Chemistry Reference
In-Depth Information
Fig. 5 (a) The structure of CdPd
3
O
4
(
a
= 5.747
˚
) to be compared with that of Li
7
VN
4
. Ca atoms -
grey
, Pd atoms -
red
, O atoms -
light blue
.(b) The structure of Li
7
VN
4
(
a
= 9.6064
˚
). V(1)N
4
tetrahedra -
grey
, Li(2)N
4
tetrahedra -
red
, Li(4)N
4
tetrahedra -
light blue
.(c) The insertion of the
A
15 CaPd
3
structure into an extended array of O atoms (
light blue cubes
) only partially occupied
by Ca atoms -
grey
O atoms, and the four remaining electrons could be then located as Lewis Pairs at
the 2
a
sites (0, 0, 0; ½, ½, ½), giving rise to the pseudo-formula
-(Os
2
W
4
)Ne
4
þ
(2e
)
2
. The two electron pairs (Lewis pairs) would occupy the 2
a
positions at
(0, 0, 0; ½, ½, ½), like the Si atoms in the Cr
3
Si. Because Pt(IV) and Pt(II) are
equivalent to
C
-[
76
Os], Pt
6
O
4
should be equivalent to an O
8
-stuffed
C
-Os
2
W
4
O
2
, a compound which has not been reported so far but which is close to
-[
74
W] and
C
C
-W
2
W
4
(2e)
2
.
An alternative explanation would imply the existence of six Pt(IV) atoms
donating a total of 24 electrons. Because only 16 can be accepted by the O atoms,
the 8 remaining electrons would play the role of either 2
C
-W
2
W
4
O
2
¼ C
-
4
Be (with a total of
C
four electrons) or 2
-Si(Ge), with four valence electrons (Si, Ge). The difficulty
of this model resides in the impossibility of locating four electrons at the same
site. However, it is noteworthy how both pseudo-elements,
C
-
4
Be and
-Si(Ge),
lead to the pseudo-formulae which agree with those of real compounds having
the
A
15 structure, i.e. Mo
3
Be and W
3
Ge, respectively.
C
C
