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.