What is the relationship between observed colour of the complex and the wavelength of light absorbed by the co

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Q. What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?

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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 observed colour of a complex is the complementary colour of the light it absorbs — if a complex absorbs light of a certain wavelength, we see the colour opposite to that wavelength on the colour wheel.

The key to understanding this lies in how our eyes perceive colour. A complex appears coloured because it absorbs some wavelengths of visible light and transmits (or reflects) the rest. The colour we see is not the absorbed colour — it is the sum of all the remaining wavelengths that reach our eyes.

Why this happens: The concept of complementary colours

When white light (which contains all visible wavelengths) falls on a complex, certain wavelengths are absorbed by the d-electrons jumping between split energy levels (crystal field splitting). The light that is not absorbed is what reaches your eye. Your brain interprets this mixture of leftover wavelengths as a single colour — the complementary colour of the absorbed light.

For example, if a complex absorbs strongly in the red region (around 650–700 nm), the transmitted light will be rich in blue-green wavelengths. So the complex appears blue-green, not red.

\text{Observed colour} = \text{Complementary colour of absorbed wavelength}

Step-by-step reasoning

White light contains all colours. When it hits a complex, some wavelengths are absorbed by the d-orbital electrons (due to crystal field splitting). The absorbed wavelength corresponds to the energy gap $\Delta$ : $\Delta = \frac{hc}{\lambda_{\text{absorbed}}}$ .
The transmitted light is white light minus the absorbed wavelengths. This is a subtractive process — the absorbed colour is removed from the spectrum.
Our eye sees the mixture of remaining wavelengths. The brain does not see individual wavelengths; it averages the remaining spectrum into a single perceived colour.
This perceived colour is the complement of the absorbed colour. On a standard colour wheel, complementary colours are opposite each other:
- Red ↔ Cyan (blue-green)
- Orange ↔ Blue
- Yellow ↔ Violet
- Green ↔ Magenta
The relationship is inverse, not direct. A common mistake is to think the complex appears the same colour as the light it absorbs. That is wrong — the complex appears the opposite colour.

Watch out

Never say "the complex is red because it absorbs red light." That is incorrect. If it absorbs red light, it appears cyan (the complement of red). The absorbed colour and observed colour are always complementary.

A concrete example

Consider the complex $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ . It absorbs light at about 500 nm (green region). The complementary colour of green is violet-red (magenta). Indeed, this complex appears violet-red in solution.

| Absorbed wavelength (nm) | Absorbed colour | Observed (complementary) colour |

|--------------------------|-----------------|----------------------------------| …

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Here are the most common mistakes students make when linking the observed colour of a coordination complex to the wavelength of light absorbed, along with clear ways to avoid each.

Mistake 1: Confusing “absorbed” with “observed” colour

The error:

Students often say “the complex is blue because it absorbs blue light.” This is wrong. A complex appears blue because it absorbs orange/red light (the complementary colour) and transmits/reflects blue light.

Why it happens:

The eye sees the light that is not absorbed. The absorbed wavelength is complementary to the observed colour.

How to avoid:

Memorise the colour wheel for $d$ -block complexes.
Use the rule:

Observed colour = complementary colour of absorbed light.
Example: If a complex absorbs at $\lambda \approx 580\ \text{nm}$ (yellow), it appears violet (complement of yellow).

Mistake 2: Forgetting the complementary colour relationship

The error:

Students try to guess the observed colour without knowing the complementary pair (e.g., green–red, blue–orange, yellow–violet).

Why it happens:

The relationship is not intuitive — it requires memorisation of the visible spectrum’s complementary pairs.

How to avoid:

Draw or visualise the colour wheel:
- Red ↔ Green
- Orange ↔ Blue
- Yellow ↔ Violet
Practice with examples:
- Absorbs blue → appears orange
- Absorbs green → appears red
- Absorbs violet → appears yellow

Mistake 3: Ignoring the wavelength range of visible light

The error:

Students treat “absorbed wavelength” as any number without checking if it falls in the visible range (400–700 nm). If a complex absorbs in the UV or IR, it appears white or colourless (no visible absorption).

Why it happens:

They assume all $d$ – $d$ transitions produce colour, but if $\Delta$ is very large or very small, absorption is outside visible range.

How to avoid:

Remember:
- $\lambda_{\text{absorbed}} < 400\ \text{nm}$ (UV) → complex appears colourless (transmits all visible light).
- $\lambda_{\text{absorbed}} > 700\ \text{nm}$ (IR) → complex appears colourless (absorbs no visible light).
Only when $\lambda_{\text{absorbed}}$ is between 400–700 nm does the complex show colour.

Mistake 4: Mixing up the effect of ligand field strength

The error:

Students think “stronger ligand → larger $\Delta$ → shorter $\lambda$ absorbed” but then incorrectly predict the observed colour.

Why it happens:

They forget that shorter $\lambda$ (higher energy) corresponds to a different complementary colour.

How to avoid:

Use the sequence:

Strong field → large $\Delta$ → absorbs shorter $\lambda$ (blue/violet) → appears orange/yellow

Weak field → small $\Delta$ → absorbs longer $\lambda$ (red/orange) → appears green/blue
Practice with examples:
- $[\text{Co(NH}_3)_6]^{3+}$ (strong field) absorbs violet → appears yellow
- $[\text{CoF}_6]^{3-}$ (weak field) absorbs red → appears green

--- …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 mark
$Q. What is the number of unpaired electrons in the complex ion [Fe(CN)6]4-?$
›Reveal solutionSolution

Fe in [Fe(CN)6]4- is Fe2+ (d6); CN- is a strong-field ligand causing all 6 d-electrons to pair up in the lower t2g set (low spin), leaving no unpaired electrons.

In [Fe(CN)6]4-, the overall charge is 4-, and CN- contributes -1 each (6 x -1 = -6), so iron must be Fe2+ (electronic configuration [Ar]3d6). CN- is a strong field ligand (high in the spectrochemical series), so it causes a large crystal field splitting (delta-o) in this octahedral complex. This favours pairing of electrons in the lower-energy t2g orbitals rather than promotion to the higher-energy eg orbitals (low-spin confi …

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Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$Q. For the complex ion [CoF6]3-, the number of unpaired electrons is -- (Co = 27)$
(a) 2
(b) 3
(c) 4
(d) 0
›Reveal solutionSolution

Co in [CoF6]3- is Co(III), a d6 ion; F- is a weak-field ligand giving a high-spin octahedral complex with 4 unpaired electrons.

Step 1 - oxidation state of Co. The complex ion has overall charge -3. Six F- ligands contribute -6. So: Co + 6(-1) = -3, giving Co = +3.

Step 2 - d-electron count. Co (Z = 27) has configuration [Ar]3d^7 4s^2. Co3+ is formed by removing 3 electrons (2 from 4s, 1 from 3d), giving 3d^6, i.e. a d6 ion.

Step 3 - field strength of F-. F- is a weak-field ligand (low in the spectrochemical series), so it cannot force pairing of electrons. The complex is therefore high-spin.

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 mark
$Q. [Ti(H2O)6]3+ is coloured, but [Sc(H2O)6]3+ is colourless — explain.$
›Reveal solutionSolution

Colour in transition-metal complexes arises from d-d electronic transitions; this requires the metal ion to have partially filled d-orbitals (neither empty nor completely full). Ti3+ (3d1) has such an electron and is coloured; Sc3+ (3d0) has none and is colourless.

[Ti(H2O)6]3+: Titanium (Z=22) has configuration [Ar]3d2 4s2; Ti3+ loses 3 electrons (2 from 4s, 1 from 3d) to give [Ar]3d1 - i.e., ONE d-electron. In the octahedral aqua complex, the surrounding water ligands split the d-orbitals into lower t2g and higher eg sets. The single d-electron can absorb a specific wavelength of visible light and get excited from t2g to eg (a d-d transition). The complementary colour of the absorbed light is what we observe (violet colour for this ion).

…

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Q.What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?

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 this happens: The concept of complementary colours

Step-by-step reasoning

A concrete example

Concept: Electronic Transitions and the Colour Wheel

Method: Complementary Colour Rule

Steps

Example …

Mistake 1: Confusing “absorbed” with “observed” colour

Mistake 2: Forgetting the complementary colour relationship

Mistake 3: Ignoring the wavelength range of visible light

Mistake 4: Mixing up the effect of ligand field strength

Q.What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?

Why this happens: The concept of complementary colours

Step-by-step reasoning

A concrete example

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