Transition Elements_part1

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.

1. Form variable Oxidation states
(a) 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.

(b) Oxidation states of the first row of transition elements.

Sc Ti V Cr Mn Fe Co Ni Cu Zn
+1

+2 +2 +2 +2 +2 +2 +2 +2 +2
+3, Sc3+ +3 +3 +3 +3 +3 +3 +3 +3

+4 +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]

2. Formation of coloured ions / complexes
(a) 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.
(b) Transition metal ions with empty or completely filled d-orbitals are colourless. This is due to no d-
d transition can occur.

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

dxy dxz dyz dx2-y2 dx2-y2 Co2+

(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].

(d) (i) 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.

(ii) 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.

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

(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
 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.

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

Blue deep blue

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

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

Element Complex Stable ion Oxidation state Colour
Scandium [Sc(H2O)6]3+ Sc3+ +3 Colourless
+3
Titanium [Ti(H2O)6]3+ T3+ +2 Violet
+3 Violet
Vanadium [V(H2O)6]2+ V2+ +2 Green
[V(H2O)6]3+ V3+ +3 Blue
+3 Green
[Cr(H2O)6]2+ Cr2+ +3 Bright Green
+3 Light green
[Cr(H2O)6]3+ Cr3+ +6 Dark green
+6 Orange
[Cr(OH)6]3- Cr3+ +2 yellow
+6 Light pink
Chromium [Cr(H2O)5Cl]2+ Cr3+ +7 green
+2 purple
[Cr(H2O)4Cl2]2+ Cr3+ +2 Green
+2 Red blood
Cr2O72- Cr6+ +3 Yellowish green
+3 Yellow
CrO42- Cr6+ +3 Yellowish green
+6 Brown
[Mn(H2O)6]2+ Mn2+ +2 Red
+2 pink
Manganese MnO42- Mn6+ +2 yellow
+3 Blue
MnO4- Mn7+ +2 Brown
+2 Light Green
[Fe(H2O)6]2+ Fe2+ +2 Light blue
+2 Green
[Fe(H2O)5(SCN)]2+ Fe2+ +2 Blue
+2 Blue
[Fe(CN)6]4- Fe2+ +2 Dark blue
+2 Green
Iron [Fe(H2O)6]3+ Fe3+ +2 Light blue
Colourless
[Fe(CN)6]3- Fe3+

[Fe(H2O)5NO]2+ Fe3+

[FeO4]2- Fe6+

[Co(H2O)6]2+ Co2+

Cobalt [Co(NH3)6]2+ Co2+
[CoCl4]2- Co2+

[Co(H2O)6]3+ Co3+

[Ni(H2O)6]2+ Ni2+

Nickel [Ni(NH3)6]2+ Ni2+
[Ni(CN)4]2- Ni2+

[Ni(edta)4]2- Ni2+

[Cu(H2O)6]2+ Cu2+

Copper [Cu(NH3)4]2+ Cu2+
[CoCl4]2- Cu2+

[Cu(edta)4]2- Cu2+

Zinc [Zn(H2O)6]2+ Zn2+

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

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

(b) 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 Cu+
+1

simple +2 Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+
Aqueous +3
Ti3+ V3+ Cr3+ Fe3+ Co3+

cations

+4 TiO2+ VO2+

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

+7 MnO4

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

(d) 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 VO2+ ; Cr2O72MnO4FeO42

 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 2-
OH
O

H2O O H2O O2-
OH2
M5+ H M5+ + 4H+

H2O OH2 H2O
H2O H2O

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+.

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
V5+

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 2-
OH
O

H2O O H2O V5+ O2-
V5+ H OH2
+ 4H+

H2O OH2 H2O
H2O H2O

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

4. Relative stabilities of 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.
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.

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.

5. 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
Cr3+ + e Ti2+ 0.41
Ti3+ + e V2+ 0.37
V3+ + e Fe2+ 0.24
Fe3+ + e Mn2+ 0.77
Mn3+ + e Co2+ 1.51
Co3+ + e 1.81

(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.

6. 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:

Ti3+ + e Ti2+ ; Eo = 0.37 V

O2 + 4H+ + 4e 2H2O ; Eo = 1.23 V

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

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,

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,

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

7. 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

through dative bond. . L

L n+ L

M

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.

(c) Example of ligands: Name of Ligand in coordination compounds / Complexes
Ligand aqua
ammine
Water, H2O carbonyl
Ammonia, NH3 ethylenediamine
Carbon monoxide, CO
cyano
ethylenediamine (en) thiocyanato
oxalato
H2NCH2CH2NH2
Cyanide, CN ethylenediaminetetraethanoate
thiocyanate, SCN fluoro
chloro
Ethanedioate ions @ bromo
Oxalate, C2O42 iodo
EDTA4 Nitro
Fluoride, F hydroxo
Chloride, Cl oxo
Bromide, Br carbonato
odide, I
Nitrite , NO2
Hydroxide, OH
Oxide, O
Carbonate ion, CO3

(d) 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  Longer 