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In this section we give you the chance to look more closely
at the lanthanide elements and their chemistry.
For now, let's shrink the main table.
First of all, the lanthanides are all metal.
Here's a selection.
Here's lanthanum.
Samarium.
Dysprosium.
If it's heated in chlorine, dysprosium forms a trichloride.
And all the other lanthanide metals do the same.
Chlorine oxidises all of them to the +3 oxidation state.
But the simplest way of reaching this oxidation state
is to dissolve the metals in dilute acid.
Here's praseodymium.
Hydrogen is evolved.
And if the reaction is done in air, then whatever the lanthanide metal,
the final product is the aqueous tripositive ion.
In this case it's green.
Compounds or ions in other oxidation states are much less stable.
For instance, the dipositive aqueous ion of samarium can be made.
It's blood red.
But it's quickly oxidised by water or hydrogen ions
to the pale yellow tripositive ion.
The highest known oxidation state of any lanthanide element is +4.
The most stable example is cerium 4 which occurs in cerium dioxide.
This is almost colourless.
Sulphuric acid converts the dioxide to a sulphate of cerium 4.
It consists of orange crystals.
When the orange sulphate is dissolved in water
the orange tetrapositive aqueous ion is formed.
But even cerium 4 species are easily reduced to the +3 state.
We'll use hydrogen peroxide as the reducing agent
and add it to the orange aqueous ion.
Oxygen is involved as the cerium 4 is reduced.
The colourless product is the aqueous tripositive ion of cerium.
So for all of the lanthanides
the +3 stage is very stable to oxidation or reduction.
The lanthanide series which begins with lanthanum
and ends with ytterbium
is a set of unusually similar elements.
Take a little time to study the pictures and information
that we've provided on the different metals.
And as you examine the reactions with acid,
ask yourself how these elements differ
from the reactions that you studied earlier
in the transition metal programme.