Answer Key
Ch 1a, Problem Set Nine
Side 1 of 8
Problem One – Oxidation Number (20 points, 5 each)
Determine the oxidation number of the metal in each of the following complexes:
a.
[Pt(Cl)
6
]
2-
Pt(IV)
b.
[Pt(Cl)
4
]
2-
Pt(IV)
c.
Mn(CO)
5
Cl,
Mn(I)
d.
[Rh(SO
4
)(NH
3
)
5
]
1+
,
Rh(III)
Problem One – Packing Efficiencies, 40 Points, 10 Each
a.
Case one – Rock salt structure (FCC)
Here ions treated as hard spheres “touch” each other along the edges of the unit cell.
That means that two radii of the corner atom (say, the anion) and one diameter of the
cation make up this edge. Therefore the side of this cube has 2
r
a
+ 2
r
c
and because
it’s FCC, there are 4 of each ion in the cell. Therefore, the packing efficiency is thus:
Packing
FCC
=
4
4
3
"
r
a
3
#
$
%
’
(
+
4
4
3
r
c
3
#
$
%
’
(
2
r
a
+
2
r
c
( )
3
=
2
3
r
a
3
+
r
c
3
( )
r
a
+
r
c
( )
3
Case two – Cesium chloride structure (BCC)
In a BCC structure, ions touch along the inner diagonal of the unit cell. From simple
geometry, a cube of side
x
has an inner diagonal of
3
x
, or if it has an inner diagonal
of
x
, it has a side of length
x
/
3
. Therefore, the length of a side is
2
r
a
+
2
r
c
( )
/
3
,
and because it’s BCC, there is only 1 of each ion in the cell…
Packing
BCC
=
4
3
r
a
3
+
4
3
r
c
3
2
r
a
+
2
r
c
( )
/ 3
[ ]
3
=
3
2
r
a
3
+
r
c
3
( )
r
a
+
r
c
( )
3
Case three – Zinc blende structure (ZnS)
This is trickier, but still possible. In zinc blende half of the tetrahedral holes are
filled. Therefore it is easier to focus on one of the tetrahedral hole regions, or a single
octant of the cube. In one octant, a cation occupies every other corner of the cube,
and the anion is in the center. Therefore the inner diagonal of this octant is 2
r
a
+ 2
r
c
and the side of this octant is
2
r
a
+
2
r
c
( )
/
3
. Again, recall that we must consider the
entire unit cell, whose sides are twice the length of the octant’s sides. Therefore, a
side of the unit cell is 2
2
r
a
+
2
r
c
( )
/
3
long, and there are 4 of each ions inside…