Amongst the following, the most stable complex is (i) [Fe(H₂O)₆]³⁺ (ii) [Fe(NH₃)₆]³⁺ (iii) [Fe(C₂O₄)₃]³⁻ (iv)

Exercises · 5.30

Q. Amongst the following, the most stable complex is

(i)

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

(ii)

[Fe(NH_3)_6]^{3+}

(iii)

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

(iv)

[FeCl_6]^{3-}

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Concept understanding — Crystal Field Stabilization Energy

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 Stabilization Energy (CFSE): Why the Formula Holds

Let's build this from first principles — not just memorizing numbers, but understanding why the energy changes happen.

1. The Core Idea: d-Orbitals Are Not All Equal in an Octahedral Field

In a free metal ion, all five d-orbitals have the same energy (degenerate). But when you place the ion inside an octahedral ligand field, something changes:

Ligands (negative point charges or dipoles) approach along the $x$ , $y$ , and $z$ axes.
Some d-orbitals point directly at the ligands → high repulsion → higher energy.
Other d-orbitals point between the ligands → less repulsion → lower energy.

Which orbitals point where?

Orbital	Lobes point toward	Repulsion with ligands?
$d_{x^2-y^2}$	Along $x$ and $y$ axes	High (directly at ligands)
$d_{z^2}$	Along $z$ axis (with a ring in $xy$ plane)	High (directly at ligands)
$d_{xy}$	Between $x$ and $y$ axes	Low (between ligands)
$d_{xz}$	Between $x$ and $z$ axes	Low
$d_{yz}$	Between $y$ and $z$ axes	Low

So the five d-orbitals split into two groups:

$e_g$ set (higher energy): $d_{x^2-y^2}$ , $d_{z^2}$
$t_{2g}$ set (lower energy): $d_{xy}$ , $d_{xz}$ , $d_{yz}$

2. The Energy Splitting: Why $\Delta_o$ and the "Barycenter" Rule

The total energy of all five d-orbitals must be conserved — it's the same as in the free ion. This is the barycenter (center of gravity) rule.

Let:

Energy of $t_{2g}$ orbitals = $-x$ (below barycenter)
Energy of $e_g$ orbitals = $+y$ (above barycenter)
The splitting energy between them = $\Delta_o$ (called $10 Dq$ in older texts)

So:

y - (-x) = \Delta_o \quad \Rightarrow \quad x + y = \Delta_o

Conservation of energy:

There are 3 $t_{2g}$ orbitals and 2 $e_g$ orbitals.
Total energy shift = $3(-x) + 2(+y) = 0$

From this:

-3x + 2y = 0 \quad \Rightarrow \quad 2y = 3x \quad \Rightarrow \quad y = \frac{3}{2}x

Substitute into $x + y = \Delta_o$ :

x + \frac{3}{2}x = \Delta_o \quad \Rightarrow \quad \frac{5}{2}x = \Delta_o \quad \Rightarrow \quad x = \frac{2}{5}\Delta_o

Then:

y = \frac{3}{2} \cdot \frac{2}{5}\Delta_o = \frac{3}{5}\Delta_o

Key result:

Each $t_{2g}$ electron is stabilized by $\mathbf{-\frac{2}{5}\Delta_o}$
Each $e_g$ electron is destabilized by $\mathbf{+\frac{3}{5}\Delta_o}$

3. The CFSE Formula for Octahedral Complexes

Let:

$n_{t_{2g}}$ = number of electrons in $t_{2g}$ orbitals
$n_{e_g}$ = number of electrons in $e_g$ orbitals

Then:

\boxed{\text{CFSE} = -\frac{2}{5}\Delta_o \cdot n_{t_{2g}} + \frac{3}{5}\Delta_o \cdot n_{e_g}}

Why this is the stabilization energy:

The negative sign means energy is lowered (stabilization).
The positive term means energy is raised (destabilization).
Net CFSE = how much more stable the complex is compared to the free ion.

4. The "Why" Behind Pairing Energy and High/Low Spin

When you add electrons beyond $d^3$ , you face a choice:

Example: $d^4$ configuration

Option A (High spin):

Put 4th electron in $e_g$ (higher energy)
Cost: $+\frac{3}{5}\Delta_o$ (destabilization)
Benefit: No pairing energy ( $P$ )

Option B (Low spin):

Pair the 4th electron in $t_{2g}$
Cost: Pairing energy $P$ (electrostatic repulsion between two electrons in same orbital)
Benefit: Avoid $+\frac{3}{5}\Delta_o$ destabilization

The decision rule:

If $\Delta_o > P$ → Low spin (pairing is cheaper than going to $e_g$ )
If $\Delta_o < P$ → High spin (going to $e_g$ is cheaper than pairing)

CFSE for low-spin $d^4$ :

4 \times \left(-\frac{2}{5}\Delta_o\right) + 0 \times \left(+\frac{3}{5}\Delta_o\right) + P = -\frac{8}{5}\Delta_o + P

CFSE for high-spin $d^4$ : …

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$[Fe(C_2O_4)_3]^{3-}$ is the most stable because oxalate is a bidentate (chelating) ligand forming three stable five-membered rings - the chelate effect gives it by far the largest formation constant. Correct option: (iii).

All four are Fe(III) ( $d^5$ ) octahedral complexes. Among the ligands, $H_2O$ , $NH_3$ and $Cl^-$ are monodentate, whereas oxalate $C_2O_4^{2-}$ is bidentate. A bidentate ligand clamps the metal into a chelate ring, and the accompanying entropy gain (the chelate effect) makes chelate complexes far more stable than comparable monodentate complexes. Three oxalate rings therefore make $[Fe(C_2O_4)_3]^{3-}$ …

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Common Mistakes in Comparing Stability of Fe3+ Complexes

Mistake 1: Ignoring the Metal's d5 Configuration

The error: Students assume all complexes have the same CFSE because they all contain Fe3+, without checking whether any ligand is actually strong enough to force pairing.

How to avoid: Always write the d-electron count first. Fe3+ = $26 - 3 = 23$ electrons, i.e. d5 configuration.

Mistake 2: Wrongly Assuming Oxalate Forces Low Spin (the key trap in this question)

The error: Students see oxalate placed above $H_2O$ / $NH_3$ in the spectrochemical series and conclude it must force a low-spin $d^5$ configuration with a large CFSE, making $[Fe(C_2O_4)_3]^{3-}$ 'win' on CFSE grounds.

Why it's wrong: For $Fe^{3+}$ ( $d^5$ ), the pairing energy is unusually high (a half-filled $t_{2g}^3 e_g^2$ arrangement is already favourable), so even oxalate is not strong enough to force pairing. $[Fe(C_2O_4)_3]^{3-}$ is experimentally high-spin ( $\mu \approx 5.9$ BM) - the same spin state as the other three complexes.

How to avoid: Don't assume higher-in-the-spectrochemical-series automatically means low spin here - for a half-filled $d^5$ ion, only very strong ligands like $CN^-$ can force low spin. Check the experimental magnetic moment when in doubt.

Mistake 3: Missing the Chelate Effect

The error: Students compare only CFSE values and, once they see (correctly or not) that CFSE is similar for all four, conclude the complexes should be similarly stable.

How to avoid: Always check ligand denticity. Oxalate ( $C_2O_4^{2-}$ ) is bidentate - it forms a 5-membered chelate ring with the metal. Replacing monodentate ligands with a chelating one is entropically favourable (more free particles released), which is the real reason $[Fe(C_2O_4)_3]^{3-}$ is far more stable - this is the chelate effect, an entropy-driven effect, not a CFSE effect.

Mistake 4: Misplacing Ligands in the Spectrochemical Series

The error: Students rank NH3 as weaker than H2O, or Cl- as stronger than H2O.

Correct order (increasing field strength):

\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^-

--- …

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Q.Amongst the following, the most stable complex is

Concept understanding — Crystal Field Stabilization Energy

Crystal Field Splitting: From Intuition to Precision

The Precise Statement

Why This Matters

Why this formula?

Crystal Field Stabilization Energy (CFSE): Why the Formula Holds

1. The Core Idea: d-Orbitals Are Not All Equal in an Octahedral Field

Which orbitals point where?

2. The Energy Splitting: Why $\Delta_o$ and the "Barycenter" Rule

Conservation of energy:

Key result:

3. The CFSE Formula for Octahedral Complexes

Why this is the stabilization energy:

4. The "Why" Behind Pairing Energy and High/Low Spin

Example: $d^4$ configuration

The decision rule:

CFSE for low-spin $d^4$ :

CFSE for high-spin $d^4$ : …

Method: Chelate-Effect Stability Comparison

Step 1: Identify the metal ion and its $d$ -electron count

Step 2: Check whether any ligand is strong enough to force low spin

Step 3: Identify the ligand that can chelate

Step 4: Apply the chelate effect (entropy argument) …

Common Mistakes in Comparing Stability of Fe3+ Complexes

Mistake 1: Ignoring the Metal's d5 Configuration

Mistake 2: Wrongly Assuming Oxalate Forces Low Spin (the key trap in this question)

Mistake 3: Missing the Chelate Effect

Mistake 4: Misplacing Ligands in the Spectrochemical Series

Q.Amongst the following, the most stable complex is

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