Chapter 7
7.3 see textbook, page 609.
ANS
Teflon is a polymer and the formal name is poly(tetrafluoroethylene). CFCs
are chlorofluorocarbons. The primary similarity is that both contain strong C-F bonds.
NaF is ionic solid and has a crystal structure similar to that of NaCl, while
Teflon is a polymer and has the structure similar to polyethylene.
The lattice energy is proportional to electrostatic interaction energy.
Uα
Z+Z-/r
7.15 see textbook, page 609.
ANS In a covalent bond, the electrons involved are shared between the nuclei,
whereas in ionic compounds, an electron is transferred from one atom to another.
Ultimately, the difference lies in the extent to which an electron is transferred during
the bond formation.
7.35 see textbook, page 610.
ANS
(a) O-H is the most polar. Since C and S have the same electronegativity value of 2.5,
C-H and S-H should have similar polarity.
(b) H-Cl is the most polar and C-Cl is the least.
(c) C-F is most polar and F-F is nonpolar.
(d) N-H is the most polar and N-Cl is the least polar.
7.57 see textbook, page 611.
ANS Overlap is the buildup or diminishment of the amplitude of a wave because of
the interaction of two or more waves. Buildup results from constructive interference,
and diminishment results from destructive interference.
7.71 see textbook, page 611.
ANS
(a) # of e-s around central atom I = 7 + 3 = 10 e-s = 5 e- pairs
In the form of AX3E2 trigonal bipyramidal arrangement, the 2 lone pair electrons
occupy two equatorial positions of trigonal bipyramid and the molecule is in T-shape.
(b) # of e-s around central atom Cl = 7 + 1 = 8 e-s = 4 e- pairs
In the form of AO3E tetrahedral arrangement, the lone pair electron occupies one
corner of the tetrahedron, and the molecule is in Trigonal pyramidal shape.
(c) # of e-s around central atom Te = 6 + 4 = 10 e-s = 5 e- pairs
In the form of AX4E trigonal bipyramidal arrangement, the lone pair electrons
occupy one equatorial position of trigonal bipyramid and the molecule is in Seesaw
shape.
(d) # of e-s around central atom Xe = 8 e-s = 4 e- pairs
In the form of AO4 Tetrahedral shape.
(a) # of e-s around central atom P = 5 + 4 -1 = 8 e-s = 4 e- pairs
In the form of AX4 Tetrahedral shape.
(b) # of e-s around central atom S = 6 + 4 = 10 e-s = 5 e- pairs
In the form of AX4O trigonal bipyramidal arrangement, the Oxygen atom
occupies one equatorial position of trigonal bipyramid.
(c) # of e-s around central atom Cl = 7 + 1 = 8 e-s = 4 e- pairs
In the form of AX2E2 tetrahedral arrangement, the two lone pair electron occupy
two corners of the tetrahedron and the molecule in Bent shape.
(d) # of e-s around central atom I = 7 + 2 + 1 = 10 e-s = 5 e- pairs
In the form of AX2E3 trigonal bipyramidal arrangement, the three lone pair
electrons occupy the three equatorial positions of trigonal bipyramid and the molecule
in Linear shape.
ANS
When a multiple bond is around the central atom, it can only occupy one
position. Therefore, a double bond will not contribute any extra electrons around the
central atom, while a triple bond has to subtract one electron from the central atom.
Examples:
For OSF4 molecule, # of electrons around central S atom are the sum of valence
electrons from S and one each from S-F single bond, and O=S bond was not added in
the counting.
For HCN molecule (C-N is a triple bond),
# of electrons around central C atom = 4 + 1 -1 = 4 => 2 e- pairs => Linear molecule
7.84.
ANS
IO3-