Ok, I've checked it out, with multiple high-level reliable
sources. I know you guys are gonna hate the findings I'm about to
reveal. (I actually love it. But then, because SEAB-Cambridge might
not be as open-minded as to see the bigger picture, meaning
in effect you guys might still lose your marks even
though your answer is arguably correct, so I can understand and
sympathize, that you might not find this quite as fascinating
as it really should be. Lastly, there's the matter of the ratio of
ions involved. If this is specified in the actual question, that
those are clues (that you ought to have taken into
consideration) leading to only 1 acceptable answer per
question part, in which case you can't say SEAB-Cambridge
wasn't fair to you).
[CuCl2]- or dichlorocopper(I) anion exists, and is linear
geometry.
But [CuCl4]3- or tetrachlorocopper(I) ion exists too, and can
exist in either tetrahedral geometry, or octahedral & square
planar geometry if it's [Cu(Cl4)(H2O)]3-. 6 ligands form an
octahedral geometry, the Cl- ligands occupy square planar while the
H2O ligands occupy top and bottom.
[CuCl4]2- or tetrachlorocopper(II) ion exists too, and
can exist in either tetrahedral geometry, or
octahedral & square planar geometry if it's
[Cu(Cl4)(H2O)2]2-). 6 ligands form an octahedral geometry, the Cl-
ligands occupy square planar while the H2O ligands occupy top and
bottom.
Since all 5 of the above ions exist, arguably SEAB-Cambridge
should accept more than one possible answer for at least some of
the questions. But whether they do, is another matter altogether.
Also, I didn't actually read the exact phrasing of the question
itself, and if the SEAB-Cambridge question specified the ratio of
Cl- ions to the metal cation in each case, then there may only be 1
acceptable answer for each question part, and you can't say
SEAB-Cambridge is being unfair to you.
A couple of points (this may go against how you're feeling now,
but Chemistry is really about learning how beautiful the Universe
is, not about scores/marks/points and hollow paper
qualifications. So take this as a post-course continuing journey of
exploration and wonder).
The more (dative aka coordinate) bonds a central metal ion can
form with ligands (electron donors), the happier it will be. Simply
because more bonds formed results in a more exothermic process,
which as far as Gibbs Free Energy is concerned, is
certainly a good, feasible thing. True, there might be a
slight decrease in entropy as a result (which is not a good,
feasible thing), but usually the favourable enthalpy change
outweighs the unfavourable entropy change. There is also electronic
stability to consider, of course.
You should be aware that Be2+ ion, being in period 2, does not
have empty 3d orbitals required to expand its octet. Consequently,
any complex ions formed with Be2+ has to be tetravalent or 4 bonds,
ie. tetrahedral geometry.
In that case, why would Cu+ or Cu2+ want to just form CuCl or
CuCl2 and stop there? Wll there's obviously the matter of
molarity and availability to consider. If there's enough Cl- ions
available, it makes sense that you would want to get [CuCl2]-
ion and (adding even even more Cl- ions) [CuCl4]3- ion, and
for Cu2+ ion, you would obtain [CuCl4]2- ion.
But if the more bonds formed the merrier, then why stop at 4 Cl-
ion ligands? Why not 6 Cl- ligands? Because Cl being in period 3,
is too huge to have 6 Cl- ligands crowding around the central metal
ion. There would be too much electronic and steric repulsion.
So we only get 4 Cl- ligands? Then obviously, we would get a
tetrahedral geometry, wouldn't we? (Notice the JC notes chillax posted
above, like all JCs' notes, are inadequate in that they just expect
students to memorize without understanding, without proper
explanation. Typical JC stuff.)
But compare the H2O molecule to Cl- anion. H2O is much smaller
(for 2 reasons - Cl- is in period 3 and it is an anion too). Hence,
the [Cu(H2O)6]2+ and [Cu(H2O)6]+ ions exist (though the process is
not as exothermic as if Cl- ligands were used, hence ligand
displacement or ligand exchange takes place).
Which means that even in tetrachlorocopper(I) and
tetrachlorocopper(II) ions, it is possible, under different
conditions, that water molecules might squeeze in between the Cl-
ligands, and you would get the octahedral geometry. 6 ligands form
an octahedral geometry, the Cl- ligands occupy square planar while
the H2O ligands occupy top and bottom. (notice that the JC notes
don't explain this? tsk tsk.)
The reason why Cl- ligands occupy square planar, is because
they're much larger, bring greater electronic and steric repulsion
as compared to the smaller H2O ligands, and hence square planar for
the 4 Cl- ligands would be furthest apart and the most stable,
while the smaller H2O ligands occupy top and bottom.
That being said, the tetrahedral [Cu(Cl)4]2- and
[Cu(Cl)4]3- complex ions also exist, just as the 'square planar'
(octahedral, really) [Cu(Cl4)(H2O)2]2-) and [Cu(Cl4)(H2O)2]3-)
complex ions also exist.
Delightful, ain't it? 
We Love Chemistry
