Chapter 9: Molecular Structures
402
EndofChapter Solutions for Chapter 9
Summary Problem
Part 1
Answer:
(a) see structure below; tetrahedral, angular, 109.5°,
sp
3
(b) see structure below; octahedral,
square planar, 90°,
sp
3
d
2
(c) see structure below; octahedral, octahedral, 90°,
sp
3
d
2
(d) see structure below;
octahedral, square planar, 90°,
sp
3
d
2
Strategy and Explanation:
Write Lewis structures for a list of formulas and identify their electron geometry, the
shape, and the hybridization of the central atom.
Follow the systematic plan for Lewis structures given in the solution to Question 8.13, then determine the number of
bonded atoms and lone pairs on the central atom, determine the designated type (AX
n
E
m
) and use Table 9.1.
Notice: It is not important to expand the octet of a central atom solely for the purposes of lowering its formal
charge, because the shape of the molecule will not change. Hence, we will write Lewis structures that follow the
octet rule unless the atom needs more than eight electrons.
Use the electron geometry to determine the number of electron pairs around the central atom. Select a hybridization
that accommodates the appropriate number of electron pairs. Use Tables 9.2 or 9.3 for reference.
(a) BrF
2
(20 e
–
)
The type is AX
2
E
2
, so the electronpair geometry is tetrahedral and the molecular
geometry is angular. The bond angle is slightly less than 109.5°. The Br atom must
be
sp
3
hybridized, according to Table 9.2, to accommodate the two bonds and two
lone pairs.
F
Br
F
.
F
Br
F
(b) BrF
4
–
(36 e
–
)
The type is AX
4
E
2
, so the electronpair geometry is octahedral and the molecular
geometry is square planar. The bond angles are 90°. The Br atom The P atom must
be
sp
3
d
2
hybridized, according to Table 9.3, to accommodate the four bonds and two
lone pairs.
.
.
..
..
.
..
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
F
Br
F
F
F
F
Br
F
F
F
.
.
(c) BrF
6
+
(48 e
–
)
The type is AX
6
E
0
, so electronpair geometry and the molecular geometry is
octahedral. The bond angles are 90°. The Br atom The P atom must be
sp
3
d
2

hybridized, according to Table 9.3, to accommodate the six bonds.
.
.
..
..
..
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
F
Br
F
F
F
F
Br
F
F
F
F
F
F
F
(d)
IBrCl
3
–
(36 e
–
)
The type is AX
4
E
2
, so the electronpair geometry is octahedral and the molecular
geometry is square planar. The bond angles are 90°. The Br atom The P atom must
be
sp
3
d
2
hybridized, according to Table 9.3, to accommodate the four bonds and two
lone pairs.
.
.
..
..
.
..
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Cl
I
Br
Cl
Cl
Cl
I
Cl
Br
Cl
.
.