d block elements.pptx

d block elements The elements in the Periodic Table which correspond to the d

sublevels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below.

The electronic structures of the d block elements shown are:

Sc [Ar] 3d14s2

Ti [Ar] 3d24s2

V [Ar] 3d34s2

Cr [Ar] 3d54s1

Mn [Ar] 3d54s2

Fe [Ar] 3d64s2

Co [Ar] 3d74s2

Ni [Ar] 3d84s2

Cu [Ar] 3d104s1

Zn [Ar] 3d104s2

You will notice that the pattern of filling isn't entirely tidy! It is broken at both chromium and copper.

This is something that you are just going to have to accept. There is no simple explanation for it which is usable at this level. Any simple explanation which is given is faulty!

People sometimes say that a half-filled d level as in chromium (with one electron in each orbital) is stable, and so it is - sometimes! But you then have to look at why it is stable. The obvious explanation is that chromium takes up this structure because separating the electrons minimises the repulsions between them - otherwise it would take up some quite different structure.

But you only have to look at the electronic configuration of tungsten (W) to see that this apparently simple explanation doesn't always work. Tungsten has the same number of outer electrons as chromium, but its outer structure is different - 5d46s2. Again the electron repulsions must be minimised - otherwise it wouldn't take up this configuration. But in this case, it isn't true that the half-filled state is the most stable - it doesn't seem very reasonable, but it's a fact! The real explanation is going to be much more difficult than it seems at first sight.

Neither can you use the statement that a full d level (for example, in the copper case) is stable, unless you can come up with a proper explanation of why that is. You can't assume that looking nice and tidy is a good enough reason!

If you can't explain something properly, it is much better just to accept it than to make up faulty explanations which sound OK on the surface but don't stand up to scrutiny!

Transition metals

• Not all d block elements count as transition metals! There are discrepancies between the various syllabuses, but the majority use the definition:

• A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.

SCANDIUM , ZINC and COPPER On the basis of this definition, scandium and zinc don't count as transition metals - even though they are members of the d block.

Scandium has the electronic structure [Ar] 3d14s2. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The Sc3+ ion has no d electrons and so doesn't meet the definition.

Zinc has the electronic structure [Ar] 3d104s2. When it forms ions, it always loses the two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d10. The zinc ion has full d orbitals and doesn't meet the definition either.

By contrast, copper, [Ar] 3d104s1, forms two ions. In the Cu+ ion the electronic structure is [Ar] 3d10. However, the more common Cu2+ ion has the structure [Ar] 3d9.

Copper is definitely a transition metal because the Cu2+ ion has an incomplete d orbitals.

Transition metal ions

You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because before filling orbitals, 4s orbitals have a lower energy than 3d orbitals.

However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital.

Transition metal ions The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons.

This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

Remember this:

When d-block elements form ions, the 4s electrons are lost first.

Transition metal ions To write the electronic structure for Co2+:

Co [Ar] 3d74s2

Co2+ [Ar] 3d7

The 2+ ion is formed by the loss of the two 4s electrons.

To write the electronic structure for V3+:

V [Ar] 3d34s2

V3+ [Ar] 3d2

The 4s electrons are lost first followed by one of the 3d electrons.

Transition metal ions To write the electronic structure for Cr3+:

Cr 1s22s22p63s23p63d54s1 Cr3+ 1s22s22p63s23p63d3

The 4s electron is lost first followed by two of the 3d electrons.

To write the electronic structure for Zn2+:

Zn 1s22s22p63s23p63d104s2 Zn2+ 1s22s22p63s23p63d10

This time there is no need to use any of the 3d electrons.

To write the electronic structure for Fe3+:

Fe 1s22s22p63s23p63d64s2 Fe3+ 1s22s22p63s23p63d5

Forming transition metal ions

The rule is quite simple. Take the 4s electrons off first, and then as many 3d electrons as necessary to produce the correct positive charge.

Variable oxidation state (number) One of the key features of transition metal chemistry is the wide range of oxidation states (oxidation numbers) that the metals can show.

It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like sulphur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals.

However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent.

Examples of variable oxidation states in the transition metals

Iron

Iron has two common oxidation states (+2 and +3) in, for example, Fe2+ and Fe3+. It also has a less common +6 oxidation state in the ferrate(VI) ion, FeO4

2-.

Manganese

Manganese has a very wide range of oxidation states in its compounds. For example:

+2 in Mn2+

+3 in Mn2O3

+4 in MnO2

+6 in MnO42-

+7 in MnO4-

Explaining the variable oxidation states in the transition metals We'll look at the formation of simple ions like Fe2+ and Fe3+.

When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound.

There are several energy terms to think about, but the key ones are:

The amount of energy needed to ionise the metal (the sum of the various ionisation energies)

The amount of energy released when the compound forms. This will either be lattice enthalpy if you are thinking about solids, or the hydration enthalpies of the ions if you are thinking about solutions.

The more highly charged the ion, the more electrons you have to remove and the more ionisation energy you will have to provide.

But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion.

Thinking about a typical non-transition metal (calcium)

Calcium chloride is CaCl2. Why is that?

If you tried to make CaCl, (containing a Ca+ ion), the overall process is slightly exothermic.

By making a Ca2+ ion instead, you have to supply more ionisation energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca2+ ions than there is if you only have a 1+ ion. The overall process is very exothermic.

Because the formation of CaCl2 releases much more energy than making CaCl, then CaCl2 is more stable - and so forms instead.

Thinking about a typical non-transition metal (calcium) What about CaCl3? This time you have to remove yet another electron from calcium.

The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionisation energy between the second and third electron removed.

Although there will be a gain in lattice enthalpy, it isn't anything like enough to compensate for the extra ionisation energy, and the overall process is very endothermic.

It definitely isn't energetically sensible to make CaCl3!

Thinking about a typical transition metal (iron)

Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion.

Fe [Ar] 3d64s2

Fe2+ [Ar] 3d6

Fe3+ [Ar] 3d5

The 4s orbital and the 3d orbitals have very similar energies. There isn't a huge jump in the amount of energy you need to remove the third electron compared with the first and second.

The figures for the first three ionisation energies (in kJ mol-1) for iron compared with those of calcium are:

metal 1st IE 2nd IE 3rd IE

Ca 590 1150 4940

Fe 762 1560 2960

There is an increase in ionisation energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons.

Thinking about a typical transition metal (iron)

However, there is much less increase when you take the third electron from iron than from calcium.

In the iron case, the extra ionisation energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made.

The net effect of all this is that the overall enthalpy change isn't vastly different whether you make, say, FeCl2 or FeCl3. That means that it isn't too difficult to convert between the two compounds.

SS CI 11.5 The d block 19

Chromium and Copper• Cr and Cu don’t fit the pattern of building up the

3d sub-shell, why?– In the ground state electrons are always arranged to

give lowest total energy– Electrons are negatively charged and repel each other– Lower total energy is obtained with e- singly in orbitals

rather than if they are paired in an orbital– Energies of 3d and 4s orbitals very close together in

Period 4

20

Chromium and Copper• At Cr

– Orbital energies such that putting one e- into each 3d and 4s orbital gives lower energy than having 2 e- in the 4s orbital

• At Cu– Putting 2 e- into the 4s orbital would give a higher energy

than filling the 3d orbitals

• On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z= 30) is not? On the basis of incompletely filled 3d orbitals in case of scandium (3d1), and completely filled in case of Zn (3d10), they are considered transition and non-transition elements respectively.


What is a transition metal?

• Transition metals [TM’s] have characteristic properties – e.g. coloured compounds, variable oxidation states

• These are due to presence of an inner incomplete d sub-shell

• Electrons from both inner d sub-shell and outer s sub-shell can be involved in compound formation



• Not all d block elements have incomplete d sub-shells – e.g. Zn has e.c. of [Ar]3d104s2, the Zn2+ ion ([Ar]

3d10) is not a typical TM ion– Similarly Sc forms Sc3+ which has the stable e.c of

Ar. Sc3+ has no 3d electrons



• For this reason, a transition metal is defined as being an element which forms at least one ion with a partially filled sub-shell of d electrons.– In period 4 only Ti-Cu are TM’s!– Note that when d block elements form ions the s

electrons are lost first


What are TM’s like?

• TM’s are metals• They are similar to each other but different from s

block metals eg Na and Mg• Properties of TM’s

– Dense metals – Have high Tm and Tb

– Tend to be hard and durable – Have high tensile strength – Have good mechanical properties


What are TM’s like?

• Properties derive from strong metallic bonding• TM’s can release e- into the pool of mobile electrons

from both outer and inner shells– Strong metallic bonds formed between the mobile pool

and the +ve metal ions– Enables widespread use of TMs!– Alloys very important: inhibits slip in crystal lattice usually

results in increased hardness and reduced malleability


TM Chemical Properties

• Typical chemical properties of the TM’s are– Formation of compounds in a variety of

oxidation states– Catalytic activity of the elements and their

compounds– Strong tendency to form complexes– Formation of coloured compounds


Variable Oxidation States

• TM’s show a great variety of oxidation states cf s block metals

• If compare successive ionisation enthalpies (Hi) for Ca and V as followsM(g) M+(g) + e- Hi(1)M+(g) M2+(g) + e- Hi(2)M2+(g) M3+(g) + e- Hi(3)M3+(g) M4+(g) + e- Hi(4)


Hi for Ca and V

Element

Ionisation Enthalpies[kJ mol-1]

Hi(1) Hi(2) Hi(3) Hi(4)

Ca [Ar]4s2 +596 +1152

+4918

+6480

V [Ar]3d34s2 +656 +142

0+283

4+451

3


Hi for Ca and V

• Both Ca & V always lose the 4s electrons• For Ca

– Hi(1) & Hi(2) relatively low as corresponds to removing outer 4s e-

– Sharp increase in Hi(3) & Hi(4) cf Hi(2) due to difficulty in removing 3p e-

• For Sc– Gradual increase from Hi(1) to Hi(4) as removing 4s then

3d e-


Oxidation States of TM’s

• In the following table– Most important OS’s in boxes– OS = +1 only important for Cu– In all others sum of Hi(1) + Hi(2) low enough for

2e- to be removed– OS = +2, where 4s e- lost shown by all except for Sc

and Ti– OS = +3, shown by all except Zn



Sc

Ti V Cr

Mn

Fe

Co

Ni

Cu

Zn

+1

+2 +2 +2 +2 +2 +2 +2 +2

+3 +3 +3 +3 +3 +3 +3 +3 +3

+4 +4 +4

+5

+6 +6 +6

+7



• No of OS’s shown by an element increases from Sc to Mn– In each of these elements highest OS is equal to no. of 3d

and 4s e-

• After Mn decrease in no. of OS’s shown by an element– Highest OS shown becomes lower and less stable– Seems increasing nuclear charge binds 3d e- more strongly,

hence harder to remove


Oxidation States of TM’s• In general

– Lower OS’s found in simple ionic compounds• E.g. compounds containing Cr3+, Mn2+, Fe3+, Cu2+ ions

– TM’s in higher OS’s usually covalently bound to electronegative element such as O or F

• E.g VO3-, vanadate(V) ion; MnO4

-, manganate(VII) ion

• Simple ions with high OS’s such as V5+ & Mn7+ are not formed


Stability of OS’s

• Change from one OS to another is a redox reaction

• Relative stability of different OS’s can be predicted by looking at Standard Electrode Potentials – E values


Stability of OS’s• General trends

– Higher OS’s become less stable relative to lower ones on moving from left to right across the series

– Compounds containing TM’s in high OS’s tend to be oxidising agents e.g MnO4

-

– Compounds with TM’s in low OS’s are often reducing agents e.g V2+ & Fe2+


Stability of OS’s• General trends (continued)

– Relative stability of +2 state with respect to +3 state increases across the series

– For compounds early in the series, +2 state highly reducing

• E.g. V2+(aq) & Cr2+(aq) strong reducing agents

– Later in series +2 stable, +3 state highly oxidising• E.g. Co3+ is a strong oxidising agent, Ni3+ & Cu3+ do not exist in

aqueous solution.

d block elements.pptx

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