Chemistry · Ch 4 — The d- and f-Block Elements

Formation of Coloured Ions

4.3.10

Formation of Coloured Ions

Why transition-metal ions are coloured

Most ions of the transition series show colour, in sharp contrast to the ions of the s- and p-block. The reason lies in the partly filled d subshell. Under the influence of the surrounding ligands (water molecules, in the simplest case of an aqueous ion), the d orbitals of the metal ion are no longer all at the same energy — they split into a set of lower-energy and higher-energy orbitals. An electron can then be excited from one of the lower-energy d orbitals to one of the higher-energy d orbitals. This is called a d-d transition.

Figure 4.5Colours of some of the first row transition metal ions in aqueous solutions. From left to right: V⁴⁺, V³⁺, Mn²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺.

Fig. 4.5 — Colours of some of the first row transition metal ions in aqueous solutions. From left to right: V⁴⁺, V³⁺, Mn²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

What the Figure Shows

The figure is a photograph of seven round-bottom flasks arranged in a row, each containing an aqueous solution of a different first-row transition-metal ion. From left to right, the solutions are:

V⁴⁺ — light blue
V³⁺ — green
Mn²⁺ — pale pink
Fe³⁺ — yellow-orange
Co²⁺ — reddish-pink
Ni²⁺ — green
Cu²⁺ — blue

There are no axes, curves, or panels — it is a direct visual demonstration of the colours produced by these ions in water.

The Physical Idea It Teaches

The colours arise from d–d transitions: when visible light falls on the solution, electrons in the partially filled $d$ orbitals of the transition-metal ion absorb specific wavelengths to jump from a lower-energy $d$ orbital to a higher-energy $d$ orbital. The colour we see is the complementary colour of the absorbed light. For example, Cu²⁺ absorbs red-orange light and appears blue.

The key point is that only ions with partially filled $d$ orbitals (i.e., $d^1$ to $d^9$ configurations) can show such colours. Ions with $d^0$ (e.g., Sc³⁺) or $d^{10}$ (e.g., Zn²⁺) configurations have no $d$ electrons to excite and are colourless.

The Key Formula Developed with This Figure

The energy difference between the split $d$ orbitals is called the crystal field splitting energy, denoted by $\Delta$ (or $10\,Dq$ ). For an octahedral complex, the wavelength $\lambda$ of light absorbed is related to $\Delta$ by:

\Delta = \frac{hc}{\lambda}

where:

$h$ = Planck’s constant ( $6.626 \times 10^{-34}\ \text{J s}$ )
$c$ = speed of light ( $3.00 \times 10^8\ \text{m s}^{-1}$ ) …

The energy needed for this excitation is small enough that it corresponds to a photon in the visible region of the spectrum:

\Delta E = h\nu

where $\Delta E$ is the energy gap between the split d orbitals, $h$ is Planck's constant, and $\nu$ is the frequency of the light absorbed.

Absorbed colour vs observed colour

When white light falls on the ion, the ion absorbs the particular frequency of visible light that matches $\Delta E$ , promoting an electron across the gap. What reaches the eye is the light that is left over — the complementary colour of the light absorbed. So the colour we see is not the colour of the light absorbed, but its complement.

The exact frequency absorbed — and hence the exact colour seen — is not fixed for a given metal. It depends on the nature of the ligand bound to the metal ion, since different ligands split the d orbitals by different amounts. The same metal ion can therefore show different colours with different ligands.

Table 4.8Colours of Some of the First Row (aquated) Transition Metal Ions

Configuration	Example	Colour
$3d^0$	$Sc^{3+}$	colourless
$3d^0$	$Ti^{4+}$	colourless
$3d^1$	$Ti^{3+}$	purple
$3d^1$	$V^{4+}$	blue
$3d^2$	$V^{3+}$	green
$3d^3$	$V^{2+}$	violet
$3d^3$	$Cr^{3+}$	violet
$3d^4$	$Mn^{3+}$	violet
$3d^4$	$Cr^{2+}$	blue
$3d^5$	$Mn^{2+}$	pink
$3d^5$	$Fe^{3+}$	yellow
$3d^6$	$Fe^{2+}$	green
$3d^6$ , $3d^7$	$Co^{3+}$ , $Co^{2+}$	blue, pink
$3d^8$	$Ni^{2+}$	green
$3d^9$	$Cu^{2+}$	blue
$3d^{10}$	$Zn^{2+}$	colourless

Which ions are colourless

An ion can only show a d-d transition, and therefore colour, if it has both an occupied d orbital for the electron to leave and a vacant d orbital for it to land in. This fails at the two extremes of the d-electron count:

$d^0$ ions (e.g. $\mathrm{Sc^{3+}}$ , $\mathrm{Ti^{4+}}$ ) have no d electron to excite in the first place. …

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