Transition Elements

Chapter 13: Transition elements
13.1 Physical Properties Of First Row Transition Elements
(A) Introduction of transition elements
1. d-block elements are elements with incompletely/ partially filled d-orbital.
2. In Period 4(First row of transition elements), there are 10 d-block elements.

d-block elements Symbol spdf notation Electronic configurations
Scandium Sc [Ar] 3d1 4s2 Orbital Diagram (valence electronic configuration)
Titanium Ti [Ar] 3d2 4s2
Vanadium V [Ar] 3d3 4s2 3d 4s
Chromium Cr [Ar] 3d5 4s1 3d 4s
Manganese Mn [Ar] 3d5 4s2 3d 4s
Iron Fe [Ar] 3d6 4s2 3d 4s
Cobalt Co [Ar] 3d7 4s2 3d 4s
Nickel Ni [Ar] 3d8 4s2 3d 4s
Copper Cu [Ar] 3d10 4s1
Zinc Zn [Ar] 3d10 4s2 3d 4s
3d 4s
3d 4s
3d 4s

3. Definition of transition element:
Transition elements are d-block elements which can form at least one simple ion with partially

filled d orbitals.

4. Thus, scandium (Sc) and Zinc (Zn) are not transition element because both scandium and zinc can
only form stable ion with only 1 oxidation state.
(a)Sc can only form Sc3+ ion with electron configuration of 1s22s22p63s23p6.
 It has no electron in the 3d orbitals.
(b)Zn can only form Zn2+ ion with electron configuration of 1s22s22p63s23p63d10.
 All the d-orbitals are completely filled.
[Note: Sc and Zn are d-block elements due to incomplete filled d-orbital/s but are not transition

element]

5. For Period 4 elements, 4s orbital is being filled first before the 3d orbital. This is because
 the empty 4s orbital has lower energy than 3d orbitals.
 Thus, according to Aufbau’s principle, the electrons must fill in 4s orbital first (lowest energy orbital ).
 However, once the 3d orbital is filled with electron, 3d orbital will have lower energy than 4s orbitals.

6. Electronic configuration of the first row of d-block elements:

Transition d-block elements Symbol Electronic configuration
elements Scandium 21Sc 1s22s22p63s23p63d14s2
Titanium 22Ti 1s22s22p63s23p63d24s2
Vanadium 23V 1s22s22p63s23p63d34s2
Chromium 24Cr 1s22s22p63s23p63d54s1
Manganese 25Mn 1s22s22p63s23p63d54s2
Iron 26Fe 1s22s22p63s23p63d64s2
Cobalt 27Co 1s22s22p63s23p63d74s2
Nickel 28Ni 1s22s22p63s23p63d84s2
Copper 29Cu 1s22s22p63s23p63d104s1
Zinc 30Zn 1s22s22p63s23p63d104s2

Questions:

1. Chromium and copper are elements in d-block of the periodic table.

(a) Write the electronic configuration of chromium and copper respectively.
Chromium : 1s22s22p63s23p63d54s1
Copper : 1s22s22p63s23p63d104s1

(b) Explain the anamolous electronic configuration in chromium and copper.
Electronic configuration of Cr is 1s22s22p63s23p63d54s1 but not according to Aufbau’s principle
(1s22s22p63s23p63d44s2). This is because the half-filled 3d orbital is more stable than partially filled
3d orbital.

Electronic configuration of Cu is 1s22s22p63s23p63d104s1 (actual electronic configuration) but not
according to Aufbau’s principle (1s22s22p63s23p63d94s2). This is because the completely filled 3d
orbital is more stable than partially filled 3d orbital.

(B) Physical properties of first row transition elements
1. All transition elements shows similarities in their physical properties (atomic radius, ionic radius
and first ionization energy).

2. Atomic radius

d-block 21Sc 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu 30Zn
elements

Atomic radius/ 162 147 134 130 135 126 125 124 128 137
pm

(a) All transition metal is smaller than the s-block metals within the same period.
 Sc to Zn is smaller than K and Ca in Period 4.

(b) Atomic radius for the first row transition elements does not change much / are about the same.

This is because from Sc to Zn,
 Nuclear charge increases from Sc to Zn,
 As the number of electrons increases from Sc to Zn, the addition electron is filled the inner 3d

orbital The added electrons shield the outer electrons from the nucleus. Thus, the screening
effect increases, attraction… weaker , size becomes bigger.
 The increase in the screening effect is cancel out by the increase in the nuclear charge. Thus,

the effective nuclear charge remain almost constant.
 Hence, the atomic size have only slightly change.

3. Melting point and boiling point

Transition elements 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu

(Transition metals) 1680 1900 1890 1240 1540 1500 1450 1080
Melting Point/ oC 3260 3400 2480 2100 3000 2900 2730 2600
Boiling Point/ oC

Transition elements have very high melting point and boiling point (even higher then Group 1 and Group
2 metals). This is because
(a)transition elements are metal (giant metallic lattice) with strong metallic bond
(b)Since the energy difference between 3d and 4s subshell are small, the valence electrons in the outer 4s

orbital and inner 3d orbital can contribute to the “sea of delocalized electrons”.
The large number of delocalized electron will accounts for stronger metallic bond for the transition
metals compared to the main group metal (Group 1 and Group 2).
Atomic size of transition metals are smaller than main group metal.

(c) Compared to 19K and 20Ca, K has only 1 valence electron per atom is delocalized to form metallic bond;
while only 2 delocalised in Ca. Thus, K and Ca have relatively lower boiling point compared to the first
row of transition elements.

(d)Melting point / boiling point of Mn and Zn is lower due to extra stability of half-filled 3d orbitals for
Mn and completely filled 3d orbital of Zn respectively. These electrons in the d-orbitals are less
available to contribute to the sea of delocalized electrons.

4. Ionisation energy (IE)

Transition elements 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu
1st I.E 658 650 653 717 759 758 737 745
2nd IE 1310 1414 1592 1509 1561 1646 1753 1958
3rd IE 2652 2828 2987 3248 2957 3232 3393 3554

(a) First Ionisation energy
(i) From Ti to Cu, first ionization energy increase slightly only.
(ii) This is because the first electron is removed from the outer 4s orbital at which the electron in 4s
orbital are shield by the inner electrons.
(iii) However, the first ionization energy have slightly increase from Ti to Cu. This is due to the
effective nuclear charge are remain almost constant (increase in nuclear charge is cancel out by
the increase in screening effect).

(b) Second Ionisation energy
(i) Successive ionization energy is increases (2nd IE is higher than 1st IE; 3rd IE is higher than 2nd
IE…) because the attraction between the nucleus to the remaining electrons becomes stronger.

More energy is needed to remove the successive electrons.

(ii) Generally, second ionization energy from the transition elements increase slightly because the
second electron is removed from the 4s orbitals. (except Cr+ and Cu+ ion).

(iii) Second ionization energy for Chromium is higher than expected.
This is because the second electron is removed from half-filled 3d orbital in Cr+ ion, which is

more stable. More energy is required to remove the electron.

Cr+ : 1s22s22p63s23p63d5

3d

(iv) Second ionization energy for copper is higher than expected.
This is because the second electron is removed from completely filled 3d orbital in Cu+ ion,

which is more stable. More energy is required to remove the electron.

Cu+ : 1s22s22p63s23p63d10

3d

(c) Third Ionisation energy

(i) Third ionization energy for manganese, Mn is higher than expected.
This is because the third electron in Mn is removed from half-filled 3d orbital in Mn2+ ion,

which is more stable. More energy is required to remove the electron.

Mn2+: 1s22s22p63s23p63d5

3d
(ii) Third ionization energy for iron, Fe is lower than expected.

This is because the third electron in Fe is removed from partially-filled 3d orbital. The paired
electrons in the same orbital in Fe2+ ion account for a greater repulsion.
Thus, less energy is needed to remove one electron from the partially 3d orbital in Fe2+ ion to
produce a more stable Fe3+ ion with half-filled 3d orbital.

Fe2+  Fe3+ + e

3d 3d
[compare between transition elements with Ca]

5. Contrast qualitatively the atomic radius, ionic radius, melting point, density, first ionisation energy and

conductivity of the first row transition elements with those of calcium as a typical s-block element

(a) Atomic radius / Ionic radius

The atomic radius of all transition elements is smaller than Ca. (Ca is bigger than transition

elements)

 This is because the electrons for the transition elements are added to the inner 3d sub-shells,

thus increase the screening effect.

 However, the additional 3d electron does not cancel out completely the increase in nuclear

charge. Thus the effective nuclear charge increase slightly. Attraction between nucleus to the

outer electron becomes stronger. Hence, transition element is smaller than Ca.

(b) Melting point
Melting point of transition element is higher than Ca.
strengthof metallicbond  number of valence electrons
Metallicradius

 Ca and transition elements/ transition metals are metal with strong metalling bond.
 Transition elements make use of the valence electrons in 3d and 4s orbitals to form metallic

bonds (due to the energy difference between the 3d and 4s orbitals are small).
 In Ca, only the electrons from 4s orbitals are delocalized to form metallic bond. (due to the

energy difference between the 4s and 3p orbitals are big).
 Since transition elements has more delocalized valence electrons than Ca,

and the atomic size of the metal (transition elements) is smaller than Calcium.
Strength of metallic bond formed in transition elements are stronger than in Ca.
Thus, transition elements have higher melting point than Ca.

(c) Density
Density of transition elements are higher than Ca. (transition elements are denser).
density, ρ  mass
volume
 Size the atomic size of transition elements are smaller than Ca atom, [Volume is smaller]
The molar mass of transition elements are higher than Ca atom.
Hence, Transition elements are denser than Ca.

(d) First ionization energy
Transition elements have higher first ionization energy than Ca.
 This is because transition elements have smaller atomic size and higher nuclear charge,
 Attraction between nucleus to the outer electron in transition elements are stronger.
 More energy is required to remove the electron in transition elements.
 Thus, transition elements has higher first ionization energy than Ca.

(e) Electrical Conductivity
Electrical conductivity of transition elements are higher than Ca.
 Both Ca and transition elements are metal with strong metallic bond.
 Transition elements make use of the valence electrons in 3d and 4s orbitals to form metallic
bonds (due to the energy difference between the 3d and 4s orbitals are small).
 In Ca, only 2 electrons per atom from the 4s orbital are used to form metallic bond.
 Thus, transition elements can contribute more delocalized electron to form metallic bond than
Ca.
 Hence, transition elements is a better electrical conductor than Ca.

13.2 Chemical properties of first row transition elements

Some of the special characteristics of transition elements compared to maun group elements:
(a)Exhibit more than one oxidation states in their compounds
(b)Form colourful ions and complex
(c)Form complex molecules or ions.
(d)Act as catalyst for chemical reactions.

13.2.1 Formation Of Variable Oxidation States

1. Transition elements can form ions with different oxidation states because the energy difference
between the inner 3d and outer 4s orbital are very small.

 Thus, both electrons in 4s orbitals and 3d orbitals can be used for chemical reaction.

2. (a) Oxidation states of the first row of transition elements.
Ti4+: [Ne]3s23p6

Sc Ti V Cr Mn Fe Co Ni Cu Zn

+1,
+2,Ti2+ +2 +2 +2 +2 +2 +2 +2, +2
+3, Sc3+ +3, Ti3+ +3 +3 +3 +3 +3 +3 +3
+4, Ti4+ +4 +4 +4 +4 +4 +4

+5 +5 +5 +5 +5

+6 +6 +6

+7

[Note: Coloured oxidation states are the most stable oxidation number]

(b) Example of some common complex and its colours:

Element Complex Stable ion Oxidation state of Colour
transition metal ion
Scandium [Sc(H2O)6]3+ @ Sc3+ Colourless
Titanium Sc3+(aq) +3
Vanadium T3+ Violet
[Ti(H2O)6]3+ @ +3
Chromium Ti3+(aq) V2+ Violet
V3+ +2 Green
Manganese [V(H2O)6]2+ Cr2+ +3 Blue
[V(H2O)6]3+ Cr3+ +2 Green
[Cr(H2O)6]2+ +3 Bright Green
[Cr(H2O)6]3+ - +3 Light green
[Cr(OH)6]3 - +3 Dark green
[Cr(H2O)5Cl]2+ - +3 Orange
[Cr(H2O)4Cl2]2+ - +6 yellow
- +6 Light pink
Cr2O72- Mn2+ +2 green
CrO42- - +6 purple
[Mn(H2O)6]2+ - +7
MnO42-
MnO4-

Iron [Fe(H2O)6]2+ Fe2+ +2 Green
[Fe(H2O)5(SCN)]2+ - +2 Red blood
Cobalt - +2 Yellowish green
Nickel [Fe(CN)6]4- - +3 Yellow
Copper [Fe(H2O)6]3+ - +3 Yellowish green
Zinc [Fe(CN)6]3- - +3 Brown
[Fe(H2O)5NO]2+ - +6 Red
+2 pink
[FeO4]2- Co2+ +2 yellow
[Co(H2O)6]2+ - +2 Blue
[Co(NH3)6]2+ - +3 Brown
+2 Light Green
[CoCl4]2- Co3+ +2 Light blue
[Co(H2O)6]3+ Ni2+ +2 Green
[Ni(H2O)6]2+ +2 Blue
[Ni(NH3)6]2+ - +2 Blue
[Ni(CN)4]2- - +2 Dark blue
[Ni(EDTA)4]2- - +2 Green
[Cu(H2O)6]2+ Cu2+ +2 Light blue
[Cu(NH3)4]2+ - +2 Colourless
-
[CoCl4]2- -
[Cu(EDTA)4]2- Zn2+
[Zn(H2O)6]2+

(c) Principal oxidation numbers of the elements in their common oxides, cations and Oxo ions.
(i) Table below shows the stable oxides of transition elements.

Oxidation Ti V Cr Mn Fe Co Ni Cu
state
+1 V2O3 Cu2O
V2O5 Copper(I)
+2 MnO FeO CoO
oxide
+3 Ti2O3 Cr2O3 Mn2O3 Fe2O3 Co2O3 NiO CuO
+4 TiO2 CrO3 MnO2 K2FeO4
+5 (copper(II)
+6 Mn2O7 oxide)
+7
NiO2

(ii)Table below shows the common aqueous cations and oxo ions of transition elements.

Oxidation Ti V Cr Mn Fe Co Ni Cu

Exist as state Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu+
simple +1 Ti3+ V3+ Cr3+ Fe3+ Co3+ Cu2+
Aqueous +2
+3

cations

+4 TiO2+ VO2+

+5 VO2+
VO3
Exist as
CrO42 MnO42 FeO42
Oxo +6 Cr2O72 Fe6+
ions
Cr6+

+7 MnO4

3. Transition metal ions with low oxidation state (+1, +2 and +3) can exist as simple aqueous cation.
Examples, [V(H2O)6]2+ , [Ni(H2O)6]2+, [Fe(H2O)6]3+
V2+(aq) , Ni2+(aq) , Fe3+(aq)

4. However, simple aqueous cations is not stable to exist for high oxidation state (+4, +5, +6 and +7).

Instead, it exist in the form of oxo ions. (oxo cations or oxo anions).

Examples, [V(H2O)4O2]2+ or VO22+(aq) ; Cr2O72aq)MnO4aq)FeO42aq)

Cr6+ Mn6+

 This is due to the transition metal ion with high oxidation state has high charge density.

 The high charge density cations will polarized the water molecules that are bonded to it and

weakens the OH bonds of the water molecules.

 Thus, H2O molecule in the simple aqueous ion will loss proton and decompose to oxo ions.

HH O2-
OH
H2O 2-
H2O O M5+
O + 4H+
M5+ H OH2
H2O
H2O OH2 H2O
H2O

Simple ion Oxo ions
[M(H2O)4O2]+
MO2+

Question:

Explain why [V(H2O)6]2+ is stable to exist while [V(H2O)6]5+ is unstable and only exist as oxo cations,
VO2+ @ [V(H2O)4O2]+.

Answer:

In [V(H2O)6]2+, oxidation number of Vanadium is +2. Thus, simple aqueous cations [V(H2O)6]2+ is stable

to exist.

H2O

H2O OH2
V2+

H2O OH2
H2O

In [V(H2O)6]5+, oxidation number of Vanadium is +5. Thus, [V(H2O)6]5+ is not stable to exist. This is

because
V5+ ion has high charge density. The electron density in the water ligand moves closer to the V5+ ion. V5+

ion will polarized the water molecules and weaken the OH bond in the H2O ligands / molecules.
Thus, H+ ion in the water molecule will be eliminated to form [V(H2O)4O2]2+ or VO2+ , which is more

stable.

HH O2-
OH
H2O 2-
H2O O V5+
V5+ H O + 4H+
OH2
H2O OH2 H2O
H2O H2O

Simple ion oxo ions

Exercise:
State the principal oxidation numbers of these elements in their common cations, oxides and oxo ions.

No. Formula of oxides/ Oxidaiton state of transition elements Formula of the
cation / oxo ions 2(Ti) + 3(2) = 0 transition metal ions
Ti = +3
1 Ti2O3 Ti3+
1(V) + 6() = +3
2 TiO2 V = +3 V3+
3 TiCl3
4 TiO2+
5 [Ti(H2O)6]2+
6 V2O3
7 V2O5
8 VO2+
9 VO2+
10 VO3
11 [V(H2O)6]3+

12 [V(H2O)6]5+
13 [VO2(H2O)4]+
14 Cr2O3

15 CrO3 1(Mn) + 4() = 1 Mn7+
16 Cr2O3 Mn = +7
17 CrO4
18 Cr2O7
19 [Cr(H2O)6]3+
20 [Cr(NH3)4Cl2]+
21 MnO

22 Mn2O3
23 MnO2
24 Mn2O7
25 MnO4
26 MnO4

27 FeO

28 Fe2O3
29 K2FeO4
30 FeO4
31 [Fe(CN)6]3
32 CoO

33 Co2O3
34 [Co(H2O)6]2+
35 [Co(NH3)6]3
36 NiO

37 NiO2
38 [Ni(NH3)6]2+
39 Cu2O
40 CuO

41 K3CuF6

(j) Relative stabilities of oxidation states (+2 and + 3 oxidation states)
(a) The most common oxidation state of transition elements are +2 and +3. However, the stability of the
high oxidation state (+3) is decrease for the elements Mn, Co, Ni and Cu.

Questions:

1. Explain why the high oxidation states become progressively less stable for transition elements on the
right of the first row of transition elements.
Answer:
Across Period from Ti to Cu, stabilities of +3 oxidation state decreases. (+2 oxidation state more
favorable)
Ti, V, Cr and Fe has a stable oxidation state of +3 but Mn, Co, Ni and Cu has stable oxidation state of +2.
This is because from Ti to Cu, nuclear charge increases and the atomic size decreases.
More energy is required to remove the third electron from the respective +2 ion.
Thus, ion with oxidation state of +3 is more difficult to form. ( more difficult to remove the third
electron from the d-orbitals)

Fe form an ion with oxidation state of +3 (Fe3+) is more stable than oxidation state of +2 (Fe2+).
This is because the half- filled 3d orbitals/subshells in Fe3+ ion is more stable than partially filled 3d
orbital in Fe2+ ion.

(k)Relative stabilities of oxidation states

(a)Relative stabilities of the oxidation state of transition elements (+2 and +3) can be explain by using
the standard reduction potential, Eoredcution.

System Cr2+ Eo / V More –ve Eo
Cr3+ + e Ti2+ 0.41 reduciton,
Ti3+ + e V2+ 0.37 oxidation
V3+ + e Fe2+ 0.24
Fe3+ + e Mn2+ 0.77 reverse rxn is
Mn3+ + e Co2+ 1.51
Co3+ + e 1.81 more favorable
+3 oxidation state

is more stable to

form.

(b). Negative Eoredcution
Positive Eoredcution  Oxidation reaction is more favorable (reverse

 reduction reaction is more favorable (forward reaction)
reaction)  +3 oxidation state is more stable

 +2 oxidation state is more stable

(c) Examples: [2]
(i) Describe the relative stability of Ti2+ and Ti3+ ion.

Answer:

Since the standard reduction potential is negative, reverse reaction (oxidation reaction) is more

favorable.
Thus, Ti3+ is more stable.

(ii) Describe the relative stability of Mn2+ and Mn3+ ion. [2]

Answer:

Since the standard reduction potential is positive, forward reaction (reduction reaction) is more

favorable.
Thus, Mn2+ is more stable.

(l) Uses of standard reduction potentials in predicting the relative stabilities of aqueous ions

Relative stabilities of the oxidation state of transition elements (+2 and +3) can be predicted by
comparing the standard reduction potential, Eoredcution of the transition elements with oxygen system.

System Cr2+ Eo / V
Cr3+ + e Ti2+ 0.41
V2+ 0.37
Ti3+ + e Fe2+ 0.24
V3+ + e Mn2+ 0.77
Fe3+ + e Co2+ 1.51
Mn3+ + e 1.81
Co3+ + e 2H2O 

O2 + 4H+ + 4e

(a) Examples:
(i) Compare the stability of Ti2+ and Ti3+ ion in aqueous solution. Explain your answer.

Answer: Ti2+ ; Eo = 0.37 V
Ti3+ + e

O2 + 4H+ + 4e 2H2O ; Eo = 1.23 V

Since the standard reduction potential of Ti3+|Ti2+ half-cell is more negative than oxygen system,

reverse reaction is more favorable (oxidation reaction).
Thus, Ti3+ ion is more stable than Ti2+ ion in aqueous solution.

(ii) Compare the stability of Fe2+ and Fe3+ ion in aqueous solution. Explain your answer.

Answer: Fe2+ ; Eo =  V
Fe3+ + e

O2 + 4H+ + 4e 2H2O ; Eo = 1.23 V

Since the standard reduction potential of Fe3+|Fe2+ half-cell is more negative than oxygen system,

reverse reaction is more favorable (oxidation reaction).
Thus, Fe3+ ion is more stable than Fe2+ ion in aqueous solution.

(iii) Compare the stability of Mn2+ and Mn3+ ion in aqueous solution. Explain your answer.

Answer:

Mn3+ + e Mn2+ ; Eo =  V

O2 + 4H+ + 4e 2H2O ; Eo = 1.23 V

Since the standard reduction potential of Mn3+|Mn2+ half-cell is more positive than oxygen

system,

forward reaction is more favorable (reduction reaction).
Thus, Mn2+ ion is more stable than Mn3+ ion in aqueous solution.

13.2.2 Nomenclature and bonding of complexes

Naming of complex ions / compounds
1. In naming of the complex,

name the ligands first, in alphabetical order, and then
name the metal atom or ion.
[Note: The metal atom or ion is written before the ligands in the chemical formula.]

Naming of Transition metal (a) If the complex is cation or neutral, the normal name of metal is used;

ion  Calculate the oxidation number of the transition metal,

1. Identify whether the  Write the oxidation number in Romanic numeral, enclosed with

complex ion is cation or bracket after the name of metal ion.

anion [Note: must be written in a word – no space in between]

Examples:

Complex  [Co(H2O)6]2+ Complex [Cr(H2O)4Cl2]Cl

Cobalt(II) Chromium(II)
Hexaaquacobalt(II) ion
hexaaquacobalt(II)

(b) If the complex ion is anion the suffix “-ate” is added to the name of

metal;

 Calculate the oxidation number of the transition metal,

 Write the oxidation number in Romanic numeral, enclosed with

bracket after the name of metal ion.
[Note: must be written in a word – no space in between]

 Change the name of the transition metal / metal as follow:

Metal Cation / Anion name
neutral

Titanium Titanium( ) Titanate

Vanadium Vanadium( ) Vanadate

Chromium Chromium( ) Chromate

Manganese Manganate

Iron iron Ferrate

Cobalt Cobalt(III) cobaltate

Nickel Nickel(III) Nickelate

Copper Cuprate

Zinc zincate

Aluminium Aluminate

Tin Stannate

Lead Plumbate

Silver Argentate

Examples: Complex  [CuCl4]2
Complex  [Fe(CN)6]4

ferrate(II) cuprate(II)

2. Ligands can be classified according to the number of lone pair electrons that can be donated to the central
metal ion. Cl- H2O
(a) Monodentate ligand : has 1 donor atom and forms only 1 dative bond with the central metal ion.

(b) Bidentate ligand : has 2 donor atoms and forms 2 dative bonds with the central metal ion.

(c) Hexadentate ligand : has only 6 donor atoms and forms 6 dative bond with the central metal ion.

Naming (a) Name of ligand
of
Classification Molecules / ions formula Name as ligand Charge
Ligands Monodentate water aqua
H2O 0
Classification Ammonia NH3 ammine 0
Bidentate Carbon monoxide CO carbonyl 0
Hexadentate Br  bromo 1
Bromide ion Cl  1
Chloride ion F chloro 1
Fluoride ion OH  fluoro 1
Hydroxide ion CN hydroxo 1
Cyanide ion SCN cyano 1
Thiocyanate O2 thiocyanato 

Oxide ion  oxo

Molecules / ions formula Name as ligand Charge
Oxalate ion C2O42 Oxalato /
(ethyldiacetate / (ethylenedizacetato 2
ethanedioate) OO / ethanedioato)

CC

OO

Ethylenediamine (en) NH2CH2CH2NH2 ethylenediamine 
diethyltriamine

Ethylenediaminetetraacetate
ion (EDTA4-) 

CH2 CH2 CH2 CH2

H2N NH NH2





O  4-
OC O
CH2 CH2  CH2 C O
OC N CH2 N O
O  CH2 C
CH2

O



Name as ligand: Ethylenediaminetetraacetato
EDTA4-

(b) If 2 or more similar ligands surrounding the center metal ion, a prefix is placed in front of

the name of the ligand.

Ni+3(-2) = -4 23456
di tri tetra penta hexa

Ni= + 2 If the name of ligand already have a prefix in its molecule, another special prefix is used to
replace di, tri, tetra…, enclosed with bracket.

23 4

bis tris terakis

Examples:
 Complex: [Ni(C2O4)3]4: tris(ethanedioato)nickelate(II)

 Trioxalatonickelate(II)

Exercises: Name
1. Name the following complexes. Hexaaquavanadium(III) ion

No. Formulae Pentaaquathiocyanatoiron(III) ion
a [V(H2O)6]3+
Dichlorocuperate(I) ion
b [Fe(H2O)5SCN]2+

c [CuCl2]

d Ni(CO)4 Tetracarbonylnickel(0)
e [V(H2O)4O2]3+ Tetraaquadioxovanadium(VII) ion

f Ni(H2NCH2CH2NH2)2Cl2 dichlorobis(ethylenediamine) nickel(II) chloride

g [Fe(CN)6]4 Hexacyanoferrate(II) ion

h [Cr(OH)4(H2O)2] Diaquatetrahydroxochromate(III) ion

i [Co(C2O4)3]3 Trioxalatocobaltate(III) ion @ tris(ethandioate)cobaltate(III)

ion

j [CoCl(NH3)5]SO4 Pentaamminechlorocobalt(III) sulphate

[CoCl(NH3)5]2+ SO42-

k Na[Pt(NH3)Cl3] Sodium amminetrichloroplatinate(II)
[Pt(NH3)Cl3] amminetrichloroplatinate(II) ion

2. Write the formulae of the complexes shown in the table below.

No. Formulae Name
a [CoCl4]2 Tetrachlorocobaltate(II) ion

b [Ag(NH3)2]2 Diamminesilver(I) ion
c [CoCl4]2 Tetrachlorocuprate(II) ion

d K3Fe(CN)6 Potassium hexacyanoferrate(III)
e [PtCl2(NH3)4]2+ Tetraamminedichloroplatinum(IV) ion
f [CrCl2(NH3)4]2+ Diamminetetrathiocyanatochromate(III) ion

g [Pt(NH3)6]Cl4 Hexaammineplatinum(IV) chloride
h [Co(H2NCH2CH2NH2)3]3+ @ Tris(ethylenediamine)cobalt(III)ion

[Co(NH2CH2CH2NH2)3]3+ Diaquadioxalatonickelate(II) ion
i [Ni(C2O4)2(H2O)2]2

j Fe(CO)5 Pentacarbonyliron(0)

13.2.3 Bond formation between ligands and the central metal atom/ ion.
(A) Complex ion formation

1. Complex ion is an ion formed when a central metal ion or atom is bonded to a group of ions or
molecules (which called ligands) through dative bond.
Thus, the compounds formed is known as coordination compounds.

2. Transition metal cations has very high tendency to form complex ion than the metal ion (Group ,Group 2

and Group13). This is due to transition metal ion

a) High charge and small size (high charge density) of the transition metal ion
b) Contain empty orbitals in the ions (3d, 4s, 4p).

(Not only d-orbital)

3. In the formation of the complex ion,
(a) the central metal ion / atom has empty orbitals in the valence shell to accept lone pair electrons
from ligands.
Hence, central metal ion is an electron pair acceptor, which known as Lewis acid.
(b) Ligand donate electron (lone pair electron) to the central metal ion to form dative bond.
Hence, Ligand is an electron pair donor, which known as Lewis base.

4. List of complex ion and its ligands

Complex ion State the Ligand Name the ligand Name the complex
[Fe(CN)6]3 CN Cyano (X) Hexacyanoferrate(III)
Cyanide ion
[Co(NH3)6]2+ NH3 Hexaamminecobalt(II)
[AlF6]3 F ammonia Hexafluoroaluminate(III)

[Cu(NH3)4]2+ NH3 Fluoride ion
[CoCl4]2 Cl
Water
[Fe(H2O)6]2+ H2O Thiocyanide ion
[Fe(H2O)5(SCN)]2+ H2O ; SCN

5. Examples of formation of coordination complex:

Central metal in / atom Ligands Coordination complex
(Lewis Acid) (complex ion or molecule)

Fe3+ + 6 CN [Fe(CN)6]3

Cu2+ + 4 NH3 [Cu(NH3)4]2+

Ni + 4 CO Ni(CO)4

6. Examples of formation of complex ions/ compounds:

(a)Formation of complex ion, hexaaquairon(III) ion, [Fe(H2O)6]3
Electronic configuration of Fe3+ ion is 1s22s22p63s23p63d5

Fe3+

3d 4s

Due to the extra stability of half-filled 3d orbital, 3d orbitals are not involved in the formation of dative

bond between Fe3+ with H2O.

Fe3+ make use of the empty 4s, 4p and 4d orbitals to form bond with H2O (accept lone pair

electron from H2O molecules). H2O is a weak ligand, does not

undergoes electron pairing.
One 4s, three 4p and two 4d orbitals are hybridized to form six sp3d2 hybridised orbitals.

Fe3+

3d 4s 4p 4d

Hybridised

Fe3+

sp3d2

Each of the sp3d2 hybridised orbital of Fe3+ ion will accept one lone pair electron from H2O.

H2O 3+ Fe + 6(0) = +3
H2O Fe = +3
OH2
Fe Name:

H2OH2O OH2 hexaaquairon(III)

(b) Formation of hexacyanoferrate(II) ion, [Fe(CN)6]4

Fe2+ + 6CN [Fe(CN)6]4

Valence Electronic configuration of Fe2+ ion:

Fe2+ : Fe
24+s 4p
3d

To form [Fe(CN)6]4 ion:
Step 1: six electrons in the Fe2+ ion undergoes pairing. (electron pairing). Because CN is a

strong ligand.

Fe2+ :

3d 4s 4p

Step 2: Fe2+ ion used two empty 3d, one empty 4s and 3 empty4p orbitals to form 6 coordinate
forms with 6 CN ion.

[Fe(CN)6]4

3d 4s 4p

form 6 coordinate with 6 CN ion.

Six sp3d2

[Note: because CN ion is a strong field ligand, thus, the electron in 3d orbitals are paired up.]
Draw the shape of [Fe(CN)6]4 ion:

Type of hybridization: sp3d2
Name of the shape: octahedral
Coordination number: 6
(depends on how many donor atoms)

(c) Formation of tetrachlorocobaltate(II) ion, [CoCl4]2

Co2+ + 4Cl [CoCl4]2

Valence Electronic configuration of Co2+ ion:

Co2+ 3d 4s 4p
:

To form [CoCl4]2ion:
Co2+ ion used one empty 4s and three empty 4p orbitals to form 4 coordinate forms with 4 Cl

ion.

[Co(Cl)4]2

3d 4s 4p

form 4 coordinate with 4 Cl ion.

[Note: because Cl ion is a weak field ligand, thus, the electron in 3d orbitals are not paired up.]
Draw the shape of [CoCl4]2ion:

Name of the shape: square planar
Coordination number: 4

(B) Geometry of complex and type of ligands

1. Geometry of complex

(a) Coordination number is the number of ligand atoms that are bonded directly to the central

metal ion in a complex ion. // number of coordinate bonds from ligands to the central metal

ion.

(b) Geometry of the complex ion depends on the coordination number and the nature of metal ion.

(c) Examples:

Coordination number Shape Have 2 ligands bonded directly to
2 Linear central metal ions

4 Square planar Have 4 ligands bonded directly to

central metal ions
Most of the d8 ions

Tetrahedral Have 4 ligands bonded directly to

central metal ions
Most of the d10 ions

6 Octahedral Have 6 monodentate ligands bonded

directly to central metal ions

Or
Have 3 bidentate ligands bonded
directly to central metal ions
Or
Have 1 hexadentate ligand bonded
directly to central metal atom

(d) Exercises

Complex ions Coordination Name Shape
number Linear Draw

[CuCl2] 2

[Ag(NH3)2] 2 Linear

[Ni(CN)4]2 4 Square planar
[Cu(NH3)4]2 4 Square planar

Complex ions Coordination Name Shape
[Pt(NH3)4]2 number Square planar Draw
4

[Cu(CN)4]3 4 Tetrahedral

[Zn(NH3)4]2 4 Tetrahedral

[CdCl4]2 4 Tetrahedral
Octahedral
[Ti(H2O)6]3 6 Octahedral
Octahedral
[V(CN)6] 6 Octahedral

[Cr(NH3)4Cl2] 6

[FeCl6]3 6

2. Types of ligands

(a) Ligand have lone pair electrons that can be donated to a metal cation to form coordinate

bond.

(b) Thus, ligand can be classified into 3 categories, according to the number of lone pairs per

ligand molecule that can be donated to central metal cation:

Monodentate Ligand Bidentate Ligand Hexadentate Ligand

Ligand that Ligand that Ligand that
 has one donor atom  has two donor atoms  has six donor atoms

 form only one  form two coordinate  form six coordinate
coordinate bond per bonds per unit ligand bonds per unit ligand

unit ligand

(c) Monodentate Ligand

i. Examples of monodentate ligands:

Monodentate Ligand Name as ligand

H2O aqua
ammine
NH3 carbonyl
bromo
CO chloro
Br  fluoro
Cl  hydroxo
F cyano
OH  thiocyanato
CN
SCN oxo
O2

ii. Draw the following Complex [V(H2O)5O]2+
[V(H2O)6]2+ Coordination number:
Coordination number:

[Fe(H2O)5SCN]2+ Ni(CO)4
Coordination number: Coordination number:

[Cr(NH3)4Cl2]+ [Co(H2O)6]Cl2
Coordination number:

(d) Bidentate Ligand

i. Examples of Bidentate ligands:

Bidentate Ligand Name as ligand

C2O42 Oxalato /
O O (ethylenedizacetato
/ ethanedioato)


C C

O O

NH2CH2CH2NH2 ethylenediamine

H2N CH2CH2 NH2

ii. Draw the following Complex
[Ni(C2O4)3]2[Name : trioxalatonikelate(IV)]
Coordination number: 6
Reason: 3 bidentate ligand. Each bidentate ligand have 2 donor atoms. Hence, it can
form 6 coordinate bond with transition metal ion. ]

Tris(ethylenediamine)cobalt(III), [Co(H2NCH2CH2NH2)3]3+ @ [Co(en)3]3+
Coordination number:

(b) Hexadentate Ligand

i. Examples of Hexadentate ligands:

Hexadentate Ligand
Name of ligand: Ethylenediaminetetraacetate ion (EDTA4)

O 4-
OC
O
OC
O CH2 CH2 CH2 CH2 C O
N N O

CH2 CH2 C

O

ii. Draw the following Complex
[Ni(EDTA)]2
Coordination number: 6

13.2.4Formation of coloured ions / complexes

1. Transition elements can form ions with different colour is due to the presence of the partially filled 3d-
orbitals which allow for the d-d transition.

2. Transition metal ions with empty or completely filled d-orbitals are colourless. This is due to no d-d
transition can occur.

3. Colour formation for transition elements Co2+
(i) In free gaseous transition metal ions, all the five 3d-orbitals are degenerate orbitals.

Energy

dxy dxz dyz dx2-y2 dx2-y2

(ii) When ligands are bind with transition metal ions in a complex, the 3d orbitals are split into 2
groups of non-degenerate 3d-orbitals with different energies.

Energy

dx2-y2 dx2-y2

dxy dxz dyz dx2-y2 dx2-y2 ΔE

dxy dxz dyz

ΔE is corresponds to the visible region of the electromagnetic spectrum.

(iii) When exposed to light, electrons at the lower energy group of d-orbitals will absorbed certain wavelength
of the visible light and get promoted to higher energy d-orbitals, which known as d-d transition.

(iv) The colour of the complex is the colour complementary to the colour absorbed.
[Note: the colour that observed is the colour that is not absorbed during the d-d transition].

4. Sc3+ ions and Ti4+ ions does not contain electrons in d-orbitals. No d-d transition occurred. Hence, the
complexes of Sc3+ and Ti4+ are colourless.
Zn2+ ion and Cu+ ion has electronic configuration of [Ar] 3d10. Thus Zn2+ and Cu+ ion has completely
filled 3d-orbitals. No d-d transition occurred. Hence, the complexes of Zn2+ and Cu+ are colourless.

5. The colour of the transition metal complex is affected by

(i) Nature of the central metal ion
Different metallic ions have different number of electrons in the d-orbital. Thus, different

ion will have different colours.
Example: Ti3+(aq) is green while Fe3+(aq) is yellow. Fe2(SO4)3

Fe2+(aq) is pale green while Cu2+(aq) is blue.

FeSO4 CuSO4 ZnSO4

(ii) the oxidation state of the transition metal ion.

Central metal ion with difference oxidation states have different numbers of electrons that

can be promoted to higher energy orbitals.
Example: Fe2+(aq) is pale green while Fe3+(aq) is yellow.

Cr2+(aq) is pale blue while Cr3+(aq) is green

(iii) the type of ligand surrounding the metal ion (Ligand Exchange)
 The stronger the ligands that surrounding the metal ion, the greater the energy difference
between the splitting of high and low 3d orbitals, the higher energy is required to promote electrons
from lower energy 3d orbital to higher energy 3d orbital.
[Stronger ligand can displace a weaker ligand from a complex.
Thus a different complex with different colour is formed]

Example: [Cu(NH3)4]2+(aq) + 2H2O (l)
[Cu(H2O)6]2+(aq) + 4 NH3(aq) deep blue

Blue

[Cu(NH3)4]2+(aq) + EDTA4-(aq) [Cu(EDTA)]2(aq) + 4 NH3(aq)
deep blue pale blue

6. Complex ions and Ligand Exchange

(a) Complex ion is an ion formed when a central metal ion is bonded to a group of ions or molecules

(Ligand) by dative bond (coordinate bond). L

L Mn+ L

L

(b) A ligand is an ion or a molecule that has lone-pair electrons that can be donated to a metal central ion

to form coordinate bonds.

 Ligands shared a pair of its electrons to the central metallic ion.

 Thus, Ligands act as Lewis Base ; Metal cation act as Lewis Acid.

(lone pair electron donor) (lone pair electron acceptor)

(c) Example of ligands:

Ligand Name of Ligand in coordination compounds / Complexes
Water, H2O aqua

Ammonia, NH3 ammine

Carbon monoxide, CO carbonyl

ethylenediamine (en) ethylenediamine

H2NCH2CH2NH2 cyano
Cyanide ion, CN thiocyanato
Thiocyanate ion, SCN

Ethanedioate ions @ oxalato
Oxalate ion, C2O42
EDTA4 ethylenediaminetetraethanoate
Fluoride ion, F fluoro
Chloride ion, Cl chloro
Bromide ion, Br bromo
odide ion, I iodo

Nitrite , NO2 Nitro
Hydroxide ion, OH hydroxo
Oxide ion, O Oxo
Carbonate ion, CO3 carbonato

Effect of Ligand Exchange

(a) Complex ion have different stabilities.
Thus, complex ion undergo ligand exchange (or substitution) reaction in solution

(b) A stronger ligand will displace a weaker ligand from a complex ion.
(c) Effect of ligand exchanged can cause the formation of complex ions and colour changes to the

solution.

Relative strength of Ligands:
Br < Cl < F < OH < C2O42- < H2O < SCN < EDTA < NH3 < en < CN < NO2 < CO

Weaker field ligand Strength of Ligand increases Stronger field ligand
Longer  Shorter 

(d) Examples: Explain the formation of complex ions and colour changes for the following cases.
(i) When anhydrous cobalt (II) chloride (blue in colour) is added to small quantity of water,
solution turned to pink. When some hydrochloric acid is added to the mixture, the solution
turned back to blue colour. When the aqueous cobalt solution is added to ammonia solution, it
turned to yellow solution; and when potassium cyanide solution is added, the mixture turned
deep red.

Explanation:

When water (ligand) is added to blue solid, Co2+ is hydrated to form [Co(H2O)6]2+ ion, which is

pink solution.

Co2+ + 6H2O  [Co(H2O)6]2+

pink

When some HCl is added to the mixture, solution turned to blue due to the formation of [CoCl4]2

[Co(H2O)6]2+ + 4 Cl [CoCl4]2

pink blue

When ammonia aqueous solution is added to [Co(H2O)6]2+ , a yellow complex ion of [Co(NH3)6]2

formed.

[Co(H2O)6]2+ + 6 NH3  [Co(NH3)6]2 

pink yellow

When KCN is added to the aqueous cobalt solution, [Co(H2O)6]2+, a complex of [Co(H2O)(CN)5]2+

which is deep red in colour.

[Co(H2O)6]2+ + 5 CN  [Co(H2O)(CN)5]2+ 

pink deep red
[Note: CNion is a stronger ligand than water. Thus can displace water to form a more stable complex

ion]

(ii) When iron (III) nitrate is added to plenty of water, yellow colour of hexaaquairon(III) ion
aqueous solution is formed. This solution is separate to two portion. First portion is added to
potassium cyanide, and change the solution to yellowish green. While the second portion is
added to potassium thiocyanate, which turned the yellow solution to blood red solution.

Explanation:

When water (ligand) is added to Iron (III) nitrate, [Fe(H2O)6]2+ ion, which is yellow colour solution

formed.

Fe3+ + 6H2O  [Fe(H2O)6]3+

yellow

When KCN is added to the iron (III) aqueous solution, a yellowish green complex ion of [Fe(CN)6]3

is formed.

[Fe(H2O)6]3++ 6 CN  [Fe(CN)6]3

yellow yellowish green

When a few drops of KSCN is added to the iron (III) aqueous solution, solution changed from yellow

to blood red solution.
[Fe(H2O)6]3++ SCN  [Fe(SCN)(H2O)5]2+

yellow blood red

[Note: SCN ion is a stronger ligand than water. Thus can displace water to form a more stable complex
ion]

Exercises:
1. When aqueous potassium thiocyanate, KSCN(aq) is added to a solution of iron(III) chloride, a blood red

colour is observed. Explain the changes of the observations.

Explanation:
An aqueous solution of iron(III) chloride contains [Fe(H2O)6]3+.
When KSCN(aq) is added, SCN ion will displace H2O ligands forming a complex ion [Fe(H2O)5SCN]2+,
which is red colour.
SCN ion is a stronger ligand than H2O.
Equation: [Fe(H2O)6]3+(aq) + SCN(aq)  [Fe(H2O)5SCN]2+(aq) + H2O(l)

This shows that SCN ion is a stronger ligand than H2O.
[Fe(H2O)5SCN]2+(aq) complex is more stable than Fe(H2O)6]3+(aq).

2. When excess ammonia solution is added to aqueous copper(II) sulphate, the solution changes from light
blue to deep blue, Subsequently, when ethylenediaminetetraacetate ions are added to the solution, the
colour changes from deep blue to pale blue. Explain the observations.

Explanation:

An aqueous solution of copper(II) sulphate contains [Cu(H2O)6]2+.
When NH3(aq) is added to Cu2+, NH3 displace H2O ligand from Cu(H2O)6]2+ to form [Cu(NH3)4]2+.

[Cu(H2O)6]2+(aq) + 4 NH3(aq)  [Cu(NH3)4]2+ (aq) + 6H2O(l)

Light blue deep blue

This shows that NH3 is a stronger ligand than H2O.

Subsequently, when ethylenediaminetetraacetate ions, EDTA4 ion will displace NH3 ligands to form
[Cu(EDTA)]2

[Cu(NH3)4]2+ (aq) + EDTA4 (aq)  [Cu(EDTA)]2 + 4 NH3(aq)

deep blue pale blue

This shows that EDTA4 is a stronger ligand than NH3

Thus, increasing relative stability of the complex is:
[Cu(H2O)6]2+(aq) < [Cu(NH3)4]2+ (aq) < [Cu(EDTA)]2

13.3.5 Catalytic properties
1. A catalyst is a substance that increase the rate of reaction by providing an alternative pathway
with lower activation energy.
2. In a chemical reaction, the catalyst may react with the reactant to form an intermediate, but it is
regenerated in a subsequent step.
Thus, catalyst takes part in the chemical reaction but does not consumed by the reaction.

3. Examples of processes catalyzed by transition elements or their compounds:

No. Industrial Process Catalyst
Fe(s) Or Fe2O3(s)
1 Haber Process in manufacture ammonia:

N2(g) + 3H2(g) 2NH3(g)

2 Ostwald Process in manufacture nitric acid: Pt(s) or Rh(s)

4NH3(g) + 5O2(g) 4NO(g) + 6H2O V2O5 (s)
Vandium(V) oxide
2NO(g) + O2(g) 2NO2(g)

3 Contact Process in manufacture sulphuric acid.

2SO2(g) + O2(g) 2SO3(g)

4 Hydrogenation of fats in the margarine industry // Pt(s) or Ni(s)
Hydrogenation of alkenes

5 Reaction of peroxodisulphate ions and iodide ions Fe2+(aq) or Fe3+(aq)
I−(aq) + S2O82aq) I2(aq) + 2 SO42aq)

4. There are 2 types of catalytic actions:
Homogeneous catalysis
Heterogeneous catalysis

(A) Homogeneous catalysis

(a) Physical states of the catalyst is the same as the reactants.

[usually are in liquid or aqueous] (Catalyst react with reactant to form an intermediate)

(b) Homogeneous catalysis can be explained by intermediate product theory.
(c) Example: Reaction between iodide ion, I− and peroxodisuphate ion, S2O82- by using iron (III)

ion as catalyst.
i. Ionic equation: I−(aq) + S2O82aq) I2(aq) + 2 SO42aq)

ii. In a single step reaction, rate of reaction for the uncatalysed reaction is very slow because

the activation energy for the reaction is very high due to it involved the reaction between 2

negatively charged.

iii. To speed up the chemical reaction, iron(III) sulphate // iron(II) sulphate is added as

catalyst.

iv. Thus, the steps of the reaction is altered to 2 steps, which can be explained by intermediate

product theory.
Step 1: Iron(III) ion, Fe3+ oxidized iodide ion, I to I2

2Fe3+ (aq) + 2I(aq)  2Fe2+(aq) + I2(aq)

Oxidising agent Intermediate

Step 2: iron(II) ion, Fe2+ reduced disulphate ion, S2O82 to SO42
2Fe2+(aq) + S2O82(aq)  Fe3+(aq) + 2SO42(aq)

Reducing agent Fe3+ is regenerated

v. Energy profile diagram:
Energy / kJ

Uncatalysed reaction

Catalysed reaction

2I + S2O82 Fe2+
(Intermediate)
I2 + 2SO42-

Reaction coordination

(B) Heterogeneous catalysis
(a) Catalyst and the reactants exist in different phases.
[The catalyst is usually solid; reactants are in liquids or gases]
(b) Heterogeneous catalysis can be explained by adsorption theory.
(c) Heterogeneous catalysis normally occur on the surface of the solid of catalyst through
adsorption process. (NOT ABSORPTION)
(d) Example: Hydrogenation of alkenes by using Ni or Pt as catalyst.

CH2 = CH2 + H2 Ni CH3 CH3

i. Nickel, Ni, provide a suitable surface for the adsorption of the reacting molecules by
making use of its available 3d orbitals.
CH2 = CH2 and H2 are adsorbed onto the surface of nickel by the formation of temporary
bonds.

ii. This weakens the covalent bonds in the CH2 = CH2 and H2 molecules, thus lowering the
activation energy for the chemical reaction.
This also hold the molecules in the correct orientation for the new bond to be formed.

iii. When the old bonds are completely break and the new bonds are completely formed,
CH3 CH3 molecules will desorb and diffuse away from the surface of catalyst.

HH H2C CH2 HH

Temporary bonds Adsorption of CH2=CH2
Nickel surface and H2 on the surface of Ni

Both Ethene and hydrogen
molecules are adsorbed onto Ni
surface.

Old Bond breaking New Bond forming

HH H2C CH2 HH

Nickel surface Bond formation and
bond breaking

H-H bonds and C=C bonds are
weaken to form bonds with metal
surface.

H H3C CH3 H

Nickel surface Desorption process

One H atom diffuses close to the
bonded carbon, C-H bond is
formed.

13.3 Uses of first row transition elements and their compounds

1. Uses of titanium:

Transition metal and Properties Uses
its compounds Aircraft body,
space capsules,
Titanium  Light nuclear reactors
 resistant to corrosion
Eyeglass frames
Titanium alloy  Light
 resistant to corrosion White pigment in paints,
Titanium (IV) oxide,  Long lasting paper
(TiO2)  Non-toxic
 High reflective index
 Does not darken when exposed to

air which containing Hydrogen
sulphide gas

2. Uses of Chromium Properties Uses
Transition metal and
its compounds  Heat resistant and does not rust Stainless steel gates,
Chromium easily stainless steel kitchen utensils
[Cr is added to iron, Fe to form
stainless steel. Cr increase the
tensile strength and increase the
steel’s resistant to corrosion by
forming an impervious chromium
oxide layer].

3. Uses of manganese and cobalt

Transition metal and Properties Uses
its compounds
Makes steel alloys
Manganese  Hard and shiny [alloys of Mn and steel is used
to make high speed cutting
Cobalt  Hard and shiny tools.]
Makes oxidising agent such as
KMnO4, MnO2

Make gemstones,
As additive to glass to produce
blue colour glass
Added to aluminium as blue
pigment of paint.