Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i)

Exercises · 5.15

Q. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(i)

[Fe(CN)_6]^{4-}

(ii)

[FeF_6]^{3-}

(iii)

[Co(C_2O_4)_3]^{3-}

(iv)

[CoF_6]^{3-}

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Concept understanding — Werner Coordination Theory

Werner Coordination Theory: The Idea That Changed Inorganic Chemistry

Imagine you're looking at a salt like cobalt(III) chloride. The formula is written as $\ce{CoCl3}$ , and when you dissolve it in water, you expect to find $\ce{Co^{3+}}$ and $\ce{Cl^-}$ ions. But something strange happens: when you add silver nitrate (which precipitates chloride ions), only some of the chlorine comes out as silver chloride. Not all of it. And the amount that precipitates depends on how you made the compound.

This was the puzzle that faced chemists in the late 1800s. Compounds like $\ce{CoCl3·6NH3}$ (orange-yellow) and $\ce{CoCl3·5NH3}$ (purple) had the same metal and the same ligands (ammonia), but different colours, different conductivities in solution, and different numbers of chloride ions that could be precipitated. The old ideas of fixed valency couldn't explain it.

Alfred Werner proposed a radical solution in 1893. He said: a metal ion has two kinds of valency.

The Core Intuition

Think of a metal ion like a king in a castle. The king has two types of relationships:

Primary valency (today: oxidation state) — this is the king's royal authority. It's fixed, non-directional, and satisfied by negative ions. For cobalt(III), this is +3. It's like the king's crown: it doesn't change.
Secondary valency (today: coordination number) — this is the king's personal bodyguard. The king can have a fixed number of guards (usually 4 or 6) who stand in specific positions around him. These guards can be neutral molecules (like ammonia) or negative ions (like chloride). The key: these guards are directly attached to the metal, forming a stable cluster called the coordination sphere.

The revolutionary idea: the chloride ions that act as bodyguards (inside the coordination sphere) do not behave like free ions. They don't precipitate with silver nitrate. They don't conduct electricity. They are "locked" to the metal.

The Precise Statement

Werner Coordination Theory (1893) Every metal atom has two types of valency : Primary valency (ionisable): corresponds to the oxidation state. It is satisfied by negative ions. These ions are outside the coordination sphere and behave as free ions in solution. Secondary valency (non-ionisable): corresponds to the coordination number. It is satisfied by neutral molecules or negative ions directly bonded to the metal. These are inside the coordination sphere and do not dissociate. The secondary valencies are directional — they point to fixed positions in space around the metal, giving the complex a definite geometry (e.g., octahedral for coordination number 6, square planar for 4). The primary valency is non-directional — it is just a number, not a spatial arrangement.

How It Explains the Puzzle

Take the compound $\ce{CoCl3·6NH3}$ (orange-yellow). Werner said:

Cobalt has primary valency +3 (needs three negative charges to satisfy it).
Cobalt has secondary valency 6 (can hold six ligands around it).
The six ammonia molecules satisfy all six secondary valencies. So the chloride ions cannot be inside the coordination sphere — they must be outside, as free ions.
Structure: $[\ce{Co(NH3)6}]\ce{Cl3}$ . All three chlorides precipitate with $\ce{AgNO3}$ .

Now take $\ce{CoCl3·5NH3}$ (purple):

Again, primary valency +3, secondary valency 6.
Five ammonia molecules satisfy five secondary valencies. One chloride ion must fill the sixth spot — it becomes a ligand inside the sphere.
The other two chlorides are outside as free ions.
Structure: $[\ce{Co(NH3)5Cl}]\ce{Cl2}$ . Only two chlorides precipitate.

Note

The number of free ions in solution determines the conductivity and the number of precipitable chlorides. Werner's theory predicted exactly these numbers — and experiments confirmed them.

The Geometry Insight …

Why this formula?

Werner Coordination Theory: Why the Key Formulas Hold

Werner Coordination Theory (1893) revolutionized inorganic chemistry by explaining how metal ions bind ligands. Let's build the reasoning from first principles — not just memorize formulas.

1. The Core Observation: Primary vs. Secondary Valence

Werner noticed that metal compounds had two types of bonding capacity:

Primary valence (now oxidation state): Satisfies the metal's charge — ionic in nature.
Secondary valence (now coordination number): Determines how many ligands attach — directional, spatial in nature.

Why this distinction?

Consider $\ce{CoCl3 \cdot 6 NH3}$ (one of Werner's classic compounds).

The compound is electrically neutral overall.
Adding $\ce{AgNO3}$ precipitates all 3 Cl⁻ as $\ce{AgCl}$ — meaning all chlorides are free ions.
Therefore, the $\ce{NH3}$ molecules must be directly bonded to $\ce{Co}$ , not the chlorides.

This forces the idea: Co has a fixed capacity for direct ligand attachment (secondary valence = 6 here), separate from its charge balance (primary valence = +3).

2. The Key Formula: Coordination Number = Number of Ligands Attached

Formula:

\text{Coordination number} = \text{number of donor atoms directly bonded to the metal}

Why this holds:

Werner's experiments showed that only a fixed number of ligands could be replaced without breaking the compound's identity.
For $\ce{CoCl3 \cdot 6 NH3}$ , adding acid doesn't remove $\ce{NH3}$ easily — they are coordinated.
The maximum number of such tightly bound ligands is the coordination number — a property of the metal ion, not the counterions.

Derivation from data:

If you have $\ce{[Co(NH3)6]Cl3}$ , conductivity measurements show 4 ions in solution ( $\ce{[Co(NH3)6]^{3+}}$ + 3 Cl⁻).

If you had $\ce{[Co(NH3)5Cl]Cl2}$ , conductivity shows 3 ions.

The number of chlorides inside the coordination sphere (non-precipitable) plus those outside must sum to the total chlorides. This gives the coordination number directly.

3. The Geometry Formula: Coordination Number Determines Shape

Werner proposed that secondary valences are directed in space — leading to specific geometries.

Coordination Number	Geometry	Why?
2	Linear	Minimizes repulsion between 2 ligands
4	Tetrahedral or Square planar	4 points in space — two arrangements possible
6	Octahedral	6 ligands at 90° angles — most symmetric

Why octahedral for 6?

6 ligands around a central atom must be placed to maximize separation.
The octahedron (6 vertices, all equidistant from center, 90° between adjacent bonds) is the only regular polyhedron with 6 vertices.
This explains why $\ce{[Co(NH3)6]^{3+}}$ is octahedral — no other arrangement gives equal bond angles and distances.

4. The Isomer Counting Formula: Why $2^n$ or $n!$ Appears

Werner used isomer counts to confirm geometry. For an octahedral complex $\ce{[Ma2b2c2]}$ :

Number of geometrical isomers = 5 (not 6, not 4)

Why this formula?

Place the two 'a' ligands: they can be cis (90°) or trans (180°).
For each, place 'b' and 'c' in remaining positions — but symmetry reduces duplicates. …

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Valence Bond Theory explains bonding in coordination compounds by considering the hybridisation of the central metal ion’s orbitals, which depends on the ligand field strength. For the given complexes: (i) $[Fe(CN)_6]^{4-}$ has $d^2sp^3$ hybridisation (inner orbital, low spin), (ii) $[FeF_6]^{3-}$ has $sp^3d^2$ hybridisation (outer orbital, high spin), (iii) $[Co(C_2O_4)_3]^{3-}$ has $d^2sp^3$ hybridisation (inner orbital, low spin), and (iv) $[CoF_6]^{3-}$ has $sp^3d^2$ hybridisation (outer orbital, high spin).

Werner Coordination Theory first suggested that metal ions have primary (ionisable) and secondary (non-ionisable) valencies. Valence Bond Theory (VBT), developed by Linus Pauling, refined this by describing how the metal ion’s vacant orbitals hybridise to accommodate lone pairs from ligands. The key insight: strong field ligands (like CN⁻, C₂O₄²⁻) cause pairing of electrons in the metal’s d-orbitals, leading to inner orbital (low spin) complexes with $d^2sp^3$ hybridisation. Weak field ligands (like F⁻) do not force pairing, giving outer orbital (high spin) complexes with $sp^3d^2$ hybridisation. The number of unpaired electrons determines magnetic behaviour.

Let’s apply this step by step to each complex.

(i) $[Fe(CN)_6]^{4-}$

Determine the oxidation state of iron.

CN⁻ is a neutral ligand (charge −1 each). Let Fe be $x$ :

$x + 6(-1) = -4 \implies x = +2$ . So, Fe is in the +2 state.
Write the electronic configuration of Fe²⁺.

Fe (atomic number 26): $[Ar] 3d^6 4s^2$ .

Fe²⁺: $[Ar] 3d^6$ (the two 4s electrons are lost first).
Identify ligand strength and decide spin state.

CN⁻ is a strong field ligand. It causes pairing of electrons in the 3d orbitals.

The six d-electrons pair up completely: three pairs occupy three d-orbitals, leaving two d-orbitals empty. This gives zero unpaired electrons (diamagnetic).
Determine hybridisation.

The metal uses two empty 3d orbitals, one 4s, and three 4p orbitals to form six equivalent $d^2sp^3$ hybrid orbitals. These accept lone pairs from six CN⁻ ligands.

This is an inner orbital (low spin) complex.

Watch out

A common mistake is to forget that CN⁻ is a strong field ligand and assume high spin for Fe²⁺. Always check the ligand series: CN⁻, CO, NH₃ (strong) vs. F⁻, Cl⁻, H₂O (weak).

(ii) $[FeF_6]^{3-}$

Oxidation state of iron.

F⁻ is −1 each. Let Fe be $x$ :

$x + 6(-1) = -3 \implies x = +3$ . So, Fe is in the +3 state.
Electronic configuration of Fe³⁺.

Fe³⁺: $[Ar] 3d^5$ (five d-electrons).
Ligand strength and spin state.

F⁻ is a weak field ligand. It does not cause pairing.

The five d-electrons occupy all five d-orbitals singly (Hund’s rule), giving five unpaired electrons (paramagnetic).
Hybridisation.

Since no d-orbitals are vacated by pairing, the metal uses outer orbitals: one 4s, three 4p, and two 4d orbitals to form six $sp^3d^2$ hybrid orbitals.

This is an outer orbital (high spin) complex.

Tip

For Fe³⁺ with weak field ligands, the maximum number of unpaired electrons is 5. This is a quick check: if you see F⁻, Cl⁻, or H₂O with Fe³⁺, expect high spin.

(iii) $[Co(C_2O_4)_3]^{3-}$

Oxidation state of cobalt.

Oxalate ion ( $C_2O_4^{2-}$ ) is a bidentate ligand with charge −2 each. Let Co be $x$ :

$x + 3(-2) = -3 \implies x = +3$ . So, Co is in the +3 state.
Electronic configuration of Co³⁺.

Co (atomic number 27): $[Ar] 3d^7 4s^2$ .

Co³⁺: $[Ar] 3d^6$ (six d-electrons).
Ligand strength and spin state.

Oxalate ( $C_2O_4^{2-}$ ) is a strong field ligand (it appears high in the spectrochemical series).

The six d-electrons pair up completely: three pairs in three d-orbitals, leaving two d-orbitals empty. Zero unpaired electrons (diamagnetic). …

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Method: Valence Bond Theory (VBT) for Coordination Compounds

Core idea: Metal ion uses hybrid orbitals to accept lone pairs from ligands. The hybridisation depends on the metal's oxidation state, coordination number, and ligand field strength (strong vs weak field ligands).

General Steps for VBT Analysis

Find oxidation state of the central metal ion.
Write electronic configuration of the metal ion (in free state).
Decide ligand field strength — strong field ligands (CN⁻, CO, en, C₂O₄²⁻) cause pairing; weak field ligands (F⁻, Cl⁻, H₂O usually) do not force pairing.
Determine number of unpaired electrons after pairing (if any).
Choose hybridisation based on coordination number and geometry:
- Coordination number 6 → $d^2sp^3$ (inner orbital, low spin) or $sp^3d^2$ (outer orbital, high spin)
Describe bonding — ligand lone pairs fill hybrid orbitals; unpaired electrons give magnetic behaviour.

(i) $[Fe(CN)_6]^{4-}$

Step 1: Oxidation state

Let Fe be $x$ : $x + 6(-1) = -4 \Rightarrow x = +2$

So, Fe(II).

Step 2: Electronic configuration

Fe (atomic no. 26): $[Ar]\,3d^6 4s^2$

Fe²⁺: $[Ar]\,3d^6$

Step 3: Ligand field

CN⁻ is a strong field ligand → causes pairing.

Step 4: Unpaired electrons after pairing

All 6 electrons pair up in 3d orbitals → 0 unpaired electrons (diamagnetic).

Step 5: Hybridisation

One 3d orbital is vacated → uses $d^2sp^3$ hybridisation (inner orbital complex).

Step 6: Bonding

Six CN⁻ donate lone pairs into six $d^2sp^3$ hybrid orbitals.

Geometry: Octahedral.

Magnetic behaviour: Diamagnetic.

(ii) $[FeF_6]^{3-}$

Step 1: Oxidation state

$x + 6(-1) = -3 \Rightarrow x = +3$

Fe(III).

Step 2: Electronic configuration

Fe³⁺: $[Ar]\,3d^5$

Step 3: Ligand field

F⁻ is a weak field ligand → no pairing.

Step 4: Unpaired electrons

All 5 electrons remain unpaired → 5 unpaired electrons (high spin).

Step 5: Hybridisation

Uses outer $sp^3d^2$ hybridisation (outer orbital complex).

Step 6: Bonding

Six F⁻ donate lone pairs into six $sp^3d^2$ hybrid orbitals.

Geometry: Octahedral.

Magnetic behaviour: Paramagnetic (5 unpaired electrons).

(iii) $[Co(C_2O_4)_3]^{3-}$

Step 1: Oxidation state

Oxalate (C₂O₄²⁻) is a bidentate ligand with charge -2 each.

$x + 3(-2) = -3 \Rightarrow x = +3$

Co(III).

Step 2: Electronic configuration

Co (atomic no. 27): $[Ar]\,3d^7 4s^2$

Co³⁺: $[Ar]\,3d^6$

Step 3: Ligand field

C₂O₄²⁻ is a strong field ligand → causes pairing.

Step 4: Unpaired electrons

All 6 electrons pair up → 0 unpaired electrons (diamagnetic).

Step 5: Hybridisation

One 3d orbital vacated → $d^2sp^3$ hybridisation (inner orbital).

Step 6: Bonding

Three oxalate ions (each bidentate, donating 2 lone pairs) occupy six coordination sites via $d^2sp^3$ hybrid orbitals.

Geometry: Octahedral. …

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Here are the most common mistakes students make when applying Valence Bond Theory (VBT) to these Werner-type coordination entities, along with how to avoid each.

Mistake 1: Forgetting to determine the correct oxidation state of the central metal ion first

Why it happens: Students jump straight to hybridization without calculating the charge on the metal. This leads to wrong $d$ -electron counts and wrong geometry.

How to avoid: Always start with the formula:

For $[Fe(CN)_6]^{4-}$ : Let oxidation state of Fe be $x$ . $x + 6(-1) = -4 \implies x = +2$ . So, Fe(II).
For $[FeF_6]^{3-}$ : $x + 6(-1) = -3 \implies x = +3$ . So, Fe(III).
For $[Co(C_2O_4)_3]^{3-}$ : Oxalate ( $C_2O_4^{2-}$ ) is bidentate. $x + 3(-2) = -3 \implies x = +3$ . So, Co(III).
For $[CoF_6]^{3-}$ : $x + 6(-1) = -3 \implies x = +3$ . So, Co(III).

Key rule: Write the charge balance equation first — never guess the oxidation state.

Mistake 2: Confusing strong-field vs weak-field ligands and their effect on pairing

Why it happens: Students memorize "CN⁻ is strong, F⁻ is weak" but forget to apply pairing logic to the $d$ -orbital configuration.

How to avoid: Use the spectrochemical series:

Strong field (low spin): CN⁻, CO, en, NH₃ (for Co³⁺), oxalate ( $C_2O_4^{2-}$ ) — causes pairing.
Weak field (high spin): F⁻, Cl⁻, H₂O (for Fe²⁺/Fe³⁺) — no pairing.

Apply to each:

$[Fe(CN)_6]^{4-}$ : Fe²⁺ has $d^6$ . CN⁻ is strong → low spin → all 6 electrons paired → $t_{2g}^6$ , $e_g^0$ .
$[FeF_6]^{3-}$ : Fe³⁺ has $d^5$ . F⁻ is weak → high spin → 5 unpaired electrons → $t_{2g}^3$ , $e_g^2$ .
$[Co(C_2O_4)_3]^{3-}$ : Co³⁺ has $d^6$ . Oxalate is strong → low spin → all paired → $t_{2g}^6$ , $e_g^0$ .
$[CoF_6]^{3-}$ : Co³⁺ has $d^6$ . F⁻ is weak → high spin → 4 unpaired electrons → $t_{2g}^4$ , $e_g^2$ .

Key rule: Pairing only happens with strong-field ligands. Weak-field ligands keep electrons unpaired.

Mistake 3: Choosing the wrong hybridization (e.g., $sp^3d^2$ vs $d^2sp^3$ )

Why it happens: Students forget that inner orbital (low spin) uses $(n-1)d$ orbitals, while outer orbital (high spin) uses $nd$ orbitals.

How to avoid:

Low spin (paired electrons) → inner $d$ -orbitals are available → $d^2sp^3$ (octahedral, inner orbital complex).
High spin (unpaired electrons) → inner $d$ -orbitals are occupied → uses outer $d$ → $sp^3d^2$ (octahedral, outer orbital complex).

Apply:

$[Fe(CN)_6]^{4-}$ : low spin → $d^2sp^3$ .
$[FeF_6]^{3-}$ : high spin → $sp^3d^2$ .
$[Co(C_2O_4)_3]^{3-}$ : low spin → $d^2sp^3$ .
$[CoF_6]^{3-}$ : high spin → $sp^3d^2$ .

Key rule: Low spin = $d^2sp^3$ (inner). High spin = $sp^3d^2$ (outer).

Mistake 4: Ignoring the magnetic property (paramagnetic vs diamagnetic)

Why it happens: Students calculate hybridization but forget to state the magnetic behavior — a common exam requirement.

How to avoid: After determining unpaired electrons:

0 unpaired → diamagnetic.
1 or more unpaired → paramagnetic.

Check:

$[Fe(CN)_6]^{4-}$ : 0 unpaired → diamagnetic.
$[FeF_6]^{3-}$ : 5 unpaired → paramagnetic.
$[Co(C_2O_4)_3]^{3-}$ : 0 unpaired → diamagnetic.
$[CoF_6]^{3-}$ : 4 unpaired → paramagnetic. …

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Q.Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

Concept understanding — Werner Coordination Theory

Werner Coordination Theory: The Idea That Changed Inorganic Chemistry

The Core Intuition

The Precise Statement

How It Explains the Puzzle

The Geometry Insight …

Why this formula?

Werner Coordination Theory: Why the Key Formulas Hold

1. The Core Observation: Primary vs. Secondary Valence

2. The Key Formula: Coordination Number = Number of Ligands Attached

3. The Geometry Formula: Coordination Number Determines Shape

4. The Isomer Counting Formula: Why $2^n$ or $n!$ Appears

Concept: Valence Bond Theory (VBT) – Hybridisation and magnetic behaviour depend on the ligand field strength (strong vs weak field ligands).

(i) $[Fe(CN)_6]^{4-}$

(ii) $[FeF_6]^{3-}$

(iii) $[Co(C_2O_4)_3]^{3-}$

Method: Valence Bond Theory (VBT) for Coordination Compounds

General Steps for VBT Analysis

(i) $[Fe(CN)_6]^{4-}$

(ii) $[FeF_6]^{3-}$

(iii) $[Co(C_2O_4)_3]^{3-}$

Mistake 1: Forgetting to determine the correct oxidation state of the central metal ion first

Mistake 2: Confusing strong-field vs weak-field ligands and their effect on pairing

Mistake 3: Choosing the wrong hybridization (e.g., $sp^3d^2$ vs $d^2sp^3$ )

Mistake 4: Ignoring the magnetic property (paramagnetic vs diamagnetic)

Q.Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(i) [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−

(ii) [FeF6]3−[FeF_6]^{3-}[FeF6​]3−

(iii) [Co(C2O4)3]3−[Co(C_2O_4)_3]^{3-}[Co(C2​O4​)3​]3−

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(i) $[Fe(CN)_6]^{4-}$

(ii) $[FeF_6]^{3-}$

(iii) $[Co(C_2O_4)_3]^{3-}$

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