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Fig. 3 (a) Corner-sharing rods of Li(2)N 4 -centred octahedra: vertices are Li(4) atoms - light blue ,
Li(1) atoms - brown , isolated V(1)Li(1) 6 octahedra - grey .(b) Corner-sharing rods of Li(2)-
centred cuboctahedra: vertices are V(2) atoms - black , Li(4)atoms - light blue , Li(1) atoms -
brown . Corner-shared V(1)Li(1) 6 octahedra - grey
such cation lies at the centre of a Li(V)N 4 tetrahedron whose vertices are common
always to eight such tetrahedra as a consequence of the anti-fluorite-type structure
of Li 7 VN 4 . The Li(2)N 4 tetrahedra, shown in Fig. 2b , are themselves each at the
centre of an octahedron of other LiN 4 tetrahedra: Fig. 3a shows only the central
Li(4), Li(3) and Li(1) atoms of these peripheral tetrahedra, which are the vertices of
the octahedra. Again, the rod structure is clearly shown, but now these octahedral
rods share corners along the three cubic axes (delineated by red, blue and green),
leaving the isolated V(1)Li(1) 6 octahedra, shown in grey, occupying the interstices
between the rods. And we can take this approach still further to consider the
polyhedron (Fig. 3b ) centred by each of the rod octahedra: it is a cuboctahedron
whose vertices are also LiN 4 tetrahedra (but not shown as such, only as the centreing
Li atoms). These cuboctahedra now share corners, completely filling the unit cell,
but leaving the V(1)Li(1) 6 octahedra in the interstices. The V(1)Li(1) 6 octahedra
(grey) are no longer isolated but share corners with the cuboctahedra. Thus, the
complete structure of Li 7 VN 4 is now represented as cuboctahedral rods with
isolated octahedra in the interstices, just as in the structure of Cr 3 Si.
Although the chemical meaning of this structure will be described in Sect. 2.3.3,
we can put forward a possible explanation as to why this structure type is chosen
by the Li(2) and V(1) atoms of g -Li 7 VN 4 (and also by the V(2) and V(1) atoms).
We treat here the respective Li(2)N 4 and V(2)N 4 tetrahedra as pseudo-atoms. If we
take the parent structure as Li 56 [V 2 Nb 6 ]N 32 rather than Li 56 [V 2 V 6 ]N 32 , then in
Fig. 4d , the grey tetrahedra are still VN 4 but the red tetrahedra are now NbN 4 .We
rewrite the parent formula as Li 56 [(V 2 N 8 )(Nb 6 N 24 )].
By applying the Zintl-Klemm concept, we first convert the N atoms to [
C
-O],
each N atom accepting one electron from a Li atom which then becomes [
C
-He]:
now we have the compound Li 24 (V 2 [
-He]. It should
be added that the 6 of the 24 Li atoms are occupying a similar position as the 6 Nb
atoms. Now we have the formula Li 18 [(V 2 [
C
-O] 8 )(Nb 6 [
C
-O] 24 )
þ
32 [
C
C
-O] 8 )(Nb 6 [
C
-O] 24 [Li 6 ])
þ
32 [
C
-He].
 
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