Transition Metals & Analytical Chemistry

The d-Block Elements

Definition of a Transition Metal

A transition metal is defined as an element that forms at least one stable ion with a partially filled d subshell. This is the IUPAC definition and the one required by A-Level examinations.

Under this definition:

• Scandium ($\mathrm{Sc}$): forms $\mathrm{Sc}^{3+}$ with configuration $[\mathrm{Ar}]\,3d^0$ -- not a transition metal (empty $d$ subshell in its common ion).
• Zinc ($\mathrm{Zn}$): forms $\mathrm{Zn}^{2+}$ with configuration $[\mathrm{Ar}]\,3d^{10}$ -- not a transition metal (full $d$ subshell in its common ion).
• Titanium through Copper (excluding Sc and Zn) are transition metals.

Electronic Configurations

The $3d$ and $4s$ subshells are close in energy. For neutral atoms, $4s$ fills before $3d$ (Aufbau principle). For cations, $4s$ electrons are removed before $3d$.

Element	$Z$	Neutral atom	$\mathrm{M}^{2+}$	$\mathrm{M}^{3+}$
Ti	22	$[\mathrm{Ar}]\,4s^2 3d^2$	$[\mathrm{Ar}]\,3d^2$	$[\mathrm{Ar}]\,3d^1$
V	23	$[\mathrm{Ar}]\,4s^2 3d^3$	$[\mathrm{Ar}]\,3d^3$	$[\mathrm{Ar}]\,3d^2$
Cr	24	$[\mathrm{Ar}]\,4s^1 3d^5$	$[\mathrm{Ar}]\,3d^4$	$[\mathrm{Ar}]\,3d^3$
Mn	25	$[\mathrm{Ar}]\,4s^2 3d^5$	$[\mathrm{Ar}]\,3d^5$	$[\mathrm{Ar}]\,3d^4$
Fe	26	$[\mathrm{Ar}]\,4s^2 3d^6$	$[\mathrm{Ar}]\,3d^6$	$[\mathrm{Ar}]\,3d^5$
Co	27	$[\mathrm{Ar}]\,4s^2 3d^7$	$[\mathrm{Ar}]\,3d^7$	$[\mathrm{Ar}]\,3d^6$
Ni	28	$[\mathrm{Ar}]\,4s^2 3d^8$	$[\mathrm{Ar}]\,3d^8$	$[\mathrm{Ar}]\,3d^7$
Cu	29	$[\mathrm{Ar}]\,4s^1 3d^{10}$	$[\mathrm{Ar}]\,3d^9$	$[\mathrm{Ar}]\,3d^8$

Variable Oxidation States

Transition metals exhibit multiple oxidation states because the $3d$ and $4s$ electrons are close in energy and can be removed in varying numbers. The range of oxidation states and their stability depends on:

1. Ionisation energies -- successive ionisation energies increase; very high states require very high energy.
2. Lattice energies / hydration enthalpies -- highly charged ions in stable lattices or strongly hydrated ions can be stabilised.
3. Electronic configuration -- half-filled ($d^5$) and fully-filled ($d^{10}$) configurations are particularly stable.

Common Oxidation States

Element	Common oxidation states	Most stable
Ti	$+2, +3, +4$	$+4$
V	$+2, +3, +4, +5$	$+5$ (as vanadate)
Cr	$+2, +3, +6$	$+3$
Mn	$+2, +3, +4, +6, +7$	$+2$ (aq), $+4$ (solid), $+7$ (oxoanion)
Fe	$+2, +3$	$+3$ (oxidising conditions), $+2$ (reducing)
Co	$+2, +3$	$+2$
Ni	$+2$	$+2$
Cu	$+1, +2$	$+2$

Manganese exhibits the widest range ($+2$ to $+7$) of any first-row transition metal.

Complex Ion Formation

Definitions

A complex ion consists of a central metal ion surrounded by ligands. A ligand is a molecule or ion that donates a lone pair of electrons to the central metal ion, forming a coordinate (dative covalent) bond.

The coordination number is the total number of coordinate bonds to the central metal ion.

Types of Ligands

Type	Description	Examples	Denticity
Monodentate	One lone pair donor	$\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, $\mathrm{Cl}^-$, $\mathrm{CN}^-$	1
Bidentate	Two lone pair donors	$\mathrm{NH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{NH}_2$ (en), $\mathrm{C}_2\mathrm{O}_4^{2-}$ (oxalate)	2
Hexadentate	Six lone pair donors	$\mathrm{EDTA}^{4-}$	6

The chelate effect: multidentate ligands form more stable complexes than an equivalent number of monodentate ligands. This is primarily an entropic effect -- one multidentate ligand replaces several monodentate ligands, increasing the number of free particles and hence the entropy of the system.

Geometry

Coordination number	Geometry	Example
4	Tetrahedral	$[\mathrm{CoCl}_4]^{2-}$
6	Octahedral	$[\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+}$, $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$
4	Square planar	$[\mathrm{PtCl}_4]^{2-}$, $[\mathrm{Ni(CN})_4]^{2-}$

Most first-row transition metal complexes are octahedral or tetrahedral. Square planar geometry is typical for $d^8$ configurations with strong-field ligands (e.g. $\mathrm{Ni}^{2+}$ with $\mathrm{CN}^-$).

Crystal Field Theory (Qualitative)

d-Orbital Splitting

In a free ion, all five $d$ orbitals are degenerate (same energy). In an octahedral complex, the six ligands approach along the $x$, $y$, and $z$ axes. The $d$ orbitals that point along these axes ($d_{z^2}$ and $d_{x^2-y^2}$, collectively the $e_g$ set) experience greater electrostatic repulsion from the ligand lone pairs than those that point between the axes ($d_{xy}$, $d_{xz}$, $d_{yz}$, collectively the $t_{2g}$ set).

This splits the $d$ orbitals into two energy levels:

• $e_g$ (higher energy): $d_{z^2}$, $d_{x^2-y^2}$ -- 2 orbitals
• $t_{2g}$ (lower energy): $d_{xy}$, $d_{xz}$, $d_{yz}$ -- 3 orbitals

The energy separation is the crystal field splitting parameter $\Delta_o$ (octahedral).

High-Spin vs Low-Spin

Electrons occupy the $d$ orbitals according to Hund's rule (maximise parallel spins) but must also respect the energy gap $\Delta_o$.

• Weak-field ligands (e.g. $\mathrm{H}_2\mathrm{O}$, $\mathrm{F}^-$): $\Delta_o$ is small. Electrons occupy all five orbitals singly before pairing (high-spin configuration). Maximum number of unpaired electrons.
• Strong-field ligands (e.g. $\mathrm{CN}^-$, $\mathrm{NH}_3$ for certain metals): $\Delta_o$ is large. Electrons pair in the lower $t_{2g}$ orbitals before occupying $e_g$ (low-spin configuration). Fewer unpaired electrons.

Spectrochemical series (weak field to strong field):

$$\mathrm{I}^- \lt \mathrm{Br}^- \lt \mathrm{Cl}^- \lt \mathrm{F}^- \lt \mathrm{OH}^- \lt \mathrm{H}_2\mathrm{O} \lt \mathrm{NH}_3 \lt \mathrm{en} \lt \mathrm{CN}^- \lt \mathrm{CO}$$

Colour of Transition Metal Complexes

Transition metal complexes are coloured because the $d$-$d$ energy gap corresponds to photon energies in the visible region of the electromagnetic spectrum ($\Delta E \approx 1.8\mathrm{--}3.1\,\mathrm{eV}$, corresponding to $\lambda \approx 400\mathrm{--}700\,\mathrm{nm}$).

When white light passes through a complex, photons of energy matching $\Delta_o$ are absorbed to promote electrons from $t_{2g}$ to $e_g$. The complementary colour of the absorbed wavelength is observed.

Complex	Colour observed	Colour absorbed
$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Blue	Orange/red
$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Pink	Green
$[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Green	Red/violet
$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}$	Violet/pale yellow	Yellow/green

Key point: $d^0$ and $d^{10}$ complexes are colourless because there are no $d$-$d$ transitions possible. This is why $\mathrm{Sc}^{3+}$ ($d^0$) and $\mathrm{Zn}^{2+}$ ($d^{10}$) complexes are colourless.

Effect of ligand substitution on colour: Changing the ligand changes $\Delta_o$, which shifts the wavelength of light absorbed, changing the observed colour.

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 4\mathrm{NH}_3 \to [\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+} + 4\mathrm{H}_2\mathrm{O}$$

Pale blue solution $\to$ deep blue solution (because $\mathrm{NH}_3$ is a stronger field ligand than $\mathrm{H}_2\mathrm{O}$, increasing $\Delta_o$).

Ligand Substitution Reactions

Ligand substitution involves the replacement of one ligand by another. The rate depends on the complex:

• Labile complexes: substitution is rapid (e.g. most first-row transition metal complexes).
• Inert complexes: substitution is slow (e.g. $\mathrm{Cr}^{3+}$ and low-spin $\mathrm{Co}^{3+}$ complexes).

Acidified Potassium Dichromate(VI)

$\mathrm{Cr}^{3+}$ (green, octahedral, $d^3$) can be oxidised to $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ (orange):

$$2[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+} + 3\mathrm{H}_2\mathrm{O}_2 + 8\mathrm{H}^+ \to \mathrm{Cr}_2\mathrm{O}_7^{2-} + 3\mathrm{H}_2\mathrm{O} + 14\mathrm{H}_2\mathrm{O}$$

The orange dichromate is reduced back to green $\mathrm{Cr}^{3+}$ by reducing agents (e.g. $\mathrm{Zn}$, $\mathrm{Fe}^{2+}$).

Cobalt Complexes

$$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+} \xrightarrow{+\mathrm{Cl}^-} [\mathrm{CoCl}_4]^{2-}$$

Pink (octahedral) $\to$ blue (tetrahedral). This equilibrium is temperature-dependent; heating shifts it towards the blue tetrahedral form.

Redox Reactions of Transition Metals

Vanadium

Vanadium exhibits oxidation states $+2$ (violet), $+3$ (green), $+4$ (blue), and $+5$ (yellow). Reduction of vanadate($\mathrm{V}$) with zinc:

$$\mathrm{VO}_2^+(\mathrm{yellow}) \xrightarrow{\mathrm{Zn/HCl}} \mathrm{VO}^{2+}(\mathrm{blue}) \xrightarrow{\mathrm{Zn/HCl}} \mathrm{V}^{3+}(\mathrm{green}) \xrightarrow{\mathrm{Zn/HCl}} \mathrm{V}^{2+}(\mathrm{violet})$$

The colour changes are striking and form a classic demonstration of variable oxidation states.

Manganese

Potassium manganate(VII) ($\mathrm{KMnO}_4$) is a powerful oxidising agent in acidic solution ($E^\circ = +1.51\,\mathrm{V}$). It is reduced to $\mathrm{Mn}^{2+}$ (pale pink, almost colourless in dilute solution):

$$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$$

In neutral/alkaline solution, it is reduced to $\mathrm{MnO}_2$ (brown precipitate) or $\mathrm{MnO}_4^{2-}$ (green).

Iron

$$\mathrm{Fe}^{3+} + e^- \rightleftharpoons \mathrm{Fe}^{2+} \quad E^\circ = +0.77\,\mathrm{V}$$

$\mathrm{Fe}^{3+}$ is a moderate oxidising agent; $\mathrm{Fe}^{2+}$ is a moderate reducing agent.

Test for $\mathrm{Fe}^{2+}$: Add $\mathrm{NaOH}$ -- green precipitate of $\mathrm{Fe}(\mathrm{OH})_2$, which turns brown on standing as it oxidises to $\mathrm{Fe}(\mathrm{OH})_3$.

Test for $\mathrm{Fe}^{3+}$: Add $\mathrm{NaOH}$ -- brown precipitate of $\mathrm{Fe}(\mathrm{OH})_3$. Add $\mathrm{KSCN}$ -- blood-red colour due to $[\mathrm{Fe}(\mathrm{SCN})(\mathrm{H}_2\mathrm{O})_5]^{2+}$.

Catalysis

Heterogeneous Catalysis

The catalyst is in a different phase from the reactants. The mechanism involves adsorption, reaction on the surface, and desorption.

Iron in the Haber process:

$$\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3$$

$\mathrm{N}_2$ and $\mathrm{H}_2$ adsorb onto the iron surface. The $\mathrm{N}\equiv\mathrm{N}$ bond is weakened, lowering the activation energy for bond breaking and subsequent reaction with $\mathrm{H}$ atoms. $\mathrm{NH}_3$ desorbs from the surface.

Vanadium(V) oxide in the Contact process:

$$2\mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{SO}_3$$

$\mathrm{V}_2\mathrm{O}_5$ provides a surface for the reaction. The mechanism involves reduction of $\mathrm{V}^{5+}$ to $\mathrm{V}^{4+}$ by $\mathrm{SO}_2$, then reoxidation of $\mathrm{V}^{4+}$ to $\mathrm{V}^{5+}$ by $\mathrm{O}_2$.

Homogeneous Catalysis

The catalyst is in the same phase as the reactants, typically forming an intermediate.

Autocatalysis: A reaction in which one of the products catalyses the reaction itself. Example: the $\mathrm{Mn}^{2+}$-catalysed reaction between oxalic acid and potassium manganate(VII). As $\mathrm{Mn}^{2+}$ is produced, it catalyses further reaction, causing an initial slow rate that accelerates.

Analytical Chemistry Summary

Tests for Metal Ions

Ion	Flame test	NaOH addition
$\mathrm{Li}^+$	Crimson red	White ppt (soluble in excess)
$\mathrm{Na}^+$	Persistent yellow	No ppt
$\mathrm{K}^+$	Lilac	White ppt
$\mathrm{Ca}^{2+}$	Brick red	White ppt (insoluble)
$\mathrm{Cu}^{2+}$	Blue-green	Blue ppt
$\mathrm{Fe}^{2+}$	--	Green ppt, turns brown
$\mathrm{Fe}^{3+}$	--	Brown ppt

Tests for Anions

Anion	Test	Observation
$\mathrm{Cl}^-$	Add dilute $\mathrm{HNO}_3$ + $\mathrm{AgNO}_3$	White ppt (soluble in $\mathrm{NH}_{3(aq)}$)
$\mathrm{Br}^-$	Add dilute $\mathrm{HNO}_3$ + $\mathrm{AgNO}_3$	Cream ppt (partially soluble in $\mathrm{NH}_{3(aq)}$)
$\mathrm{I}^-$	Add dilute $\mathrm{HNO}_3$ + $\mathrm{AgNO}_3$	Yellow ppt (insoluble in $\mathrm{NH}_{3(aq)}$)
$\mathrm{SO}_4^{2-}$	Add $\mathrm{HCl}$ + $\mathrm{BaCl}_2$	White ppt
$\mathrm{CO}_3^{2-}$	Add dilute acid	Effervescence, gas turns limewater milky

For further analytical techniques (IR, MS, NMR, chromatography), see Organic Chemistry.

Magnetism in Transition Metal Complexes

The magnetic properties of transition metal complexes arise from the spin of unpaired electrons. A complex is paramagnetic if it contains unpaired electrons (attracted into a magnetic field) and diamagnetic if all electrons are paired (weakly repelled by a magnetic field).

Magnetic Moments

The magnetic moment $\mu$ (in Bohr magnetons, $\mu_\mathrm{BM}$) is related to the number of unpaired electrons $n$ by the spin-only formula:

$$\mu = \sqrt{n(n+2)}\,\mu_\mathrm{BM}$$

Unpaired electrons ($n$)	$\mu_\mathrm{calc}$ ($\mu_\mathrm{BM}$)	Typical ion ($d^n$)
0	0.00	$\mathrm{Zn}^{2+}$ ($d^{10}$), $\mathrm{Cu}^+$ ($d^{10}$)
1	1.73	$\mathrm{Cu}^{2+}$ ($d^9$)
2	2.83	$\mathrm{V}^{3+}$ ($d^2$)
3	3.87	$\mathrm{Cr}^{3+}$ ($d^3$), $\mathrm{V}^{2+}$ ($d^3$)
4	4.90	$\mathrm{Mn}^{2+}$ ($d^5$, high-spin), $\mathrm{Cr}^{2+}$ ($d^4$, high-spin)
5	5.92	$\mathrm{Mn}^{2+}$ ($d^5$, high-spin), $\mathrm{Fe}^{3+}$ ($d^5$, high-spin)

Experimental magnetic moments may differ slightly from spin-only values due to orbital contribution (spin-orbit coupling), which adds a small positive correction. For first-row transition metals, the spin-only approximation is usually adequate.

Using Magnetism to Distinguish High-Spin and Low-Spin

Magnetic measurements provide experimental evidence for high-spin vs low-spin configurations. Consider $\mathrm{Fe}^{2+}$ ($d^6$):

• High-spin (weak field, e.g. $\mathrm{H}_2\mathrm{O}$): $t_{2g}^4\,e_g^2$, 4 unpaired electrons, $\mu \approx 4.90\,\mu_\mathrm{BM}$.
• Low-spin (strong field, e.g. $\mathrm{CN}^-$): $t_{2g}^6\,e_g^0$, 0 unpaired electrons, $\mu \approx 0\,\mu_\mathrm{BM}$.

The difference is dramatic and unambiguous. A $\mathrm{Fe}^{2+}$ complex with $\mu \approx 5\,\mu_\mathrm{BM}$ must be high-spin; one with $\mu \approx 0$ must be low-spin.

Worked Example. $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ has $\mu = 5.1\,\mu_\mathrm{BM}$. $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ has $\mu = 0\,\mu_\mathrm{BM}$. Explain.

$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$: $\mathrm{Fe}^{2+}$ is $d^6$. $\mathrm{H}_2\mathrm{O}$ is a weak-field ligand, so $\Delta_o$ is small. The electrons occupy the $t_{2g}$ set singly first ($t_{2g}^4$), then the $e_g$ set ($e_g^2$), giving 4 unpaired electrons (high-spin). The spin-only value for $n = 4$ is $\sqrt◆LB◆4 \times 6◆RB◆ = 4.90\,\mu_\mathrm{BM}$, close to the experimental $5.1\,\mu_\mathrm{BM}$ (the slight excess is from orbital contribution).

$[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$: $\mathrm{CN}^-$ is a strong-field ligand, so $\Delta_o$ is large. All 6 electrons pair in the $t_{2g}$ set ($t_{2g}^6\,e_g^0$), giving 0 unpaired electrons (low-spin, diamagnetic).

Tetrahedral Splitting

In a tetrahedral complex, the splitting is the inverse of the octahedral case. The $d$ orbitals pointing between the axes ($t_2$: $d_{xy}$, $d_{xz}$, $d_{yz}$) are closer to the ligands and are higher energy. The orbitals pointing along the axes ($e$: $d_{z^2}$, $d_{x^2-y^2}$) are lower energy. The splitting parameter $\Delta_t \approx \frac{4}{9}\Delta_o$.

Because $\Delta_t$ is much smaller than $\Delta_o$, tetrahedral complexes are almost always high-spin. Low-spin tetrahedral complexes are essentially unknown for first-row transition metals.

Detailed Colour and Spectroscopy

Calculating the Energy of Absorption

The energy of the photon absorbed in a $d$-$d$ transition can be calculated from the wavelength of absorption:

$$\Delta E = \frac◆LB◆hc◆RB◆◆LB◆\lambda◆RB◆$$

where $h = 6.626 \times 10^{-34}\,\mathrm{J\,s}$, $c = 3.00 \times 10^8\,\mathrm{m/s}$, and $\lambda$ is the wavelength in metres.

Worked Example. $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ absorbs at $\lambda = 510\,\mathrm{nm}$ (green). Calculate $\Delta_o$ in $\mathrm{kJ/mol}$.

$$\Delta E = \frac◆LB◆hc◆RB◆◆LB◆\lambda◆RB◆ = \frac◆LB◆(6.626 \times 10^{-34})(3.00 \times 10^8)◆RB◆◆LB◆510 \times 10^{-9}◆RB◆ = 3.90 \times 10^{-19}\,\mathrm{J}$$

Per mole:

$$\Delta_o = 3.90 \times 10^{-19} \times 6.022 \times 10^{23} = 234,800\,\mathrm{J/mol} = 235\,\mathrm{kJ/mol}$$

Colour Wheel and Complementary Colours

The observed colour of a complex is the complementary colour of the light absorbed:

Wavelength absorbed (nm)	Colour absorbed	Colour observed
400--435	Violet	Yellow-green
435--480	Blue	Yellow
480--490	Green-blue	Orange
490--500	Blue-green	Red
500--560	Green	Purple/magenta
560--580	Yellow-green	Violet
580--595	Yellow	Blue
595--605	Orange	Green-blue
605--700	Red	Blue-green

Charge Transfer Transitions

In addition to $d$-$d$ transitions, some complexes exhibit charge transfer transitions, which involve the transfer of an electron between the metal and the ligand. These are generally much more intense than $d$-$d$ transitions.

• Ligand-to-metal charge transfer (LMCT): An electron moves from a ligand-based orbital to a metal-based orbital. Common in oxoanions: $\mathrm{MnO}_4^-$ (deep purple, $\mathrm{Mn}^{7+}$, $d^0$ -- no $d$-$d$ transitions possible; the colour is entirely due to LMCT), $\mathrm{CrO}_4^{2-}$ (yellow).
• Metal-to-ligand charge transfer (MLCT): An electron moves from the metal to the ligand. Less common at A-Level.

Charge transfer transitions explain why some $d^0$ complexes (like $\mathrm{MnO}_4^-$) are intensely coloured despite having no $d$-$d$ transitions.

The Chelate Effect in Detail

Quantitative Treatment

The chelate effect can be demonstrated quantitatively by comparing stability constants:

$$[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 6\mathrm{NH}_3 \rightleftharpoons [\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+} + 6\mathrm{H}_2\mathrm{O} \quad \log\beta_6 \approx 8.6$$ $$[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 3\mathrm{en} \rightleftharpoons [\mathrm{Ni}(\mathrm{en})_3]^{2+} + 6\mathrm{H}_2\mathrm{O} \quad \log\beta_3 \approx 18.4$$

Both reactions involve 6 coordinate bonds to nitrogen donors, yet the tris(en) complex is approximately $10^{10}$ times more stable than the hexammine complex. This enormous difference is primarily entropic:

• When 6 $\mathrm{NH}_3$ molecules coordinate, 6 free particles are lost (net change: $-6$ particles).
• When 3 en molecules coordinate, only 3 free particles are lost (net change: $-3$ particles).

The greater increase in entropy for the bidentate case makes $\Delta G$ more negative, favouring complex formation.

Biological Significance

The chelate effect is central to biochemistry. Haemoglobin uses a porphyrin ring (tetradentate) to bind $\mathrm{Fe}^{2+}$ extremely tightly. EDTA is used in medicine to treat heavy metal poisoning: it chelates toxic metal ions (e.g. $\mathrm{Pb}^{2+}$) and allows them to be excreted. EDTA forms exceptionally stable complexes because it is hexadentate.

Transition Metal Chemistry of Specific Elements

Chromium

Chromium has two important oxidation states in aqueous chemistry: $\mathrm{Cr}^{3+}$ ($d^3$) and $\mathrm{Cr}^{6+}$ (as $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ or $\mathrm{CrO}_4^{2-}$).

Species	Oxidation state	Colour	$d$ electrons
$[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+}$	$+3$	Violet (green in impure form)	$d^3$
$\mathrm{Cr}_2\mathrm{O}_7^{2-}$	$+6$	Orange	$d^0$
$\mathrm{CrO}_4^{2-}$	$+6$	Yellow	$d^0$

The dichromate-chromate equilibrium is pH-dependent:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons 2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+$$

In acid (low pH): equilibrium shifts left, favouring orange $\mathrm{Cr}_2\mathrm{O}_7^{2-}$.

In alkali (high pH): equilibrium shifts right, favouring yellow $\mathrm{CrO}_4^{2-}$.

Worked Example. Write equations for the oxidation of $\mathrm{Cr}^{3+}$ to $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ using hydrogen peroxide in alkaline solution, followed by acidification.

In alkaline solution, $\mathrm{Cr}^{3+}$ is first oxidised to chromate:

$$2[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+} + 3\mathrm{H}_2\mathrm{O}_2 + 10\mathrm{OH}^- \to 2\mathrm{CrO}_4^{2-} + 8\mathrm{H}_2\mathrm{O}$$

On acidification, chromate converts to dichromate:

$$2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+ \to \mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O}$$

Observation: green solution turns yellow (chromate formation), then orange (dichromate on acidification).

Cobalt

Cobalt(II) complexes provide classic examples of both octahedral and tetrahedral geometry:

Complex	Geometry	Colour	Explanation
$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	Octahedral	Pink	$d^7$, high-spin, 3 unpaired electrons
$[\mathrm{CoCl}_4]^{2-}$	Tetrahedral	Blue	$d^7$, always high-spin (tetrahedral)

The equilibrium between these two forms is temperature-dependent:

$$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 4\mathrm{Cl}^- \rightleftharpoons [\mathrm{CoCl}_4]^{2-} + 6\mathrm{H}_2\mathrm{O} \quad \Delta H \gt 0$$

The forward reaction is endothermic, so heating shifts equilibrium to the right (blue). Cooling shifts it to the left (pink). Adding concentrated $\mathrm{HCl}$ (increasing $[\mathrm{Cl}^-]$) also shifts it right. Adding water shifts it left. Adding $\mathrm{AgNO}_3$ removes $\mathrm{Cl}^-$ as $\mathrm{AgCl}$ precipitate, shifting it left.

Common Pitfalls

1. Defining transition metals incorrectly. The IUPAC definition requires a partially filled $d$ subshell in at least one stable ion, not in the neutral atom. Sc and Zn are not transition metals under this definition.

2. Writing wrong electron configurations for ions. Remove $4s$ electrons before $3d$ when forming cations.

3. Confusing $e_g$ and $t_{2g}$. The $e_g$ set ($d_{z^2}$, $d_{x^2-y^2}$) is higher energy in octahedral complexes. The $t_{2g}$ set is lower energy.

4. Stating that colour is due to $d$-$d$ transitions without specifying the mechanism. Be precise: photons are absorbed when electrons are promoted from lower-energy $d$ orbitals to higher-energy $d$ orbitals. The observed colour is the complementary colour of the absorbed light.

5. Forgetting that $d^0$ and $d^{10}$ complexes are colourless. There are no $d$-$d$ transitions if there are no $d$ electrons or the $d$ subshell is full.

6. Assuming tetrahedral complexes can be low-spin. Because $\Delta_t \approx \frac{4}{9}\Delta_o$, tetrahedral complexes are almost always high-spin for first-row transition metals.

7. Confusing LMCT with $d$-$d$ transitions. $\mathrm{MnO}_4^-$ is coloured ($d^0$) due to ligand-to-metal charge transfer, not $d$-$d$ transitions. Always check the $d$ count first.

8. Stating that the chelate effect is due to stronger bonds. The chelate effect is primarily an entropic effect, not an enthalpic one. The bonds are of similar strength; the stability gain comes from the increased number of free particles.

Complex Ion Formation and Stability Constants

Stepwise and Overall Formation Constants

For the formation of $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$:

$$\mathrm{Cu}^{2+} + \mathrm{NH}_3 \rightleftharpoons [\mathrm{Cu}(\mathrm{NH}_3)]^{2+} \quad K_1$$ $$[\mathrm{Cu}(\mathrm{NH}_3)]^{2+} + \mathrm{NH}_3 \rightleftharpoons [\mathrm{Cu}(\mathrm{NH}_3)_2]^{2+} \quad K_2$$ $$[\mathrm{Cu}(\mathrm{NH}_3)_2]^{2+} + \mathrm{NH}_3 \rightleftharpoons [\mathrm{Cu}(\mathrm{NH}_3)_3]^{2+} \quad K_3$$ $$[\mathrm{Cu}(\mathrm{NH}_3)_3]^{2+} + \mathrm{NH}_3 \rightleftharpoons [\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+} \quad K_4$$

The overall formation constant is:

$$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$$

Typically $K_1 \gt K_2 \gt K_3 \gt K_4$ because each successive ligand added experiences increasing steric and electrostatic repulsion from the ligands already present.

Haemoglobin and Oxygen Transport

Haemoglobin contains $\mathrm{Fe}^{2+}$ ions coordinated to a porphyrin ring (a tetradentate ligand) and a protein chain (globin). The sixth coordination site binds $\mathrm{O}_2$ reversibly:

$$[\mathrm{Hb-Fe}^{2+}] + \mathrm{O}_2 \rightleftharpoons [\mathrm{Hb-Fe}^{2+}-\mathrm{O}_2]$$

Carbon monoxide poisoning: $\mathrm{CO}$ is a stronger ligand than $\mathrm{O}_2$ (greater $d$-orbital overlap due to $\pi$-backbonding). It binds irreversibly to haemoglobin, blocking $\mathrm{O}_2$ transport:

$$[\mathrm{Hb-Fe}^{2+}] + \mathrm{CO} \to [\mathrm{Hb-Fe}^{2+}-\mathrm{CO}] \quad \text{(essentially irreversible)}$$

The equilibrium constant for $\mathrm{CO}$ binding is approximately 200 times that for $\mathrm{O}_2$ binding.

Qualitative Analysis of Transition Metal Ions

Standard Inorganic Tests

Ion	Test	Observation
$\mathrm{Cu}^{2+}$	Add $\mathrm{NaOH}$	Blue precipitate of $\mathrm{Cu(OH)}2$; dissolves in excess $\mathrm{NH}_3$ to deep blue $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$
$\mathrm{Fe}^{2+}$	Add $\mathrm{NaOH}$	Green precipitate of $\mathrm{Fe(OH)}_2$; turns brown on standing (oxidised to $\mathrm{Fe(OH)}_3$)
$\mathrm{Fe}^{3+}$	Add $\mathrm{NaOH}$	Red-brown precipitate of $\mathrm{Fe(OH)}_3$
$\mathrm{Fe}^{3+}$	Add $\mathrm{KSCN}$	Blood red solution of $[\mathrm{Fe}(\mathrm{SCN})(\mathrm{H}_2\mathrm{O})_5]^{2+}$
$\mathrm{Mn}^{2+}$	Add $\mathrm{NaOH}$, then $\mathrm{H}_2\mathrm{O}_2$	White precipitate darkens to brown $\mathrm{MnO}_2$
$\mathrm{Cr}^{3+}$	Add $\mathrm{NaOH}$ then $\mathrm{H}_2\mathrm{O}_2$, heat	Green solution turns yellow ($\mathrm{CrO}_4^{2-}$)
$\mathrm{Ni}^{2+}$	Add dimethylglyoxime + $\mathrm{NH}_3$	Bright red precipitate
$\mathrm{Co}^{2+}$	Add $\mathrm{SCN}^-$ in acetone	Blue organic layer of $[\mathrm{Co}(\mathrm{SCN})_4]^{2-}$

Using Flame Tests for Transition Metals

Transition metal ions produce characteristic flame colours (though these are less commonly used than for Group 1 and 2):

Ion	Flame colour
$\mathrm{Cu}^{2+}$	Blue-green
$\mathrm{Li}^+$ (Group 1, for comparison)	Crimson red

Most transition metal flame colours are not distinctive enough for reliable identification; precipitation tests are preferred.

Practice Problems

Problem 1

Explain why $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is purple but $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{4+}$ is colourless.

Solution:

$\mathrm{Ti}^{3+}$ has the electron configuration $[\mathrm{Ar}]\,3d^1$. The single $d$ electron can be promoted from a $t_{2g}$ orbital to an $e_g$ orbital by absorbing a photon of visible light. The energy gap corresponds to the green-yellow region of the spectrum, so the complementary colour (purple) is transmitted.

$\mathrm{Ti}^{4+}$ has the configuration $[\mathrm{Ar}]\,3d^0$. With zero $d$ electrons, there are no $d$-$d$ transitions possible, so no visible light is absorbed and the complex is colourless.

Problem 2

Write equations for the reactions of $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ with (a) excess $\mathrm{NH}_3$, (b) concentrated $\mathrm{HCl}$, and (c) $\mathrm{NaOH}$. State any observations.

Solution:

(a) With excess $\mathrm{NH}_3$:

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 4\mathrm{NH}_3 \to [\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+} + 4\mathrm{H}_2\mathrm{O}$$

Pale blue $\to$ deep blue solution.

(b) With concentrated $\mathrm{HCl}$:

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 4\mathrm{Cl}^- \rightleftharpoons [\mathrm{CuCl}_4]^{2-} + 6\mathrm{H}_2\mathrm{O}$$

Pale blue $\to$ yellow solution (tetrahedral complex).

(c) With $\mathrm{NaOH}$:

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 2\mathrm{OH}^- \to \mathrm{Cu}(\mathrm{OH})_2(s) + 6\mathrm{H}_2\mathrm{O}$$

Blue solution $\to$ blue precipitate.

Problem 3

$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is diamagnetic. Explain this observation and predict the colour of the complex relative to $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$.

Solution:

$\mathrm{Co}^{3+}$ has the electron configuration $[\mathrm{Ar}]\,3d^6$. In $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, $\mathrm{NH}_3$ is a strong-field ligand (high in the spectrochemical series). The large $\Delta_o$ means it is energetically favourable for all 6 electrons to pair in the $t_{2g}$ orbitals ($t_{2g}^6\,e_g^0$) rather than promoting electrons to the $e_g$ set. With zero unpaired electrons, the complex is diamagnetic ($\mu = 0\,\mu_\mathrm{BM}$).

Since $\mathrm{NH}_3$ is a stronger field ligand than $\mathrm{H}_2\mathrm{O}$, $\Delta_o$ is larger for $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ than for $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$. A larger $\Delta_o$ means absorption of higher-energy (shorter wavelength) photons. The absorbed light shifts from the green region (observed: pink) to the blue/violet region, so $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ appears yellow-orange.

Problem 4

Explain why $\mathrm{MnO}_4^-$ is an intense purple colour despite $\mathrm{Mn}^{7+}$ having a $d^0$ configuration.

Solution:

$\mathrm{Mn}^{7+}$ has the configuration $[\mathrm{Ar}]\,3d^0$. With zero $d$ electrons, no $d$-$d$ transitions are possible. The intense purple colour of $\mathrm{MnO}_4^-$ is due to ligand-to-metal charge transfer (LMCT). An electron is transferred from a lone pair on one of the oxygen atoms into an empty $d$ orbital on the manganese. This charge transfer transition involves a much larger change in dipole moment than a $d$-$d$ transition, giving it a much higher extinction coefficient (more intense colour).

By contrast, $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ($\mathrm{Mn}^{2+}$, $d^5$, high-spin) is a very pale pink because $d$-$d$ transitions are Laporte-forbidden (centrosymmetric complexes have very weak $d$-$d$ absorption) and because the high-spin $d^5$ configuration has all $d$ orbitals singly occupied, so every $d$-$d$ transition requires an electron to change spin as well as orbital, making it additionally spin-forbidden.

Problem 5

Explain the observations when aqueous ammonia is added dropwise to a solution containing $\mathrm{Cu}^{2+}$ ions until in excess.

Solution:

Initial addition (limited $\mathrm{NH}_3$): A pale blue precipitate of copper(II) hydroxide forms:

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 2\mathrm{NH}_3 \to \mathrm{Cu(OH)}_2(s) + 2\mathrm{NH}_4^+ + 4\mathrm{H}_2\mathrm{O}$$

The ammonia acts as a base, accepting protons from the water ligands and raising the pH until $\mathrm{Cu}(\mathrm{OH})_2$ precipitates.

Excess $\mathrm{NH}_3$: The precipitate dissolves to form a deep blue solution of tetraamminecopper(II):

$$\mathrm{Cu(OH)}_2(s) + 4\mathrm{NH}_3 \to [\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+} + 2\mathrm{OH}^-$$

The deep blue colour of $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ is due to $d$-$d$ transitions. Ammonia is a stronger field ligand than water (higher in the spectrochemical series), so the splitting is larger and the absorption shifts to shorter wavelengths (higher energy), transmitting blue light.

Stability constant: The formation of $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ has a large overall stability constant ($\beta_4 \approx 10^{13}$), confirming that the complex is very stable and the equilibrium lies far to the right.

Problem 6

Explain the role of $\mathrm{V}_2\mathrm{O}_5$ in the Contact process and describe the mechanism of catalysis.

Solution:

$\mathrm{V}_2\mathrm{O}_5$ catalyses the oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$:

$$2\mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{SO}_3$$

Mechanism (simplified):

1. $\mathrm{SO}_2$ adsorbs onto the $\mathrm{V}_2\mathrm{O}_5$ surface.
2. The $\mathrm{V}_2\mathrm{O}_5$ is reduced: $\mathrm{V}^{5+} \to \mathrm{V}^{4+}$ as oxygen is transferred to $\mathrm{SO}_2$ to form $\mathrm{SO}_3$.
3. $\mathrm{O}_2$ from the gas phase re-oxidises the catalyst: $\mathrm{V}^{4+} \to \mathrm{V}^{5+}$.

The catalyst provides an alternative pathway with lower activation energy. The oxidation state of vanadium cycles between +4 and +5. The catalyst is heterogeneous (solid surface), so it provides adsorption sites that weaken the S--O bonds in $\mathrm{SO}_2$ and the O=O bond in $\mathrm{O}_2$, facilitating the reaction.

Crystal Field Theory and Colour in Detail

Crystal Field Splitting

When a transition metal ion is surrounded by six ligands in an octahedral arrangement, the five degenerate $d$-orbitals split into two energy levels:

• $t_{2g}$ orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$): Lower energy. These orbitals point between the ligands and experience less repulsion.
• $e_g$ orbitals ($d_{x^2-y^2}$, $d_{z^2}$): Higher energy. These orbitals point directly at the ligands and experience more repulsion.

The energy difference between the two sets is the crystal field splitting energy, $\Delta_o$ (or $10\mathrm{Dq}$).

Crystal Field Stabilisation Energy (CFSE)

The CFSE is the net energy lowering of the $d$-electrons relative to the hypothetical spherical field (where all five $d$-orbitals are degenerate). Each $t_{2g}$ electron stabilises the complex by $-0.4\Delta_o$, and each $e_g$ electron destabilises it by $+0.6\Delta_o$.

Worked Example: CFSE of $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$

$\mathrm{Fe}^{2+}$ has the electron configuration $[\mathrm{Ar}]\,3d^6$.

$\mathrm{H}_2\mathrm{O}$ is a weak-field ligand, so the complex is high-spin:

$t_{2g}^4\,e_g^2$

$\text{CFSE} = 4(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$

For $\mathrm{H}_2\mathrm{O}$, $\Delta_o \approx 10400\,\mathrm{cm}^{-1}$:

$\text{CFSE} = -0.4 \times 10400 = -4160\,\mathrm{cm}^{-1}$

The negative sign indicates stabilisation.

Worked Example: CFSE of $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$

$\mathrm{CN}^-$ is a strong-field ligand, so the complex is low-spin:

$t_{2g}^6\,e_g^0$

$\text{CFSE} = 6(-0.4\Delta_o) + 0(0.6\Delta_o) = -2.4\Delta_o$

For $\mathrm{CN}^-$, $\Delta_o \approx 33000\,\mathrm{cm}^{-1}$:

$\text{CFSE} = -2.4 \times 33000 = -79200\,\mathrm{cm}^{-1}$

The much larger CFSE for the cyanide complex explains why low-spin complexes are more stable with strong-field ligands: the pairing energy cost is more than compensated by the large $\Delta_o$.

The Spectrochemical Series

Ligands are arranged in order of increasing crystal field splitting energy:

$\mathrm{I}^- \lt \mathrm{Br}^- \lt \mathrm{Cl}^- \lt \mathrm{F}^- \lt \mathrm{OH}^- \lt \mathrm{H}_2\mathrm{O} \lt \mathrm{NH}_3 \lt \text{en} \lt \mathrm{NO}_2^- \lt \mathrm{CN}^- \lt \mathrm{CO}$

Weak-field ligands (left) give small $\Delta_o$ and high-spin complexes. Strong-field ligands (right) give large $\Delta_o$ and low-spin complexes.

Colour and Absorption of Light

Transition metal complexes are coloured because they absorb visible light. The absorbed photon promotes an electron from a $t_{2g}$ orbital to an $e_g$ orbital:

$E = h\nu = hc/\lambda = \Delta_o$

The colour observed is the complementary colour of the light absorbed.

Worked Example: $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ absorbs light at $\lambda = 510\,\mathrm{nm}$ (green). The transmitted/reflected light is purple/violet, which is the observed colour.

$\Delta_o = \frac◆LB◆hc◆RB◆◆LB◆\lambda◆RB◆ = \frac◆LB◆6.626 \times 10^{-34} \times 2.998 \times 10^8◆RB◆◆LB◆510 \times 10^{-9}◆RB◆ = 3.89 \times 10^{-19}\,\mathrm{J}$

Converting to $\mathrm{cm}^{-1}$: $\frac◆LB◆1◆RB◆◆LB◆\lambda◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆510 \times 10^{-7}\,\mathrm{cm}◆RB◆ = 19608\,\mathrm{cm}^{-1}$

Complex	$\Delta_o$ ($\mathrm{cm}^{-1}$)	$\lambda_{\max}$ (nm)	Colour observed
$[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$	20,300	493	Purple
$[\mathrm{V}(\mathrm{H}_2\mathrm{O})_6]^{3+}$	17,800	562	Green
$[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{3+}$	17,400	575	Violet
$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	9,300	1075	Pink
$[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$	8,500	1176	Green

Why $\mathrm{Zn}^{2+}$ and $\mathrm{Sc}^{3+}$ Complexes Are Colourless

$\mathrm{Zn}^{2+}$ has the electron configuration $[\mathrm{Ar}]\,3d^{10}$ (full $d$ subshell). All $d$-orbitals are fully occupied, so no $d$-$d$ transitions are possible. The complexes are colourless.

$\mathrm{Sc}^{3+}$ has the electron configuration $[\mathrm{Ar}]$ (empty $d$ subshell). There are no $d$-electrons to promote, so no $d$-$d$ transitions occur. The complexes are colourless.

Ligand Exchange Reactions

Ligand exchange involves the substitution of one ligand for another in a transition metal complex:

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}(aq) + 4\mathrm{NH}_3(aq) \rightleftharpoons [\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l)$

This reaction is easily observed: the pale blue $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ ion turns deep blue when $\mathrm{NH}_3$ is added.

Factors Affecting Ligand Exchange

1. Ligand field strength: Stronger ligands (higher in the spectrochemical series) displace weaker ligands. $\mathrm{CN}^-$ displaces $\mathrm{H}_2\mathrm{O}$, but $\mathrm{H}_2\mathrm{O}$ does not displace $\mathrm{CN}^-$.

2. Concentration: Adding a large excess of the incoming ligand drives the equilibrium towards the substituted complex.

3. Steric effects: Bulky ligands may be unable to coordinate due to steric hindrance, even if they are stronger field ligands.

4. Entropy: If the exchange increases the number of particles (e.g. bidentate replacing two monodentate ligands), the entropy change favours the substitution (chelate effect).

The Chelate Effect

Bidentate and multidentate ligands form more stable complexes than monodentate ligands with the same donor atoms. This is the chelate effect.

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + \text{en} \to [\mathrm{Cu}(\text{en})(\mathrm{H}_2\mathrm{O})_4]^{2+} + 2\mathrm{H}_2\mathrm{O}$

(en = ethane-1,2-diamine, a bidentate ligand)

The chelate effect is partly entropic: one bidentate ligand replacing two monodentate ligands increases the number of free particles in solution, increasing entropy.

Worked Example: Stepwise Ligand Exchange

When concentrated $\mathrm{HCl}$ is added to $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}(aq)$, the following stepwise substitutions occur:

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + \mathrm{Cl}^- \rightleftharpoons [\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}]^+ + \mathrm{H}_2\mathrm{O}$

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}]^+ + \mathrm{Cl}^- \rightleftharpoons [\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2] + \mathrm{H}_2\mathrm{O}$

$\ldots \to [\mathrm{CuCl}_4]^{2-} + 6\mathrm{H}_2\mathrm{O}$

The colour changes from pale blue ($\mathrm{H}_2\mathrm{O}$ ligands) through green to yellow ($\mathrm{Cl}^-$ ligands). The yellow colour of $[\mathrm{CuCl}_4]^{2-}$ indicates a smaller $\Delta_o$ for $\mathrm{Cl}^-$ than for $\mathrm{H}_2\mathrm{O}$.

Variable Oxidation States in Detail

Vanadium Oxidation States

Vanadium exhibits oxidation states from +2 to +5. The colours of vanadium ions in aqueous solution:

Oxidation state	Ion	Colour
+2	$\mathrm{V}^{2+}$	Violet
+3	$\mathrm{V}^{3+}$	Green
+4	$\mathrm{VO}^{2+}$	Blue
+5	$\mathrm{VO}_2^+$	Yellow

These can be interconverted by reduction with zinc and acid:

$\mathrm{VO}_2^+ \xrightarrow{\mathrm{Zn},\,\mathrm{H}^+} \mathrm{VO}^{2+} \xrightarrow{\mathrm{Zn},\,\mathrm{H}^+} \mathrm{V}^{3+} \xrightarrow{\mathrm{Zn},\,\mathrm{H}^+} \mathrm{V}^{2+}$

Chromium Oxidation States

Chromium exhibits oxidation states from +2 to +6:

Oxidation state	Ion/Compound	Colour
+2	$\mathrm{Cr}^{2+}(aq)$	Blue
+3	$\mathrm{Cr}^{3+}(aq)$	Green/violet
+6	$\mathrm{CrO}_4^{2-}(aq)$	Yellow
+6	$\mathrm{Cr}_2\mathrm{O}_7^{2-}(aq)$	Orange

Dichromate-chromate equilibrium:

$\mathrm{Cr}_2\mathrm{O}_7^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons 2\mathrm{CrO}_4^{2-}(aq) + 2\mathrm{H}^+(aq)$

Adding acid shifts the equilibrium to the left (orange dichromate). Adding alkali shifts it to the right (yellow chromate).

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Explain why $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ is blue but $[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ is colourless.

Mark Scheme:

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ contains $\mathrm{Cu}^{2+}$ with the electron configuration $[\mathrm{Ar}]\,3d^9$ (1 mark). The $d$-orbitals are split by the octahedral crystal field of water ligands (1 mark). An electron can be promoted from a $t_{2g}$ orbital to an $e_g$ orbital by absorbing visible light (1 mark). The complementary colour (blue) is transmitted/reflected.

$[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ contains $\mathrm{Zn}^{2+}$ with the electron configuration $[\mathrm{Ar}]\,3d^{10}$ (1 mark). All $d$-orbitals are fully occupied, so no $d$-$d$ electron transitions are possible, and no visible light is absorbed (1 mark).

Q2 (6 marks)

Describe the colour changes observed when excess concentrated ammonia solution is added to a solution containing $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}(aq)$ ions. Write equations for the reactions occurring.

Mark Scheme:

Initial solution: Pale blue (1 mark).

On adding a few drops of $\mathrm{NH}_3$: A pale blue precipitate of $\mathrm{Cu}(\mathrm{OH})_2$ forms (1 mark):

$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+} + 2\mathrm{NH}_3 \to \mathrm{Cu}(\mathrm{OH})_2(s) + 2\mathrm{NH}_4^+ + 4\mathrm{H}_2\mathrm{O}$

(1 mark for equation.)

On adding excess $\mathrm{NH}_3$: The precipitate dissolves to form a deep blue solution (1 mark):

$\mathrm{Cu}(\mathrm{OH})_2(s) + 4\mathrm{NH}_3 \to [\mathrm{Cu}(\mathrm{NH}_3)_4(\mathrm{H}_2\mathrm{O})_2]^{2+} + 2\mathrm{OH}^-$

(1 mark for equation.)

The deep blue colour is due to the tetraamminecopper(II) complex, which has a different crystal field splitting than the hexaaqua complex, absorbing different wavelengths of light (1 mark for explanation of colour change).

Q3 (4 marks)

Explain what is meant by the term ligand. Give one example of a bidentate ligand and explain why it forms more stable complexes than monodentate ligands.

Mark Scheme:

A ligand is a molecule or ion that can donate a lone pair of electrons to a transition metal ion to form a coordinate (dative covalent) bond (1 mark).

Example of a bidentate ligand: ethane-1,2-diamine ($\mathrm{H}_2\mathrm{NCH}_2\mathrm{CH}_2\mathrm{NH}_2$, "en") or oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$) (1 mark).

Bidentate ligands form more stable complexes because of the chelate effect: the entropy change is more favourable (one ligand replaces two, releasing two molecules) and the chelate ring structure provides additional stability through the macrocyclic effect (1 mark for entropy, 1 mark for ring structure).

Q4 (5 marks)

The complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is diamagnetic, whereas $[\mathrm{CoF}_6]^{3-}$ is paramagnetic with four unpaired electrons. Explain these observations using crystal field theory.

Mark Scheme:

$\mathrm{Co}^{3+}$ has the electron configuration $[\mathrm{Ar}]\,3d^6$ (1 mark).

$\mathrm{NH}_3$ is a strong-field ligand (high in the spectrochemical series), so $\Delta_o$ is large (1 mark). The pairing energy is less than $\Delta_o$, so all six $d$-electrons pair in the $t_{2g}$ orbitals: $t_{2g}^6\,e_g^0$ (low-spin, diamagnetic) (1 mark).

$\mathrm{F}^-$ is a weak-field ligand, so $\Delta_o$ is small (1 mark). The pairing energy is greater than $\Delta_o$, so the electrons occupy both $t_{2g}$ and $e_g$ orbitals following Hund's rule: $t_{2g}^4\,e_g^2$ (high-spin, paramagnetic with four unpaired electrons) (1 mark).

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tip

Diagnostic Test Ready to test your understanding of Transition Metals? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Transition Metals with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.