[Al(H2O)6]3+ is formed from AlCl3 in excess water, because Al is
in period 3, there's energetically accessible vacant d-orbitals ,
so Al can expand it's octet, is it?
Why doesnt it form [Al(H2O)4]3+?
[Be(H2O)4]2+ is formed instead of [Be(H2O)6]2+ from BeCl2 in
excess water, because period 2 elements are unable to expand their
octet is it?
Yes correct. Even if Al can accept just 4 H2O ligands, but it'll
prefer 6, because bond forming (including dative bonds) are
exothermic and hence thermodynamically favorable.
Hi Ultima, may i ask you,
AJC /2014/P3/2(b)
State the shape and bond angle of [Pt(NH3)4]2+
I think the challenge to this question is, to determine the
number of valence electrons of centre atom Pt right?
This is what i did : i do not know how to determine valence e-
of Pt, as we did not learn how to write electronic configuration
beyond period 3 (is it?) .
So i just assume Pt has 2 valence e- , so lost 2 valence e- ,
and 4 NH3 ligands datively bonded to Pt, i get a tetrahedral shape,
bond angle 109°
The answer wrote :
Shape : square planar
Bond angle : 90°
But, accept tetrahedral 109°
How did they get sq planar or assumed pt has 6 valence
e-?
Did a google search for electronic config of Pt , it's complex
and their still debating abt it
May i ask u for guidance for the thought process towards this
qn?
For transition metal cations, you can ignore any lone pairs from
the metal atom itself, and deduce its geometry solely based on the
coordinate dative bonds accepted from its ligands. If the ligands
are huge, eg. Cl- ligands, then to avoid excessive steric strain
from van der Waals repulsions between the ligands, there will only
be 4 instead of 6 ligands. When you know there are 4 ligands, then
whether the geometry of the coordination complex is tetrahedral or
square planar depends on the Jahn–Teller
effect (which is of course way beyond the A level H2 syllabus),
as well as other conditions that favor either stereoisomer (some
coordination complexes can exist as an equilibrium between
tetrahedral and square planar, with the position of equilibrium
affected by other factors beyond A levels).
For A level purposes, unless the coordination complex is
well-known and student familiarity is expected (eg. such as Tetraamminecopper(II)
sulfate and Cisplatin), Cambridge
(and AJC here) will accept either square planar or tetrahedral, as
long as the corresponding bond angles are correct.
Thank you.
Yeah it's perplexing, AJ wrote the answer square planar, and
then a second statement below
also accept tetrahedral,
Its like suggesting sq planar should come to mind first rather
than tetrahedral.
But after your input, i feel that tetrahedral is more
straightforward.
I was thinking, is it becos they just assume structure
consisting of 4 bond pairs? If they did this, then there's a
possibility of seesaw/distorted tetrahedral.
Why then, distorted tetrahedral not accepted/wrong?
Or is it acceptable?
As I said, ignore lone pairs from the transition metal itself, take
it that the only electron charge clouds present are the dative
bonds from the ligands. And the AJC qn was closely related to
Cisplatin, which is a
famous chemotherapy drug and hence you should be aware that it's
square planar and has cis-trans isomerism (although it's not in the
Singapore H2 syllabus, but it's in the IB and other A level Chem
syllabuses, and if you're serious about A grade, you should expose
yourself to stuff like this beyond your own A level syllabus).
