CHE
211
Answers
in
BOLD
RED
EXAM
3
(Ch.
8­10)
KEY
On
this
mastery­based
exam,
you
must
score
at
least
a
70%
to
complete
the
exam.
If
your
score
is
less
than
70%
you
will
have
to
do
a
retake
of
one
or
two
chapters’
material
(normally
those
on
which
you
did
worst).
Any
new
points
beyond
what
you
score
this
time
on
those
chapters
will
then
be
added
to
your
score
up
to
a
maximum
of
75%.
Multiple
Choice
(3%
each)
and
Problems
(6
or
7%
each,
partial
credit
possible)
grouped
by
chapter
Please
select
the
BEST
answer
to
each
multiple­choice
question
and
mark
it
clearly
on
your
answer
sheet.
Problems
should
be
answered
in
the
space
immediately
after
each.
SHOW
ALL
WORK
(if
any)!
Thank
you.
CHAPTER
8
27
question
points
+
6
problem
points
=
33
points
1.
Which
type
of
orbital’s
electrons
best
shield
the
nuclear
charge
from
the
outer
shells?
a.
f
b.
p
c.
s
d.
d
2.
B
and
F
both
lose
an
outer
p
electron
if
ionized
to
a
+1
ion.
Why
is
the
energy
required
for
F
much
greater?
a.
F
has
more
electrons
b.
F
has
a
higher
nuclear
charge
c.
B
is
a
metalloid
d.
F
is
a
nonmetal
3.
Which
of
these
elements
is
a
Transition
metal?
a.
Ge
b.
Re
4.
Which
of
these
is
likely
to
have
the
most
endothermic
first
Ionization
Energy
value?
a.
K
b.
S
c.
Na
d.
Al
5.
Based
upon
the
normal
rules
of
orbital
filling,
how
many
unpaired
electrons
does
a
normal
As
atom
have?
a.
3
b.
2
c.
1
d.
0
6.
Which
sequence
of
three
successive
ionization
energies
is
most
likely
to
be
that
for
Ca?
(all
are
in
megaJoules)
a.
1.10,
1.90,
2.91
b.
1.52,
2.67,
3.93
c.
0.42,
3.05,
4.41
d.
0.59,
1.14,
4.91
7.
How
many
outer
electrons
does
S
have?
a.
4
8.
Which
of
these
isoelectronic
ions
will
be
the
smallest?
a.
S2–
b.
P3–
c.
K+
9.
Which
of
the
following
oxides
is
most
likely
to
produce
a
basic
solution
when
dissolved
in
water?
a.
CO
b.
SrO
c.
SO2
d.
Al2O3
c.
Ba
d.
Be
b.
1
c.
6
d.
14
d.
Cl–
PROBLEM
Using
spectroscopic
notation
(1s2…,
etc)
AND
ALSO
orbital
diagrams
(lines
or
boxes
with
arrows
=
electrons)
give
the
proper
electron
filling
for
the
ground
state
atoms
or
ions
listed
below:
Ca
Mn
Co3+
ion
Ca
1s22s22p63s23p64s2
Mn
1s22s22p63s23p64s23d5
Co+3
1s22s22p63s23p64s03d6
[or
Ca[Ar]4s2
here
and
on
others]
methods
of
modeling
can
use
either
lines,
boxes,
or
circles
for
the
following
4s
4s
3s
3p
2s
2p
3s
3p
2s
2p
1s
1s
3d
4s
3d
3s
3p
2s
2p
1s
3d
CHAPTER
9
(BONDING
MODELS)
10.
27
question
points
+
7
problem
points
=
34
points
Which
of
the
following
is
the
wrong
Lewis
dot
symbol
(model)
for
the
element
indicated?
Silicon
(answer
a)
Si
Cl
Mg
a.
b.
c.
d.
Ga
11.
The
reaction
2
Mg(s)
+
O2(g)

2
MgO(s)
has
∆H°RXN
=
‐1202.4
kJ.
If
the
values
of
the
key
terms
(all
per
mol)
for
Mg
IE1
=
737.7
kJ,
IE2
=
1450.7
kJ
and
Mg(s)

Mg(g)
=
127.6
kJ
with
O
EA1
=
–141
kJ
,
EA2
=
+844
kJ
and
½
O2(g)

O(g)
=
249
kJ,
what
is
the
value
of
the
lattice
energy
that
can
be
determined
from
all
these?
a.
–3728
kJ/mol
b.
–3869
kJ/mol
c.
–4470
kJ/mol
d.
–2667
kJ/mol
12.
Which
of
these
substances
is
most
likely
to
be
an
ionic
substance
with
the
greatest
lattice
energy?
a.
MgO
b.
CsSe
c.
SiC
d.
SF4
13.
Which
bond
is
the
shortest?
a.
C≡C
14.
An
unknown
substance
is
a
soft
solid
&
does
not
dissolve
in
water;
it
melts
at
125°C.
Its
bonding
is
most
likely
a.
ionic
b.
covalent
c.
metallic
d.
unpredictable
15.
Using
the
values
of
average
bond
energies
listed
in
the
data
sheet’s
table,
determine
∆HRXN
of
this
reaction:
H
a.
–209
kJ
O
O
O
C
b.
–1143
kJ
H
N
+
c.
–212
kJ
+ N
O
+ 3 O O
2 H
H
C
H
d.
–2635
kJ
O
H
b.
C=C
c.
C–C
H
H
d.
all
are
equal
16.
Based
on
normal
electronegativity
values,
SiH4
(all
H
on
Si)
will
be
______________,
while
MgSi
will
be____________.
a.
nonpolar
covalent;
polar
covalent
b.
polar
covalent;
ionic
b.
nonpolar
covalent;
ionic
d.
ionic;
polar
covalent
17.
When
you
are
forming
a
covalent
bond,
energy
is
released
as
the
two
atoms
approach
each
other…
a.
until
the
nuclei
touch
b.
unless
they
are
the
same
element
c.
until
the
nuclear
charges
repel
enough
18.
When
two
elements
of
unknown
oxidation
state
react,
we
assign
a
negative
oxidation
state
a.
to
the
least
electronegative
b.
to
the
more
electronegative
one
c.
to
the
later
group
member
PROBLEM
State
the
typical
properties
of
an
ionic
substance
and
of
a
metal;
then
explain
these
using
the
nature
of
the
bonding
in
each.
In
other
words,
use
the
differences
in
bonding
of
an
ionic
and
a
metallic
substance
to
explain
the
normal
properties
of
each
type
of
material.
Include
their
responses
to
a
sharp
hit
with
a
hammer,
their
ability
to
conduct
electricity
or
not,
and
their
normal
melting
and
boiling
temperatures.
Type
Ionic
Structure
A
lattice
of
oppositely‐charged
ions
in
a
3D
array
A
series
of
metal
cations
with
their
former
valence
Metallic
electrons
in
a
loosely‐
held
‘sea’
of
delocalized
electrons
Hammer
hit…
Conductivity…
Melt/Boil
Point…
Does
not
conduct
when
Solid
will
break.
solid
because
ions
Both
are
high
temperatures
The
movement
of
any
of
cannot
move.
When
because
the
lattice
energy
the
ions
will
cause
like‐
melted
or
dissolved
in
must
be
overcome
to
allow
charged
ions
to
come
into
water,
ions
can
separate
many
movement
of
the
contact,
repel,
and
crack
and
become
mobile,
ions.
apart.
conducting
charge.
Dislocation
of
the
metal
cations
is
easy
because
the
electron
sea
will
simply
‘flow’
with
them,
allowing
changes
in
shape
Delocalized
electrons
can
readily
flow
through
the
solid,
making
it
conductive.
Melting
points
are
fairly
high
because
the
metal
ions
and
electrons
have
a
strong
mutual
attraction
that
must
be
overcome.
Boiling
points
are
VERY
high
because
forces
must
be
totally
overcome
to
boil.
CHAPTER
10
(MOLECULAR
SHAPE)
27
question
points
+
6
problem
points
=
33
points
19.
Which
compound
cannot
meet
the
octet
rule
due
to
it
having
an
odd
number
of
valence
electrons?
a.
NH3
b.
CF4
c.
BF3
d.
NO2
20.
How
many
nonbonding
pairs
(‘lone
pairs’)
are
there
altogether
in
the
compound
H‐O‐Cl
(hypochlorous
acid)?
a.
2
b.
3
c.
5
d.
10
21.
Which
of
the
following
has
a
trigonal
planar
molecular
geometry
based
on
normal
bonding
and
VSEPR
class?
a.
NH3
b.
BCl3
c.
H2O
d.
BrF3
22.
In
the
Lewis
structure
for
SOF2,
each
F
has
a
single
bond
to
S
while
the
O
has
a
double
bond
to
S.
What
is
S’s
formal
charge
in
this
structure
(S
has
a
lone
pair)?
a.
+2
b.
+1
c.
–1
d.
0
23.
Based
on
the
electronic
&
molecular
geometries
that
VSEPR
predicts
for
BrF5,
what
is
the
F–Br–F
bond
angle?
a.
≈
90°
b.
≈
109°
c.
≈
120°
d.
≈
45°
24.
What
molecular
geometry
name
is
assigned
to
the
predicted
VSEPR
geometry
of
BrF5
in
question
23?
a.
trigonal
bipyramidal
b.
square
planar
c.
tetrahedral
d.
square
pyramidal
25.
Resonance
will
be
necessary
to
adequately
describe
the
true
bonding
character
of
a.
CO2
b.
NH4+
c.
H‐CN
d.
CO32–
26.
Xe
forms
XeF2
and
XeF4.
Assume
Xe
has
electronegativity
≈
2.5.
Which
Xe
compound
has
a
dipole
moment?
a.
XeF2
b.
XeF4
c.
both
d.
neither
27.
Which
of
the
following
would
be
predicted
to
NOT
have
a
dipole
moment?
a.
CO
b.
PF5
c.
CHF3
d.
NF3
PROBLEM
DRAW
Lewis
structures
for
PCl3
and
for
HN3
(in
which
all
the
atoms
are
in
line:
H‐N‐N‐N).
Then
name
the
electronic
geometry
and
the
molecular
geometry
of
the
P
atom
in
PCl3
and
the
second
N
(bold
above)
in
HN3.
P
Cl
Cl
H
Cl
Tetrahedral
electron
geometry
Trigonal
pyramidal
molecular
geometry
N
N
N
Linear
geometry
(both
categories)
Note
the
HN3
molecule
has
several
alternative
ways
it
can
be
drawn;
this
is
the
most
ideal
but
I
will
accept
a
few
others
that
are
almost
as
good
as
long
as
there
are
no
formal
charges
of
over
±1
EXTRA
CREDIT
(can
only
receive
five
points
if
the
problem
above
is
done)
Redraw
your
HN3
structure
and
determine
the
formal
charges
on
each
N
atom.
SHOW
YOUR
WORK!
5
–
2
–
6/2
=
0
H
N
N
N
5
–
4
–
4/2
=
–1
5
–
0
–
8/2
=
+1