6b transition elements
TRANSCRIPT
S. R. Chase 1
TRANSITION METAL CHEMISTRY
Study to show yourself worthy of the grade. To whom much is given much is expected!
“ Good as any but better than many”
I
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• Transition metals• Characteristics of transition metals• Electron configuration of the metals and their
ions• Atomic radii, ionic radii and ionisation
energies• Formation of coloured ions• Experiments
to show the variation in oxidation states of vanadium
TABLE OF CONTENTS
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TABLE OF CONTENTS
• Properties of transition elements when compared to those of calcium as a typical s-block element
• Predict the shapes of complexes of transition elements
• Discuss the use of the following as redox systems
• Explain the principle of Ligand exchange• Perform experiments on ligand exchange
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1st Row Transition Metals
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Ener
gy
Overlap of subshells results in the 4s
subshell having lower energy than the 3d
subshell
Energy levels in the various subshells in a many-electron atom
2
8
18
32
No. of electrons in energy level = 2n2
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[Ar] 3d14s2
[Ar] 3d24s2
[Ar] 3d34s2
[Ar] 3d54s1
[Ar] 3d54s2
[Ar] 3d64s2
[Ar] 3d74s2
[Ar] 3d84s2
[Ar] 3d104s1
[Ar] 3d104s2
Electronic configuration
(z=21)
(z=22)
(z=23)
(z=24)
(z=25)
(z=26)
(z=27)
(z=28)
(z=29)
(z=30)
**
**
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What is a transition element?
Transition elements are elements that form one or more ions with a partially filled d subshell.
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Co
ScTi V
Cr
Mn Fe
NiCu
Zn
TRANSITION METALS
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General uses of transition elements
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TRANSITION METALS
First Row TS elements Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
21, 22, 23, 24, 25, 26. 27, 28, 29, 30
Some [Sc]Thing [Ti]Very [V]Crazy [Cr]
Make [Mn]I [Fe]
Could [Co]Not [Ni]Cut [Cu]Zinc [Zn]
Can you make up a pneumonic to remember this
order?
12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn0
200
400
600
800
1000
1200
1400
1600
1800
2000
m.p./ C of some elements
m.p./ C
Mel
ting
poin
t (
C)
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OBJECTIVE 5.1
• Variations in oxidation number• Complex formation• Coloured compounds• Catalytic activity• Magnetic properties
Describe the characteristics of transition elements
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There are several common characteristic properties of transition elements:
They often form coloured compounds.
They can have a variety of different oxidation states.
At least one of their compounds has an incomplete d-electron subshell.
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They are often good catalysts.
They are silvery-blue at room temperature (except copper, iron and gold)
They are solids at room temperature (except mercury).
They form complex ions (aqua ions included).
They are often paramagnetic.
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Patterns in oxidation state emerge across the period of transition elements:
The number of oxidation states of each ion increases up to Mn, after which they decrease
When the elements are in lower oxidation states, they can be found as simple ions. However, transition metals in higher oxidation states are usually bonded covalently to electronegative elements like oxygen or fluorine, forming polyatomic ions such as chromate (CrO4
2-, vanadate VO2- , or permanganate (MnO4
-).
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Other properties with respect to the stability of oxidation states:
Ions in higher oxidation states tend to make good oxidizing agents, whereas elements in low oxidation states become reducing agents. The 2+ ions across the period start as strong reducing agents and become more stable. The 3+ ions start stable and become more oxidizing across the period.
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Variable oxidation statesAs opposed to group 1 and group 2 metals, ions of the transition elements may have multiple oxidation states, since they can lose d electrons without a high energetic penalty. Manganese, for example has two 4s electrons and five 3d electrons, which can be removed. Loss of all of these electrons leads to a +7 oxidation state.
Mn4s 3d 4p
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Lower Oxidation States in Ionic Compounds
There is a wide range of oxidation states in the transition metals. Contrast this with Group 1 and Group 2 elements, each of which has only one oxidation state (+1 and +2 for group 1 and 2 respectively.) The difference is explained by the closeness of the 4s and 3d energy levels. In Ca the two 4s electron are easily removed, but to
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remove another electron means breaking into the 3p subshell which requires much more energy. This is not the case with the transition metals. Since the first electrons to be lost in forming a transition metal ion are the 4s electrons, which makes the +2 oxidation state common, as in Fe2+ and Co2+ . However the 3d electrons may also be lost which means the ions such as V3+ , Fe3+ , and Cr3+ are also common.
Lower Oxidation States in Ionic Compounds
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1000
2000
3000
4000
5000
6000
Ionisation Energy of Selected ElementsIE (1) / kJ mol-1IE (2) / kJ mol-1IE (3) / kJ mol-1
Ioni
satio
n En
ergi
es (k
J mol
-1)
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A solid circle represents a common oxidation state, and a ring represents a less common
(less energetically favorable) oxidation state.
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Pale pink
brown
black
green
purple
Mn 2+
Mn 3+
Mn 4+
Mn 6+
Mn 7+
Common oxidation states on manganese
Manganese shows most oxidation numbers from +2 to +7 inclusive. Notice the maximum oxidation number is equal to the total number of 3d and 4s electrons. Compounds containing higher ox. states tend to be covalent e.g. MnO4
- while those with lower ox. states tend to be ionic e.g. Mn2+
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Purple
green
blue
yellow
V +2
V +3
V +4
V +5
Common oxidation states on Vanadium
• Vanadium shows oxidation numbers from +2 to +5 inclusive. Notice the maximum oxidation number is equal to the total number of 3d and 4s electrons.
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COMPLEXES
A complex is formed when a central metal atom or ion is surrounded by species that donate lone pairs of electrons. A species that donates a lone pair of electrons is called a ligand.
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Ligands are usually molecules such as water or negative ions such as the chloride ion Cl-.
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Many complexes are positively or negatively charged, but some are neutral. In the formula, the central metal atom is written first followed by the ligands. The charge on the complex is the charge of the central metal ion and the charges of the surrounding ligands added together.
The overall charge of [Cu(H2O)6] 2+
results from the 2+ charge of Cu2+ , since the water molecules carry no charge.
[Cu(H2O)6] 2+
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In the case of [CuCl4]2- there are four Cl- ions to give a charge of 4-, which when added to the 2+ charge of Cu2+ , gives an overall charge of 2-.The charge on a complex ion is delocalised over the whole ion and is usually shown outside square brackets.
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What is the charge on the central metal ion in the following complexes?
(a) [Ag(NH3)2]+
(b) [Fe(H2O)6]3+
(c) [Ni(CN)4]2-
(d) [NiCl4]2-
(e) [Fe(CN)6]3-
1+3+2+2+3+
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LIGANDS
Ligands such as H2O, NH3, Cl- and CN- donate one pair of electrons to the central metal atom and are called monodentate ligands. Those ligands which form two dative covalent bonds to the metal ion are called bidentate ligands. When these ligands attach to the central metal ion, they form five-membered rings and the complexes are called chelates.
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1,2-diaminoethane – NH2CH2CH2NH2
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CHELATION
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LIGAND - EDTA
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USES OF EDTA
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USES OF EDTA
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Ingredients:Vegetable oil (canola, and/or soybean), water, sugar, distilled vinegar, Vidalia Onions (5%), mustard (vinegar, water, mustard seed, salt, tumeric, paprika, spices, garlic powder), salt, dried onions, natural spices, xanthan gum, sodium benzoate, potassium sorbate and calcium disodium EDTA (preservatives), yellow 5 (color).
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Multimol-IronMultimol Iron- ferrous EDTA 12%, is sprayed in soil deficient in iron content to increase productivity and soil fertility. Composition: iron content percentage by weight minimum in Ferrous EDTA 12%.Dosage: 0.5 to 1.0 g/Litre
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Contains Tetrasodium Edta to aid in removing Ca ions from the water preventing scum formation
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NAMING COMPLEX IONS
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NAMING COMPLEX IONS
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HETEROGENEOUS AND HOMOGENEOUS CATALYSIS
A catalyst is a substance that alters the rate of a chemical reaction without becoming permanently involved in the reaction. A catalyst works by providing an alternate reaction pathway with a lower energy of activation.
There are two types of catalyst. Homogeneous catalysts are in the same phase as the reactants they catalyse, while heterogeneous catalysts are in a different phase from the reactants they catalyse.
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HETEROGENEOUS CATALYST
Nickel catalyses the hydrogenation of carbon-carbon double bonds, an important reaction in the production of fats from oil.
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HETEROGENEOUS CATALYSTThe Haber process for the production of ammonia uses finely divided iron to catalyse the reaction between nitrogen and ammonia:
This is one of the most important industrial reactions because ammonia is used to produce nitrogenous fertilizers.
N2 (g) + 3H2 (g) 2NH3 (g)Fe catalyst
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HETEROGENEOUS CATALYST
The contact process, which leads to the production of sulphuric acid is catalysed by vanadium pentoxide, V2O5.
2SO2 (g) + O2 (g) 2SO3 (g)V2O5 catalyst
A heterogeneous catalyst works by providing a surface on which the reactants can bond. These are called active sites. It is the availability of the 3d and 4s electrons coupled with the ability to use variable oxidation states that makes transition metals and their compounds such good catalysts.
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A heterogeneous catalyst works by providing a surface on which the reactants can bond. These are called active sites
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HETEROGENEOUS CATALYST
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HETEROGENEOUS CATALYST
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HOMOGENEOUS CATALYST
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MAGNETIC PROPERTIES
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MAGNETIC PROPERTIESDue to the presence of unpaired electrons in the d orbitals most of the transition metal ions and their compounds are paramagnetic i.e. they are attracted by the magnetic field. As the number of unpaired electrons increases from 1 to 5, the magnetic moment and hence paramagnetic character also increases.
Those transition elements which have paired electrons are diamagnetic i.e. they are repelled by the magnetic field.
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Metals like Fe, Co and Ni possess very high (special form) paramagnetism where they obtain permanent magnetic moment and are referred to as ferromagnetic
Ferromagnetism occurs when individual atoms are paramagnetic.
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MAGNETISMAlnico which is an alloy of Al, 12%, Ni, 20%, Co, 50% and 18% Fe is used to make permanent magnets.
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MAGNETISMWhen an object is placed in a strong magnetic field one of three things happen: it could
• experience no force whatsoever (non-magnetic)• be repelled from the magnetic field (diamagnetic)• be attracted weakly to the field (paramagnetic)
Ferromagnetic materials are strongly attracted by a magnetic force. The elements iron (Fe), nickel (Ni), and cobalt (Co) are such materials.
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Paramagnetism is a form of magnetism which occurs only in the presence of an externally applied magnetic field. Paramagnetic materials are attracted to magnetic fields Paramagnetism is associated with unpaired electrons.
PARAMAGNETISM
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PARAMAGNETISMParamagnetism is associated with unpaired electrons. Many transition element complexes are often paramagnetic because they contain unpaired d electrons. It is found that the strength of the paramagnetism shown by complexes is dependent on the total number of unpaired electrons.
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Ion Sc3+
Ti4+Ti3+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
Elec. Config. d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
No. of unpaired electrons
0 1 2 3 4 5 4 3 2 1 0
Magnetic moment
0 1.73 2.83 3.87 4.90 5.92 4.90 3.87 2.83 1.73 0
Effective Magnetic Moment of some Transition Metal ions found in transition
metal compounds
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E
3d-Free gas phase atom
3d-Octahedral complex
Paramagnetism in Octahedral complex
In the isolated gas phase atoms, all five orbitals have the same energy, but in complexes they split into two groups with an energy gap between them. Strong ligands cause a large split and weak ligands cause a small split.
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Paramagnetism in Octahedral complex
Strong ligands, e.g. CN-, cause a large split and weak ligands , e.g. H2O, cause a small split.
When the split is large the electrons tend to pair up in the lower level before they fill the upper level. When the gap is small they tend to fill up both the upper and lower orbitals singly before pairing up.
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Many TS elements and their compounds are paramagnetic, which means that they are attracted to a magnetic field.(i) Explain the origin
of paramagnetism in TS elements
(ii) Suggest the explanations for the relative
Formula of Complex
Relative Paramagnetism
[V(H2O)6]2+ 3[V(H2O)6]3+ 2
[Fe(H2O)6]2+ 4[Fe(H2O)6]3+ 5
[Fe(CN)6]4- 0
[Fe(CN)6]3- 1
paramagnetism of the following complexes, explaining your reasoning.
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Formula of Complex
3d ELECTRONS
Relative Paramagnetism
[V(H2O)6]2+ 3d3 3[V(H2O)6]3+ 3d2 2[Fe(H2O)6]2+ 3d6 4[Fe(H2O)6]3+ 3d5 5[Fe(CN)6]4- 3d6 0[Fe(CN)6]3- 3d5 1
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(a) (i) Transition elements with unpaired 3d-electrons are paramagnetic. In the presence of a magnetic field, these odd electrons align themselves in such a way that they are attracted to the field.
(ii) The strength of the paramagnetism shown by complexes depends on the total number of unpaired electrons
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The strength of paramagnetism shown by complexes depends on the total number of unpaired electrons, In [V(H2O)6]3+ and [V(H2O)6]2+, there are 3 and 2 odd (unpaired) electrons, Hence, their relative paramagnetism are 3 and 2 respectively.
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[Fe(H2O)6]2+ (3d6) [Fe(H2O)6]3+ (3d5)
In both [Fe(H2O)6]2+ and [Fe(H2O)6]3+ the water ligands are weak ligands and they cause only a small splitting in the 3d orbitals. Hence all the 3d-electrons occupy singly first before pairing up i.e. there are 4 and 5 odd electrons. Hence, there are 4 and 5 odd electrons respectively and their relative paramagnetism are 4 and 5 respectively.
small splitting in the 3d-orbitals
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[Fe(CN)6]4- (3d6) [Fe(CN)6]3- (3d5)
CN- are strong ligands and they cause a big splitting in the 3d-orbitals in both [Fe(CN)6]4- and [Fe(CN)6]3- . Hence the 3d-electrons paires up first before occupying the higher energy 3d orbitals. Hence, there are 0 and 1 electron respectively and their relative paramagnetism are 0 and 1 respectively.
big splitting in the 3d-orbitals
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(b) Describe in each case one test you would carry out on a solution to discover whether it contained
(i) Fe2+ (aq) ions
(ii) Fe3+ (aq) ions
(i) Add potassium hexacyanoferrate(III), K3[Fe(CN)6]. If Fe2+ is present, a blue precipitate would be seen.
(ii) Add potassium hexacyanoferrate(II), K4[Fe(CN)6]. If Fe3+ is present, a blue precipitate would be seen.
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Iron(II)ions reacts with potassium
hexacyanoferrate(III) solution.
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Powdered potassium ferrocyanide (left) and
ferricyanide (right), showing their different
colours
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The weak field and strong field ligands are based on their capability of splitting. Below is a derived list of some ligands ordered by the size of the splitting Δ that they produce. weakI− < Br− < S2− < SCN− < Cl− < NO3
− < N3− < F− < OH− <
C2O42− < H2O < NCS− < CH3CN < NH3 < en < phen
(1,10-phenanthroline) < NO2− < CN− < CO
strong(Order of Increasing splitting weak field to strong field)
WEAK FIELD AND STRONG FIELD LIGANDS
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Unlike ferromagnets, paramagnets do not retain any magnetization in the absence of an externally applied magnetic field, because thermal motion causes the spins to become randomly oriented without it. Thus the total magnetization will drop to zero when the applied field is removed. Even in the presence of the field there is only a small induced magnetization because only a small fraction of the spins will be oriented by the field.
PARAMAGNETISM
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Which of the following compounds would you expect to be paramagnetic? Arrange the paramagnetic ones in order of increasing magnetic moment.NH3, NO2, TiCl3, TiO2, VCl3, MnCl2
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Determine the electronic configuration of the first row transition elements and of
their ions
• -mention changes in oxidation number.
OBJECTIVE 5.2
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Ener
gy
Overlap of subshells results in the 4s
subshell having lower energy than the 3d
subshell
Energy levels in the various subshells in a many-electron atom
2
8
18
32
No. of electrons in energy level = 2n2
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For each element from scandium to zinc, the number of protons increases by one. This increases the positive charge on the nucleus. However because the electrons are added to the inner 3d subshell, the outer 4s electrons tend to be shielded from the increasing charge. This partly explains the similarity in physical and chemical properties. Also, because the 3d and 4s subshells have similar energies, the electrons from both can take part in bonding which gives rise to some characteristic transition metal properties.
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Electrons occupy first the orbitals of lowest energy. The orbitals in a subshell are first occupied singly by electrons that spin in the same direction, which helps to minimize the interelectron repulsion. Only when all the d orbitals in the d subshell are singly occupied do electrons pair up.
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The table shows the electron configurations of the first-row transition elements. Notice that the 4s subshell has already been filled because this is at a lower energy than the 3d subshell. The electron configuration of argon (z=18) is shown below and is often abbreviated [Ar]
Electron configurations
The full electron configuration of titanium (Z=22) is: 1s2 2s2 2p6 3s2 3p6 3d2 4s2
or abbreviated [Ar] 3d2 4s2
[Ar] = 1s2 2s2 2p6 3s2 3p6
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[Ar] 3d14s2
[Ar] 3d24s2
[Ar] 3d34s2
[Ar] 3d54s1
[Ar] 3d54s2
[Ar] 3d64s2
[Ar] 3d74s2
[Ar] 3d84s2
[Ar] 3d104s1
[Ar] 3d104s2
Electronic configuration
(z=21)
(z=22)
(z=23)
(z=24)
(z=25)
(z=26)
(z=27)
(z=28)
(z=29)
(z=30)
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Oxidation state
0
+2
+3
0
+2
+3 *
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OBJECTIVE 5.3
• Explain the relatively small changes in atomic radii, ionic radii and ionisation energies of the elements across the period
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[Ar] 3d14s2
[Ar] 3d24s2
[Ar] 3d34s2
[Ar] 3d54s1
[Ar] 3d54s2
[Ar] 3d64s2
[Ar] 3d74s2
[Ar] 3d84s2
[Ar] 3d104s1
[Ar] 3d104s2
Electronic configuration
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Sc Ti V Cr Mn Fe Co Ni Cu Zn0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18 Metallic or Atomic Radius / nm
Metallic Radius / nm
Transition Metals
Ioni
c Rad
ius /
nm
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Sc Ti V Cr Mn Fe Co Ni Cu Zn0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17Metallic Radius / nm
Metallic Radius / nm
Transition Metals
Ioni
c Rad
ius /
nm
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Sc Ti V Cr Mn Fe Co Ni Cu Zn0
500
1000
1500
2000
2500
3000
3500
4000
4500
Trend in IE of Transition Metals
IE (1) / kJ mol-1IE (2) / kJ mol-1IE (3) / kJ mol-1
Transition Metals
Ioni
zatio
n En
ergy
/kJ m
ol-1
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Sc Ti V Cr Mn Fe Co Ni Cu Zn0
500
1000
1500
2000
2500
3000
3500
4000
Trend in m.p. and b.p. of Transition Metals
m.p./ Cb.p. / C
Transition Metals
Tem
pera
ture
/ C
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Melting and Boiling points
Compared to the s-block elements, the melting points of the d-block elements are high. This suggest strong bonding in transition elements.
Metallic bonding involves delocalisation of electrons and the availability for this purpose of the 3d electrons, in addition to the 4s electrons, offers a feasible explanation of the strong bonding.
The reason for the general similarity of the 10 elements is the relatively small difference in effective nuclear charge across the period.
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OBJECTIVE 5.4
• Explain the formation of coloured ions by transition elements
d orbital separation of energy in octahedral complexes
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The Formation of coloured ions
The formation of coloured ions is another characteristic property of transition metals. Most of their compounds are coloured and they form coloured solutions. The colours of nine transition metal ions in aqueous solution are given in the table. Scandium and zinc ions are also included.
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The Formation of coloured ions
Being colourless is another reason why Sc3+(aq) and Zn2+
(aq) are not usually regarded as transition metal ions. A coloured ion results from an incomplete d subshell, which neither of these ions has.
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Colors of representative compounds of the
Period 4 transition metals.
titanium oxide
sodium chromate
potassium ferricyanide
nickel(II) nitrate hexahydrate
zinc sulfate heptahydrate
scandium oxide
vanadyl sulfate dihydrate
manganese(II) chloride
tetrahydrate cobalt(II) chloride hexahydrate
copper(II) sulfate pentahydrate
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Vanadium pentoxide - V2O5
Vanadyl sulfate - VOSO4·xH2O
Potassium dichromate
- K2Cr2O7
Potassium chromate
- K2CrO4
Potassium permanganate
- KMnO4
Manganese dioxide - MnO2
Manganese violet -
Mn NH4 P2O7
Manganous sulfate -
MnSO4·H2O
Ferric oxide - Fe2O3
"Ferric ammonium citrate" -
Fe3+/NH4+/C6H5O7
3-
Nickel sulfate - NiSO4·6H2O
Nickel carbonate basic -
NiCO3 · x Ni(OH)2
Colors of representative compounds of the Period 4 transition metals
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Colors of representative compounds of the Period 4 transition metals
Copper sulfate - CuSO4·5H2O
Copper carbonate basic - CuCO3·Cu(OH)2
Cupric chloride - CuCl2·2H2O Cobaltous chloride -
CoCl2·6H2OCobaltous chloride anhydrous - CoCl2
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Colour in TS compoundsIn the isolated gas phase atoms, all five orbitals have the same energy, but in complexes they split into two groups with an energy gap between them.
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3d Orbitals
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E
3d-Free gas phase atom
3d-Octahedral complex
Colour in Octahedral complex
In the isolated gas phase atoms, all five orbitals have the same energy, but in complexes they split into two groups with an energy gap between them.
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Transition metal compounds are frequently coloured, both in the solid state and in solution. Their colour is entirely due to the uneven absorption of light in the visible region of the electromagnetic spectrum. If energy is absorbed in one part of the visible region, then a colour, which is the complement of the colours absorbed, is seen
High energy
Lowenergy
An artist’s colour wheel
The color of [Ti(H2O)6]3+
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Regions of the electromagnetic spectrum.
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In most cases the colour is caused by electrons absorbing energy as they move from one d-orbital to a higher one. Remember that in a complex ion, the d- orbitals are not all of the same energy, but are split into two groups. Such d-->d transitions are responsible for the colour on many transition metal compounds. For example , blue Cu2+
(aq) ions undergo the following electron transition when they absorb light.
[Ar] 3d104s1
4s 3d
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The colour of a complex can, therefore, be seen to depend on the energy gap between the two groups of d-orbitals. This in turn depends on
(a) The nature of the metal and its oxidation state
Taking the +2 and +3 oxidation states of chromium and iron as examples: Cr2+
(aq) is sky-blue (E =190 kJ mol-1), whereas Cr3+(aq) is
green (E =205 kJ mol-1). Fe2+(aq) is pale green, whereas Fe3+
(aq) is red brown.
High energy
Lowenergy
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(b) The nature of the ligand
Different ligands have different effects on the relative energies of d orbitals of a particular ion. For example ammonia ligands cause a larger difference than water molecules in the relative energies of the two groups of d orbitals in the ligand field. This factor results in the colour change from blue to blue violet when ammonia is added to aqueous solution of a copper(II) salt
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OBJECTIVE 5.5
• Use of an acidified solution of ammonium vanadate(V) and grannulated zinc.
• Refer to Eo values
Perform experiments to show the variation in oxidation states of vanadium
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Distinctive colors reveal the oxidation states of vanadium (V) in certain aqueous solutions: V+2 (violet), V+3 (green), V+4 (blue) and V+5
(yellow)
States of Vanadium
V+2
V+3
V+4
V+5
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Vanadium: V(V), V(IV), V(III), AND V(II)
A striking illustration of variable oxidation states is shown by the chemistry of vanadium. The highest oxidation states occur when the elements combine with the highly electronegative element oxygen. Vanadium shows its oxidation number +5 in the vanadium(V) ion. When zinc reduces ammonium vanadate(V) in concentrated hydrochloric acid, the solution changes from yellow to blue-green, and eventually violet as the oxidation changes from +5 through to +2.
VO2+
(aq) VO2+ (aq) V3+
(aq) V2+ (aq)
yellow blue blue-green violet
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To prepare a transition metal in a low oxidation state, the general process is:1. Add an acid2. Add a reductant ( a reducing agent)
Zinc is the reductant in this case.
VO2+
(aq) VO2+ (aq) V3+
(aq) V2+ (aq)
yellow blue blue-green violet
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A comparison of the four oxidation states of VVO2
+ (aq) VO2+
(aq) V3+ (aq) V2+
(aq)
yellow blue blue-green violet
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VO2+
(aq) VO2+ (aq) V3+
(aq) V2+ (aq)
yellow blue blue-green violet
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A comparison of the four oxidation states of V
V+2V+3V+4V+5
VO2+
(aq) VO2+ (aq) V3+
(aq) V2+ (aq)
yellow blue blue-green violet
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VO2+ to VO2+
(yellow to blue) Eo = 1.0 V
VO2+ to V3+
(blue to green) Eo = .33 V
V3+ to V2+
(green to violet) Eo = -.26 V
V2+ to V Eo = -1.18 V
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Vanadium Chemistry
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Question
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Solution
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Solution
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*
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OBJECTIVE 5.6
Discuss qualitatively the properties of transition elements when compared to those of calcium as a typical s-block element
• Melting point• Density• Atomic radius• Ionic radius• First ionisation energy• Conductivity
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Transition elements tend to have high tensile strength, density and melting and boiling points. As with many properties of transition metals, this is due to d orbital electrons' ability to delocalise within the metal lattice. In metallic substances, the more electrons shared between nuclei, the stronger the metal.
S. R. Chase 135K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn0
200
400
600
800
1000
1200
1400
1600
1800
2000
m.p./ C of some elements
m.p./ C
Mel
ting
poin
t (
C)
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K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn0123456789
10 Density / g cm -3
Density / g cm -3
Dens
ity (g
cm-3
)
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0
50
100
150
200
250
Atomic and Ionic Radius /pmAtomic radius /pmM2+ ionic radius/pm
Atom
ic/io
nic R
adiu
s/pm
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0
1000
2000
3000
4000
5000
6000
Ionisation Energy of Selected Elements
IE (1) / kJ mol-1
IE (2) / kJ mol-1
IE (3) / kJ mol-1
Ioni
satio
n En
ergi
es (k
J mol
-1)
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K Sc V Mn Co Cu0
0.1
0.2
0.3
0.4
0.5
0.6
Electrical conductivity x 10 -8 k/S m-1
Electrical conductivity x 10 -8 k/S m-1
Elements
Elec
tric
al co
nduc
tivity
x 1
0 -8
k/S
m-1 Cu and Ag are the best
metallic conductors at room temperature
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CONDUCTION OF ELECTRICITY
The conductivity of metals results from delocalization of the outer electrons in the metallic bond, which makes the electrons free to move under the influence of a potential difference.
In calcium for example only the 4s electrons are delocalised but for the transition metal atoms the 4s as well as the 3d electrons are delocalised as part of the metallic bond, thus they are better electrical conductors than Ca.
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K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Electrical conductivity x 10 -8 k/S m-1
Electrical conductivity x 10 -8 k/S m-1
Elements
Elec
tric
al co
nduc
tivity
x 1
0 -8
k/S
m-1
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OBJECTIVE 5.7
• Octahedral• Tetrahedral• Square planar
Predict the shapes of complexes of transition elements
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OBJECTIVE 5.8
• Fe3+(aq) / Fe2+
(aq)
• MnO4-(aq) /Mn2+
(aq)
• Cr2O72-
(aq) / Cr3+(aq)
Discuss the use of the following as redox systems
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Fe2+(aq) / Fe3+
(aq) oxidation RA
• Fe2+(aq) Fe3+
(aq) + e-
MnO4-(aq) /Mn2+
(aq) reduction OA
• MnO4-(aq) + 5e- + 8H+ (aq)Mn2+
(aq) + 4H2O (l)
Cr2O72-
(aq) / Cr3+(aq) reduction OA
• Cr2O72-
(aq) + 14H+ (aq) + 6e- 2Cr3+(aq)+ 7H2O (l)
Redox Systems
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Acidified potassium manganate
MnO4- Mn2+ Cr2O7
2- Cr3+
Acidified potassium dichromate
Oxidizing Agents/Oxidant
• MnO4-(aq) + 5e- + 8H+
(aq) Mn2+(aq) +
4H2O (l) • Cr2O7
2-(aq) + 14H+ (aq) + 6e- 2Cr3+
(aq)+ 7H2O (l)
+7 +2 +6 +3
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Ppt. formed with NaOH
Fe2+ Fe3+(Fe2+)Reducing Agent/
Reductant
+2 +3
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+ -
NaOHFe2+
Iron(II) salt is a known
Reducing agent
The pale green Fe2+ ion is oxidized to the Fe3+ which is yellow to
orange/brown in solution
Sodium hydroxide is added to confirm the
presence of the iron(III) ion(brown ppt)
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Iron(II) sulphate
A B C D E
H2O2 C6H12O6 H2O NaClO NaNO2
After a while, notice the precipitates settled to the bottom of the test tubes
H2O2 C6H12O6H2O NaClO NaNO2
OA OA OANot-OA
Sodium hydroxide added
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KMnO4 /H+
Testing for Reducing agents in the lab
Potassium manganate(VII)
is a known oxidizing agent
The purple MnO4- ion is
reduced to the colourless Mn2+ ion
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A B C D E
Acidified potassium manganateIs a known oxidising agent
H2O2 C6H12O6 H2O NaClO NaNO2
RA RA RANot - RA
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Potassium manganate(VII) is a powerful oxidising agent used extensively in the determination of the concentration of a wide range of reducing agents and also for various quantitative analyses. The solid compound and its aqueous solution are intensely purple coloured, hence external indicators are not needed to signal the end-point of the titration. The end-point of titrations involving additions of aqueous potassium manganate(VII) from a burette is taken as the first permanent pink colouration.
Most redox reactions using Potassium manganate(VII) are carried out under acidic conditions (only H2SO4 suitable)to facilitate reduction to the very pale pink (almost colourless) Mn2+according to the equation
MnO4-(aq) + 5e- + 8H+ (aq)Mn2+
(aq) + 4H2O (l)
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Potassium manganate (VII) oxidizes iron(II) ions to iron(III) such that:
1mol MnO4- (aq) 5 mol Fe2+
(aq)
MnO4-(aq) + 5Fe2+
(aq) + 8H+ (aq)Mn2+(aq) + 5Fe3+
(aq) + 4H2O (l)
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Testing for Reducing agents in the lab
K2Cr2O7 /H+
Potassium dichromate(VI)
is a known oxidizing agent
The Orange Cr2O72- ion is
reduced to the green/blue Cr3+ ion
S. R. Chase 161After about 2 hr
Acidified potassium dichromate
H2O2 C6H12O6 H2O NaClO NaNO2
RA RANot - RA Not - RA
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Potassium dichromate(VI) , K2Cr2O7, is a reasonably good oxidizing agent, reacting with reducing agents according to the following half-equation:
• Cr2O72-
(aq) + 14H+ (aq) + 6e- 2Cr3+(aq)+ 7H2O (l)
It is however less versatile than the more powerful oxidizing agent Potassium manganate(VII).
½ Cr2O72-
(aq) + 7H+ (aq) + 3e- Cr3+(aq)+ 3½ H2O (l) E = + 1.33 V
MnO4-(aq) + 5e- + 8H+ (aq)Mn2+
(aq) + 4H2O (l) E = + 1.52 V
*
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OBJECTIVE 5.9
• Stability constants• The CO/O2 haemoglobin and
• NH3(aq)/Cu2+ (aq) systems
Explain the principle of Ligand exchange
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Stability constants
The course of ligand exchange reactions may be considered in terms of competition for sites on the central ion by the ligands present. The value of a stability constant is an indication of the inherent tendency of a particular ligand to replace water in aqua-complex. The stability of chelate complexes , involving polydentate ligands, is usually greater than that of complexes based on monodentate ligands.
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Ligand exchange using the metal ion Cu2+ and the ligands ; water, ammonia and chloride
H2O Cl- Cl- Cl- NH3 EDTA
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Stability constants Stability constants vary in magnitude over several
powers of 10, and it is often more convenient to use the logarithm of the stability constant (lg K) to allow an easier comparison.
The larger the value of lg K, the more stable the complex is relative to the ‘simple’ aqueous ion and the more readily it will be formed.
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Stability constants
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Stability constants
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The CO/O2 haemoglobin Ligand Exchange
Haemoglobin, a protein molecule, is a vital component of blood. It transports oxygen from the lungs around the body via the arteries. This large molecule is made up of four protein groups (globulin groups). In the centre of each globulin group is the ion Fe 2+ bound in a complex. Four sites of attachment around the Fe2+ are occupied by a planar ring structure, called a porphyrin .
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The iron-porphyrin part of the complex is called the haem group. The fifth ligand site is occupied by a dative covalent bond to the protein globin, while molecular oxygen loosely attaches to the sixth site.
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Hemoglobin and the octahedral complex in haem.
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When oxygen is bonded to the Fe2+, the whole complex, called oxyhaemoglobin, takes on a red colour. Once the oxygen is removed, it is replaced by a water ligand, which changes the colour of the complex to blue. The complex is now called deoxyhaemoglobin. Deoxyhaemoglobin gives the blood in our veins its blue colour as it returns to the lungs to pick up more oxygen.
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oxyhaemoglobin, takes on a red colour. Once the oxygen is removed, it is replaced by a water ligand, which changes the colour of the complex to blue. The complex is now called deoxyhaemoglobin. Deoxyhaemoglobin gives the blood in our veins its blue colour as it returns to the lungs to pick up more oxygen.
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Carbon monoxide gas is sometimes referred to as the silent killer. The CO molecule has a lone pair and can form dative covalent bonds with Fe2+ in the same place where oxygen bonds with haemoglobin, to form carboxyhaemoglobin. As the bonds formed by CO are stronger than those formed by oxygen the uptake of oxygen is prevented, which is why carbon monoxide is so deadly.
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*
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OBJECTIVE 5.10
Include • Cu2+
(aq)
• Co2+(aq)
Perform experiments to demonstrate Ligand exchange
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Cobalt Chemistry
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Cobalt Chemistry
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Replacing the water in the hexaaquacobalt(II) ionIf you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions (for example, cobalt(II) chloride solution), the solution turns from its original pink colour to a dark rich blue. The six water molecules are replaced by four chloride ions.
The reaction taking place is reversible.
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Concentrated hydrochloric acid is used as the source of chloride ions because it provides a very high concentration compared to what is possible with, say, sodium chloride solution. Concentrated hydrochloric acid has a chloride ion concentration of approximately 10 mol dm-3.
The high chloride ion concentration pushes the position of the equilibrium to the right according to Le Chatelier's Principle.
Notice the change in the co-ordination of the cobalt. Chloride ions are bigger than water molecules, and there isn't room to fit six of them around the central cobalt ion.
Cobaltous chloride - CoCl2·6H2O
The anhydrous form of cobaltous chloride. This salt is soluble in water very well and gives pink solutions, just like the hydrated form of this compound. When this comes in contact with water, it becomes purple at once, due to formation of the hydrate.
Cobaltous chloride anhydrous - CoCl2
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Cobalt Chemistry
Addition of water to the blue solution produces a new equilibrium mixture in which the pink species predominates
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Replacing the water in the hexaaquacobalt(II) ionThis time, all the water molecules get replaced.
The straw coloured solution formed changes colour very rapidly on standing to a deep reddish brown. The hexaamminecobalt(II) ions are oxidised by the air to hexaamminecobalt(III) ions. However, that is a quite separate reaction, and isn't a part of the ligand exchange reaction.
S. R. Chase 2010 194H2O Cl-
NH3A few dropsNH3excess
pink blue green ppt.
straw colour
Ligand exchange using the metal ion Co2+ and the ligands ; water, ammonia and chloride After a
while the Co2+
is oxidised to Co3+
Cobalt(II) Ligand Exchange
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Cobalt(II) Ligand Exchange
195
NH3A few drops
After a while
the Co2+
is oxidised to Co3+
NH3excess
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Copper Chemistry
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CuCl2 Cu(NO3)2 Cu CuSO4 CuSO4hydrated anhydrous
CuCO3 CuO
Which of these copper compounds are soluble in water and which are not?
Adding water to the anhydrous copper sulphate can restore the water of crystallisation and it can regain its crystalline, blue structure.
CuSO4
hydrated
CuSO4
anhydrous
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Copper Chemistry
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Cu2+ in solutionCu2+ + a few drops aqueous ammonia
Cu2+ + excess aqueous ammonia
[Cu(NH3)4](OH)2(aq)
Complex soluble hydroxide
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Ligand exchange using the metal ion Cu2+ and the ligands ; water, ammonia and chloride
H2O Cl- Cl- Cl-
lg K =5.6NH3
lg K =13.1
EDTAlg K = 18.8
Copper(II) Ligand Exchange
Colour change
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Now 6B try
these Gp 7
questions
? ?? ?? ? ?
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