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

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

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

Fe is in +2 state ($3d^6$). CN⁻ is a strong field ligand, causing pairing. Hybridisation: $d^2sp^3$ (inner orbital). All electrons paired → diamagnetic.

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

Fe is in +3 state ($3d^5$). F⁻ is a weak field ligand, no pairing. Hybridisation: $sp^3d^2$ (outer orbital). Five unpaired electrons → paramagnetic.

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

Co is in +3 state ($3d^6$). Oxalate ($C_2O_4^{2-}$) is a strong field ligand, causing pairing. Hybridisation: $d^2sp^3$ (inner orbital). All electrons paired → diamagnetic.

(iv) $[CoF_6]^{3-}$ …

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

1. 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.

2. 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).

3. 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).

4. 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.

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

1. 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.

2. Electronic configuration of Fe³⁺.

   Fe³⁺: $[Ar] 3d^5$ (five d-electrons).

3. 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).

4. 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.

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

1. 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.

2. Electronic configuration of Co³⁺.

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

   Co³⁺: $[Ar] 3d^6$ (six d-electrons).

3. 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). …

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

1. Find oxidation state of the central metal ion.
2. Write electronic configuration of the metal ion (in free state).
3. Decide ligand field strength — strong field ligands (CN⁻, CO, en, C₂O₄²⁻) cause pairing; weak field ligands (F⁻, Cl⁻, H₂O usually) do not force pairing.
4. Determine number of unpaired electrons after pairing (if any).
5. 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)
6. 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. …

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. …