Chemistry Reference
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8.509 ˚ ). Ru(1)O 4 tetrahedra - brown ,Ru(2)O 4 tetrahedra -
Fig. 8 (a) The structure of RuO 4 ( a
¼
8.595 ˚ ). Os(2)F 4 tetrahedra - grey , Os(1)(F/O) 4 tetrahedra
- red , F atoms - green , F/O atoms - yellow
pink . O atoms - blue .(b)OsF 2 O 2 ( a
¼
-He] 56 and [(Nb 3 ) 2 (Nb 3 ) 6 ][
to each of the eight Nb atoms to give [
C
C
-O] 32 .
Nb 3 is
-Ru(2).
For the OsF 2 O 2 structure, we take as parent the hypothetical Li 56 [Re 2 Re 6 ]N 32 .
Justification for this is that the compound Li 56 [Mn 2 Mn 6 ]N 32 does exist, and Re is in
the same group. We require 48 electrons to convert the 32 N atoms to 16 [
C
-Ru, and so we have [
C
-RuO 4 ] with both
C
-Ru(1) and
C
C
-O]
and16 [
C
-F] atoms and the remaining 8 from the 56 Li atoms to convert the Re to
C
-[Os(F 2 )(O 2 )]. The resulting structure
involves a body-centred cube comprising two Os(2)F 4 tetrahedra, the inscribed
icosahedron being formed by six Os(1)(F/O) 4 tetrahedra.
-Os. The nett result is [
C
-He] 56 plus
C
3.2.3 Sodalite and Danalite
Another substructure of Li 56 [V(1) 2 V(2) 6 ]N 32 is that of the sodalite type typified by
Na 8 [Al 6 Si 6 O 24 ](ClO 4 ) 2 [ 30 ] , derived from the hypothetical parent Na 56 [V(1) 2
V(2) 6 ]N 32 , where the V(2)N 4 tetrahedra correspond with SiO 4 tetrahedra, Na(2)N 4
tetrahedra with AlO 4 tetrahedra and V(1)N 4 tetrahedra with ClO 4 tetrahedra
(Fig. 9a, b ) .
In rationalizing this substructure, we begin with the parent structure,
Na 56 [P(1) 2 P(2) 6 ]N 32 (note the substitution of Li atoms by the Na atoms of the
original sodalite). First, we convert the 32 N atoms to 32[
-O] by the acceptance of
32 electrons from 32 Na atoms, leaving 24 Na atoms plus 32 [ C -Ne]. The formula
becomes Na 24 [
C
-O] 32 . Within the [P(1) 2 P(2) 6 ] group, we
now have 2P(1) atoms accept 2e each from Na to form 2[
C
-Ne] 32 [P(1) 2 P(2) 6 ][
C
C
-Cl], and we then
associate with 2[
C
-Cl] eight of the 32[
C
-O] atoms to form 2[
C
-ClO 4 ]: the pseudo-
formula is now Na 20 [
C
-Ne] 36 [[
C
-ClO 4 ] 2 P(2) 6 ][
C
-O] 24 , which we rewrite as
Na 14 [
-ClO 4 ] 2 . We need now to convert [Na 6 P(2) 6 ]
to [Al 6 Si 6 ], and this requires that the 6 P(2) atoms donate 1e each and the six Na
atoms accept 2e each, a net acceptance of 6e for the group, leading to a final
pseudo-formula Na 8 [
C
-He] 36 [Na 6 P(2) 6 ][
C
-O] 24 [
C
C
-Ne] 42 [Al 6 Si 6 ][
C
-O] 24 [
C
-ClO 4 ] 2 , a neon-stuffed sodium
perchlorate sodalite.
 
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