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Transcript
Physical properties of transition metals.
Definition
An element that has its ‘d’ orbitals partly filled, in some of its compounds.
Strictly speaking this means for example in the first row of transition elements we include titanium to copper.
‘d’ block elements in say the first row (period 4) include scandium to zinc.
In some books the phrases ‘transition metal’ and ‘d’ block are synonymous. The transition metals resemble each other closely
in their physical properties. They are all hard, lustrous, weakly electropositive metals with high melting and boiling points.
Melting points
1. All transition metals have melting points above 1000oC. This suggests metallic bonding.
2. 3d as well as 4s electrons available for delocalisation is the best explanation for this, since the more electrons within the
‘sea’ of electrons the greater the electrostatic attraction, and this determines the melting point of the metal.
Density
1. Much denser than ‘s’ block elements.
2. Sc – Cu gradual increase in density, this is due to an increase in nuclear charge.
3. Metallic radius generally decreases across a period due to the increasing nuclear charge. At the same time the Relative
Atomic Mass (RAM) is increasing.
Electronic configuration
Filling of 3d and 4s orbitals
In ‘d’ block, electrons are added to an inner ‘d’ orbital and this shields the outer ‘s’ electrons from the increased nuclear charge. Therefore, the
atomic radius decreases only slightly and electronegativity and ionisation energies increase only slightly.
The difference in energy between the second and third shells is less than between the first and second. By the time the fourth shell is reached,
there is, in fact an overlap between the third and fourth shells. In other words, from scandium onwards, the orbitals of highest energy in the
third shell (the 3d orbital) have higher energy than those of the lowest energy in the fourth shell (the 4s orbital). Hence, when writing the
electronic configurations of these ‘d’ block elements we fill the 4s then the 3d orbitals.
Exceptions do occur:
Cr [Ar] 4s13d5
and
Cu [Ar] 4s13d10
This can be explained by the extra stability offered by full and half-filled ‘d’ orbitals.
This property can be illustrated by the examples below:
Fe2+ (3d6) is readily oxidised to Fe3+ (3d5)
Mn2+ (3d5) is not readily oxidised to Mn3+ (3d4)
When transition metals form ions, electrons are lost first from the 4s sub-shell rather than the 3d sub-shell. Thus Fe2+ions
have the electronic structure [Ar]3d6 rather than [Ar]4s23d4.
This occurs due to the presence of electrons in the 3dlevel, these repel the 4s electrons even further from the nucleus.
Therefore, the 4s electrons are pushed to a higher energy level, higher than 3d. Consequently, when transition atoms
become ions, the electrons from the 4s level before the 3d.
This means that all transition metals will have similar chemical
properties which are dictated by the behaviour of the 4s outer
electrons.
Properties All transition metals except Sc are capable of bivalency. The 4s electrons are first used and then 3d electrons.
There is a great variety of oxidation states but patterns can be found.
Reason: Close similarity in energy of 4s and 3d electrons. Calcium, for example, only has oxidation state number +2 in
compounds due to ease at which electrons are lost from 4s, but any further loss would need much greater energy since the
third electron is to be found in an inner shell.
Known Oxidation states can be summarised by the table below,
Note: Mn can have an oxidation state of +7 due to the hypothetical loss of 7 electrons (4s2 3d5) – after this nuclear charge
binds electrons more strongly. Mn has the maximum number of unpaired electrons available for bond formation. The number
of unpaired electron decreases steadily on either side of Mn. Hence, the pattern shown here.
Sc
Ti
Oxidation
numbers that
occurs in
compounds
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
7
3
6
6
6
5
5
5
5
5
4
4
4
4
4
4
4
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
All show oxidation state +2 (except Sc) due to loss of two 4s electrons.
2
All show +3, but rare in Ni and Cu.
Since, Transition metal ions are small they have a high charge density, therefore, display similar properties to Aluminium.
Stability of oxidation states
1. Stability of higher oxidation states decreases from left to right. A possible reason is the increase in nuclear charge.
2. Relative Stability of +2 and +3 state
a) The increasing stability of +2 across the period is caused by the greater difficulty of removing a third electron as nuclear
charge increases.
b) Mn2+/Mn3+ and Fe2+/Fe3+ have stabilities that do not fit in this pattern. This can be explained by the stability of 3d5 found
in Fe3+ and Mn2+.
Symbol
Electronic structure of Atom
Commonion
Electronic configuration of ion
Potassium
K
(Ar) 4s1
K+
(Ar)
Calcium
Ca
(Ar) 4s2
Ca2+
(Ar)
candium
Sc
(Ar) 3d14s2
Sc3+
(Ar)
Titanium
Ti
(Ar) 3d24s2
Ti4+
(Ar)
Vanadium
V
(Ar) 3d34s2
V3+
(Ar) 3d2
Chromium
Cr
(Ar) 3d54s1
Cr3+
(Ar) 3d3
Manganese
Mn
(Ar) 3d54s2
Mn2+
(Ar) 3d5
Iron
Fe
(Ar) 3d64s2
Fe2+
(Ar) 3d6
Fe3+
(Ar) 3d5
Cobalt
Co
(Ar) 3d74s2
Co2+
(Ar) 3d7
Nickel
Ni
(Ar) 3d84s2
Ni2+
(Ar) 3d8
Copper
Cu
(Ar) 3d104s1
Cu+
(Ar) 3d10
Cu2+
(Ar) 3d9
Zn2+
(Ar) 3d10
Zinc
Zn
(Ar) 3d104s2
Formation
1. Substances that are coloured will absorb part of the electromagnetic spectrum and reflect another.
Non-transition metal solutions tend to be colourless suggesting they absorb no part of the spectrum.
2. Compounds that are coloured have electrons promoted from a ground state to an excited state. This energy transition must
coincide with the frequency of light absorbed.
For Na+ to excite electrons they must be promoted from their outer 2p orbital to a 3s orbital. This energy gap is too big.
3. In the case of Ti(H2O)63+ ions there is a narrow gap in electron energy levels which matches the energy in part of the visible
spectrum.
This electron is not promoted to 4s since the gap is too large. Instead the presence of the water ligands causes the energy
symmetry of the 5 3’d’ orbitals to distort and split into two levels.
The new energy gap matches part of the visible frequency (green/yellow) therefore the resulting solution appears
purple.
The principle is the same in other systems but details more complicated since there are more electrons involved.
Different ligands affect the energetic symmetry of ‘d’ orbitals differently therefore, producing different colours.
Different oxidation states will vary colour greatly due to different numbers of ‘d’ electrons available.
Paramagnetism
Most substances are weakly repelled by a strong magnetic field while others are weakly attracted to it. When they are
repelled, it is known as diamagnetism and when attracted it is known as paramagnetism.
Paramagnetism arises because electrons can be regarded as spinning on their axes and, just as an electric current flowing
through a wire generates a magnetic moment, so does the spinning electron. Electrons that occupy the same orbital (i.e.
have opposite spins), have, of course, zero magnetic moment since the two contributions cancel each other.
Heterogenous catalysis
A catalyst enables a reaction to take a different reaction pathway or mechanism of a lower overall activation energy.
Catalysts are regenerated at the end of a reaction therefore a small amount of catalyst will catalyse an infinite amount of
reactants.
Catalysts do take part in a reaction but are not ‘consumed’.
Transition metals are considered to be good catalysts, due to their ability to form a variety of different oxidation states and ‘d’
electrons allow adsorption onto their surface. This weakens bonds in reactant molecules.
Transition metals tend to be heterogenous catalysts meaning, they are in a different state (usually solid) than the products.
Example:Nickel powder is used in the hydrogenation of C=C, as in ethene. Ni atoms provide sites for holding C=C and H2 in
correct orientation for bonding.
Complex ions
Tendency to form complex ions
Complexes are atoms or more ions or compounds in which an ion is surrounded by a number of molecules or oppositely
charged ions called ligands.
The ligands are joined to the central atom or ion by co-ordinate bonds. The pair of electrons been donated by the ligand. The
donor atoms are usually oxygen, nitrogen and halogens.
The transition elements with vacant ‘d’ orbitals readily form complexes but, is not limited to these elements.
The most common ligands in complexes are:
1. Water: most ions form hydrated ions in solution. For example:
[Cu(H20)6]2+ hexaaqua copper (II) ion
2. Ammonia: the co-ordinate bond is formed between the nitrogen atom and the metal ion, the nitrogen atom donating the
lone pair of electrons. For example:
[Co(NH3)6]3+ hexaammine colbalt (III) ion.
3. The Cyanide ion: the lone pair of carbon atoms is used to form co-ordinate bonds. For example: [Fe(CN)6]4hexacyano ferrate (II) ion
4. The Halide ion: the halide provides the lone pair of electrons. For example: [CuCl4]3- tetrachloro cuprate (II) ion
Naming complex ions
1. Number ligands using mono-, di, tri, tetra, penta, hexa2. Identify ligands using names ending in –o for anions.
F- fluoro, CN- cyano, Cl- chloro, OH- hydroxo (H2O is aqua, NH3 is ammine).
3. Name the cation using English name in +ve complex but latinized name in –ve complex with suffix –ate.
aluminate, plumbate, cuprate, ferrate, zincate, nickelate.
4. Indicate oxidation number of central cation using I, II, III etc.
Common complex ion reactions: These reactions typically involve the replacement of a ligand by another ligand and are Ligand
Replacement Reactions.
A solution of copper (II) sulphate is blue. This solution contains hydrated Cu2+ ions. The addition of ammonia changes the
solution to a dark blue containing the ion tetraammine copper (II).
The reaction can be summarised as show: [Cu(H20)6]2+(aq) + 4NH3(aq)  [Cu(NH3)4(H20)2]2+(aq) + 4H20(l)When potassium
thiocyanate is added to a pale yellow solution of iron (III) ions, a deep blood red solution is formed, containing the ion
pentaaquathiocyanato iron (III).
[Fe(H20)6]3+(aq) + SCN-(aq) [Cu(SCN)(H2O)5]2+(aq) + H2O(l)
There are many more examples of ligand replacement
(or exchange) reactions. Check and learn the ones for your syllabus.