ ( $3 d^{3}$ ) uses two empty 3d orbitals: $d^{2} s p^{3}$ .
Q2 · 2 marks · Short answerBoard level
Name the hybridisation and shape for coordination number 4 and for coordination number 6 complexes according to VBT, and state which of the octahedral types is an inner orbital complex.
Hint
Two choices for each coordination number.
Answer
CN 4: $s p^{3}$ tetrahedral or $d s p^{2}$ square planar. CN 6: $d^{2} s p^{3}$ or $s p^{3} d^{2}$ , both octahedral. $d^{2} s p^{3}$ (using inner 3d orbitals) is the inner orbital type.
Q3 · 2 marks · Short answerHarder board level
Predict the hybridisation, shape and magnetic behaviour of $[A g (N H_{3})_{2}]^{+}$ (Ag Z = 47).
Hint
Ag⁺ is $4 d^{10}$ .
Answer
Ag⁺ is $4 d^{10}$ with empty 5s and 5p. Two NH₃ lone pairs enter 5s and one 5p: $s p$ , linear, no unpaired electrons, diamagnetic.
Q4 · 3 marks · Short answerHarder board level
Using VBT, compare $[T i (H_{2} O)_{6}]^{3 +}$ and $[V (H_{2} O)_{6}]^{3 +}$ : give the hybridisation, number of unpaired electrons and spin-only magnetic moment of each. (Z: Ti 22, V 23)
Hint
Both ions have fewer than four d electrons.
Answer
Ti³⁺ $3 d^{1}$ : $d^{2} s p^{3}$ , 1 unpaired, 1.73 BM. V³⁺ $3 d^{2}$ : $d^{2} s p^{3}$ , 2 unpaired, 2.83 BM. Both are inner orbital octahedral complexes because two 3d orbitals stay empty.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $[N i (C O)_{4}]$ is tetrahedral and diamagnetic.
Reason (R): In the zero oxidation state, the 4s electrons of nickel move into 3d under the influence of CO, giving $3 d^{10}$ , and the empty 4s and 4p orbitals form $s p^{3}$ hybrids.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count electrons in 3d after rearrangement.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
State three limitations of valence bond theory as applied to coordination compounds.
Hint
Think about colour, magnetism and ligand strength.
Answer
Any three: it involves several assumptions; it gives no quantitative interpretation of magnetic data; it does not explain colour; it does not explain thermodynamic or kinetic stability quantitatively; it cannot reliably predict tetrahedral versus square planar for CN 4; it does not distinguish weak and strong ligands.
Q7 · 3 marks · Short answerHOTS / competency
A four-coordinate complex ion $[C o X_{4}]^{2 -}$ of cobalt(II) has a magnetic moment of about 3.9 BM. Deduce the number of unpaired electrons and use VBT to decide whether it is tetrahedral or square planar. (Co Z = 27)
Hint
What would $d s p^{2}$ need?
Answer
$n (n + 2) \approx 15$ , so $n = 3$ . Co²⁺ is $3 d^{7}$ with 3 unpaired electrons when no pairing occurs; $s p^{3}$ uses 4s and 4p and leaves the 3d set untouched, so it fits. For $d s p^{2}$ one 3d orbital must be emptied, which would force seven electrons into four orbitals and leave only 1 unpaired electron (about 1.73 BM). So the complex is tetrahedral ( $s p^{3}$ ).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student measures the magnetic moments of two manganese(III) complexes (Mn Z = 25): $[M n (C N)_{6}]^{3 -}$ gives about 2.8 BM and $[M n C l_{6}]^{3 -}$ gives about 4.9 BM.
(i) Find the number of unpaired electrons in each. [1]
(ii) Give the VBT hybridisation of each and classify it as inner or outer orbital. [1]
(iii) Which ligand causes pairing of the 3d electrons? [1]
Hint
Mn³⁺ is $3 d^{4}$ .
Answer
(i) 2.8 BM: 2 unpaired; 4.9 BM: 4 unpaired. (ii) $[M n (C N)_{6}]^{3 -}$ : two of the four electrons pair, freeing two 3d orbitals, $d^{2} s p^{3}$ , inner orbital. $[M n C l_{6}]^{3 -}$ : no pairing, $s p^{3} d^{2}$ , outer orbital. (iii) CN⁻.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Hybridisation and shape of $[N i (C N)_{4}]^{2 -}$ ?Show answer
$d s p^{2}$ , square planar, diamagnetic.
2.Hybridisation of $[C o F_{6}]^{3 -}$ ?Show answer
$s p^{3} d^{2}$ , outer orbital, 4 unpaired electrons.
3.What is an inner orbital complex?Show answer
One that uses inner (n−1)d orbitals in $d^{2} s p^{3}$ hybridisation; usually low spin.
4.Shape of $[F e (C O)_{5}]$ ?Show answer
Trigonal bipyramidal ( $d s p^{3}$ ).
5.Magnetic moment of $[C r (N H_{3})_{6}]^{3 +}$ ?Show answer
3.87 BM (3 unpaired electrons).
6.Name one thing VBT cannot explain.Show answer
The colour of complexes.

Valence Bond Theory of Coordination Compounds: frequently asked questions

What is the difference between inner orbital and outer orbital complexes?

Inner orbital complexes use (n−1)d orbitals (d^2sp^3) and are usually low spin; outer orbital complexes use nd orbitals (sp^3d^2) and are high spin.

How does VBT predict the magnetic behaviour of a complex?

By showing how many 3d electrons remain unpaired after the metal makes its hybrid orbitals available; any unpaired electrons make the complex paramagnetic.

Why is [Ni(CN)₄]²⁻ square planar but [NiCl₄]²⁻ tetrahedral?

CN⁻ forces the eight 3d electrons to pair, freeing one 3d orbital for dsp^2 (square planar); Cl⁻ does not, so sp^3 (tetrahedral) is used and 2 electrons stay unpaired.

What are the limitations of valence bond theory?

It cannot explain colour, gives no quantitative magnetic data, cannot distinguish strong and weak ligands, and cannot reliably predict tetrahedral versus square planar geometry.

Do hybrid orbitals actually exist?

No. Hybridisation is a mathematical way of combining atomic orbital wave functions to describe bonding.