The transition metals

Class 12 Chemistry, unit 6, book pages 94–127

Practise 58 MCQs Take the unit quiz Review 38 cards

Transition metals such as iron, copper and titanium sit in the d-block at the centre of the periodic table. Because their 3d and 4s electrons are close in energy, they show variable oxidation states, act as catalysts, form complex ions with ligands and give coloured compounds. This unit explains these properties through electron configurations and d-orbital splitting, and applies them to naming complexes, their shapes and isomers, ligand exchange, stability constants and redox titrations.

Learning outcomes (28)

Identify the general physical and chemical properties of the first row of transition elements, titanium to copper.
Define a transition element as a d-block element which forms one or more stable ions with incomplete d orbitals.
Sketch the shape of the $3 d_{x y}$ and $3 d_{z^{2}}$ orbitals.
Identify the properties of transition elements: variable oxidation states, catalytic activity, formation of complex ions and formation of coloured compounds.
Explain why transition elements have variable oxidation states in terms of the similarity in energy of the 3d and 4s sub-shells.
Explain why transition elements behave as catalysts in terms of having more than one stable oxidation state, and vacant d orbitals that are energetically accessible and can form dative bonds with ligands.
Explain why transition elements form complex ions in terms of vacant d orbitals that are energetically accessible.
Explain the reaction of transition elements with ligands to form complexes, including complexes of copper(II) and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions.
Define the term ligand as a species that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom or ion.
Use the term monodentate ligand, including water, ammonia, chloride and cyanide; bidentate ligand, including 1,2-diaminoethane and the ethanedioate ion; and polydentate ligand, including EDTA.
Define the term complex as a molecule or ion formed by a central metal atom or ion surrounded by one or more ligands.
Describe the geometry (shape and bond angles) of transition element complexes which are linear, square planar, tetrahedral or octahedral.
State what is meant by coordination number.
Predict the formula and charge of a complex ion, given the metal ion, its charge or oxidation state, the ligand and its coordination number or geometry.
Explain qualitatively that ligand exchange can occur, including the complexes of copper(II) and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions.
Predict, using $E^{\circ}$ values, the feasibility of redox reactions involving transition elements and their ions.
Analyse reactions involving $MnO X_{4} X^{-} / C X_{2} O X_{4} X^{2 -}$ , $MnO X_{4} X^{-} / Fe X^{2 +}$ and $Cu X^{2 +} / I X^{-}$ in solution given suitable data (including describing the reaction and doing calculations).
Perform calculations involving the colour of complexes.
Use the terms degenerate and non-degenerate d orbitals.
Describe the splitting of degenerate d orbitals into two non-degenerate sets and use $Δ E$ : in octahedral complexes two higher and three lower d orbitals; in tetrahedral complexes three higher and two lower d orbitals.
Explain why transition elements form coloured compounds in terms of the frequency of light absorbed as an electron is promoted between two non-degenerate d orbitals.
Describe, in qualitative terms, the effects of different ligands on $Δ E$ , the frequency of light absorbed, and hence the complementary colour that is observed.
Use the complexes of copper(II) and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions as examples of ligand exchange affecting the colour observed.
Describe the types of stereoisomerism shown by complexes, including those associated with bidentate ligands: geometrical isomerism, for example square planar $[Pt (NH X_{3}) X_{2} Cl X_{2}]$ and octahedral $[Co (NH X_{3}) X_{4} (H X_{2} O) X_{2}] X^{2 +}$ and $[Ni (H X_{2} NCH X_{2} CH X_{2} NH X_{2}) X_{2} (H X_{2} O) X_{2}] X^{2 +}$ ; optical isomerism, for example $[Ni (H X_{2} NCH X_{2} CH X_{2} NH X_{2}) X_{3}] X^{2 +}$ and $[Ni (H X_{2} NCH X_{2} CH X_{2} NH X_{2}) X_{2} (H X_{2} O) X_{2}] X^{2 +}$ .
Deduce the overall polarity of complexes.
Define the stability constant ( $K_{stab}$ ) of a complex as the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.
Write an expression for $K_{stab}$ of a complex ( $[H X_{2} O]$ is not included) and use $K_{stab}$ expressions to perform calculations.
Explain ligand exchanges in terms of $K_{stab}$ values and understand that a large $K_{stab}$ is due to the formation of a stable complex ion.

6Transition elements: definition and positionOften examined

Look at the centre of the periodic table. From period 4 onwards there are ten columns between group 2 (the s-block) and group 13 (the p-block). In these elements the last electron enters a d sub-shell, so they are called the d-block elements. The first row (period 4) runs from scandium ( $Z = 21$ ) to zinc ( $Z = 30$ ).

These metals are vital in daily life. Copper is used in wiring, coins and plumbing; iron in bridges, vehicle parts and buildings; chromium for plating plumbing fixtures; gold and silver in jewellery and electronics; platinum in catalytic converters; titanium in bicycles, aircraft and artificial joints; and nickel, vanadium, molybdenum and tantalum in industry and medicine. Transition metals also form alloys such as brass (copper and zinc) and bronze (copper and tin).

d-block element: An element in which the last (differentiating) electron enters a d sub-shell. In period 4 these are Sc to Zn.
Transition element: A d-block element which forms one or more stable ions with an incomplete (partly filled) d sub-shell. The book also states it as: elements which have an incomplete d sub-shell in their most stable ionic states.

The word 'transition' means change. Across period 4 these elements show a gradual change in properties from the very reactive group 2 metal calcium to the much less reactive group 13 element gallium.

The first row of the d-block. Titanium to copper are the transition elements studied in this unit. Zinc is a d-block element but not a transition element.

Why zinc is not a transition element

Zinc has the configuration $[Ar] 3 d^{10} 4 s^{2}$ and forms only one ion, $Zn X^{2 +}$ , which is $[Ar] 3 d^{10}$ . The d sub-shell is complete both in the atom and in its only stable ion, so zinc does not satisfy the definition. Zinc does form complexes, such as $[Zn (NH X_{3}) X_{4}] X^{2 +}$ , but forming complexes alone is not enough to make an element a transition element.

Because its d sub-shell is full, no d-d transition is possible, so zinc compounds are white or colourless.
Zinc shows only one oxidation state (+2), not variable oxidation states.
Zinc has the lowest melting point of the row, because its full 3d and 4s electrons take little part in metallic bonding.

Why scandium was controversial

The common ion of scandium, $Sc X^{3 +}$ , is $[Ar]$ : its 3d sub-shell is empty, so $Sc X^{3 +}$ compounds are colourless. For this reason scandium was not always counted as a transition element. It was later included by the IUPAC system when compounds with scandium in the +1 and +2 oxidation states, such as $CsScCl X_{3}$ , were made. In these compounds scandium has a partly filled 3d sub-shell ( $Sc X^{2 +}$ is $[Ar] 3 d^{1}$ ).

This unit mainly deals with the first row, from titanium to copper. To understand their physical and chemical properties we first need their electronic configurations.

6.1Electron configurations of the d-block elementsOften examined

The electrons of the d-block atoms fill orbitals according to the Aufbau principle: lower-energy sub-shells fill first. In potassium and calcium the 4s sub-shell is lower in energy than 3d, so 4s fills first. From scandium to zinc each new electron enters the 3d sub-shell, while the 4s sub-shell stays $4 s^{2}$ . The inner core $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ is written $[Ar]$ .

Electronic configurations of the first-row d-block elements and their ions (book table, with a few common ions added).
Element (Z)	Atom	Common ions and their configurations
Sc (21)	$[Ar] 3 d^{1} 4 s^{2}$	$Sc X^{3 +}$ : $[Ar]$ ; $Sc X^{2 +}$ : $[Ar] 3 d^{1}$ ; $Sc X^{+}$ : $[Ar] 3 d^{1} 4 s^{1}$
Ti (22)	$[Ar] 3 d^{2} 4 s^{2}$	$Ti X^{2 +}$ : $[Ar] 3 d^{2}$ ; $Ti X^{3 +}$ : $[Ar] 3 d^{1}$ ; $Ti X^{4 +}$ : $[Ar]$
V (23)	$[Ar] 3 d^{3} 4 s^{2}$	$V X^{2 +}$ : $[Ar] 3 d^{3}$ ; $V X^{3 +}$ : $[Ar] 3 d^{2}$ ; $V X^{4 +}$ : $[Ar] 3 d^{1}$
Cr (24)	$[Ar] 3 d^{5} 4 s^{1}$	$Cr X^{2 +}$ : $[Ar] 3 d^{4}$ ; $Cr X^{3 +}$ : $[Ar] 3 d^{3}$
Mn (25)	$[Ar] 3 d^{5} 4 s^{2}$	$Mn X^{2 +}$ : $[Ar] 3 d^{5}$ ; $Mn X^{3 +}$ : $[Ar] 3 d^{4}$
Fe (26)	$[Ar] 3 d^{6} 4 s^{2}$	$Fe X^{2 +}$ : $[Ar] 3 d^{6}$ ; $Fe X^{3 +}$ : $[Ar] 3 d^{5}$
Co (27)	$[Ar] 3 d^{7} 4 s^{2}$	$Co X^{2 +}$ : $[Ar] 3 d^{7}$ ; $Co X^{3 +}$ : $[Ar] 3 d^{6}$
Ni (28)	$[Ar] 3 d^{8} 4 s^{2}$	$Ni X^{2 +}$ : $[Ar] 3 d^{8}$
Cu (29)	$[Ar] 3 d^{10} 4 s^{1}$	$Cu X^{+}$ : $[Ar] 3 d^{10}$ ; $Cu X^{2 +}$ : $[Ar] 3 d^{9}$
Zn (30)	$[Ar] 3 d^{10} 4 s^{2}$	$Zn X^{2 +}$ : $[Ar] 3 d^{10}$

The two exceptions: chromium and copper

Chromium and copper break the Aufbau pattern. Chromium is $[Ar] 3 d^{5} 4 s^{1}$ , not $3 d^{4} 4 s^{2}$ , and copper is $[Ar] 3 d^{10} 4 s^{1}$ , not $3 d^{9} 4 s^{2}$ . One 4s electron moves into the 3d sub-shell. The book gives these reasons:

A half-filled ( $3 d^{5}$ ) or completely filled ( $3 d^{10}$ ) d sub-shell is extra stable (lower in energy).
Moving one electron out of 4s avoids the inter-electronic repulsion between the two paired electrons in the 4s orbital.
In $3 d^{5}$ and $3 d^{10}$ the electron cloud is symmetrical, and a symmetrical cloud shields the nucleus in a more effective way.

Electrons-in-boxes for chromium and copper. The observed configurations have a half-filled or a completely filled 3d sub-shell and only one 4s electron.

Configurations of ions

Writing the configuration of an ion

Write the electronic configurations of

Cr X^{3 +}

Mn X^{2 +}

and

Cu X^{2 +}

, and state the number of unpaired electrons in each.

Chromium. Start from the atom and remove the 4s electron first, then two 3d electrons.
$Cr : [Ar] 3 d^{5} 4 s^{1} ⟶ Cr X^{3 +} : [Ar] 3 d^{3}$
The three 3d electrons occupy separate orbitals (Hund's rule): 3 unpaired electrons.
Manganese. Remove the two 4s electrons only.
$Mn : [Ar] 3 d^{5} 4 s^{2} ⟶ Mn X^{2 +} : [Ar] 3 d^{5}$
Half-filled 3d: 5 unpaired electrons.
Copper. Remove the single 4s electron, then one 3d electron.
$Cu : [Ar] 3 d^{10} 4 s^{1} ⟶ Cu X^{2 +} : [Ar] 3 d^{9}$
Four pairs and one single electron: 1 unpaired electron.

Cr X^{3 +} : [Ar] 3 d^{3} (3), Mn X^{2 +} : [Ar] 3 d^{5} (5), Cu X^{2 +} : [Ar] 3 d^{9} (1)

Relative energies of the 4s and 3d sub-shells

The energy gap between 3d and 4s changes across the period. In calcium the 4s orbital has lower energy than 3d, because the 4s electron penetrates close to the nucleus and feels almost the full attraction of the unscreened nucleus. This is why 4s fills before 3d.

After calcium, electrons are added to the inner 3d sub-shell. The 3d electrons shield the 4s electrons, so the 4s electrons are held only a little more strongly as the nuclear charge rises. As a result the first and second ionisation energies, which remove the two 4s electrons, rise only slightly from scandium to zinc. The 3d electrons themselves are poorly shielded, feel a higher effective nuclear charge and are strongly attracted by the nucleus, so the third ionisation energy, which removes a 3d electron, rises much more steeply across the series.

6.2, 6.2.1Physical properties of transition elementsOften examined

The transition elements are much more like each other than the elements across period 3. Across period 3 the elements change from metals (Na, Mg, Al) to a metalloid (Si), then to non-metals (P, S, Cl) and a noble gas (Ar). All the transition elements are metals. Their atomic radii also change much less than in period 3, because each new electron goes into an inner (3d) sub-shell. Because of these similarities, transition metals show similar physical properties.

General physical properties

High melting and boiling points: strong metallic bonding, because electrons of both the 3d and 4s sub-shells are delocalised into the metallic bond.
High density: the atoms are heavy and small and are closely packed in the metal lattice.
Good conductors of electricity: they have free (delocalised) electrons that move through the lattice when a voltage is applied.
Good conductors of heat: the free electrons and the closely packed vibrating ions carry thermal energy.
Shiny metallic lustre: the free electrons at the surface absorb and re-emit light.
Malleable and ductile: layers of metal ions can slide over one another without breaking the metallic bond, so the metals can be hammered into sheets and drawn into wires.
Hard and strong: this makes them suitable for construction and manufacturing (iron in steel, titanium in aircraft).
Magnetic properties: many transition metals and their compounds are paramagnetic, and iron, cobalt and nickel are ferromagnetic. These properties are used in many applications, including magnetic resonance imaging (MRI).

How transition metals differ from s-block metals.
Property	Group 1 and 2 metals (K, Ca)	Transition metals (Fe, Cu)
Melting point	low (K $63^{\circ} C$ , Ca $842^{\circ} C$ )	high (Fe $1538^{\circ} C$ , Ti $1668^{\circ} C$ )
Density	low (K $0.86$ , Ca $1.55 g c m^{- 3}$ )	high (Fe $7.87$ , Cu $8.96 g c m^{- 3}$ )
Hardness	soft (potassium can be cut with a knife)	hard and strong
Reactivity	very reactive (react quickly with water)	much less reactive
Oxidation states	one (+1 or +2)	several (variable)
Colour of compounds	white or colourless	usually coloured
Complex formation	weak, few complexes	form many stable complexes
Catalytic activity	little	good catalysts

6.2.2 (1)Density

Density is the mass per unit volume of a substance. For metals it is usually expressed in grams per cubic centimetre ( $g c m^{- 3}$ ). It is an important physical property because it shows how closely the metal atoms are packed in the metallic lattice.

ρ = \frac{m}{V}

$ρ$ : density ( $g c m^{- 3}$ ), $m$ : mass (g), $V$ : volume ( $c m^{3}$ ).

Moving through the first row from scandium to copper, the density generally increases. There are two reasons:

The atomic mass increases across the period, so each atom is heavier.
The atomic radius decreases across the period, because the increasing nuclear charge pulls the electron cloud closer. Each atom takes up less volume.

Density is directly related to mass and inversely related to volume, so more mass in less volume gives a higher density.

Density of the first-row d-block metals

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Z

ρ / g c m^{- 3}

2.99

4.51

6

7.15

7.3

7.87

8.86

8.9

8.96

7.14

density

Data-book values at room temperature (the book gives no numbers). The general rise has two exceptions: manganese (lower than the trend) and zinc (a sharp drop).

There are exceptions to the general trend. Manganese is less dense than expected, because it has an unusual, loosely packed crystal structure. Zinc is much less dense than copper: its full $3 d^{10}$ sub-shell takes little part in metallic bonding, so the zinc atoms are larger and less tightly packed in the metallic lattice, which reduces the density.

6.2.2 (2)Atomic radiiOften examined

The size of an atom depends on two opposing effects: the nuclear charge, which pulls the electrons inwards, and the shielding effect of the inner electrons together with electron-electron repulsion, which lets the outer electrons move outwards.

Fig 6.1: Atomic radii of the first-row transition metals

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

100

120

140

160

180

Z

r / pm

162

147

134

128

127

126

125

124

128

134

atomic radius

The radius falls quickly at first (Sc to Cr), stays almost constant in the middle (Cr to Ni) and then rises again (Cu, Zn).

Explaining the trend

Scandium to chromium: the radius decreases. The nuclear charge increases by one unit at each step and this effect overcomes the shielding effect, so the electrons are pulled closer (Sc 162, Ti 147, V 134, Cr $128 pm$ ).
Chromium to nickel: the radius is almost constant. Each added electron goes into the inner 3d sub-shell. These 3d electrons shield the outer 4s electrons and repel them, which cancels the effect of the extra nuclear charge. The two factors counterbalance each other (Mn 127, Fe 126, Co 125, Ni $124 pm$ ).
Nickel to copper to zinc: the radius increases. The 3d sub-shell becomes full. A full $3 d^{10}$ sub-shell shields strongly and the electron-electron repulsion is large, which pushes the 4s electrons away from the nucleus (Cu $128 pm$ ). In zinc the additional electron in the 4s sub-shell increases the repulsion and shielding further, which outweighs the effect of the higher nuclear charge (Zn $134 pm$ ).

6.2.2 (3)Melting and boiling pointsOften examined

The melting and boiling points of the transition elements are generally much higher than those of the s- and p-block elements (for example calcium melts at 842\,^\circ\mathrm{C} but titanium at 1668\,^\circ\mathrm{C}). They have strong metallic bonding, because electrons of both the 3d and 4s sub-shells are involved in the metallic bond.

Fig 6.2: Melting and boiling points of the first-row d-block metals

Values from the book (in ^\circ\mathrm{C}). Both curves rise at first, dip at manganese, and fall to their lowest values at zinc.

Explaining the shape of the graph

Sc to V and Cr: increase. The number of unpaired 3d electrons increases (Sc 1, Ti 2, V 3, Cr 5). More electrons take part in the metallic bond, so the bond gets stronger. The melting point reaches its highest values at vanadium (1910\,^\circ\mathrm{C}) and chromium (1907\,^\circ\mathrm{C}).
A sharp minimum at manganese. Manganese, [\mathrm{Ar}]\,3d^{5}4s^{2}, has a stable half-filled 3d sub-shell and a full 4s sub-shell. These electrons are held firmly and are not readily available for metallic bonding, so its metallic bond is weaker (1246\,^\circ\mathrm{C}). Both graphs show a minimum at manganese.
Fe to Cu: gradual decrease. After the middle of the series the 3d electrons begin to pair up. Paired electrons are less available for metallic bonding, so the melting point falls gradually (Fe 1538, Co 1495, Ni 1455, Cu 1085\,^\circ\mathrm{C}).
Zinc is the lowest. In zinc, [\mathrm{Ar}]\,3d^{10}4s^{2}, both sub-shells are full and cannot take part in metallic bonding effectively, so zinc has the lowest melting point (419\,^\circ\mathrm{C}) and boiling point (907\,^\circ\mathrm{C}). For the same reason mercury, which belongs to the zinc family (full 5d and 6s sub-shells), is a liquid at room temperature.

6.2.2 (4)Magnetic properties

Transition metals and their compounds show magnetic properties, which give information about the chemical bonding in them and have important uses in modern technology. An electron spinning about its own axis generates a magnetic field, just as an electric current flowing through a wire does. So a single unpaired electron acts like a small magnet. In a filled orbital the two electrons have opposite spins, and their magnetic fields cancel. The magnetic behaviour of a substance therefore depends on its unpaired electrons.

Paired electrons cancel (diamagnetic). Unpaired spins point at random unless a field lines them up (paramagnetic). In a ferromagnetic solid the spins of neighbouring atoms line up together in domains.

Paramagnetic substance: A substance that is weakly attracted into a magnetic field because it has unpaired electrons. The unpaired electrons on different atoms or ions are not influenced by one another, so their magnetic moments point randomly. In a magnetic field they line up parallel to the field and a weak attraction is observed. When the magnet is removed, the alignment and the paramagnetism are lost. Examples: aluminium, sodium, and ions such as \ce{Cu^2+}, \ce{Fe^3+} and \ce{Mn^2+}.
Ferromagnetic substance: A substance that is very strongly attracted by a magnet (nearly a million times more than a paramagnetic substance), for example iron, cobalt and nickel. Their unpaired electrons are influenced by the electrons of neighbouring atoms, so their spins line up in the same direction. They keep their magnetic moment even after the field is removed, so they can be made into permanent magnets.
Diamagnetic substance: A substance that is weakly repelled by a magnet because all its electrons are paired. The paired electrons cancel each other's magnetic moments, so the net magnetic moment is zero. Examples: zinc, copper metal, and ions with d^{0} or d^{10} configurations such as \ce{Sc^3+}, \ce{Ti^4+}, \ce{Zn^2+} and \ce{Cu+}.

Unpaired electrons in simple ions (each d orbital is singly filled before pairing starts).
Ion	d electrons	Unpaired electrons	Magnetic behaviour
\ce{Sc^3+}, \ce{Ti^4+}	d^{0}	0	diamagnetic
\ce{Ti^3+}	d^{1}	1	paramagnetic
\ce{V^3+}	d^{2}	2	paramagnetic
\ce{Cr^3+}	d^{3}	3	paramagnetic
\ce{Mn^2+}, \ce{Fe^3+}	d^{5}	5	strongly paramagnetic
\ce{Fe^2+}	d^{6}	4	paramagnetic
\ce{Co^2+}	d^{7}	3	paramagnetic
\ce{Ni^2+}	d^{8}	2	paramagnetic
\ce{Cu^2+}	d^{9}	1	paramagnetic
\ce{Cu+}, \ce{Zn^2+}	d^{10}	0	diamagnetic

6.2.2 (5)Alloys

An alloy is a mixture of metals (sometimes with a little of a non-metal such as carbon) that has more useful properties than the pure metals. Transition metals form alloys easily because their atoms have similar atomic sizes (radii 124 to 134\,\mathrm{pm} from chromium to zinc). Atoms of one metal can therefore substitute for atoms of another in the metallic lattice without disturbing it much.

Common alloys of transition metals.
Alloy	Composition	Useful properties
Steel (stainless steel)	iron with chromium, nickel and manganese (and a little carbon)	hard, strong, resists corrosion and stays shiny: cutlery, surgical tools, sinks
Brass	copper and zinc	harder than copper, golden colour, easily shaped: taps, fittings, musical instruments
Bronze	copper and tin	hard and resistant to corrosion: statues, medals, bearings

6.2.3 (1)Variable oxidation statesOften examined

Across the first row the physical properties are similar, but the chemical properties show much less similarity. The most important chemical property is variable oxidation states: a transition element can form compounds in which it has different oxidation states, for example iron is +2 in \ce{FeCl2} and +3 in \ce{FeCl3}, and manganese is +2 in \ce{MnSO4}, +4 in \ce{MnO2} and +7 in \ce{KMnO4}.

Why the oxidation state can vary

The 4s and 3d sub-shells have very similar energies. After the two 4s electrons are lost, one or more 3d electrons can also be lost (or used in covalent bonds) without a very large jump in energy. So different numbers of electrons can be used in bonding, which gives different oxidation states. In calcium, by contrast, a third electron would have to come from the inner 3p sub-shell, which needs far more energy, so calcium shows only +2.

The common oxidation state is +2, obtained by the loss of the two 4s electrons to form an \ce{M^2+} ion. The maximum stable oxidation state frequently corresponds to the total number of 3d and 4s electrons available for bonding. For example manganese, [\mathrm{Ar}]\,3d^{5}4s^{2}, has a maximum oxidation state of +7 (in \ce{MnO4^-}).

Oxidation states of the first-row d-block elements.
Element	Outer configuration	3d + 4s electrons	Oxidation states (book table)
Sc	3d^{1}4s^{2}	3	+3
Ti	3d^{2}4s^{2}	4	+2, +3, +4
V	3d^{3}4s^{2}	5	+2, +3, +4, +5
Cr	3d^{5}4s^{1}	6	+2, +3, +6
Mn	3d^{5}4s^{2}	7	+2, +3, +4, +6, +7
Fe	3d^{6}4s^{2}	8	+2, +3
Co	3d^{7}4s^{2}	9	+2, +3
Ni	3d^{8}4s^{2}	10	+2
Cu	3d^{10}4s^{1}	11	+1, +2
Zn	3d^{10}4s^{2}	12	+2

Oxidation states across the first row

Dots mark the oxidation states in the book's table. The highest state rises by one at each step up to manganese (+7) and then falls.

Patterns to remember

Up to manganese, the maximum oxidation state equals the number of 3d + 4s electrons: Sc +3, Ti +4, V +5, Cr +6, Mn +7.
After manganese the 3d electrons are held more strongly by the increasing nuclear charge and begin to pair, so high oxidation states become harder to reach. The maximum falls to +3 (Fe, Co) and then +2.
The stability of the +2 state relative to the +3 state increases from left to right, because it becomes increasingly difficult to remove a 3d electron as the nuclear charge increases.
The stability of the higher oxidation states decreases from left to right across the series.
Simple ions exist in low oxidation states (such as \ce{Mn^2+}, \ce{Fe^2+}), while high oxidation states are stable in complex ions and oxo-anions (such as \ce{MnO4^-}, \ce{Cr2O7^2-}), where the metal is covalently bonded to oxygen.

Stability of half-filled sub-shells: manganese and iron

\ce{Mn^2+} is [\mathrm{Ar}]\,3d^{5} (half-filled), so it is more stable than \ce{Mn^3+}, [\mathrm{Ar}]\,3d^{4}. \ce{Fe^3+} is [\mathrm{Ar}]\,3d^{5} (half-filled), so it is more stable than \ce{Fe^2+}, [\mathrm{Ar}]\,3d^{6}. This is why \ce{Fe^2+} is easily oxidised to \ce{Fe^3+}, and why pale green iron(II) solutions slowly turn yellow-brown in air.

Finding an oxidation state from a formula

Find the oxidation state of manganese in \ce{KMnO4} and of chromium in \ce{K2Cr2O7}.

The sum of the oxidation states in a neutral compound is zero. K is +1 and O is -2. Let Mn be x.
(+1)+x+4(-2)=0\ \Longrightarrow\ x=+7
For dichromate, there are two chromium atoms, each y.
2(+1)+2y+7(-2)=0\ \Longrightarrow\ 2y=+12,\ y=+6

\ce{Mn}=+7,\qquad \ce{Cr}=+6

6.2.3 (2)Catalytic activityOften examined

Catalyst: A substance that increases the rate of a chemical reaction without itself being used up; it is chemically unchanged at the end of the reaction. It works by providing an alternative pathway (mechanism) with a lower activation energy.
Activation energy, E_a: The minimum energy that colliding particles must have for a reaction to take place.

Energy profile of an exothermic reaction with and without a catalyst. The catalyst lowers the activation energy; the enthalpy change \Delta H stays the same.

The catalysed path has a lower peak, so at the same temperature a much larger fraction of collisions has energy equal to or greater than E_a, and the rate increases.
The energies of the reactants and products are not changed, so \Delta H is the same with or without a catalyst.
The catalyst lowers the activation energy of the forward and the backward reactions equally, so it does not change the position of equilibrium; equilibrium is only reached faster.

Why transition metals are good catalysts

Transition elements or their compounds are used as catalysts in many reactions. Two unique properties are responsible for their catalytic activity:

They can exist in more than one stable oxidation state. The metal can change easily between oxidation states, such as \ce{Fe^2+} and \ce{Fe^3+} or \ce{Cu+} and \ce{Cu^2+}. During the reaction it temporarily gains or loses electrons and so helps the reactants change into products. At the end it regains its original oxidation state and is ready to act as a catalyst again.
They have vacant d orbitals that are energetically accessible. These vacant orbitals can accept electron pairs from reactant molecules (acting as ligands) and form temporary dative bonds. The dative bonds hold the reactants on the surface of the metal and bring them close together. As a result the bonds within the reactant molecules weaken and break, and products form on the surface. After the reaction the metal releases the products and binds new reactants.

Heterogeneous catalysis

In heterogeneous catalysis the catalyst and the reactants are in different physical states, usually a solid catalyst with gaseous reactants. The reaction takes place on the catalyst surface.

Haber process (manufacture of ammonia): iron adsorbs nitrogen and hydrogen molecules on its surface, weakens their bonds and makes them react faster.
Contact process (manufacture of sulphuric acid): vanadium(V) oxide, \ce{V2O5}, catalyses the oxidation of sulphur dioxide by a reversible change in the oxidation state of vanadium between +5 and +4.

\ce{N2(g) + 3H2(g) <=>[Fe][450^\circ C,\ 200\ atm] 2NH3(g)}

\ce{2SO2(g) + O2(g) <=>[V2O5][450^\circ C] 2SO3(g)}

\begin{aligned}&\ce{V2O5 + SO2 -> V2O4 + SO3}&&(\text{V: }+5\to+4)\\&\ce{V2O4 + 1/2O2 -> V2O5}&&(\text{V: }+4\to+5)\end{aligned}

Adding the two steps gives \ce{SO2 + 1/2O2 -> SO3}; the \ce{V2O5} is regenerated.

Homogeneous catalysis

In homogeneous catalysis the catalyst and the reactants are in the same physical state, for example all in aqueous solution. An example is the oxidation of iodide ions by peroxodisulphate(VI) ions:

\ce{2I^-(aq) + S2O8^2-(aq) -> 2SO4^2-(aq) + I2(aq)}

This reaction is slow, because both reactants are negatively charged ions that repel each other, so the activation energy is high. It is catalysed by adding d-block metal ions such as \ce{Fe^2+(aq)}, which change their oxidation state reversibly between +2 and +3:

\begin{aligned}&\text{Step 1:}\quad \ce{2Fe^2+(aq) + S2O8^2-(aq) -> 2Fe^3+(aq) + 2SO4^2-(aq)}\\&\text{Step 2:}\quad \ce{2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(aq)}\end{aligned}

Both steps involve a reaction between oppositely charged ions, which attract each other, so each step is faster than the uncatalysed reaction. Adding the two steps gives back the overall equation, and the \ce{Fe^2+} used in step 1 is regenerated in step 2.

With \ce{Fe^2+} the reaction goes by two steps, each with a lower activation energy than the direct reaction between two negative ions.

6.2.3LigandsOften examined

A ligand is an atom, ion or molecule which is electron-rich and has at least one lone pair of electrons that it donates to a transition metal atom or ion, forming a dative covalent (coordinate) bond with it. Ligands may be neutral molecules or negative ions made of non-metal atoms. Because they donate electron pairs, ligands are Lewis bases, and they are also nucleophiles.

Monodentate ligands

A monodentate ligand (Latin: 'one-toothed') forms only one dative bond with the metal atom or ion. It may have more than one lone pair, but it can donate only one of them to the metal. Examples: aqua (\ce{H2O}), ammine (\ce{NH3}), halo (\ce{X^-}, such as chloro \ce{Cl^-}), hydroxo (\ce{OH^-}) and cyano (\ce{CN^-}).

Polydentate ligands

A polydentate ligand ('many-toothed') can donate more than one lone pair and so forms more than one dative bond with the same metal ion. A bidentate ligand forms two bonds. 1,2-diaminoethane (\ce{H2NCH2CH2NH2}, also called ethylenediamine and written en) bonds through its two N atoms, and the ethanedioate (oxalate) ion (\ce{C2O4^2-}) bonds through two O atoms.

1,2-diaminoethane (en)

ethanedioate (oxalate) ion

EDTA (the anion of ethylenediaminetetraacetic acid) is a hexadentate ligand. It donates six lone pairs, two from its N atoms and four from its \ce{O^-} atoms, and wraps around one metal ion to form very stable 1:1 complexes such as \ce{[Cu(EDTA)]^2-}.

EDTA (tetra-anion) The two N atoms and the four O atoms carrying the negative charges are the six donor atoms.

Common ligands.
Ligand	Formula	Charge	Donor atom(s)	Denticity	Name in a complex
water	\ce{H2O}	0	O	mono	aqua
ammonia	\ce{NH3}	0	N	mono	ammine
carbon monoxide	\ce{CO}	0	C	mono	carbonyl
chloride	\ce{Cl^-}	-1	Cl	mono	chloro
bromide	\ce{Br^-}	-1	Br	mono	bromo
hydroxide	\ce{OH^-}	-1	O	mono	hydroxo
cyanide	\ce{CN^-}	-1	C	mono	cyano
nitrite	\ce{NO2^-}	-1	N	mono	nitro
1,2-diaminoethane (en)	\ce{H2NCH2CH2NH2}	0	2 N	bi	1,2-diaminoethane
ethanedioate (oxalate)	\ce{C2O4^2-}	-2	2 O	bi	ethanedioato (oxalato)
EDTA	\ce{EDTA^4-}	-4	2 N, 4 O	hexa	ethylenediaminetetraacetato

6.3Formation of complex compoundsOften examined

A complex compound (coordination compound) is formed when ligands donate lone pairs of electrons into vacant orbitals of a transition metal atom or ion. The ligand is the electron-pair donor, so it is a Lewis base. The metal accepts the electron pairs, so it is a Lewis acid. The bonds formed are coordinate covalent (dative) bonds.

Complex: A molecule or ion formed by a central metal atom or ion surrounded by one or more ligands bonded to it by dative covalent bonds. Examples: \ce{[Cu(H2O)6]^2+}, \ce{[CuCl4]^2-}, \ce{[Ni(CO)4]}.
Complex ion: A complex that carries an overall charge, positive (such as \ce{[Co(NH3)6]^3+}) or negative (such as \ce{[Fe(CN)6]^4-}). A complex with zero charge, such as \ce{[Ni(CO)4]}, is a neutral complex.

Why transition metals form complexes

Small size and high charge. Transition metal ions are small and carry a charge of +2 or +3, so they have a high charge density. They attract the lone pairs of ligands strongly.
Vacant orbitals of low energy. The ions have vacant 3d orbitals (and empty 4s, 4p and 4d orbitals) that are close in energy and energetically accessible. Ligands can easily donate lone pairs into them to form dative bonds.

Hybridisation in the book's model

The book explains complex formation in three steps. The metal atom first loses electrons to become a cation. The vacant orbitals of the cation then hybridise (mix) to give the same number of new, equivalent orbitals of equal energy. Finally each ligand donates one lone pair into one hybrid orbital. Six hybrid orbitals (sp^{3}d^{2}) give an octahedral complex; four (sp^{3}) give a tetrahedral complex.

Complexes of copper(II) with water and chloride ligands

The copper(II) ion is [\mathrm{Ar}]\,3d^{9}. Its 3d orbitals hold nine electrons, so the vacant 4s, three 4p and two 4d orbitals are used. They hybridise into six sp^{3}d^{2} orbitals, and six water molecules each donate a lone pair to form the octahedral hexaaquacopper(II) ion.

\ce{Cu^2+(aq) + 6H2O(l) -> [Cu(H2O)6]^2+(aq)}

Formation of \ce{[Cu(H2O)6]^2+}: six vacant orbitals (4s, 4p, 4d) hybridise to sp^{3}d^{2} and accept six lone pairs from water.

With chloride ions, copper(II) mixes only its four vacant 4s and 4p orbitals to give four sp^{3} hybrid orbitals. Four chloride ligands bond to them, giving the tetrahedral tetrachlorocuprate(II) ion. Only four chloride ions fit because the chloride ion is large (Cl is in period 3) and six of them cannot surround the small copper(II) ion.

\ce{Cu^2+(aq) + 4Cl^-(aq) -> [CuCl4]^2-(aq)}

Formation of \ce{[CuCl4]^2-}: four vacant orbitals (4s, 4p) hybridise to sp^{3} and accept four lone pairs from chloride ions.

Complexes of cobalt(II) with water and chloride ligands

The cobalt(II) ion is [\mathrm{Ar}]\,3d^{7}. With water it forms the octahedral, pink hexaaquacobalt(II) ion, \ce{[Co(H2O)6]^2+}: six empty orbitals (4s, 4p, 4d) hybridise to sp^{3}d^{2} and accept six lone pairs from six water molecules. With four chloride ions it forms the tetrahedral, blue tetrachlorocobaltate(II) ion, \ce{[CoCl4]^2-}, using four sp^{3} hybrid orbitals.

Formation of \ce{[Co(H2O)6]^2+} from \ce{Co^2+} (3d^{7}) using six sp^{3}d^{2} hybrid orbitals.

Formation of \ce{[CoCl4]^2-} from \ce{Co^2+} using four sp^{3} hybrid orbitals.

6.3Terminology of complex compoundsOften examined

Parts of the complex compound \ce{[Co(NH3)6]Cl3}.

Central metal atom or ion: The transition metal atom or ion to which the ligands are bonded, for example \ce{Co^3+} in \ce{[Co(NH3)6]Cl3}. It acts as a Lewis acid.
Ligand: A species (atom, ion or molecule) that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom or ion.
Coordination number: The number of coordinate (dative) covalent bonds formed by the ligands with the central metal atom or ion. The most common coordination numbers are 4 and 6; coordination number 2 is also known.
Coordination sphere: The central metal atom or ion together with its ligands, written inside square brackets. It may be cationic, anionic or neutral.
Counter ions: The simple ions outside the square brackets that balance the charge of the coordination sphere, for example the three \ce{Cl^-} ions in \ce{[Co(NH3)6]Cl3}.

Coordination numbers of some complexes.
Complex	Ligands	Dative bonds	Coordination number
\ce{[Cu(NH3)4]^2+}	4 \ce{NH3}	4	4
\ce{[Ni(CO)4]}	4 \ce{CO}	4	4
\ce{Na4[Fe(CN)6]}	6 \ce{CN^-}	6	6
\ce{[Co(NO2)3(NH3)3]}	3 \ce{NO2^-} + 3 \ce{NH3}	6	6
\ce{[Ni(en)3]^2+}	3 en (bidentate)	6	6
\ce{[Ag(NH3)2]+}	2 \ce{NH3}	2	2

Charge on the coordination sphere

The charge on the coordination sphere is the sum of the charges on the central metal ion and on all the ligands. It is equal in size but opposite in sign to the total charge of the counter ions outside the brackets.

Compound	Coordination sphere	Type
\ce{K4[Fe(CN)6]}	\ce{[Fe(CN)6]^4-}	anionic
\ce{[Cu(NH3)4]SO4}	\ce{[Cu(NH3)4]^2+}	cationic
\ce{[Ni(CO)4]}	\ce{[Ni(CO)4]}	neutral

Charge on the coordination sphere

Find the charge on the coordination sphere in (i) \ce{Na4[Fe(CN)6]} and (ii) \ce{[Cu(NH3)4]SO4}.

(i) Iron is in the +2 oxidation state. Each cyanide ligand carries -1.
\text{charge}=(+2)+6(-1)=-4\ \Rightarrow\ \ce{[Fe(CN)6]^4-}
Check: four \ce{Na+} counter ions give +4, which balances -4.
(ii) Copper is +2 and ammonia is a neutral ligand.
\text{charge}=(+2)+4(0)=+2\ \Rightarrow\ \ce{[Cu(NH3)4]^2+}
One \ce{SO4^2-} counter ion balances it.

\ce{[Fe(CN)6]^4-}\ \text{and}\ \ce{[Cu(NH3)4]^2+}

6.3.1Naming complex compounds (IUPAC)Often examined

IUPAC gives each complex a systematic name built from its parts. Learn the six rules below; then practise with the examples, because naming questions are common in FBISE papers.

Cation before anion. Name the cation first and then the anion, whether each is a simple ion or a complex ion (as in sodium chloride). Example: in \ce{K4[Fe(CN)6]}, potassium comes first.
Ligands first, then metal. Inside the coordination sphere, name the ligands first, in alphabetical order, and then the central metal.
Anionic complex: '-ate'. If the coordination sphere is an anion, the metal name ends in -ate: ferrate (iron), cuprate (copper), argentate (silver), cobaltate (cobalt), nickelate (nickel), zincate (zinc), platinate (platinum), chromate (chromium). For a cationic or neutral sphere the metal keeps its normal name.
Number prefixes. Use di, tri, tetra, penta, hexa to show the number of each simple ligand. For ligands whose name already contains a number, such as 1,2-diaminoethane, use bis (2), tris (3), tetrakis (4), pentakis (5) and put the ligand name in brackets.
Ligand names. Anionic ligands end in -o: chloro (\ce{Cl^-}), bromo (\ce{Br^-}), hydroxo (\ce{OH^-}), cyano (\ce{CN^-}), nitro (\ce{NO2^-}), carbonato (\ce{CO3^2-}), ethanedioato (\ce{C2O4^2-}). Neutral ligands usually keep their names, with special names for aqua (\ce{H2O}), ammine (\ce{NH3}) and carbonyl (\ce{CO}).
Oxidation state. Give the oxidation state of the central metal in Roman numerals in brackets straight after the metal name, for example iron(III), nickel(0).

Examples of complex names (the first eight are from the book).
Formula	IUPAC name
\ce{[Pt(NH3)4BrCl]Cl}	tetraamminebromochloroplatinum(IV) chloride
\ce{[Cu(NH3)2]+}	diamminecopper(I) ion
\ce{K4[Fe(CN)6]}	potassium hexacyanoferrate(II)
\ce{[Cr(H2O)6]Cl3}	hexaaquachromium(III) chloride
\ce{[Ni(CO)4]}	tetracarbonylnickel(0)
\ce{[Co(NO2)3(NH3)3]}	triamminetrinitrocobalt(III)
\ce{[Cr(en)2Cl2]Cl}	dichlorobis(1,2-diaminoethane)chromium(III) chloride
\ce{K2[PtCl6]}	potassium hexachloroplatinate(IV)
\ce{[Co(NH3)6]Cl3}	hexaamminecobalt(III) chloride
\ce{K3[Fe(CN)6]}	potassium hexacyanoferrate(III)
\ce{[CuCl4]^2-}	tetrachlorocuprate(II) ion
\ce{[Ag(NH3)2]+}	diamminesilver(I) ion

Naming a complex step by step

Name \ce{[Pt(NH3)4BrCl]SO4}.

Identify the ions. The complex is the cation and \ce{SO4^2-} (sulphate) is the anion, so the name starts with the complex.
Find the oxidation state of Pt. The sphere has charge +2 (it balances one \ce{SO4^2-}). \ce{NH3} is 0, \ce{Br^-} and \ce{Cl^-} are -1 each.
x+4(0)+(-1)+(-1)=+2\ \Rightarrow\ x=+4
Name the ligands in alphabetical order: ammine (a), bromo (b), chloro (c), with prefixes.
\text{tetraammine}+\text{bromo}+\text{chloro}
The sphere is a cation, so the metal keeps its name. Add the oxidation state, then the anion.

\begin{gathered}\text{tetraamminebromochloroplatinum(IV)}\\\text{sulphate}\end{gathered}

Concept-Assessment Exercise 6.1, question 2

Name the following: (i) \ce{[Co(NH3)4Cl2]Cl} (ii) \ce{[Zn(OH)4]^2-} (iii) \ce{[CrCl2(NH3)4]+}

(i) The sphere is +1 (balances one \ce{Cl^-}). x+0-2=+1, so Co is +3. Ligands: ammine before chloro.
tetraamminedichlorocobalt(III) chloride
(ii) The ion is an anion, so zinc becomes zincate. x+4(-1)=-2, so Zn is +2.
tetrahydroxozincate(II) ion
(iii) Cation with charge +1. x+2(-1)+0=+1, so Cr is +3. Alphabetical order: ammine, chloro.
tetraamminedichlorochromium(III) ion

6.3.2Working out the charge and formula of a complexOften examined

The total charge on a complex ion depends on the oxidation state of the metal ion and on the number and charges of the ligands. Any one of these can be found if the others are known.

\text{charge on complex}=\text{oxidation state of metal}+\sum(\text{charges of ligands})

\text{oxidation state of metal}=\text{charge on complex}-\sum(\text{charges of ligands})

Charges of common ligands.
Ligand	\ce{H2O}, \ce{NH3}, \ce{CO}, en	\ce{Cl^-}, \ce{Br^-}, \ce{OH^-}, \ce{CN^-}, \ce{NO2^-}	\ce{C2O4^2-}	\ce{EDTA^4-}
Charge	0	-1	-2	-4

Book worked example 1

Deduce the charge on the tetraamminedichlorocobalt(III) complex ion. What is its formula?

The name shows two chloride ligands (-1 each) and four ammonia molecules (0). Cobalt is +3.
\text{charge}=(+3)+2(-1)+4(0)=+1

\ce{[CoCl2(NH3)4]+}

Book worked example 2

Deduce the oxidation state of copper in the \ce{[CuCl4]^2-} complex ion. What is its name?

The net charge is -2 and the four chloride ligands contribute -4.
x+4(-1)=-2\ \Rightarrow\ x=+2
The complex is an anion, so copper becomes cuprate.

\text{Cu}=+2;\ \text{tetrachlorocuprate(II) ion}

Concept-Assessment Exercise 6.1, question 1

Determine the oxidation states of the transition metals in: (i) \ce{[Co(NH3)6]Cl3} (ii) \ce{Na2[MnCl5]} (iii) \ce{[Co(en)2(H2O)2]} (iv) \ce{[CuCl4]^2-} (v) \ce{Na4[Fe(CN)2(OH)4]}

(i) Three \ce{Cl^-} counter ions, so the sphere is +3; \ce{NH3} is neutral.
x=+3
(ii) Two \ce{Na+}, so the sphere is -2.
x+5(-1)=-2\ \Rightarrow\ x=+3
(iii) en and \ce{H2O} are neutral. As printed the complex has no charge, so cobalt would be 0.
x=0
The book has probably left out the charge; the real ions are \ce{[Co(en)2(H2O)2]^2+} (Co +2) and \ce{[Co(en)2(H2O)2]^3+} (Co +3).
(iv)
x+4(-1)=-2\ \Rightarrow\ x=+2
(v) Four \ce{Na+}, so the sphere is -4; two \ce{CN^-} and four \ce{OH^-} give -6.
x-6=-4\ \Rightarrow\ x=+2

+3,\ +3,\ 0\ (\text{as printed}),\ +2,\ +2

Predicting the formula from the metal ion, ligand and coordination number

To write the formula of a complex ion, put the metal and the correct number of ligands in square brackets, then add the charges. The coordination number (or the shape) tells you how many ligands: octahedral 6, tetrahedral or square planar 4, linear 2.

Metal ion	Ligand	Coordination number / shape	Formula and charge
\ce{Fe^3+}	\ce{CN^-}	6, octahedral	(+3)+6(-1)=-3: \ce{[Fe(CN)6]^3-}
\ce{Co^3+}	\ce{NH3}	6, octahedral	(+3)+0=+3: \ce{[Co(NH3)6]^3+}
\ce{Co^2+}	\ce{Cl^-}	4, tetrahedral	(+2)+4(-1)=-2: \ce{[CoCl4]^2-}
\ce{Ni^2+}	\ce{CN^-}	4, square planar	(+2)+4(-1)=-2: \ce{[Ni(CN)4]^2-}
\ce{Ni^2+}	en	6, octahedral	3 en give 6 bonds: \ce{[Ni(en)3]^2+}
\ce{Ag+}	\ce{NH3}	2, linear	(+1)+0=+1: \ce{[Ag(NH3)2]+}

6.3.3Coloured complexesOften examined

When white light falls on a transition metal complex (solid or solution), part of the light is absorbed and the rest is transmitted or reflected. What we see depends on what is absorbed:

If all the visible light is absorbed, the substance looks black.
If some of the light is absorbed, the substance looks coloured.
If none is absorbed (all is reflected or transmitted), it looks white (solid) or colourless (solution).

Transition metal ions are coloured because they absorb part of visible light. Light is absorbed only by ions with an incomplete 3d sub-shell. The copper(II) ion ([\mathrm{Ar}]\,3d^{9}) forms blue compounds, but the zinc ion ([\mathrm{Ar}]\,3d^{10}) forms white or colourless compounds because its d sub-shell is complete.

Shapes of the d orbitals

A d sub-shell has five orbitals. In a free ion (no ligands) all five have the same energy: they are degenerate orbitals. They differ only in the directions of their lobes:

d_{xy}, d_{yz} and d_{xz}: four lobes lying between the axes, in the xy, yz and xz planes.
d_{x^2-y^2}: four lobes lying along the x and y axes.
d_{z^2}: two lobes along the z axis with a doughnut-shaped ring around the middle in the xy plane.

Shapes of the five 3d orbitals. Three have lobes between the axes; d_{x^2-y^2} and d_{z^2} have lobes along the axes.

Splitting of the d orbitals by ligands

In an octahedral complex six ligands approach the metal ion along the x, y and z axes. The lone pairs of the ligands repel the d electrons. The two orbitals that point along the axes (d_{x^2-y^2} and d_{z^2}) point straight at the ligands and are repelled more, so they rise more in energy. The three orbitals that point between the axes (d_{xy}, d_{yz}, d_{xz}) are repelled less. The d sub-shell therefore splits into two sets of non-degenerate orbitals: a higher set of two, called e_g, and a lower set of three, called t_{2g}. Both sets are higher in energy than the 3d orbitals of the free ion.

Splitting of the five degenerate 3d orbitals by six ligands in an octahedral complex. \Delta E is the splitting energy.

d-d transitions and colour

The energy gap \Delta E between the two sets equals the energy of a photon of visible light. When white light passes through the complex, a photon of exactly this energy is absorbed and an electron is promoted from the lower set (t_{2g}) to the higher set (e_g). This is called a d-d electronic transition. The absorbed colour is removed from the white light, and we see the complementary colour of the light absorbed.

\Delta E=h\nu=\dfrac{hc}{\lambda}

h: Planck's constant 6.626\times10^{-34}\,\mathrm{J\,s}; \nu: frequency of the absorbed light (Hz); c: speed of light 3.00\times10^{8}\,\mathrm{m\,s^{-1}}; \lambda: wavelength (m).

The colour wheel. The colour seen is the one opposite (complementary to) the colour absorbed.

Complementary colours.
Colour absorbed	Colour observed
red	green
orange	blue
yellow	violet
yellow-green	red-violet
green	red
blue	orange
violet	yellow

For example, \ce{[Ni(H2O)6]^2+} absorbs red light and appears green. \ce{[Cu(H2O)6]^2+} absorbs orange (and red) light and appears blue.

How \ce{Cu^2+(aq)} gets its colour: orange light is removed from white light, and the transmitted light looks blue.

Why d^{0} and d^{10} ions are colourless

A d-d transition needs an electron in the lower set and a vacancy in the upper set. \ce{Sc^3+} and \ce{Ti^4+} (d^{0}) have no d electrons to promote. \ce{Zn^2+} and \ce{Cu+} (d^{10}) have a completely filled upper set, so there is no space to promote an electron into. No visible light is absorbed, so their compounds are white or colourless.

In \ce{Cu^2+} (d^{9}) the upper set has a vacancy, so a d-d transition is possible. In \ce{Zn^2+} (d^{10}) both sets are full and no transition can occur.

Calculations with the colour of complexes

Energy of the d-d transition

\ce{[Ti(H2O)6]^3+} absorbs most strongly at a wavelength of 500\,\mathrm{nm}. Calculate (a) the frequency of the absorbed light, (b) the splitting energy \Delta E for one ion and (c) for one mole of ions. (d) Predict the colour of the solution.

Given

\lambda = 500\,\mathrm{nm}=5.00\times10^{-7}\,\mathrm{m}
h = 6.626\times10^{-34}\,\mathrm{J\,s}
c = 3.00\times10^{8}\,\mathrm{m\,s^{-1}}
N_A = 6.022\times10^{23}\,\mathrm{mol^{-1}}

To find

\nu
\Delta E

(a) Frequency from c=\nu\lambda.
\nu=\dfrac{c}{\lambda}=\dfrac{3.00\times10^{8}}{5.00\times10^{-7}}=6.00\times10^{14}\,\mathrm{Hz}
(b) Energy of one photon equals \Delta E for one ion.
\Delta E=h\nu=6.626\times10^{-34}\times6.00\times10^{14}=3.98\times10^{-19}\,\mathrm{J}
(c) Multiply by the Avogadro constant and convert to kJ.
\Delta E=3.98\times10^{-19}\times6.022\times10^{23}=2.39\times10^{5}\,\mathrm{J\,mol^{-1}}=239\,\mathrm{kJ\,mol^{-1}}
(d) 500\,\mathrm{nm} lies in the green (yellow-green) region of the spectrum. The complementary colour is red-violet, which is the colour of \ce{Ti^3+(aq)}.

\nu=6.00\times10^{14}\,\mathrm{Hz},\ \Delta E=3.98\times10^{-19}\,\mathrm{J}=239\,\mathrm{kJ\,mol^{-1}}

6.3.4d-d splitting in octahedral and tetrahedral complexesOften examined

The pattern of splitting depends on where the ligands are compared with the lobes of the d orbitals. In an octahedral complex the six ligands sit on the axes. In a tetrahedral complex the four ligands sit at alternate corners of a cube, between the axes.

Positions of the ligands. Octahedral: on the x, y and z axes. Tetrahedral: at four alternate corners of a cube, nearer to the lobes of d_{xy}, d_{yz} and d_{xz}.

Octahedral complexes: two higher, three lower

Take the octahedral hexaaquatitanium(III) ion, \ce{[Ti(H2O)6]^3+}. \ce{Ti^3+} is [\mathrm{Ar}]\,3d^{1}. The three t_{2g} orbitals (d_{xy}, d_{yz}, d_{xz}) lie between the six ligands and are repelled less, so they have lower energy. The two e_g orbitals (d_{x^2-y^2}, d_{z^2}) point directly at the ligands and are repelled more, so they have higher energy. In the ground state the single electron is in the t_{2g} set. White light excites it to the e_g set by absorbing yellow-green light, so the complex looks red-violet.

The d-d transition in octahedral \ce{[Ti(H2O)6]^3+}: the single electron moves from t_{2g} to e_g.

Tetrahedral complexes: three higher, two lower

In a tetrahedral complex the pattern is reversed. The three orbitals d_{xy}, d_{yz}, d_{xz} point nearer to the four ligands, are repelled more and form the higher set (called t_2; the book writes t_{2g}). The two orbitals d_{x^2-y^2} and d_{z^2} point between the ligands, are repelled less and form the lower set (e; the book writes e_g). Example: tetrachlorocuprate(II), \ce{[CuCl4]^2-}, with \ce{Cu^2+} [\mathrm{Ar}]\,3d^{9}. Four electrons fill the lower e set and five are in the upper set. Light promotes an electron from e to the single vacancy in t_2.

The d-d transition in tetrahedral \ce{[CuCl4]^2-}: an electron moves from the lower pair to the upper three.

Octahedral and tetrahedral splitting side by side. The tetrahedral splitting is smaller, about \tfrac49 of the octahedral value for the same metal ion and ligand.

Comparison of the two splitting patterns.
Feature	Octahedral	Tetrahedral
Number of ligands	6	4
Ligand positions	on the x, y, z axes	between the axes (alternate corners of a cube)
Higher energy set	2 orbitals: e_g (d_{x^2-y^2}, d_{z^2})	3 orbitals: t_2 (d_{xy}, d_{yz}, d_{xz})
Lower energy set	3 orbitals: t_{2g} (d_{xy}, d_{yz}, d_{xz})	2 orbitals: e (d_{x^2-y^2}, d_{z^2})
Size of splitting	larger, \Delta_o	smaller, \Delta_t\approx\tfrac49\Delta_o
Examples	\ce{[Ti(H2O)6]^3+}, \ce{[Cu(H2O)6]^2+}	\ce{[CuCl4]^2-}, \ce{[CoCl4]^2-}

6.3.5Effect of ligands on the colour of complexesOften examined

When ligands approach a metal ion, their electric fields interact with the d orbitals and split them. Different ligands split the d orbitals by different amounts, so they have different field strengths. The book's order of field strength is:

\ce{CN^-} > \ce{NH3} > \ce{H2O} > \ce{OH^-} > \ce{Cl^-} > \ce{Br^-}

strong field ligands (large splitting) on the left, weak field ligands (small splitting) on the right

A strong field ligand splits the d orbitals more and gives a larger \Delta E, so the complex absorbs light of higher frequency (shorter wavelength). A weak field ligand gives a smaller \Delta E, so light of lower frequency is absorbed. Because a different colour is absorbed, the complementary colour we see also changes.

A strong field ligand gives a larger splitting \Delta E than a weak field ligand.

Example: copper(II) with water and with ammonia

Ammonia is a stronger field ligand than water. When ammonia solution is added in excess to a solution of copper(II) ions, four ammonia molecules replace four water molecules. The splitting \Delta E increases, the absorption moves to higher frequency, and the colour changes from pale blue to deep violet-blue.

\underset{\text{pale blue}}{\ce{[Cu(H2O)6]^2+(aq)}}+\ce{4NH3(aq)}\ce{->}\underset{\text{deep violet-blue}}{\ce{[Cu(NH3)4(H2O)2]^2+(aq)}}+\ce{4H2O(l)}

The same metal ion gives different colours with different ligands.
Copper(II) complex	Ligands	Shape	Colour
\ce{[Cu(H2O)6]^2+}	6 \ce{H2O}	octahedral	pale blue
\ce{[Cu(H2O)4(OH)2]}	4 \ce{H2O}, 2 \ce{OH^-}	octahedral	pale blue precipitate
\ce{[Cu(NH3)4(H2O)2]^2+}	4 \ce{NH3}, 2 \ce{H2O}	octahedral	deep blue (violet-blue)
\ce{[CuCl4]^2-}	4 \ce{Cl^-}	tetrahedral	yellow (green when mixed with the blue aqua ion)

Concept Assessment Exercise 6.2

Degenerate and non-degenerate orbitals

What are degenerate and non-degenerate 3d orbitals?

Degenerate orbitals have the same energy: the five 3d orbitals of a free metal ion (no ligands) are degenerate. Non-degenerate orbitals have different energies: in a complex the ligands split the 3d orbitals into two sets of different energy, such as t_{2g} (lower) and e_g (higher) in an octahedral complex.

Why splitting happens in some complexes

Why do degenerate 3d orbitals split into two sets in some complexes but not in others?

The d orbitals split only when ligands approach and repel them unequally. Orbitals pointing towards the ligands are repelled more than those pointing between them. Splitting is therefore seen in complexes of ions whose d orbitals interact with the ligands. It shows up as colour only when the d sub-shell is partly filled (d^{1} to d^{9}). For d^{0} ions (\ce{Sc^3+}, \ce{Ti^4+}) and d^{10} ions (\ce{Zn^2+}, \ce{Cu+}) no d-d transition is possible, so no effect of splitting is observed.

Different ligands, different splitting

Which different ligands cause different splitting energy \Delta E? Give one example.

Strong field ligands such as \ce{CN^-} and \ce{NH3} cause a large \Delta E; weak field ligands such as \ce{Cl^-} and \ce{Br^-} cause a small \Delta E. Example: replacing water by ammonia in copper(II) increases \Delta E, and the colour changes from pale blue \ce{[Cu(H2O)6]^2+} to deep violet-blue \ce{[Cu(NH3)4(H2O)2]^2+}.

6.3.6Geometries of complexesOften examined

The shape of a complex can be predicted with VSEPR theory, as for simple molecules. The dative bond pairs arrange themselves as far apart as possible to keep repulsion to a minimum. The shape depends mainly on the coordination number.

Coordination number 6: octahedral

Six donor atoms sit at the six corners (vertices) of an octahedron (Greek octa: eight, hedron: faces). All the angles between neighbouring ligands are 90^\circ (and 180^\circ between opposite ligands). Octahedral complexes are the most common and most stable type; almost all transition metals form them, with monodentate ligands (as in \ce{[Ni(H2O)6]^2+}) or polydentate ligands (as in \ce{[Ni(en)3]^2+}, where each curved line is the \ce{-CH2CH2-} chain of one en).

Octahedral complexes. Wedges point towards the reader and dashed bonds point away. In \ce{[Ni(en)3]^2+} each curved line joins the two N atoms of one 1,2-diaminoethane ligand.

Copper(II) bonds to six water or six ammonia ligands but to only four chloride ligands. The chloride ion is bigger than a water or ammonia molecule, because chlorine is in period 3 while O and N are in period 2. There is not enough space around the small copper(II) ion for six chloride ions.

Coordination number 4: tetrahedral or square planar

Four ligands can be arranged in two ways. Tetrahedral complexes follow VSEPR theory: the four ligands are as far apart as possible, at 109.5^\circ. They are less common than octahedral complexes and are often formed with chloride ligands, as in \ce{[CuCl4]^2-} and \ce{[CoCl4]^2-}. They are also common for d^{10} ions such as \ce{Cu+} and \ce{Zn^2+}: \ce{[Cu(CN)4]^3-}, \ce{[Zn(NH3)4]^2+}.

Square planar complexes have the four ligands at the corners of a square with the metal at the centre, all in one plane, at 90^\circ to each other. This arrangement does not follow simple VSEPR theory. They are less common than tetrahedral complexes and are mostly formed by d^{8} ions such as \ce{Ni^2+}, \ce{Pt^2+} and \ce{Au^3+}: \ce{[Ni(CN)4]^2-}, \ce{[AuCl4]^-}, \ce{[Pt(NH3)4]^2+} and diamminedichloroplatinum(II), \ce{[Pt(NH3)2Cl2]}.

The two shapes for coordination number 4.

Coordination number 2: linear

Two monodentate ligands lie on opposite sides of the metal at 180^\circ, giving a linear complex. Linear complexes are less common and less stable; they are typical of d^{10} ions such as \ce{Ag+}, \ce{Cu+} and \ce{Au+}. Examples: diamminesilver(I) \ce{[Ag(NH3)2]+} (the active species in Tollens' reagent), \ce{[CuCl2]^-} and \ce{[Au(CN)2]^-}.

The linear diamminesilver(I) ion.

Shapes of complexes.
Coordination number	Shape	Bond angle	Hybrid orbitals	Examples
2	linear	180^\circ	sp	\ce{[Ag(NH3)2]+}, \ce{[CuCl2]^-}, \ce{[Au(CN)2]^-}
4	tetrahedral	109.5^\circ	sp^{3}	\ce{[CuCl4]^2-}, \ce{[CoCl4]^2-}, \ce{[Zn(NH3)4]^2+}
4	square planar	90^\circ	dsp^{2}	\ce{[Pt(NH3)2Cl2]}, \ce{[Ni(CN)4]^2-}, \ce{[AuCl4]^-}
6	octahedral	90^\circ	sp^{3}d^{2} or d^{2}sp^{3}	\ce{[Cu(H2O)6]^2+}, \ce{[Ni(en)3]^2+}, \ce{[Fe(CN)6]^3-}

6.4Ligand substitution (exchange) reactionsOften examined

Transition metal complexes take part in redox, deprotonation and ligand substitution reactions. A ligand exchange (ligand substitution) reaction is one in which one ligand in a complex is replaced by another. It takes place when the new complex is more stable than the old one. The exchange may be complete (all ligands replaced) or partial (only some replaced). Because the colour of a complex depends on its ligands, ligand exchange usually causes a colour change. The book uses copper(II) and cobalt(II) complexes with water, ammonia, hydroxide and chloride ligands.

Copper(II) complexes

1. With sodium hydroxide. Adding \ce{NaOH(aq)} to blue copper(II) sulphate solution gives a pale blue precipitate. Two hydroxide ligands replace two water ligands and the neutral complex precipitates.

\underset{\text{blue solution}}{\ce{[Cu(H2O)6]^2+(aq)}}+\ce{2OH^-(aq)}\ce{->}\underset{\text{pale blue ppt}}{\ce{[Cu(H2O)4(OH)2](s)}}+\ce{2H2O(l)}

The hydroxide ions do not substitute the water ligands directly. They act as a base and remove protons (\ce{H+}) from two coordinated water molecules, leaving \ce{OH^-} groups bonded to the copper. This is called deprotonation.

2. With a little aqueous ammonia. Ammonia is also a base, so it gives the same pale blue precipitate by deprotonation:

\ce{[Cu(H2O)6]^2+(aq) + 2NH3(aq) -> [Cu(H2O)4(OH)2](s) + 2NH4+(aq)}

3. With excess ammonia. The precipitate dissolves to give a deep blue solution. Four ammonia molecules replace two water and two hydroxide ligands; this is a true ligand exchange.

\underset{\text{pale blue ppt}}{\ce{[Cu(H2O)4(OH)2](s)}}+\ce{4NH3(aq)}\ce{->}\underset{\text{deep blue solution}}{\ce{[Cu(NH3)4(H2O)2]^2+(aq)}}+\ce{2H2O(l) + 2OH^-(aq)}

4. With concentrated hydrochloric acid. Added dropwise, it exchanges the six water ligands for four chloride ligands, giving yellow tetrachlorocuprate(II). The reaction is reversible; the mixture of blue and yellow species looks green. Adding water reverses it.

\underset{\text{blue}}{\ce{[Cu(H2O)6]^2+(aq)}}+\ce{4Cl^-(aq)}\ce{<=>}\underset{\text{yellow}}{\ce{[CuCl4]^2-(aq)}}+\ce{6H2O(l)}

Cobalt(II) complexes

The pink hexaaquacobalt(II) ion, \ce{[Co(H2O)6]^2+}, gives these reactions:

\underset{\text{pink}}{\ce{[Co(H2O)6]^2+(aq)}}+\ce{2OH^-(aq)}\ce{->}\underset{\text{blue ppt}}{\ce{[Co(H2O)4(OH)2](s)}}+\ce{2H2O(l)}

with sodium hydroxide: blue precipitate of cobalt(II) hydroxide (it turns pink on standing)

\underset{\text{pink}}{\ce{[Co(H2O)6]^2+(aq)}}+\ce{6NH3(aq)}\ce{->}\underset{\text{brown (straw)}}{\ce{[Co(NH3)6]^2+(aq)}}+\ce{6H2O(l)}

with concentrated ammonia added slowly: complete exchange of all six water ligands

\underset{\text{pink}}{\ce{[Co(H2O)6]^2+(aq)}}+\ce{4Cl^-(aq)}\ce{<=>}\underset{\text{blue}}{\ce{[CoCl4]^2-(aq)}}+\ce{6H2O(l)}

with concentrated hydrochloric acid: octahedral pink to tetrahedral blue

Summary of ligand exchange reactions.
Complex ion	Reagent	Product	Observation
\ce{[Cu(H2O)6]^2+} (blue)	\ce{NaOH} or a little \ce{NH3}	\ce{[Cu(H2O)4(OH)2]}	pale blue precipitate
\ce{[Cu(H2O)4(OH)2]}	excess \ce{NH3}	\ce{[Cu(NH3)4(H2O)2]^2+}	precipitate dissolves, deep blue solution
\ce{[Cu(H2O)6]^2+}	conc. \ce{HCl}	\ce{[CuCl4]^2-}	yellow (green mixture)
\ce{[Co(H2O)6]^2+} (pink)	\ce{NaOH}	\ce{[Co(H2O)4(OH)2]}	blue precipitate
\ce{[Co(H2O)6]^2+}	conc. \ce{NH3}	\ce{[Co(NH3)6]^2+}	brown (straw-coloured) solution
\ce{[Co(H2O)6]^2+}	conc. \ce{HCl}	\ce{[CoCl4]^2-}	blue solution

6.4.1Stability of complexes and stability constantOften examined

Ligand exchange reactions are reversible, so they reach equilibrium and can be described by an equilibrium constant, just like K_c in chemical equilibrium. Consider the competition between water and chloride ligands for cobalt(II):

\underset{\text{pink}}{\ce{[Co(H2O)6]^2+(aq)}}+\ce{4Cl^-(aq)}\ce{<=>}\underset{\text{blue}}{\ce{[CoCl4]^2-(aq)}}+\ce{6H2O(l)}

The equilibrium lies on the side of the more stable complex. Adding water shifts it to the left (pink colour returns); adding concentrated \ce{HCl} shifts it to the right (blue). Chloride and water ligands compete for the cobalt(II) ion.

Stability constant, K_{\mathrm{stab}}: The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. In water, [\ce{H2O}] is not included in the expression.

Derivation K_{\mathrm{stab}} expression for \ce{[CoCl4]^2-}

Aim. Write the stability constant for the formation of tetrachlorocobaltate(II) in water.

Write K_c using the law of mass action: products over reactants, each concentration raised to the power of its coefficient.
K_c=\dfrac{[\ce{[CoCl4]^2-}][\ce{H2O}]^{6}}{[\ce{[Co(H2O)6]^2+}][\ce{Cl^-}]^{4}}
Water is the solvent and is present in large excess, so its concentration stays practically constant. It is combined with K_c into a new constant and is not shown.
K_{\mathrm{stab}}=\dfrac{K_c}{[\ce{H2O}]^{6}}
This gives the stability constant.
K_{\mathrm{stab}}=\dfrac{[\ce{[CoCl4]^2-}]}{[\ce{[Co(H2O)6]^2+}][\ce{Cl^-}]^{4}}
Units: put \mathrm{mol\,dm^{-3}} for every concentration.
\dfrac{\mathrm{mol\,dm^{-3}}}{(\mathrm{mol\,dm^{-3}})(\mathrm{mol\,dm^{-3}})^{4}}=(\mathrm{mol\,dm^{-3}})^{-4}=\mathrm{dm^{12}\,mol^{-4}}

ResultK_{\mathrm{stab}}=\dfrac{[\ce{[CoCl4]^2-}]}{[\ce{[Co(H2O)6]^2+}][\ce{Cl^-}]^{4}}\quad(\mathrm{dm^{12}\,mol^{-4}})

The larger K_{\mathrm{stab}}, the further the equilibrium lies to the right and the more stable the complex. The unit is \mathrm{dm^{3n}\,mol^{-n}} when n ligand particles are used up.

What the values tell us

Stability constants show how firmly a ligand is bonded to the central metal ion. Generally, donor atoms of lower electronegativity (such as N in \ce{NH3}) donate their lone pair more readily and bond more firmly than donor atoms of higher electronegativity (such as O in \ce{H2O}). So ammonia complexes have higher K_{\mathrm{stab}} values than aqua complexes.

Stability constants of some copper(II) complexes. The values are so large that they are given on a \log_{10} scale, which has no units.
Complex	\log_{10}K_{\mathrm{stab}}
\ce{[CuCl4]^2-}	5.6
\ce{[Cu(NH3)4(H2O)2]^2+}	13.1
\ce{[Cu(EDTA)]^2-}	18.8

Complexes of polydentate ligands have much higher stability constants than those of monodentate ligands (the chelate effect). One EDTA ion replaces six water molecules, so the number of free particles in solution increases, and the complex is very stable. The same happens with en:

\ce{[Cu(H2O)6]^2+(aq) + 3en(aq) -> [Cu(en)3]^2+(aq) + 6H2O(l)}

Using K_{\mathrm{stab}} to predict ligand exchange

A ligand that forms a complex with a larger K_{\mathrm{stab}} will replace a ligand whose complex has a smaller K_{\mathrm{stab}}; the equilibrium shifts towards the more stable complex. For example, excess ammonia added to yellow \ce{[CuCl4]^2-} gives a deep blue solution, because \log K_{\mathrm{stab}} of the ammonia complex (13.1) is much larger than that of the chloride complex (5.6):

\ce{[CuCl4]^2-(aq) + 4NH3(aq) + 2H2O(l) -> [Cu(NH3)4(H2O)2]^2+(aq) + 4Cl^-(aq)}

Book worked example 1

A mixture of aqueous tetrachlorocuprate(II) and hexaaquacopper(II) ions forms when concentrated hydrochloric acid is added to aqueous copper(II) sulphate. (i) Write the expression for the stability constant of tetrachlorocuprate(II). (ii) What are its units?

\ce{[Cu(H2O)6]^2+(aq) + 4Cl^-(aq) <=> [CuCl4]^2-(aq) + 6H2O(l)}

(i) Leave out water, the solvent.
K_{\mathrm{stab}}=\dfrac{[\ce{[CuCl4]^2-}]}{[\ce{[Cu(H2O)6]^2+}][\ce{Cl^-}]^{4}}
(ii) Substitute units.
\dfrac{\mathrm{mol\,dm^{-3}}}{\mathrm{mol\,dm^{-3}}\,(\mathrm{mol\,dm^{-3}})^{4}}=\mathrm{mol^{-4}\,dm^{12}}

\mathrm{dm^{12}\,mol^{-4}}

A calculation with K_{\mathrm{stab}}

For \ce{[Cu(H2O)6]^2+ + en <=> [Cu(H2O)4(en)]^2+ + 2H2O}, K_{\mathrm{stab}}=3.98\times10^{10}\,\mathrm{dm^{3}\,mol^{-1}}. At equilibrium [\mathrm{en}]=1.0\times10^{-3}\,\mathrm{mol\,dm^{-3}}. Find the ratio of the concentrations of the en complex and the aqua complex.

Write the expression (water omitted) and rearrange for the ratio.
K_{\mathrm{stab}}=\dfrac{[\ce{[Cu(H2O)4(en)]^2+}]}{[\ce{[Cu(H2O)6]^2+}][\mathrm{en}]}\ \Rightarrow\ \dfrac{[\ce{[Cu(H2O)4(en)]^2+}]}{[\ce{[Cu(H2O)6]^2+}]}=K_{\mathrm{stab}}[\mathrm{en}]
Substitute.
=3.98\times10^{10}\times1.0\times10^{-3}=3.98\times10^{7}

Even at this low concentration of en, there are about 40 million en complexes for every aqua complex: practically all the copper is in the more stable complex.

\text{ratio}\approx4.0\times10^{7}

Concept Assessment Exercise 6.3

Order of stability constants

Arrange the ligands \ce{CN^-}, \ce{OH^-}, \ce{NH3}, \ce{H2O}, \ce{Br^-} in increasing order of stability constants in ligand substitution reactions of iron(II) complexes.

Stronger ligands form more stable complexes and displace weaker ones, so the order follows the book's ligand strength series, from weakest to strongest:

\ce{Br^-} < \ce{OH^-} < \ce{H2O} < \ce{NH3} < \ce{CN^-}

Why ammonia replaces water

Given the stability constants, why can ammonia exchange for water in the copper(II) complex?

The ammonia complex, \ce{[Cu(NH3)4(H2O)2]^2+}, has a much larger stability constant (\log K_{\mathrm{stab}}=13.1) than the aqua complex. The nitrogen of \ce{NH3} is less electronegative than the oxygen of \ce{H2O}, so it donates its lone pair more readily and forms a stronger dative bond. The equilibrium therefore lies far to the side of the more stable ammonia complex, and ammonia replaces water.

6.5Redox reactions of transition metal ionsOften examined

Transition elements exist in several oxidation states, and in redox reactions their oxidation states change. Reduction is a decrease in oxidation state (gain of electrons); oxidation is an increase in oxidation state (loss of electrons). An oxidising agent accepts electrons and is itself reduced; a reducing agent loses electrons and is itself oxidised.

Predicting feasibility with E^\circ values

Write both half-equations as reductions with their standard electrode (reduction) potentials E^\circ, as IUPAC recommends.
The half-reaction with the more positive E^\circ goes forward as a reduction; its oxidised species is the oxidising agent.
Reverse the half-equation with the less positive E^\circ (it becomes the oxidation) and change the sign of its E^\circ.
Multiply the half-equations so that electrons lost equal electrons gained. E^\circ values are not multiplied, because they do not depend on the coefficients.
Add the half-equations and the E^\circ values. If E^\circ_{\text{cell}} is positive, the reaction is feasible; if it is negative, it is not feasible under standard conditions.

E^\circ_{\text{cell}}=E^\circ_{\text{reduction}}-E^\circ_{\text{oxidation half (as a reduction)}}

Standard electrode potentials used in this unit.
Half-equation (reduction)	E^\circ/\mathrm{V}
\ce{Zn^2+ + 2e- <=> Zn}	-0.76
\ce{2CO2 + 2e- <=> C2O4^2-}	-0.49
\ce{Cr^3+ + e- <=> Cr^2+}	-0.41
\ce{Cu^2+ + e- <=> Cu+}	+0.15
\ce{I2 + 2e- <=> 2I^-}	+0.54
\ce{Fe^3+ + e- <=> Fe^2+}	+0.77
\ce{Cu^2+ + I^- + e- <=> CuI(s)}	+0.86
\ce{Cr2O7^2- + 14H+ + 6e- <=> 2Cr^3+ + 7H2O}	+1.33
\ce{MnO4^- + 8H+ + 5e- <=> Mn^2+ + 4H2O}	+1.52
\ce{S2O8^2- + 2e- <=> 2SO4^2-}	+2.01

(i) Acidified manganate(VII) with ethanedioate ions

\begin{aligned}&\ce{2CO2(g) + 2e- -> C2O4^2-(aq)}&&E^\circ=-0.49\,\mathrm{V}\\&\ce{MnO4^-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}&&E^\circ=+1.52\,\mathrm{V}\end{aligned}

Which way will the reaction go? The manganate(VII) half-reaction has the more positive E^\circ, so \ce{MnO4^-} is reduced to \ce{Mn^2+} (it is the oxidising agent), and ethanedioate is oxidised to carbon dioxide (it is the reducing agent). Reverse the ethanedioate half-equation and balance electrons (10 each):

\begin{array}{ll}\ce{2MnO4^- + 16H+ + 10e- -> 2Mn^2+ + 8H2O}&E^\circ=+1.52\,\mathrm{V}\\\ce{5C2O4^2- -> 10CO2 + 10e-}&E^\circ=+0.49\,\mathrm{V}\\\hline \ce{2MnO4^-(aq) + 5C2O4^2-(aq) + 16H+(aq) -> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)}&E^\circ_{\text{cell}}=+2.01\,\mathrm{V}\end{array}

The large positive E^\circ_{\text{cell}} shows that the reaction is feasible. The purple colour of \ce{MnO4^-} disappears as it reacts, so \ce{KMnO4} is its own indicator. In practice the solution is warmed to about 60^\circ\mathrm{C} because the reaction is slow at first; the \ce{Mn^2+} formed then catalyses it (autocatalysis).

(ii) Acidified manganate(VII) with iron(II) ions

\begin{aligned}&\ce{MnO4^-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l)}&&E^\circ=+1.52\,\mathrm{V}\\&\ce{Fe^3+(aq) + e- -> Fe^2+(aq)}&&E^\circ=+0.77\,\mathrm{V}\end{aligned}

Manganate(VII) has the higher E^\circ and is reduced. Reverse the iron half-equation (now E^\circ=-0.77\,\mathrm{V} as written for oxidation) and multiply it by 5:

\begin{array}{ll}\ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O}&+1.52\,\mathrm{V}\\\ce{5Fe^2+ -> 5Fe^3+ + 5e-}&-0.77\,\mathrm{V}\\\hline \ce{MnO4^-(aq) + 5Fe^2+(aq) + 8H+(aq) -> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l)}&E^\circ_{\text{cell}}=+0.75\,\mathrm{V}\end{array}

The positive value shows the reaction is feasible. The mole ratio \ce{MnO4^-}:\ce{Fe^2+}=1:5 is used in titration calculations. The acid must be dilute sulphuric acid; hydrochloric acid is not used because \ce{MnO4^-} would oxidise \ce{Cl^-} to chlorine.

(iii) Copper(II) ions with iodide ions

When excess potassium iodide is added to copper(II) sulphate solution, a white precipitate of copper(I) iodide forms in a red-brown solution of iodine:

\ce{2Cu^2+(aq) + 4I^-(aq) -> 2CuI(s) + I2(aq)}

Split into half-equations: \ce{I2 + 2e- -> 2I^-} (E^\circ=+0.54\,\mathrm{V}) and \ce{Cu^2+ + e- -> Cu+} (E^\circ=+0.15\,\mathrm{V}). On these values alone, iodine should oxidise \ce{Cu+}, not the other way round:

\begin{array}{ll}\ce{2I^-(aq) -> I2(aq) + 2e-}&-0.54\,\mathrm{V}\\\ce{2Cu^2+(aq) + 2e- -> 2Cu+(aq)}&+0.15\,\mathrm{V}\\\hline \ce{2Cu^2+(aq) + 2I^-(aq) -> 2Cu+(aq) + I2(aq)}&E^\circ_{\text{cell}}=-0.39\,\mathrm{V}\end{array}

Yet the reaction does happen. The reason is that \ce{Cu+} is removed as soon as it forms, as the insoluble precipitate \ce{CuI}. By Le Chatelier's principle the continuous removal of product pulls the equilibrium to the right. In E^\circ terms, the relevant half-equation is \ce{Cu^2+ + I^- + e- -> CuI(s)} with E^\circ=+0.86\,\mathrm{V}, which gives E^\circ_{\text{cell}}=0.86-0.54=+0.32\,\mathrm{V} (feasible).

The iodine formed is measured by titration with sodium thiosulphate, using starch as indicator (blue-black to colourless at the end point):

\ce{I2(aq) + 2S2O3^2-(aq) -> 2I^-(aq) + S4O6^2-(aq)}

Acidified dichromate(VI) as an oxidising agent

Potassium dichromate(VI), \ce{K2Cr2O7}, in dilute sulphuric acid is another common oxidising agent. Orange \ce{Cr2O7^2-} (Cr +6) is reduced to green \ce{Cr^3+} (Cr +3). With iron(II) ions (for example from Mohr's salt, \ce{FeSO4.(NH4)2SO4.6H2O}) and with ethanedioic (oxalic) acid:

\begin{aligned}&\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 6Fe^3+ + 7H2O}&&E^\circ_{\text{cell}}=1.33-0.77=+0.56\,\mathrm{V}\\&\ce{Cr2O7^2- + 8H+ + 3H2C2O4 -> 2Cr^3+ + 6CO2 + 7H2O}&&\end{aligned}

Redox titration: iron(II) with manganate(VII)

Practical activity 6.1. The \ce{MnO4^-}/\ce{Fe^2+} reaction is used to find the amount of iron in a sample such as an iron tablet. A known volume (for example 25.0\,\mathrm{cm^{3}}) of the \ce{Fe^2+(aq)} solution, acidified with dilute sulphuric acid, is placed in a conical flask. Potassium manganate(VII) solution of known concentration (0.05\,\mathrm{mol\,dm^{-3}}) is run in from a burette. The purple colour disappears as it reacts. The end point is the first permanent pale pink colour, when all the \ce{Fe^2+} has reacted and \ce{MnO4^-} is in slight excess. No separate indicator is needed.

Redox titration of iron(II) with potassium manganate(VII).

Titration calculation (MnO4-/Fe2+)

25.0\,\mathrm{cm^{3}} of an acidified \ce{Fe^2+} solution needs 12.5\,\mathrm{cm^{3}} of 0.050\,\mathrm{mol\,dm^{-3}} \ce{KMnO4} to reach the end point. Find the concentration of \ce{Fe^2+}.

Moles of manganate(VII) used: moles = concentration × volume (in \mathrm{dm^{3}}).
n(\ce{MnO4^-})=0.050\times\dfrac{12.5}{1000}=6.25\times10^{-4}\,\mathrm{mol}
From the equation, 1\,\mathrm{mol}\ \ce{MnO4^-} reacts with 5\,\mathrm{mol}\ \ce{Fe^2+}.
n(\ce{Fe^2+})=5\times6.25\times10^{-4}=3.125\times10^{-3}\,\mathrm{mol}
Divide by the volume of the iron solution.
c(\ce{Fe^2+})=\dfrac{3.125\times10^{-3}}{0.0250}=0.125\,\mathrm{mol\,dm^{-3}}

c(\ce{Fe^2+})=0.125\,\mathrm{mol\,dm^{-3}}

Worked example 6.1 (book)

An excess of potassium iodide solution was added to 30.00\,\mathrm{cm^{3}} of copper(II) sulphate solution. The mixture was titrated against 0.05\,\mathrm{mol\,dm^{-3}} sodium thiosulphate solution using starch indicator; 10\,\mathrm{cm^{3}} was needed. Calculate the concentration of the copper(II) sulphate solution.

Given

V(\ce{Cu^2+}) = 30.00\,\mathrm{cm^{3}}=0.0300\,\mathrm{dm^{3}}
V(\ce{S2O3^2-}) = 10\,\mathrm{cm^{3}}=0.010\,\mathrm{dm^{3}}
c(\ce{S2O3^2-}) = 0.05\,\mathrm{mol\,dm^{-3}}

To find

c(\ce{Cu^2+})

Moles of thiosulphate used.
n(\ce{S2O3^2-})=0.05\times0.010=5.0\times10^{-4}\,\mathrm{mol}
Link the two equations: \ce{2Cu^2+ -> I2} and \ce{I2 -> 2S2O3^2-}, so \ce{Cu^2+}:\ce{S2O3^2-}=1:1.
n(\ce{Cu^2+})=5.0\times10^{-4}\,\mathrm{mol}
Concentration of copper(II).
c=\dfrac{5.0\times10^{-4}}{0.0300}=0.0167\,\mathrm{mol\,dm^{-3}}

c(\ce{CuSO4})=0.0167\,\mathrm{mol\,dm^{-3}}

Titration calculation (MnO4-/C2O4 2-)

25.0\,\mathrm{cm^{3}} of 0.0500\,\mathrm{mol\,dm^{-3}} sodium ethanedioate, acidified and warmed, needs 25.0\,\mathrm{cm^{3}} of potassium manganate(VII). Find the concentration of the \ce{KMnO4}.

Moles of ethanedioate.
n(\ce{C2O4^2-})=0.0500\times0.0250=1.25\times10^{-3}\,\mathrm{mol}
Ratio from the equation: \ce{2MnO4^-}:\ce{5C2O4^2-}.
n(\ce{MnO4^-})=\tfrac25\times1.25\times10^{-3}=5.00\times10^{-4}\,\mathrm{mol}
Concentration.
c=\dfrac{5.00\times10^{-4}}{0.0250}=0.0200\,\mathrm{mol\,dm^{-3}}

c(\ce{KMnO4})=0.0200\,\mathrm{mol\,dm^{-3}}

Disproportionation

A species disproportionates when it is oxidised and reduced at the same time. Use E^\circ in the same way: for \ce{3Cr^2+ -> 2Cr^3+ + Cr}, E^\circ_{\text{cell}}=E^\circ(\ce{Cr^2+}/\ce{Cr})-E^\circ(\ce{Cr^3+}/\ce{Cr^2+})=-0.91-(-0.41)=-0.50\,\mathrm{V}, so chromium(II) does not disproportionate (see long question i).

6.6Stereoisomerism in coordination compoundsOften examined

Stereoisomers have the same formula and the same bonds but a different arrangement of atoms or ligands in space. Complexes show two types: geometric (cis-trans) isomerism and optical isomerism.

(i) Geometric isomerism

Geometric isomers differ in the positions of ligands relative to each other around the central metal ion. They are also called cis-trans isomers or diastereomers. In the cis isomer two identical ligands are next to each other (90^\circ apart); in the trans isomer they are opposite each other (180^\circ apart).

Square planar complexes

Diamminedichloroplatinum(II), \ce{[Pt(NH3)2Cl2]}, is square planar and exists as two isomers:

cis- and trans-diamminedichloroplatinum(II). cis-platin is an anticancer drug; trans-platin is not.

Octahedral complexes

Octahedral complexes of the type \ce{MA4B2} also show cis-trans isomerism. In tetraamminediaquacobalt(III), \ce{[Co(NH3)4(H2O)2]^3+}, the cis isomer has the two water ligands next to each other and the trans isomer has them on opposite sides.

cis- and trans-\ce{[Co(NH3)4(H2O)2]^3+}.

Complexes with bidentate ligands can also have cis and trans forms, for example bis(1,2-diaminoethane)diaquanickel(II), \ce{[Ni(en)2(H2O)2]^2+}:

cis- and trans-\ce{[Ni(en)2(H2O)2]^2+}. Each curve joins the two N atoms of one en ligand; en always spans two cis positions.

Polarity of cis and trans isomers

Geometric isomers have different symmetry and so different polarity. In a cis isomer such as cis-\ce{[Co(NH3)4(H2O)2]^3+} the two water ligands are on the same side. Oxygen is more electronegative than nitrogen, so electron density is pulled towards the water side, which becomes partially negative, while the ammonia side becomes partially positive. The two bond dipoles add up to a net dipole: the cis isomer is polar. In the trans isomer the two water ligands are opposite, their dipoles act in opposite directions and cancel: the trans isomer is symmetric and non-polar. The same reasoning applies to cis- and trans-\ce{[Pt(NH3)2Cl2]}.

(ii) Optical isomerism

Optical isomers (enantiomers) are non-superimposable mirror images of each other. They have identical physical properties except their effect on plane-polarised light: one rotates it clockwise and the other anticlockwise by the same angle. In complexes, optical isomerism is shown by octahedral complexes containing bidentate ligands:

Two bidentate + two monodentate ligands, \ce{[M(AA)2B2]}: only the cis isomer is optically active. Example: cis-\ce{[Ni(en)2(H2O)2]^2+}. The trans isomer has a mirror plane and is not chiral.
Three bidentate ligands, \ce{[M(AA)3]}: example \ce{[Ni(en)3]^2+}, which exists as a pair of mirror images that cannot be superimposed.

The two optical isomers of cis-\ce{[Ni(en)2(H2O)2]^2+}.

The two optical isomers of tris(1,2-diaminoethane)nickel(II), \ce{[Ni(en)3]^2+}.

Which complexes show stereoisomerism.
Complex type	Example	Geometric isomers	Optical isomers
tetrahedral \ce{MA2B2}	\ce{[Zn(NH3)2Cl2]}	no	no
square planar \ce{MA2B2}	\ce{[Pt(NH3)2Cl2]}	cis and trans	no
octahedral \ce{MA6}	\ce{[Cu(H2O)6]^2+}	no	no
octahedral \ce{MA4B2}	\ce{[Co(NH3)4(H2O)2]^3+}	cis and trans	no
octahedral \ce{M(AA)2B2}	\ce{[Ni(en)2(H2O)2]^2+}	cis and trans	cis form only
octahedral \ce{M(AA)3}	\ce{[Ni(en)3]^2+}	no	yes

Formula sheet

Density

\rho=\dfrac{m}{V}

m: mass, V: volume

Unit: \mathrm{g\,cm^{-3}}

Charge on a complex ion

\text{charge}=\text{O.S. of metal}+\sum\text{charges of ligands}

O.S.: oxidation state of the central metal; neutral ligands (\ce{H2O}, \ce{NH3}, \ce{CO}, en) count 0, \ce{Cl^-}, \ce{CN^-}, \ce{OH^-} count -1

Unit: none

Coordination number

\text{C.N.}=\sum(\text{number of ligands}\times\text{denticity})

denticity: number of dative bonds formed by one ligand (1 for H2O, 2 for en, 6 for EDTA)

Unit: none

Energy of absorbed light (splitting energy)

\Delta E=h\nu=\dfrac{hc}{\lambda}

h=6.626\times10^{-34}\,\mathrm{J\,s}, \nu: frequency, c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}, \lambda: wavelength

Units: J per ion; multiply by N_A for \mathrm{J\,mol^{-1}}

Tetrahedral and octahedral splitting

\Delta_t\approx\tfrac49\Delta_o

\Delta_t: tetrahedral splitting, \Delta_o: octahedral splitting (same metal ion and ligand)

Unit: \mathrm{kJ\,mol^{-1}}

Stability constant (example)

K_{\mathrm{stab}}=\dfrac{[\ce{[CoCl4]^2-}]}{[\ce{[Co(H2O)6]^2+}][\ce{Cl^-}]^{4}}

square brackets outside the formula: concentration in \mathrm{mol\,dm^{-3}}; [\ce{H2O}] is left out

\mathrm{dm^{3n}\,mol^{-n}} for n ligands taken up (here \mathrm{dm^{12}\,mol^{-4}}); \log K_{\mathrm{stab}} has no unit

Standard cell potential (feasibility)

E^\circ_{\text{cell}}=E^\circ_{\text{reduced}}-E^\circ_{\text{oxidised}}

both E^\circ values are standard reduction potentials; the reaction is feasible when E^\circ_{\text{cell}}>0

Unit: V

Moles from a titre

n=c\times V

c: concentration in \mathrm{mol\,dm^{-3}}, V: volume in \mathrm{dm^{3}} (cm^3 ÷ 1000)

Unit: mol

Titration mole ratios

\begin{aligned}\ce{MnO4^-}:\ce{Fe^2+}&=1:5\\\ce{MnO4^-}:\ce{C2O4^2-}&=2:5\\\ce{Cu^2+}:\ce{S2O3^2-}&=1:1\\\ce{Cr2O7^2-}:\ce{Fe^2+}&=1:6\end{aligned}

from the balanced redox equations

Unit: none

Key terms

d-block element: An element in which the last electron enters a d sub-shell; in period 4, Sc to Zn.
Transition element: A d-block element that forms one or more stable ions with an incomplete d sub-shell (Sc to Cu in period 4; not Zn).
Paramagnetic: Weakly attracted by a magnetic field because of unpaired electrons whose spins line up only while the field is applied.
Ferromagnetic: Very strongly attracted by a magnet; unpaired spins of neighbouring atoms line up in domains and stay aligned after the field is removed (Fe, Co, Ni).
Diamagnetic: Weakly repelled by a magnet because all electrons are paired.
Alloy: A mixture of metals (sometimes with a non-metal such as carbon) with more useful properties than the pure metals, such as steel, brass and bronze.
Variable oxidation states: The ability of a transition element to form compounds in several oxidation states, because 3d and 4s electrons have similar energies.
Catalyst: A substance that increases the rate of a reaction by providing a path of lower activation energy, and is chemically unchanged at the end.
Homogeneous catalysis: Catalysis in which the catalyst and reactants are in the same physical state, for example \ce{Fe^2+(aq)} in the \ce{S2O8^2-}/\ce{I^-} reaction.
Heterogeneous catalysis: Catalysis in which the catalyst and reactants are in different physical states, for example solid Fe in the Haber process.
Ligand: A species with a lone pair of electrons that forms a dative covalent bond to a central metal atom or ion; a Lewis base.
Monodentate ligand: A ligand that forms one dative bond with the metal, such as \ce{H2O}, \ce{NH3}, \ce{Cl^-}, \ce{CN^-}.
Bidentate ligand: A ligand that forms two dative bonds with the same metal ion, such as 1,2-diaminoethane (en) and ethanedioate.
Polydentate ligand: A ligand that forms more than one dative bond with the same metal ion; EDTA is hexadentate.
Chelate: A complex in which a polydentate ligand forms a ring with the metal ion; chelates are very stable.
Complex: A molecule or ion formed by a central metal atom or ion surrounded by one or more ligands.
Coordination number: The number of dative bonds formed between the ligands and the central metal atom or ion.
Coordination sphere: The central metal with its ligands, written in square brackets.
Degenerate orbitals: Orbitals of equal energy, such as the five 3d orbitals of a free ion.
Non-degenerate orbitals: Orbitals of different energy, such as the t_{2g} and e_g sets in a complex.
\Delta E (splitting energy): The energy gap between the two sets of split d orbitals; equal to h\nu of the light absorbed.
d-d transition: Promotion of an electron from the lower set of split d orbitals to the higher set by absorbing visible light.
Complementary colour: The colour seen when a particular colour is absorbed from white light; it lies opposite on the colour wheel.
Strong field ligand: A ligand that splits the d orbitals by a large \Delta E, such as \ce{CN^-} and \ce{NH3}.
Ligand exchange (substitution): A reaction in which one ligand in a complex is replaced by another.
Stability constant, K_{\mathrm{stab}}: The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
Disproportionation: A reaction in which the same species is both oxidised and reduced.
Geometric (cis-trans) isomers: Stereoisomers that differ in the relative positions of ligands: cis (next to each other) and trans (opposite).
Optical isomers (enantiomers): Non-superimposable mirror images that rotate plane-polarised light in opposite directions.

Multiple choice questionsp. 125–126

Which one of the following species has d^{10} sub-shell?

A\ce{Cu^2+}
B\ce{Zn^2+}Answer
CMn
DCr

Why

\ce{Zn} is [\mathrm{Ar}]\,3d^{10}4s^{2}; losing the two 4s electrons gives \ce{Zn^2+}, [\mathrm{Ar}]\,3d^{10}.

\ce{Cu^2+} is 3d^{9}, Mn is 3d^{5}4s^{2} and Cr is 3d^{5}4s^{1}.

Which one is correct statement about zinc element? (1) It has complete 3d sub-shell. (2) It is transition metal. (3) It forms colourless complexes. (4) It can exist in one stable oxidation state.

A1 and 2 only
B1, 3 and 4 onlyAnswer
C1 and 4 only
Dall are correct

Why

Zinc has a complete 3d^{10} sub-shell in the atom and in its only ion \ce{Zn^2+}, so its complexes are colourless and it shows only the +2 state. Statements 1, 3 and 4 are correct.

Statement 2 is wrong: zinc is a d-block element but not a transition element, because its ion has no incomplete d sub-shell.

iii

The electronic configuration of iron cation in the complex, \ce{[Fe(CN)6]^3-}, is:

A[\mathrm{Ar}]\,3d^{6}
B[\mathrm{Ar}]\,3d^{5}Answer
C[\mathrm{Ar}]\,3d^{6}4s^{1}
D[\mathrm{Ar}]\,3d^{6}4s^{2}

Why

Oxidation state of Fe: x+6(-1)=-3, so x=+3. \ce{Fe} ([\mathrm{Ar}]\,3d^{6}4s^{2}) loses both 4s electrons and one 3d electron, giving \ce{Fe^3+} [\mathrm{Ar}]\,3d^{5}.

Option A is \ce{Fe^2+}; options C and D keep 4s electrons, which are always lost first.

Note. The book labels the options a, b, b, d; the third option is taken as C.

Why is the hexaaquacopper(II) ion blue in colour?

AThe d-d electronic transition absorbs blue wavelength of visible spectrum
BThe d-d electronic transition absorbs all wavelengths except blue
CThe d-d electronic transition transmits all wavelengths of white light except blue
DThe d-d electronic transition absorbs red wavelength and transmit the remaining lightAnswer

Why

\ce{[Cu(H2O)6]^2+} absorbs light at the red-orange end of the spectrum in a d-d transition. The remaining light is transmitted and we see its complementary colour, blue.

A would make the ion look orange-yellow; B is wrong because only one region is absorbed, not everything except blue; C describes an ion that absorbs blue.

The cyanide ligand (\ce{CN^-}) can form two complexes with iron with formulae \ce{[Fe(CN)6]^3-} and \ce{[Fe(CN)6]^4-}. What is the oxidation state of iron in them?

A+2 and +3 respectively
B+4 and -2 respectively
C+2 and +4 respectively
D+3 and +2 respectivelyAnswer

Why

x+6(-1)=-3 gives x=+3; x+6(-1)=-4 gives x=+2. So iron is +3 in \ce{[Fe(CN)6]^3-} and +2 in \ce{[Fe(CN)6]^4-}.

Transition metals are different from alkali metals and alkaline earth metals in many ways. Which one is incorrect statement about transition metals when compared with group 1 and group 2 metals?

Athey form coloured compounds
Bthey show variable oxidation states
Cthey have high melting points
Dthey are more reactiveAnswer

Why

Transition metals are much less reactive than group 1 and 2 metals (compare iron or copper with sodium or calcium in water). The other three statements are true.

vii

The species that can act as ligand in transition metal complex is:

A\ce{CH4}
B\ce{C6H6}
C\ce{Na}
D\ce{C6H4(OH)2}Answer

Why

A ligand must have a lone pair of electrons to donate. The O atoms of the two \ce{-OH} groups in \ce{C6H4(OH)2} each have two lone pairs (benzene-1,2-diol can even act as a bidentate ligand).

\ce{CH4} and \ce{C6H6} have no lone pairs, and sodium is a metal atom that loses electrons rather than donating pairs.

viii

Transition metals can show different oxidation states. Which one is the most common oxidation state in first row transition elements?

A+2Answer
B+3
C+4
D+7

Why

Almost all the first-row elements (Ti to Zn) form an \ce{M^2+} ion by losing their two 4s electrons, so +2 is the most common oxidation state.

Which one of the following properties of transition metals is held responsible for their catalytic behaviour?

Athey form coloured complexes
Bthey show variable oxidation statesAnswer
Cthey have high melting points
Dthey have high density

Why

A transition metal catalyst can change its oxidation state (for example \ce{Fe^2+ <=> Fe^3+}) while it transfers electrons to and from reactants, and then return to its original state. Vacant d orbitals that form temporary bonds are the other reason.

The coordination number of nickel in the complex \ce{[Ni(en)2(OH2)2]^2+} is:

A2
B4
C6Answer
D8

Why

Each en is bidentate (2 bonds) and each water is monodentate: 2\times2+2\times1=6 dative bonds, so the coordination number is 6 (octahedral).

Which one of the following complexes cannot show stereoisomerism? (1) \ce{[Cu(H2O)6]^2+} (2) \ce{[CuCl4]^2-} (3) \ce{[Cu(NH3)4(OH2)2]^2+} (4) \ce{[Ni(en)3]^2+}

A1, 2 and 3 only
B1, 3 and 4 only
C1 and 2 onlyAnswer
D1 and 4 only

Why

\ce{[Cu(H2O)6]^2+} has six identical ligands and tetrahedral \ce{[CuCl4]^2-} has four identical ligands: neither has isomers.

\ce{[Cu(NH3)4(OH2)2]^2+} has cis and trans forms, and \ce{[Ni(en)3]^2+} has two optical isomers.

xii

The weakest field ligand among the following is:

A\ce{H2O}
B\ce{CN^-}
C\ce{NH3}
D\ce{OH^-}Answer

Why

In the series \ce{CN^-} > \ce{NH3} > \ce{H2O} > \ce{OH^-} > \ce{Cl^-} > \ce{Br^-}, hydroxide comes after the other three, so it causes the smallest splitting.

xiii

The highest oxidation state shown by first row transition metals is shown by:

Acopper
BmanganeseAnswer
Cnickel
Dcobalt

Why

Manganese, [\mathrm{Ar}]\,3d^{5}4s^{2}, can use all seven 3d and 4s electrons and reaches +7, as in \ce{KMnO4}. Cu, Ni and Co reach only +2 or +3 in common compounds.

Short answer questionsp. 126

i3 marks

The melting point of titanium is higher than calcium in the same period. Justify this statement.

Solution

The melting point of a metal depends on the strength of its metallic bonding, that is, on how many electrons each atom gives to the sea of delocalised electrons and how strongly the nuclei attract them.

Calcium ([\mathrm{Ar}]\,4s^{2}) can release only its two 4s electrons into the metallic bond. Titanium ([\mathrm{Ar}]\,3d^{2}4s^{2}) releases its two 4s electrons and its two 3d electrons, because 3d and 4s are very close in energy. Titanium also has a higher nuclear charge and smaller atoms. So its metallic bonding is much stronger, and more energy is needed to break it.

\text{m.p.: Ti } 1668^\circ\mathrm{C}\ \gg\ \text{Ca } 842^\circ\mathrm{C}

ii3 marks

The first-row elements of d-block exist in more than one oxidation state. However, zinc shows only +2 oxidation state in its complexes. Why?

Solution

Most first-row d-block elements show variable oxidation states because their 4s and 3d electrons have similar energies, so different numbers of them can be lost or used in bonding.

Zinc is [\mathrm{Ar}]\,3d^{10}4s^{2}. It loses its two 4s electrons easily to form \ce{Zn^2+}, [\mathrm{Ar}]\,3d^{10}. The remaining 3d sub-shell is completely filled and very stable, and it is held strongly by the high nuclear charge. Removing a third electron from it would need far too much energy, which is never recovered in bonding. So zinc shows only the +2 oxidation state.

iii3 marks

Define the terms (a) ligand (b) coordination number (c) complex ion.

Solution

(a) Ligand: An atom, ion or molecule that contains at least one lone pair of electrons and donates it to a central metal atom or ion, forming a dative covalent bond. Examples: \ce{H2O}, \ce{NH3}, \ce{Cl^-}, \ce{CN^-}. A ligand is a Lewis base.
(b) Coordination number: The number of dative (coordinate) bonds formed between the ligands and the central metal ion. Example: 6 in \ce{[Cu(H2O)6]^2+}, 4 in \ce{[CuCl4]^2-}, 6 in \ce{[Ni(en)3]^2+} (three bidentate ligands).
(c) Complex ion: A charged species formed by a central metal ion surrounded by ligands bonded to it by dative bonds, for example \ce{[Co(NH3)6]^3+} (cationic) or \ce{[Fe(CN)6]^4-} (anionic).

iv3 marks

Explain why hydrated \ce{Ti^4+} complexes are colourless and hydrated \ce{Ti^3+} complexes are coloured.

Solution

Colour in transition metal complexes comes from a d-d transition: ligands split the 3d orbitals into two sets separated by \Delta E, and an electron in the lower set absorbs visible light of energy \Delta E=h\nu to jump to the upper set. The complementary colour is seen.

\ce{Ti^3+} is [\mathrm{Ar}]\,3d^{1}. In \ce{[Ti(H2O)6]^3+} the single electron in t_{2g} absorbs yellow-green light and moves to e_g, so the ion looks red-violet.

\ce{Ti^4+} is [\mathrm{Ar}]\,3d^{0}: it has no d electron to promote, so no visible light is absorbed and its compounds are colourless (for example white \ce{TiO2}).

v3 marks

Why do the melting points of first-row transition metals increase up to the middle and then decrease? Comment.

Solution

Melting point depends on the strength of metallic bonding, which depends mainly on the number of unpaired 3d and 4s electrons available for the metallic bond.

From Sc to V and Cr the number of unpaired 3d electrons increases (1, 2, 3, then 5), so the bonding gets stronger and the melting point rises to about 1900^\circ\mathrm{C}. After the middle the 3d electrons start to pair up (Fe 3d^{6} to Cu 3d^{10}); paired electrons are less available for bonding, so the melting point falls (Cu 1085^\circ\mathrm{C}, Zn 419^\circ\mathrm{C}).

Manganese is an exception with a low value (1246^\circ\mathrm{C}): its stable half-filled 3d^{5} and full 4s^{2} sub-shells hold their electrons firmly.

vi3 marks

Why are transition metals used as catalysts in industries for performing different reactions? Give one example.

Solution

Transition metals and their compounds are good catalysts for two reasons:

They have variable oxidation states, so they can gain and lose electrons during a reaction (for example \ce{V^5+ <=> V^4+}, \ce{Fe^2+ <=> Fe^3+}) and then return to their original state.
They have vacant d orbitals that accept lone pairs from reactant molecules, forming temporary dative bonds. This holds the reactants close together on the surface and weakens their bonds, giving a path of lower activation energy.

Example: iron is the catalyst in the Haber process, \ce{N2 + 3H2 <=>[Fe] 2NH3} (or \ce{V2O5} in the Contact process, \ce{2SO2 + O2 <=>[V2O5] 2SO3}).

vii3 marks

The blue cobalt chloride paper is used to test the presence of water. If water is present, the paper turns pink. This is because six water ligands exchange for four chloride ligands present in cobalt chloride, \ce{[CoCl4]^2-}. Write the equation to show the ligand substitution reaction when the test is positive.

Solution

\underset{\text{blue (tetrahedral)}}{\ce{[CoCl4]^2-}}+\ce{6H2O}\ce{->}\underset{\text{pink (octahedral)}}{\ce{[Co(H2O)6]^2+}}+\ce{4Cl^-}

The dry paper contains the blue tetrachlorocobaltate(II) ion. When water touches it, six water molecules replace the four chloride ligands. The coordination number changes from 4 to 6, the shape from tetrahedral to octahedral, and the colour from blue to pink (hexaaquacobalt(II)). The reaction is reversible: heating the paper drives off the water and restores the blue colour.

viii3 marks

Octahedral complexes of copper(II) ion have different colours. Explain why.

Solution

In an octahedral copper(II) complex (3d^{9}) the ligands split the 3d orbitals into a lower t_{2g} set and a higher e_g set. The colour comes from a d-d transition that absorbs light of energy \Delta E=h\nu; we see the complementary colour.

The size of \Delta E depends on the ligand. A strong field ligand (\ce{NH3}) splits the orbitals more than a weaker one (\ce{H2O}). So each complex absorbs a different frequency and shows a different colour: \ce{[Cu(H2O)6]^2+} is pale blue, \ce{[Cu(H2O)4(OH)2]} is a pale blue precipitate and \ce{[Cu(NH3)4(H2O)2]^2+} is deep violet-blue (larger \Delta E).

Long answer questionsp. 126–127

i7 marks

The reduction potential for \ce{Cr^2+} ions being reduced to chromium metal is -0.91\,\mathrm{V} and that for \ce{Cr^3+} ions being reduced to \ce{Cr^2+} ions is -0.41\,\mathrm{V} as shown. Predict whether the chromium(II) ion will disproportionate to chromium(III) ion and chromium metal. If yes, then write the possible ionic equation.

\begin{aligned}\ce{Cr^2+ + 2e- &<=> Cr}&\quad E^\circ&=-0.91\,\mathrm{V}\\\ce{Cr^3+ + e- &<=> Cr^2+}&\quad E^\circ&=-0.41\,\mathrm{V}\end{aligned}

Solution

What disproportionation means

In a disproportionation reaction the same species is oxidised and reduced at the same time. For chromium(II) this would mean some \ce{Cr^2+} ions are reduced to Cr metal (+2 to 0) while others are oxidised to \ce{Cr^3+} (+2 to +3).

The two half-reactions

Reduction half: \ce{Cr^2+} gains two electrons and becomes Cr. Its E^\circ is used as it stands.
\ce{Cr^2+ + 2e- -> Cr}\qquad E^\circ=-0.91\,\mathrm{V}
Oxidation half: reverse the second equation and change the sign of E^\circ. Multiply by 2 so that 2 electrons are lost, matching the 2 gained. E^\circ is not multiplied.
\ce{2Cr^2+ -> 2Cr^3+ + 2e-}\qquad E^\circ=+0.41\,\mathrm{V}
Add the half-equations and the potentials.
\ce{3Cr^2+(aq) -> 2Cr^3+(aq) + Cr(s)}\qquad E^\circ_{\text{cell}}=-0.91+0.41=-0.50\,\mathrm{V}

Prediction

E^\circ_{\text{cell}} is negative (-0.50\,\mathrm{V}), so the reaction is not feasible under standard conditions. Chromium(II) ions will not disproportionate into chromium(III) ions and chromium metal. The reason is that E^\circ for \ce{Cr^2+}/\ce{Cr} (-0.91\,\mathrm{V}) is more negative than for \ce{Cr^3+}/\ce{Cr^2+} (-0.41\,\mathrm{V}): \ce{Cr^2+} is a poorer oxidising agent than \ce{Cr^3+}, so it cannot oxidise another \ce{Cr^2+} ion.

The equation that would have represented disproportionation is \ce{3Cr^2+ -> 2Cr^3+ + Cr}, but it does not occur. The reverse reaction has E^\circ=+0.50\,\mathrm{V} and is feasible: chromium metal reduces chromium(III) ions to chromium(II).

\ce{Cr(s) + 2Cr^3+(aq) -> 3Cr^2+(aq)}\qquad E^\circ_{\text{cell}}=+0.50\,\mathrm{V}

E^\circ_{\text{cell}}=-0.50\,\mathrm{V}:\ \ce{3Cr^2+ -> 2Cr^3+ + Cr}\ \text{is not feasible}

ii2026 Annual I, Set D, Q.47 marks

Explain the difference in the splitting pattern of the 3d atomic orbitals in the two copper(II) complexes, \ce{[Cu(H2O)6]^2+} and \ce{[CuCl4]^2-}.

Solution

Degenerate orbitals and why they split

In a free \ce{Cu^2+} ion ([\mathrm{Ar}]\,3d^{9}) the five 3d orbitals are degenerate (equal in energy). When ligands approach, their lone pairs repel the d electrons. Orbitals whose lobes point towards the ligands are repelled more and rise more in energy, so the 3d orbitals split into two non-degenerate sets separated by \Delta E. The pattern depends on the shape of the complex.

\ce{[Cu(H2O)6]^2+}: octahedral

Six water ligands approach along the x, y and z axes. d_{x^2-y^2} and d_{z^2} point straight at the ligands and form the higher e_g set (two orbitals). d_{xy}, d_{yz} and d_{xz} point between the ligands and form the lower t_{2g} set (three orbitals). The nine electrons are arranged t_{2g}^{6}e_g^{3}. The splitting \Delta_o is large.

\ce{[CuCl4]^2-}: tetrahedral

Four large chloride ligands sit at alternate corners of a cube, between the axes. Now d_{xy}, d_{yz} and d_{xz} point nearer to the ligands and form the higher set (t_2, three orbitals), while d_{x^2-y^2} and d_{z^2} form the lower set (e, two orbitals). The electrons are arranged e^{4}t_2^{5}. The splitting \Delta_t is smaller, about \tfrac49\Delta_o, because there are only four ligands and none points directly at a d orbital (chloride is also a weaker field ligand than water).

Diagram

3d splitting in the two copper(II) complexes, with the nine d electrons filled in.

Summary of differences

	\ce{[Cu(H2O)6]^2+}	\ce{[CuCl4]^2-}
Shape, coordination number	octahedral, 6	tetrahedral, 4
Higher set	2 orbitals (e_g)	3 orbitals (t_2)
Lower set	3 orbitals (t_{2g})	2 orbitals (e)
Splitting	larger \Delta_o	smaller \Delta_t\approx\tfrac49\Delta_o
Colour	blue (absorbs orange-red)	yellow (absorbs at a different, lower-energy region)

Because \Delta E is different, the two complexes absorb different frequencies of light in their d-d transitions and so have different colours: blue for the aqua complex and yellow for the chloro complex.

iii7 marks

The following scheme shows some reactions of copper(II) ion. (a) Suggest the formulae of A, B and C. (b) State the colours of solutions B and C. (c) Name the type of reaction occurring when C is heated with copper. Also state the role of copper in this reaction. (d) When the solution of complex \ce{[CuCl2]^-} is poured into water, which colour precipitate is formed? Justify its colour.

Reaction scheme from the book.

Solution

(a) Formulae of A, B and C.

A is the pale blue precipitate formed when a base (\ce{NaOH} or a little \ce{NH3}) removes protons from two water ligands: \ce{[Cu(H2O)4(OH)2]} (copper(II) hydroxide, \ce{Cu(OH)2}).

\ce{[Cu(H2O)6]^2+ + 2OH^- -> [Cu(H2O)4(OH)2](s) + 2H2O}

B forms when excess ammonia dissolves A by ligand exchange: \ce{[Cu(NH3)4(H2O)2]^2+}, tetraamminediaquacopper(II) ion.

\ce{[Cu(H2O)4(OH)2](s) + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 2H2O + 2OH^-}

C forms when concentrated \ce{HCl} replaces the water ligands by chloride: \ce{[CuCl4]^2-}, tetrachlorocuprate(II) ion.

\ce{[Cu(H2O)6]^2+ + 4Cl^- <=> [CuCl4]^2- + 6H2O}

(b) Colours of solutions B and C.

Solution B, \ce{[Cu(NH3)4(H2O)2]^2+}, is deep blue (violet-blue). Solution C, \ce{[CuCl4]^2-}, is yellow; with some \ce{[Cu(H2O)6]^2+} still present the mixture usually looks yellow-green.

\ce{[CuCl4]^2-(aq) + Cu(s) ->[heat] 2[CuCl2]^-(aq)}

This is a redox reaction of the type called comproportionation (the reverse of disproportionation): copper in the +2 state (in C) and copper in the 0 state (the metal) both end up in the same +1 state in \ce{[CuCl2]^-}. Copper metal is oxidised (0 to +1), so it acts as the reducing agent; it reduces copper(II) to copper(I). The excess chloride ions stabilise copper(I) as the linear complex \ce{[CuCl2]^-}.

(d) Colour of the precipitate when \ce{[CuCl2]^-} is poured into water.

\ce{[CuCl2]^-(aq) ->[\text{water}] CuCl(s) + Cl^-(aq)}

Dilution lowers the chloride concentration, the complex breaks up and a white precipitate of copper(I) chloride forms. It is white because copper(I) is [\mathrm{Ar}]\,3d^{10}: the 3d sub-shell is completely filled, so no d-d transition is possible, no visible light is absorbed, and the compound looks white.

iv7 marks

Transition metal complexes such as \ce{[Cu(H2O)4(NH3)2]^2+} and \ce{Pt(NH3)2Cl2} show same type of stereoisomerism. (a) Name the type of isomerism. (b) Draw three-dimensional structures for the given two isomers, two for each.

Solution

(a) Type of isomerism.

Both show geometric (cis-trans) isomerism, a type of stereoisomerism. The isomers have the same formula and the same bonds, but the positions of the ligands differ: in the cis isomer the two identical ligands (\ce{NH3}) are next to each other at 90^\circ; in the trans isomer they are opposite each other at 180^\circ.

\ce{[Cu(H2O)4(NH3)2]^2+} is octahedral (\ce{MA4B2}) and \ce{[Pt(NH3)2Cl2]} is square planar (\ce{MA2B2}); both types can have two different arrangements.

(b) Three-dimensional structures, two for each.

cis- and trans-diamminetetraaquacopper(II). Wedges come towards the reader and dashed bonds go behind the page.

cis- and trans-diamminedichloroplatinum(II), square planar.

The cis isomers are polar (the bond dipoles do not cancel) and the trans isomers are non-polar (the dipoles cancel). cis-\ce{[Pt(NH3)2Cl2]} (cisplatin) is an anticancer drug; the trans isomer is not. Neither pair shows optical isomerism, because each structure has a plane of symmetry.

v7 marks

Aqueous copper(II) ion can form complexes with the ligands ammonia and 1,2-diaminoethane as shown. (a) Write an expression for the stability constant for both equations and state their units. (b) Of the three complexes in the above two equations, state the formula of the complex that is the most stable. Give a reason also.

\begin{aligned}&\ce{[Cu(H2O)6]^2+(aq) + en(aq) <=> [Cu(H2O)4(en)]^2+(aq) + 2H2O(l)}&&K_{\mathrm{stab}}=3.98\times10^{10}\\&\ce{[Cu(H2O)6]^2+(aq) + 2NH3(aq) <=> [Cu(H2O)4(NH3)2]^2+(aq) + 2H2O(l)}&&K_{\mathrm{stab}}=5.01\times10^{7}\end{aligned}

Solution

(a) Stability constant expressions and units.

The stability constant is the equilibrium constant for forming the complex in water. Water is the solvent, in large excess, so [\ce{H2O}] is left out.

K_{\mathrm{stab}}(1)=\dfrac{[\ce{[Cu(H2O)4(en)]^2+}]}{[\ce{[Cu(H2O)6]^2+}][\mathrm{en}]}

\text{units}=\dfrac{\mathrm{mol\,dm^{-3}}}{(\mathrm{mol\,dm^{-3}})(\mathrm{mol\,dm^{-3}})}=\mathrm{dm^{3}\,mol^{-1}}

K_{\mathrm{stab}}(2)=\dfrac{[\ce{[Cu(H2O)4(NH3)2]^2+}]}{[\ce{[Cu(H2O)6]^2+}][\ce{NH3}]^{2}}

\text{units}=\dfrac{\mathrm{mol\,dm^{-3}}}{(\mathrm{mol\,dm^{-3}})(\mathrm{mol\,dm^{-3}})^{2}}=\mathrm{dm^{6}\,mol^{-2}}

K_{\mathrm{stab}}(1):\ \mathrm{dm^{3}\,mol^{-1}};\quad K_{\mathrm{stab}}(2):\ \mathrm{dm^{6}\,mol^{-2}}

(b) Most stable complex and reason.

The most stable complex is \ce{[Cu(H2O)4(en)]^2+}.

Both reactions have very large K_{\mathrm{stab}} values, so both products are more stable than \ce{[Cu(H2O)6]^2+}: the equilibria lie far to the right, and the aqua complex is the least stable of the three.
The en complex has the larger constant (3.98\times10^{10} against 5.01\times10^{7}), about 800 times larger, so its equilibrium lies furthest to the right. Both new ligands bond through N atoms, so the difference comes from the chelate effect.
en is bidentate: one en molecule forms two dative bonds and makes a stable five-membered ring with copper. One en particle replaces two water molecules, so the number of free particles increases (2 on the left, 3 on the right). This increase in entropy makes chelate complexes more stable than complexes of similar monodentate ligands such as \ce{NH3}.

\ce{[Cu(H2O)4(en)]^2+}

Note. The book prints the aqua ion as '\ce{[Cu(H2O)6]2+}' with a misplaced charge; it is \ce{[Cu(H2O)6]^2+}.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 51 mark

Ligands form bonds with transition metals by:

AAccepting electrons
BForming ionic bonds only
CLosing protons
DDonating lone pairs of electronsAnswer

Why

A ligand has a lone pair of electrons and donates it into a vacant orbital of the metal ion, forming a dative (coordinate) bond. The ligand is a Lewis base.

It does not accept electrons (A), the bonds are not ionic (B), and losing protons (C) describes an acid, not a ligand.

2026 Annual I, Set B, Q.2 (iv) OR1+1+1 = 3 marks

Predict the shape, approximate bond angles and coordination number of \ce{[Ag(NH3)2]+}.

Solution

Shape (1 mark): linear. The two \ce{NH3} ligands lie on opposite sides of the silver ion. Two bond pairs repel each other least when they are as far apart as possible (VSEPR).
Bond angle (1 mark): \ce{H3N-Ag-NH3} angle =180^\circ.
Coordination number (1 mark): 2, because two monodentate \ce{NH3} ligands each form one dative bond to \ce{Ag+}.

The linear diamminesilver(I) ion.

Silver is +1 ([\mathrm{Kr}]\,4d^{10}) and ammonia is neutral, so the ion carries a charge of +1. It is the active species in Tollens' reagent.

2026 Annual I, Set B, Q.2 (v) OR3 marks

Why transition elements with d^{10} configuration are usually colourless?

Solution

The colour of a transition metal compound comes from a d-d transition. Ligands split the 3d orbitals into a lower and a higher set separated by \Delta E. An electron in the lower set absorbs a photon of visible light with energy \Delta E=h\nu and jumps into a vacant place in the higher set; we see the complementary colour.

In a d^{10} ion, such as \ce{Zn^2+}, \ce{Cu+} or \ce{Ag+}, all five d orbitals are completely filled. There is no vacancy in the upper set, so no electron can be promoted and no visible light is absorbed. All the light is transmitted or reflected, so the compounds are white or colourless (for example \ce{ZnSO4}, \ce{CuI}).

d^{9} has a vacancy in the upper set (coloured); d^{10} has none (colourless).

2026 Annual I, Set B, Q.2 (xiv)1+2 = 3 marks

What is a catalyst? Explain its action with a general graphical representation.

Solution

Catalyst (1 mark): a substance that increases the rate of a chemical reaction without being used up; it is chemically unchanged at the end of the reaction.

Action (2 marks): a catalyst provides an alternative pathway (mechanism) with a lower activation energy E_a. At the same temperature many more colliding particles now have energy equal to or greater than E_a, so more collisions are successful and the reaction is faster. The energies of reactants and products do not change, so \Delta H is the same, and the equilibrium position is not changed (it is only reached sooner).

Energy profile with and without a catalyst: the catalysed path has a lower peak, \Delta H is unchanged.

Transition metals are good catalysts because they have variable oxidation states and vacant d orbitals that form temporary bonds with reactants, for example iron in the Haber process.

2026 Annual I, Set B, Q.41+3+3 = 7 marks

What are transition elements? Write trend of melting point and atomic radii from Ti to Cu in 3d series.

Solution

Transition elements (1 mark)

Transition elements are d-block elements which form one or more stable ions with an incomplete (partly filled) d sub-shell. In period 4 they are Sc to Cu (Ti to Cu in the book's study). Zinc is a d-block element but not a transition element, because \ce{Zn^2+} is 3d^{10}.

Trend in melting point, Ti to Cu (3 marks)

	Ti	V	Cr	Mn	Fe	Co	Ni	Cu
m.p. / ^\circ\mathrm{C}	1668	1910	1907	1246	1538	1495	1455	1085

Melting points of the first-row d-block metals

Melting points are high because of strong metallic bonding: both 3d and 4s electrons are delocalised into the metallic bond.
Ti to V (and Cr): the melting point rises (1668\to1910^\circ\mathrm{C}) because the number of unpaired 3d electrons available for bonding increases.
Mn: a sharp dip (1246^\circ\mathrm{C}). Its stable half-filled 3d^{5} and full 4s^{2} sub-shells hold their electrons firmly, so fewer electrons take part in metallic bonding.
Fe to Cu: the melting point falls gradually (1538\to1085^\circ\mathrm{C}) because the 3d electrons pair up and become less available for bonding.

Trend in atomic radius, Ti to Cu (3 marks)

	Ti	V	Cr	Mn	Fe	Co	Ni	Cu
radius / pm	147	134	128	127	126	125	124	128

Atomic radii of the first-row d-block metals

Ti to Cr: decreases (147 to 128 pm). The nuclear charge rises by one at each step and outweighs the extra shielding, pulling the electrons closer.
Cr to Ni: almost constant (128 to 124 pm). Each new electron enters the inner 3d sub-shell; its shielding of the outer 4s electrons and electron-electron repulsion balance the increase in nuclear charge.
Ni to Cu: slight increase (124 to 128 pm). The 3d sub-shell becomes full; the full 3d^{10} shields strongly and repels the 4s electron, pushing it outwards.

2026 Annual I, Set D, MCQ 21 mark

Catalysts accelerate reactions by:

AIncreasing temperature only
BChanging reaction enthalpy
CIncreasing reactant concentration
DLowering activation energyAnswer

Why

A catalyst provides an alternative pathway with a lower activation energy, so more collisions are successful. It does not change the enthalpy change (B) or the concentrations (C), and it works without raising the temperature (A).

2026 Annual I, Set D, MCQ 61 mark

Zinc is not considered as a transition element because:

Ad-orbitals are completely filledAnswer
BIt is a non-metal
CIt has no d-orbitals
DIt is highly reactive

Why

Zinc ([\mathrm{Ar}]\,3d^{10}4s^{2}) and its only ion \ce{Zn^2+} (3d^{10}) have completely filled d orbitals, so zinc has no stable ion with an incomplete d sub-shell.

Zinc is a metal (not B), it does have d orbitals (not C), and it is not highly reactive (not D).

2026 Annual I, Set D, Q.2 (iv) OR1.5+1.5 = 3 marks

Write the IUPAC name of the following coordination compounds: (a) \ce{[Pt(NH3)4BrCl]SO4} (b) \ce{K4[Fe(CN)6]}

Solution

(a) 1.5 marks

The complex is the cation; \ce{SO4^2-} is the anion, so the sphere has charge +2. Find Pt:
x+4(0)+(-1)+(-1)=+2\ \Rightarrow\ x=+4
Ligands in alphabetical order: ammine, bromo, chloro. The sphere is a cation, so the metal keeps its name.

tetraamminebromochloroplatinum(IV) sulphate

(b) 1.5 marks

Potassium is the cation; the complex \ce{[Fe(CN)6]^4-} is the anion. Find Fe:
x+6(-1)=-4\ \Rightarrow\ x=+2
Six cyano ligands; the sphere is an anion, so iron becomes ferrate.

potassium hexacyanoferrate(II)

2026 Annual I, Set D, Q.2 (v) OR1x3 = 3 marks

Write any three physical properties of transition elements.

Solution

High melting and boiling points (for example Fe 1538^\circ\mathrm{C}), because strong metallic bonding uses both 3d and 4s electrons.
High density (for example Cu 8.96\,\mathrm{g\,cm^{-3}}), because the atoms are heavy, small and closely packed in the lattice.
Good conductors of heat and electricity, because they have delocalised (free) electrons.

Other acceptable properties: hard and strong; shiny metallic lustre; malleable and ductile; paramagnetic or ferromagnetic (Fe, Co, Ni); form alloys because of similar atomic sizes.

2026 Annual I, Set D, Q.43.5+3.5 = 7 marks

Compare the arrangement of d-orbitals in octahedral and tetrahedral complexes. (Support your answer by drawing their diagrams of d-d splitting pattern.)

Solution

In a free metal ion the five 3d orbitals are degenerate (same energy). When ligands approach, their lone pairs repel the d electrons. Orbitals whose lobes point towards the ligands are repelled more and are raised more in energy, so the d orbitals split into two non-degenerate sets. The pattern depends on where the ligands are.

Octahedral ligands lie on the axes; tetrahedral ligands lie between the axes.

Octahedral complexes (3.5 marks)

Six ligands approach along the x, y and z axes.
d_{x^2-y^2} and d_{z^2} point directly at the ligands, are repelled more and form the higher e_g set (2 orbitals).
d_{xy}, d_{yz}, d_{xz} point between the ligands, are repelled less and form the lower t_{2g} set (3 orbitals).
The gap \Delta_o is large. Example: \ce{[Ti(H2O)6]^3+}, \ce{[Cu(H2O)6]^2+}.

Tetrahedral complexes (3.5 marks)

Four ligands sit at alternate corners of a cube, between the axes.
d_{xy}, d_{yz}, d_{xz} point nearer to the ligands and form the higher t_2 set (3 orbitals).
d_{x^2-y^2} and d_{z^2} point away from the ligands and form the lower e set (2 orbitals).
The pattern is the reverse of octahedral, and the gap is smaller, \Delta_t\approx\tfrac49\Delta_o, because there are fewer ligands and none points straight at an orbital. Example: \ce{[CuCl4]^2-}, \ce{[CoCl4]^2-}.

d-d splitting patterns: octahedral (two higher, three lower) and tetrahedral (three higher, two lower).

Because \Delta E differs, octahedral and tetrahedral complexes of the same ion absorb different frequencies of light and have different colours, for example pink \ce{[Co(H2O)6]^2+} and blue \ce{[CoCl4]^2-}.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 4, old syllabus1 mark

In the compound \ce{[Co(NH3)6]Cl3}, the coordination number of cobalt is:

A3
B4
C6Answer
D9

Why

Six \ce{NH3} ligands each form one dative bond to cobalt, so the coordination number is 6. The three \ce{Cl^-} ions are counter ions outside the coordination sphere, so 9 is wrong; 3 is the oxidation state of cobalt, not its coordination number.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 5, old syllabus1 mark

The electronic configuration of \ce{Fe^2+} ion is:

A[\mathrm{Ar}]\,3d^{6}Answer
B[\mathrm{Ar}]\,3d^{5}4s^{1}
C[\mathrm{Ar}]\,3d^{4}4s^{2}
D[\mathrm{Ar}]\,3d^{7}

Why

\ce{Fe} is [\mathrm{Ar}]\,3d^{6}4s^{2}. The two 4s electrons are lost first, giving \ce{Fe^2+} [\mathrm{Ar}]\,3d^{6}. Options B and C wrongly remove 3d electrons before 4s; D has one electron too many.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i) OR, old syllabus1.5+1.5 = 3 marks

Write the chemical equation for the reaction of \ce{K2Cr2O7} (a) with Mohr's salt and (b) with oxalic acid in acidic medium.

Solution

In dilute sulphuric acid, orange dichromate(VI) is reduced to green chromium(III) (Cr: +6 to +3, 6 electrons per \ce{Cr2O7^2-}).

(a) With Mohr's salt, \ce{FeSO4.(NH4)2SO4.6H2O}

Iron(II) is oxidised to iron(III) (1 electron each), so 6 \ce{Fe^2+} are needed per \ce{Cr2O7^2-}:

\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 6Fe^3+ + 7H2O}

\begin{aligned}&\ce{K2Cr2O7 + 6FeSO4.(NH4)2SO4.6H2O + 7H2SO4}\\&\qquad\ce{-> K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 6(NH4)2SO4 + 43H2O}\end{aligned}

(b) With oxalic acid, \ce{H2C2O4}

Oxalic acid is oxidised to carbon dioxide (C: +3 to +4, 2 electrons per molecule), so 3 molecules are needed per \ce{Cr2O7^2-}:

\ce{Cr2O7^2- + 8H+ + 3H2C2O4 -> 2Cr^3+ + 6CO2 + 7H2O}

\ce{K2Cr2O7 + 3H2C2O4 + 4H2SO4 -> K2SO4 + Cr2(SO4)3 + 6CO2 + 7H2O}

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii) OR, old syllabus1.5+1.5 = 3 marks

Write IUPAC name for the coordination compounds \ce{[Fe(CN)6]^3-} and \ce{[Co(NH3)4Cl2]Cl}.

Solution

\ce{[Fe(CN)6]^3-} (1.5 marks)

Oxidation state: x+6(-1)=-3, so iron is +3. The complex is an anion, so iron is named ferrate. Six cyano ligands give 'hexacyano'.

hexacyanoferrate(III) ion

\ce{[Co(NH3)4Cl2]Cl} (1.5 marks)

The sphere \ce{[Co(NH3)4Cl2]+} is the cation (one \ce{Cl^-} counter ion). Oxidation state: x+4(0)+2(-1)=+1, so cobalt is +3. Ligands in alphabetical order: ammine, then chloro. The cation is named first, then the chloride anion.

tetraamminedichlorocobalt(III) chloride

Exam focus

Define a transition element and describe the trends in melting point and atomic radius from Ti to Cu. (long)
Asked as a full Section C question in 2026 (Set B Q.4, 1+3+3).
Go to answer
Compare d-orbital splitting in octahedral and tetrahedral complexes with diagrams. (long)
2026 Set D Q.4 (7 marks) and book long question ii; draw both diagrams with labels.
Go to answer
Write IUPAC names of complexes such as \ce{[Pt(NH3)4BrCl]SO4}, \ce{K4[Fe(CN)6]}, \ce{[Co(NH3)4Cl2]Cl}. (short)
Naming appeared in both the new (Set D) and old 2026 papers.
Go to answer
Why are d^{0} and d^{10} compounds colourless while d^{1} to d^{9} compounds are coloured? (short)
2026 Set B Q.2 (v) OR and book short question iv.
Go to answer
What is a catalyst? Explain why transition metals are good catalysts, with an energy profile. (short)
2026 Set B Q.2 (xiv) and two 2026 MCQs; book short question vi.
Go to answer
Shape, bond angle and coordination number of complexes such as \ce{[Ag(NH3)2]+}, \ce{[CuCl4]^2-}, \ce{[Ni(H2O)6]^2+}. (short)
2026 Set B Q.2 (iv) OR; a direct SLO.
Go to answer
Ligand exchange reactions of copper(II) and cobalt(II) with \ce{OH^-}, \ce{NH3} and \ce{Cl^-}, with colours. (long)
Book long question iii and short question vii are built on these reactions.
Go to answer
Write K_{\mathrm{stab}} expressions with units and use them to explain which complex is more stable. (long)
Book long question v; the chelate effect is a favourite reasoning question.
Go to answer
Use E^\circ values to predict feasibility (\ce{MnO4^-}/\ce{Fe^2+}, \ce{MnO4^-}/\ce{C2O4^2-}, \ce{Cu^2+}/\ce{I^-}, disproportionation of \ce{Cr^2+}). (long)
SLO with calculations; book long question i.
Go to answer
Equations for acidified \ce{K2Cr2O7} with Mohr's salt and with oxalic acid. (short)
Asked in the 2026 old-syllabus paper; also useful for titration questions.
Go to answer
Draw cis and trans isomers of \ce{[Pt(NH3)2Cl2]} and octahedral complexes, and the optical isomers of \ce{[Ni(en)3]^2+}. (long)
Book long question iv and MCQ xi; diagrams earn most of the marks.
Go to answer
Electronic configurations of Cr, Cu and ions such as \ce{Fe^2+}, \ce{Fe^3+}, \ce{Cu^2+}. (mcq)
2026 old-paper MCQ 5 and book exercise MCQs i and iii.
Go to answer