Pissed off losing marks at the deductive question 3.
EDIT: Lost 22 marks minimum in total. Need to work for the other
papers to get the 2.5% back
EDIT again: By the way for the CuO and NH3 question, it is in
part (c). part (b) already has a question which tests on Cu2+
dissolving in NH3 (dropwise to excess) with equations +
observations. So unless they are that kind to repeat the same
answering content (which I highly doubt), the colour of soln should
not be deep blue again.
Yes, I agree that these 3 questions (and the other killer
questions of the paper, if they be of similar difficulty) are
indeed unfairly difficult for H2 'A' levels standard.
-------------------------------------
Structure of NO2 :
NO2 has an unpaired electron (free radical) on N, which
is doubly bonded to an O atom, and singly datively
bonded to another O atom. The N atom has a +ve charge, the singly
bonded O atom a -ve charge, and the doubly bonded O atom has no
charge. N lacks a stable octet in NO2, and the molecule is a free
radical.
-------------------------------------
Why FO2 doesn't exist :
Reason #1
The problem with FO2, is that F (being in period
2), cannot expand its octet (since it doesn't have vacant,
energetically accessible 3d orbitals to use). Hence, it cannot
form a similar structure as ClO2, which has expanded its octet
(4 bond pairs + 1.5 lone pairs on Cl = 11 electrons in terms of an
expanded octet).
Reason #2
For FO2 to have single bonds between the F and two O
atoms, F will gain a 2+ charge and each O atom with gain
a -ve charge, and this separation of charge is
destabilizing, since F is more electronegative than O.
Reason #3
In addition, in such a model, F atom would have an unpaired
electron, making the molecule an unstable free radical.
Reason #4
And furthermore, the F atom wouldn't have a stable octet in
such a model, which defeats the purpose of forming the
structure in the 1st place.
-------------------------------------
CuO reacting with NH3 :
The oxide anion (ie. the base) abstracts a proton from NH3
molecule (ie. the acid), generating first the OH- ion and NH2- ion.
These two anions, together with four other NH3 molecules, act as
ligands to donate dative bonds to the Cu2+ ion, generating an
octahedral complex ion.
Another possibility is that the OH- abstracts a further proton from
another NH3 molecule (since liquid ammonia used means NH3 is
present in large excess), generating H2O and another NH2- ligand.
In such an octahedra complex ion, two ligands are NH2-, one is the
H2O generated, and 3 are NH3 molecules.
The resulting complex ion should still be deep blue in colour, or a
slight variant (eg. cyan or violet) of the deep blue colour of the
usual tetraaminecopper(II) complex ion, since the Cu2+ ion remains
as Cu2+, and the ligands are similar.
CuO + 6NH3 --> Cu(NH2)(OH)(NH3)4
or
CuO + 6NH3 --> Cu(NH2)2(H2O)(NH3)3
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