Unit 5.2: Transition metal chemistry:
derive the electronic configurations of atoms of the d-block elements (Sc to Zn), and their simple ions, from their position in the
Periodic Table
recall the transition metals as d-block elements forming one or more stable ions which have incompletely filled d-orbitals
Transition metal – D block elements which form ions which have partially filled d-orbitals
Scandium and zinc are not transition elements
Element
atomic Symbol Electronic structure of atom
number
Common ion(s) Electronic
structure of
ion
Scandium
21
Sc
(Ar)3d14s2
Sc3+
(Ar)
Titanium
22
Ti
(Ar)3d24s2
Ti4+
Ti3+
(Ar)
(Ar)3d1
Vanadium
23
V
(Ar)3d34s2
V3+
(Ar)3d2
Chromium 24
Cr
(Ar)3d54s1
Cr3+
(Ar)3d3
Manganese 25
Mn
(Ar)3d54s2
Mn2+
(Ar)3d5
Iron
26
Fe
(Ar)3d64s2
Fe2+
Fe3+
(Ar)3d6
(Ar)3d5
Cobalt
27
Co
(Ar)3d74s2
Co2+
(Ar)3d7
Nickel
28
Ni
(Ar)3d84s2
Ni2+
(Ar)3d8
Copper
29
Cu
(Ar)3d104s1
Cu+
Cu2+
(Ar)3d10
(Ar)3d9
Zinc
30
Zn
(Ar)3d104s2
Zn2+
(Ar)3d10
recall appropriate parts of the chemistry of chromium, iron and copper to illustrate the properties of transition elements described
in c, d and e.
recall characteristic properties of the transition elements, such as
(i) the formation of coloured aqueous ions, and other complex ions
Transition metal compounds are coloured, d orbitals are normally of similar energy but when surrounded by ligands the orbitals
are split into higher and lower energy sets.
 absorbs light (in visible region)  d-orbitals split by ligands  electron moves to a higher energy level
[Cr(H2O)6]3+(aq) blue hexaaquachromium(III)
dichromate Cr2O72-(aq) orange in acid
chromate [CrO4]2-(aq) yellow in alkali
Fe2+(aq) green
Fe3+(aq) yellow/brown
Cu(H2O)6]2+(aq) blue hexaaquacopper(II)
Managnate(VII) MnO4-(aq) purple
Tetraamminecopper(II) Cu(NH3)42+(aq) deep blue/purple
Tetrachlorocuprate(II) CuCl42-(aq) yellow
hydrated thiocyanate complex of iron (III) [Fe(SCN)(H2O)5]2+ blood red
[Co(H2O)6]2+(aq) pink hexaaquacobalt(II)
[CoCl4]2-(aq) blue tetrachlorocobalte(II)
(ii)
the formation of a range of compounds in which they are present in different stable oxidation states
Transition metals show the following characteristic properties:
1. Variable oxidation states:Transition metals have electrons of similar energy in both the 3d and 4s levels. This means that one particular element
can form ions of roughly the same stability by losing different numbers of electrons. Thus, all transition metals from
titanium to copper can exhibit two or more oxidation states in their compounds.
When Transition Metals form positive ions they loose their electrons from the 4s sub-shell first, then the 3d sub-shell.
Oxidation states of some Transition Metals:
Titanium- +2, +3, +4
Vanadium- +2, +3, +4, +5
Chromium- +2, +3, +6
Manganese- +2, +3, +4, +5, +6, +7
Iron- +2, +3
Cobalt- +2, +3
2.
3.
4.
Nickel- +2, +3, +4
Copper- +1, +2
Formation of complex ions:- As a lot of the transition metals have some empty spaces in their 3d-orbitals, they can
receive lone pairs of electrons and form dative covalent bonds thus producing complex compounds.
Coloured compounds:- When electrons move from a d-orbital (with lower energy) to another d-orbital (with higher
energy), energy is taken in. This energy in the form of light is missing from the reflected light thus producing coloured
compounds.
Catalytic properties:- As Transition Metals have variable oxidation states, they tend to have catalytic properties. The
small differences in ionisation energies make variable oxidation numbers possible. The reversible redox reactions
involving transition metal ions make lower energy routes for other reactions.
Understand the bonding, shape and cause of colour in complex ions and aqua ions
stereoisomerism in such complex ions will not be examined
colour should be related to a simple transfer of electrons between d-orbitals
Bonding between the ligand and the metal ion is DATIVE COVALENT and this CAUSES A SPLITTING OF D-ORBITALS
Complex ions (cationic or anionic) – A central metal ion surrounded by ligands/anions (species donating electron(s)
In transition metal complexes, non-bonded pairs of electrons on the ligand form co-ordinate/dative covalent bonds to the central
ion by donating these unshared electron pairs into vacant orbitals
Co-ordination number
Shape
Octahedral
tetrahedral
Linear
The number of co-ordinate bonds from ligands to the central ion
[Fe(H2O)6]3+
[Ni(CO)4]2+
[CuCl2]-
Octahedron
[CrCl(H2O)5]+2 pentaaquachlorochromium (III)
[NiCl4]2- tetrachloronickelate(II)
[Cr(OH)6]3- hexahydroxochromate(III)
Writing formulae:
Central atom first, then negative ions, then follow any neutral molecules. Everything is then put in square brackets and the
charge added. For example, tetraaquachlorocopper (II) would be written as [CuCl(H 2O)4]+
Naming:
For a positive complex ion: prefix, ligands (alphabetical order), then metal, then oxidation number.
For a negative complex ion: the name of the metal is changed. Iron-ferrate, copper-cuprate, chromium-chromate, cobaltcobalate, nickel-nickelate, zinc-zincate
Di
Tri
Tetra
Penta
Hexa
Name of molecule or ion Symbol Ligand name
Water
H 2O
Aqua
Ammonia
NH3
Ammine
Chloride ion
Cl-
Chloro
Hydroxide ion
OH-
Hydroxo
Cynide ion
CN-
Cyano
Ethane- 1,2- diammine
EN
Ethane- 1,2- diammine
EDTA
EDTA EDTA
understand simple ligand exchange processes
the concepts of deprotonation and ligand exchange should be applied to these reactions
recall colours of the OH ppts and complexes on addition of excess alkali or ammonia solution, limited to Cr3+, Mn2+, Fe2+, Fe3+,
Co2+, Ni2+, Cu2+, Zn2+
Ligand exchange- This involves the replacement of one ligand by another
[ Cu(H2O)4 ]2+(aq) + 4Cl- <----> [ CuCl4 ]2-(aq) + 4H20(l)
blue
green
colour changes as the different ligands create a different amount of splitting
between d orbitals in the copper ion.
Deprotonation H+ removed by (NH3/OH-) from (H2O) ligands

[Cu(H2O)6]2+ + OH-
[Cu(OH)(H2O)5]+ +
H2O
[Ni(H2O)6]2+ + 2OH
Zn(OH)2(H2O)2(s) + 2OH-(aq) 
Ni(OH)2 + 2H2O
[Zn(OH)4]2-(aq) +
2H2O
Cr(OH)3(H2O)3(s) + 3OH-(aq) ---> [Cr(OH)6]3-(aq) +
3H2O
Deprotonation
with NaOH (aq)
Or ligand
exchange with
NH3 (aq)
Cr3+
Cr(OH)3(s)
grey green
Mn2+
Mn(OH)2(s)
white/brown
Fe3+
Fe(OH)2(s)
muddy/green
Fe3+
Fe(OH)3(s)
rust brown
Co2+
Co(OH)2(s)
light blue
Ni2+
Ni(OH)2(s)
green
Cu2+
Cu(OH)2
light
blue
Zn2+
Zn(OH)2
white
For Vanadium, ox state[+2, +3, +4, +5] in its compounds,
Recall colours of aqueous solutions,
And given E values, work out the reagents for the interconversion of the metal ions, oxo anions and oxo cations
Describe reactions for the interconversion of the oxidation states of vanadium in aqueous solution
Oxidation state
+5
+4
+3
+2
Colour in aqueous
solution
yellow
blue
green
violet
Ion
VO3- or
VO2+
V3+
V2+
Name
vanadate(V) or
dioxovanadium(V)ion
oxovanadium (IV) ion
vanadium (III)
vanadium (II)
VO2+
Vanadium oxidised from (+2 to +5) V2+ to VO2+, add acidified manganate (VII) to the vanadium (II) V 2+ -----> VO2+
Vanadium is reduced from (+5 to +2) VO2+ to V2+, add zinc to the dioxovanadium
VO2+ -----> V2+
Vanadium can be reduced from the +5 states right through to the +2 state by zinc.
recall that transition elements and their compounds are important catalysts in industrial processes, and that their catalytic activity
is often associated with the variable oxidation states of the elements
recall examples of catalytic action by vanadium, iron and nickel and/or their compounds.
2. (a) Complete the following electronic configurations.
Ni
Ni2+
(b) Explain why nickel is classified as a transition metal.
Forms ions which have partially filled d-orbitals
(c) Consider the following reaction scheme. (i) What types of bond are present in the [Ni(H2O) 6] 2+ ion?
Dative covalent/co-ordinate
(ii) Write an equation for the formation of the pale green precipitate.
OR
OR
OR
(iii) Explain why Step 1 is a deprotonation reaction.
NH3 OR OH-)  from (H2O) ligands
 H+ removed (by
(iv) Name the type of reaction occurring in Step 2.
exchange/replacement/substitution
Ligand
(v) Give an equation for the reaction in Step 2.
OR
OR
 d-orbitals split (in energy) by
(d) Explain why the hexaaquanickel(II) ion, [Ni(H2O) 6] 2+ is coloured.
ligands  absorbs light (in visible region)
 electron moves to a higher
energy level
3. (a) The elements from scandium to zinc belong to the d-block. Some, but not all, of these elements are transition elements.
(i) What is meant by the term transition element?
Forms ions which have partially filled d-orbitals
(ii) Which of the elements, from scandium to zinc inclusive, are in the d-block but are not transition elements?
Scandium / Sc and Zinc / Zn
(b) (i) Complete the electronic configurations of the Fe 2+ and Mn2+ ions
2+
Fe [Ar] 3d
6
2+
Mn [Ar] 3d
5
(ii) Suggest why Fe2+ ions are readily oxidised to Fe3+ ions, but Mn2+ ions are not readily oxidised to Mn3+ ions.
3+
5
 Fe is 3d / half
filled d-subshell which is more stable than 3d
6
2+
 Mn is (already) 3d
5
4
(which is more stable than 3d )
(c) Draw a diagram to show the three-dimensional structure of the [Fe(CN)6]4– complex ion.
(d) A solution of potassium manganate(VII), KMnO 4, can be standardised by titration with arsenic(III) oxide, As 2O3. In this
reaction, 5 mol of arsenic(III) oxide are oxidised to arsenic(V) oxide, As 2O5, by 4 mol of manganate(VII) ions, MnO4– Calculate
the final oxidation number of the manganese
 Two As atoms oxidised from
−
+3 to +5 per mole of As O (loss of 4e )
2
3
 ∴ if 5 moles oxidised, total
−
20e lost / change in oxidation no. = 20
−
-
 ∴ 4 moles MnO reduced, total 20e gained / change in oxidation no. 20 ∴ each
4
−
Mn(VII) gains 5e /change in oxidation no. 5
 ∴ Mn(II) / Mn2+
(e) Ammonium vanadate(V), NH4VO3, reacts with dilute sulphuric acid to form a solution containing yellow VO2 + ions.
(i) Write an ionic equation for the reaction of the anion in NH4VO3 with dilute sulphuric acid.
VO3- + 2H+/2H3O+  VO2+ + H2O/3H2O
(ii) Is the reaction in (i) a redox reaction?
VO2+ / Oxidation no. of V unchanged (at +5)
No because oxidation no. of V is +5 in
(iii) Addition of zinc to the solution containing VO2+ ions causes the colour to change from yellow to green then to blue, followed
by green again and finally violet. State the formulae of the ions responsible for each of these colours.
The first green colour
The second green colour
The violet colour
+
First green colour : VO and VO
2
2+
Violet colour : V / [V(H O) ]
2
6
2+
2+
3+
Second green colour : V / [V(H O) ]
2
6
3+