The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the follow

NCERT Exemplar · Q6

Q. The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?

(i)

[Fe(CO)_5]

(ii)

[Fe(CN)_6]^{3-}

(iii)

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

(iv)

[Fe(H_2O)_6]^{3+}

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Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.

Now imagine that six negative ions (or the negative ends of polar molecules) march in from the $x$ , $y$ , and $z$ axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.

This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.

The Precise Statement

d

For the most common geometry — octahedral — here is what happens:

Six ligands sit at the corners of an octahedron, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ , $-z$ axes.
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point their lobes directly along these axes. These are the $e_g$ set. They feel maximum repulsion → higher energy.
The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point their lobes between the axes (into the octahedral faces). These are the $t_{2g}$ set. They feel less repulsion → lower energy.

The energy gap between these two sets is denoted by $\Delta_o$ (or $10\,Dq$ ). The $t_{2g}$ set drops by $0.4\Delta_o$ and the $e_g$ set rises by $0.6\Delta_o$ , keeping the average energy unchanged (the "barycentre" rule).

Note

The labels $e_g$ and $t_{2g}$ come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.

Why This Matters

Crystal field splitting explains three things you will see repeatedly:

Colour — electrons can jump from $t_{2g}$ to $e_g$ by absorbing visible light. The gap $\Delta_o$ determines the colour you see.
Magnetism — if $\Delta_o$ is large, electrons pair up in the lower $t_{2g}$ set (low spin). If $\Delta_o$ is small, electrons spread out (high spin). This changes the number of unpaired electrons. …

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.

1. The Starting Point: Degenerate d-Orbitals

In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ — lobes lie between the x, y, z axes (called $t_{2g}$ set in octahedral symmetry)
$d_{x^2-y^2}$ , $d_{z^2}$ — lobes point directly along the x, y, z axes (called $e_g$ set)

Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

Their lobes point directly at the ligands.
The negatively charged ligands repel the electron density in these orbitals.
This repulsion raises the energy of these orbitals — they become less stable (higher energy).

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

Their lobes point between the axes (e.g., $d_{xy}$ lobes lie in the xy-plane but at 45° to x and y).
They avoid the ligands — less repulsion.
Their energy is lower than the $e_g$ set.

The Splitting Pattern

\Delta_{\text{oct}} = E(e_g) - E(t_{2g})

Where:

$E(e_g)$ = energy of $d_{x^2-y^2}$ and $d_{z^2}$ (higher)
$E(t_{2g})$ = energy of $d_{xy}$ , $d_{xz}$ , $d_{yz}$ (lower)
$\Delta_{\text{oct}}$ is called the crystal field splitting energy (CFSE)

Why the name? The $e_g$ orbitals are "doubly degenerate" (2 orbitals), $t_{2g}$ are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.

3. The Energy Conservation Rule

The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:

The center of gravity (average energy) of the split set equals the original degenerate energy.
For octahedral splitting:
- 2 $e_g$ orbitals go up by $+0.6\Delta_{\text{oct}}$ each
- 3 $t_{2g}$ orbitals go down by $-0.4\Delta_{\text{oct}}$ each

Check:

2 \times (+0.6\Delta) + 3 \times (-0.4\Delta) = 1.2\Delta - 1.2\Delta = 0

This conservation of energy is a fundamental constraint — the splitting is not arbitrary.

4. The Tetrahedral Case: Why It's Opposite and Smaller

In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:

The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals now point closer to the ligands (more repulsion).
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point away from ligands (less repulsion).

Result:

$e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower energy
$t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher energy

The splitting is inverted compared to octahedral.

Magnitude:

\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}

Why smaller?

Only 4 ligands (vs. 6) → less total repulsion.
Ligands are not directly along axes → weaker interaction. …

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The chelate effect makes multidentate ligands form more stable complexes than monodentate ones. Among the options, $[Fe(C_2O_4)_3]^{3-}$ has three bidentate oxalate ligands, giving the highest chelation and thus the greatest stability. The most stable complex is (iii).

Why the chelate effect decides stability

The chelate effect is the extra thermodynamic stability gained when a polydentate ligand (one that binds through multiple donor atoms) wraps around a metal ion, forming a ring. This happens because:

Entropy favours chelation: One chelating ligand replaces several monodentate ligands, releasing more free molecules into solution. More particles mean higher entropy, which drives the equilibrium toward the complex.
Kinetic and thermodynamic synergy: Chelating ligands also dissociate more slowly because breaking one bond still leaves the ligand attached — it has to break all bonds at once to leave.

So, the more chelate rings a complex has, the more stable it tends to be. Let’s examine each option.

Step-by-step analysis

Identify the ligand denticity in each complex
- (i) $[Fe(CO)_5]$ — CO is monodentate (binds through carbon only). No chelation.
- (ii) $[Fe(CN)_6]^{3-}$ — CN⁻ is monodentate (binds through carbon). No chelation.
- (iii) $[Fe(C_2O_4)_3]^{3-}$ — Oxalate ( $C_2O_4^{2-}$ ) is bidentate (binds through two oxygen atoms). Each oxalate forms a 5-membered chelate ring. Three oxalates give three chelate rings.
- (iv) $[Fe(H_2O)_6]^{3+}$ — H₂O is monodentate. No chelation.
Compare the chelate effect strength

Only option (iii) has any chelation. The other three rely solely on monodentate ligands, so they lack the entropy advantage. Even if we compare bond strengths, the chelate effect typically outweighs differences in ligand field strength for overall stability — a chelated complex is almost always more stable than an analogous monodentate one.

Tip

A quick way to spot the chelate effect in a multiple-choice question: look for the complex with a polydentate ligand. Here, oxalate is the only bidentate ligand among the options. …

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Common Mistakes & How to Avoid Them

Mistake 1: Assuming the strongest-field ligand automatically gives the most stable complex

The error: Students recall that $CN^-$ and $CO$ are very strong-field ligands (large $\Delta_o$ , large CFSE) and jump to the conclusion that $[Fe(CO)_5]$ must be the most stable species in the list.

Why it's wrong: Overall thermodynamic stability of a complex is not decided by crystal-field splitting alone — when a polydentate (chelating) ligand is present, the chelate effect is usually the dominant contribution, and it easily outweighs a purely ligand-field argument. $[Fe(CO)_5]$ is also a fundamentally different kind of species (zero-valent iron, trigonal bipyramidal, stabilised by synergic $\pi$ -back-bonding) — it isn't on the same footing as the three octahedral $Fe^{3+}$ complexes the question is really comparing.

How to avoid: Whenever a polydentate ligand (like oxalate, en, EDTA) appears among the options, check for chelation first — in "which is most stable" questions, it is very often the deciding factor.

Mistake 2: Forgetting that oxalate ( $C_2O_4^{2-}$ ) is bidentate

The error: Treating oxalate as a simple monodentate anion, the way $Cl^-$ or $NO_3^-$ behaves.

Why it's wrong: Oxalate binds through two oxygen atoms (one from each carboxylate group), forming a stable five-membered chelate ring with the metal. $[Fe(C_2O_4)_3]^{3-}$ has three such rings around the iron centre.

How to avoid: Always check ligand denticity before comparing "stability" across options. $NH_3$ , $H_2O$ , $CN^-$ , $CO$ are monodentate; oxalate, ethylenediamine (en), and EDTA are polydentate.

Mistake 3: Confusing ligand-field strength with overall thermodynamic stability

The error: Assuming $[Fe(CN)_6]^{3-}$ must be the "most stable" option simply because $CN^-$ sits at the strong-field end of the spectrochemical series.

Why it's wrong: A large CFSE does add to a complex's stability, but the chelate effect — driven largely by the favourable entropy of releasing several monodentate ligands (or solvent molecules) for one chelating ligand — typically gives an even bigger stability boost. This is exactly why $Fe^{3+}$ –oxalate complexes are so much more stable in practice than a simple ligand-field comparison would suggest.

How to avoid: Don't treat "strong-field ligand" and "most stable complex" as synonyms. Chelation is a separate — and often larger — contribution to stability.

Mistake 4: Ignoring that $[Fe(CO)_5]$ is not a comparable species

The error: Trying to rank $[Fe(CO)_5]$ (iron in the 0 oxidation state) directly against the three $Fe^{3+}$ complexes. …

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Q.The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?

Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

The Precise Statement

Why This Matters

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

1. The Starting Point: Degenerate d-Orbitals

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

The Splitting Pattern

3. The Energy Conservation Rule

4. The Tetrahedral Case: Why It's Opposite and Smaller

Result:

Magnitude:

Why the chelate effect decides stability

Step-by-step analysis

Method: Chelate Effect & Crystal Field Stabilisation Energy (CFSE) Analysis

Step 1 — Identify the central metal ion and its oxidation state

Step 2 — Determine ligand field strength and geometry

Step 3 — Apply the chelate effect

Common Mistakes & How to Avoid Them

Mistake 1: Assuming the strongest-field ligand automatically gives the most stable complex

Mistake 2: Forgetting that oxalate ( $C_2O_4^{2-}$ ) is bidentate

Mistake 3: Confusing ligand-field strength with overall thermodynamic stability

Mistake 4: Ignoring that $[Fe(CO)_5]$ is not a comparable species

Q.The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?

Why the chelate effect decides stability

Step-by-step analysis

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