13.1 First Row d-block Elements

The Periodic Table.

There is nothing “God-given” about the Periodic Table.

We use the current shape because it is convenient.

Sixth form students think of it in blocks, according to the type of orbital in the outer electrons are in (mostly).

We generally put the f-block at the bottom, mostly because it’s out of the way – and many of the elements are so radioactive that they’re essentially useless.

We name the elements either Lanthanides (or Lanthanoids) and Actinides (Actinoids) after the first element in that row of the f-block

To be consistent with what we did with the d-block, we should put the f-block in as below.

But you very rarely see the table in this form

(note H can be placed in more than one spot. Why?)

Not much.

Most d-block elements are Transition elements but we exclude Zn and Sc because they don’t have the usual of Transition Element properties.

1

2

3

4

Fill in the electron box for common ions of the top row d-block elements

Zn²⁺ is the only ion that Zinc forms.

Sc³⁺ is (almost) the only ion that Scandium forms

We need to exclude these elements from the list of Transition Elements because they have the wrong properties so we say:

“A Transition element is a ____-block element that forms ions with an _____________________ d sub-shell”

You don’t have to learn these but…

Notice that on the left (before the half-way point) the most stable Oxidation states are usually the highest available.

But on the right, low oxidation states are most common and that this is +2 – which all Transition elements can do.

High oxidation state Mn/Cr compounds make good oxidising agents because they can fall to a stable low oxidation state

Eg: MnO₄^- --> Mn²⁺

Oxidation State = _____ --> Oxidation State = _____

Eg: Cr₂O₇^2- --> Cr³⁺

Oxidation State = _____ --> Oxidation State = _____

1. Why does Aluminium need to lose 3 electrons?

2. Look at the diagram above, why is it unlikely to lose even more electrons?

3. Look at the diagram below. How does it show why Transition elements can form variable oxidation states?

1. Why might there be a smaller difference between successive IEs for d-block elements? (think about which shell the next electron comes from)

Because Zn & Sc are not Transition elements, we expect their compounds to be ______________ when solid & to dissolve to form _______________ solutions.

You don’t need to be able to explain how the colour is formed in detail yet but you should know the basic Physics.

When complex ions form, the d-orbitals are not at the same energy.

Energy is ____________ to promote electrons from lower to higher orbitals

Energy is ________________ to frequency.

So different energy gaps absorb different frequencies

1. If a compound absorbed the blue end of the spectrum what colour would it be?

2. The electronic Configuration of Zn²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d¹⁰. Why aren’t Zn (II) solutions colourful?

3. The electronic Configuration of Sc³⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁰. Why aren’t Sc (III) solutions colourful?

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A Complex consists of:

1. A central metal ion

2. Co-ordinately bonded Ligands

A co-ordinate bond is a covalent bond in which _________________________

________________________________________________________________

So, to be a ligand a species needs to possess a _______________________.

It can be a neutral compound (NH₃, H₂O) or an ion (Cl^-, OH^-)

The co-ordination number of these complexes is 6 (6 coordinate bonds)

A bidentate ligand should be able to __________________________________

________________________________________________________________

To do this it would need _________ lone pairs, generally on atoms that are separated by 2 Carbon atoms.

The two most common bidentate ligands

You would assume that a complex with a coordination number of 4 would be tetrahedral – and many of them are.

You can fit 6 water ligands round a metal ion but only 4 Chloride ions

Why? ___________________________________________________________

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Why are tetrahedral complexes most common for complexes with a coordination number of 4? _________________________________________

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But Platinum and Nickel generally form square planar complexes

Only Cisplatin has cancer-therapy uses. Why doesn’t transplatin have the same effect? __________________________________________________________

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You should know some catalysts from GCSE and from earlier topics;

• The catalyst for the Haber Process: _____________________________

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

• The catalyst for the Hydrogenation: _____________________________

C₂H_4(g) + H_2(g) ⇌ C₂H_6(g)

• The catalyst decomposing peroxide: _____________________________

2 H₂O_2(aq) ⇌ 2H₂O_(l) + O_2(g)

• The catalyst in catalytic converters: _____________________________

2NO_(g) + 2CO_(g) ⇌ N_2(g) + 2CO_2(g)

These are heterogeneous catalysts – what does this mean?

________________________________________________________

A paramagnetic material is ________________________ to magnetic fields.

A diamagnetic material is ________________________ by magnetic fields.

A paramagnetic material contains some ______________ electrons.

A diamagnetic material does not.

These ions would show ______magentic behaviour

You’d expect Nickel to be paramagnetic but the effect of ligands could make it diamagnetic