CHAPTER 9: ANSWERS TO ASSIGNED PROBLEMS
Hauser- General Chemistry I
revised 8i03l08
9.19 Give the molecular geometries of a molecule that has the following electron
domains on its central atom:
(a) four bonding domains and no non-bonding domains
I
X
TETRA I+€DR/+L
//:
(b) ttnee bonding domains and two nonbonding domains
aa
fr
T-
-X-
S HAPEO
I
(c) five bonding domains and one nonbonding domain
>r(
leu,ARE py nnm tDA L
(e) four bonding domains and two nonbonding
SQUNR€
domains
.
)X i
?MNA{I
"\
9-21 Give the molecular geometries for the following molecules
(a) HCN
G)
so32-
L\NEAR
Tftt boAJAL a{ff1,rvrrDA
(c)SF+ See rA^J
Fl-
t ..
o
:
and ions:
r ,. ]a-
C=tl:
[-ljro']
F:
:
;F>
i,
I
t.Ft
r.
lFl
t
(d)
*
L
r
ci
tI
I J
NH3cr* TETRAI+E Dffl.pF6-
o cTA
t+
€DR/+
:
IJ
I
(e)
(r)Nr- L\NeftR
L-''r
[..
l-<+
l:w-N=Nll
L.
J
F
,.
\ b-E
-'ljl-r\ F;
r0
rrJ
CQSo"tq,ncg
I
I
I
I
I
I
9,29 (a) Explain why BrF+- is square planar, whereas BF+- is tetrahedral.
_-a
rFj
l;){Gl
:F
bl
1.(,'f,'
'rJ
_
I
I
TETRAHEDRAL
SQUARE PLAIIAR
WIIILE BOTH STRUCTURES HAVE 4 BONDS TO THE CENTRAL ATOM,
TTIE BFI'ALSO HAS TWO LONE PAIRS TIIAT IMPACT THE GEOMETRY.
(b) Water, HzO, is a bent molecule. Predict the shape of the molecular ion formed from
the water molecule if you were able to remove four electrons to make
tt
rt'
Ato.u
H
- O-
Fl
TIIE (H2o)4* LACKs rHE LoNE pArRs rHAT
FoRCED A BENT GEoMETRy
FOR WATER, THEREFORE IT HAS A LIIYEAR GEOMETRY.
9.76 The molecules SiF+, SF+, and XeF+ have molecular formulas of the type AFa, but
the molecules have different molecular geometries. Predict the shape of each molecule,
and explain why the shapes differ.
,.
.I.
tf'
)_.
'.:F{i'
:.f.' :!:,'
,
lF-f
rt
:5'r I
tF,. t
,*o.) i
lF.
'
TETRAHEDRAL
.o
ro
\.
/Y.i
,ir, r {' \.-,
'..
'
SQUARE PLANAR
SEESAW
WHILE EACH STRUCTURE CONTAINS 4 BONDS ON THE CENTRAL ATOM,
THE PRESENCE OF LONE PAIRS ALTERS THE GEOMETRY.
9.31 (a) Does SCl2 have a dipole moment? If so, in which direction does the net dipole
point? DIPOLE MOMENT: BENT GEOMETRY LACKS SYMMETRY
r.
:cQ
-
at ^t'
)
DrPor€
rt
-c-Q-l
T,uu/A{tD
r Ay OeIruE
l\s(E ENJ (A
(b) Does BeClz have a dipole moment? If so, in which direction does the net dipole
point? NO DIPOLE MOMENT: EQUAL AND OPPOSITE PULLS FOR THIS
LIN.EAR MOLECULB.
t.
r.
i CA-l'
B?--C-Q.:
r r
Poanoxl
Predict whether each of the following molecules is polar or nonpolar:
9.35
(a) IF
,{-f;
LIN/EFIR
It
S=-C=S
(b) cs2
NONPOLAR
aa
Lrlve$R
l'
I o'.
IS\
.:/
ie ,g;
(c) So:
NONPOLAR (double bond not a factor here)
(d) PCl3
T1tl rooru&u
P
uAlAa
P
../r\
j9',,1:
:*"
POLAR
(e) SFo
T-R\bCIN,ftL PYRXtnt0.Fu
.:.
.. .F'
.E r \- ti:
O cxfir+qDRflL
:F- ,f'it
NONPOLAR
:Fr
I-i;
..\
F'
.l
!t
(0IFs
POLAR
-l--
5 OUftfl,A P YRAh\OAL
i? -. I J-- F.,
.,
rl, F'
r |
'v..
9.81
The PFr molecule has a dipole moment of 1.03 D, but BFr has a dipole moment
zero. How can you explain the difference?
of
1F;
-p
r.Ffi,,-ir
.r
;
Fz
h
"- Fi
THE PFT MOLECULE IS TRIGONAL PYRAMIDAL, WHILE THE BF3 HAS
THE SYMMETRY OF TRIGONAL PLAI\AR GEOMETRY WITH IDENTICAL
SUBSTITUENTS.
9.7 The orbital diagram below presents the final step in the formation of hybrid orbitals
by a silicon atom.
-:i:..
-'!-
What type of hybrid orbital is produced in this hybridization? Sp3 HYBRID
ORBTTALS FORM FROM oNE "s" AND THREE "p" ATOMIC ORBITALS.
9.41
Consider the bonding in an MgHr molecule.
(a) Draw a Lewis structure for the molecule, and predict its molecular geometry.
H-Mg-H
LINEAR
(b) What hybridization is used in MgH2?
TWO "REGIONS" CALL FOR TWO sp IIYBRID ORBITALS.
9.47 Indicate the hybridizationof the central atom
(a)
BCl:
(c)
CSz
in
{
,l
!- .,!
lCI.- -el
sp2
S:- c=5
fo
aa
sp
I I
jFit
(d) KrFz
l-
drpt
(e)
&i
F'[,
PFo-
dttpt
rO
kc -
rl
,s: ,..] -
tc
l*
F-"
t
e
I
ADDITIONAL EXERCISE #7
*r
#7
For the ion NOz
a) Draw three resonance forms (use nitrogen as central atom).
-i']
['o=
N
: Ie= g=nJ: Fr'.-
b)
Predict the geometry. LINEAR (use any form)
c)
Predict the hybridization of the central atom. sp (use any form)
9.79 From their Lewis structures, determine the number of @ and
the following molecules or ions:
(a)COz
3er,2.1y
(b) thiocyanate iorq NCS-
]o, alf
fI
xr=o:J
bonds in each
O= C:? e> ,^O - C 3o:6.'O=
[-"
lN:
L"
at
C=S
ll
I
e
C-
fdDno.nce
of
t
lol
ll
(c) formaldehyde, I{2CO
c
30 andlII
HH
(d) formic acid, HCOOH, which has one H and two O atoms attached to C.
40
t?,
andtfl
f{ - c-
,.
g-bl
ADDITIONAL EXERCISE #8
If a carbon atom is "sp" hybridized, what is the maximum number of
zr
bonds that
this atom could be involved in?
SINCE TWO p ORBITALS ARE UNHYBRIDIZED AND AVAILABLE
FOR n BONDING, TWO II BONDS COULD FORIVI.
ADDITIONAL EXERCISE #9
For the following structure
.....
H
I
H-c
tltll
HHHHH
.How many atoms atoms are spt hybridized? FOUR
.How many pi (n) bonds between two carbon atoms exist in this structure? TWO
9.65
(a)
What are the relationships among bond order, bond length, and bond energy?
HIGHER BOND ORDER =+ SHORTER BOND LENGTH
HIGHER BOND ENERGY
==>
(b) According to molecular orbital theory, would either Bez or Bez+be expected to exist?
Explain.
o* 2p
r*
2p
o2p
rr 2p
I
o* 2s
I
I
o2s
I
I
ENERGY
Use for Li. Be. B. C. and N species
Bez has 4 valence e- Diagram shows (2 BONDING
wlLL NOT EXIST
Bez* has 3 valence e- Diagram
shows (2 BONDING
CAN EXIST
-
-
2 ANTIBONDING)
/2=0
1 ANTIBONDINGI I
2= 0.5
9.67 (a) What does the term diamagnetism mean? (b) How does a diamagnetic substance
respond to a magnetic field?
SUBSTAI\CES WITH NO T]NPAIRED ELECTRONS ARE NOT ATTRACTED
BY A MAGNETIC FIELD.
(c) Which of the following ions would you expect to be diamagnetic: Nr'- , Or2- ,8e22*
STRATEGY: USE THE PROPER FRAMEWORK. COUNT VALENCE
ELECTRONS WITH ADJUSTMENT FOR IONS. AI\ TJIIPAIRED ELECTRON
,
Cz-?
LEADS TO DIAMAGNETISM.
o* 2p
o* 2p
rr* 2p
fi*
o2p
rr 2p
rt 2p
ozp
o* 2s
o2s
U,se
I
I
for Li. Be. B. C. and N species
Nr2- 5+5+2=l2valence
Or'' 6+6+2=l4valence
Be22* 2+2 -Z=Zvalence
Cz- 4+4+1=9valence
2p
o*2s
<r2s
Use for O. F. and Ne species
ENERGY
paramagnetic
diamagnetic
diamagnetic
paramagnetic
9.69 Using Figures 9.37 and 9.45 as guides, draw the molecular orbital electron
configuration for the following. In each case indicate whether the addition of an electron
to the ion would increase or decrease the bond order of the species.
Brn
3 + 3 -1 = 5 valence gives paramagnetic with BO = 0.5
An extra electron will INCREASE the bond order.
(b)Liz*
1+ 1- 1= l valence givesparamagneticwith BO:0.5
An extra electron will INCREASE the bond order.
(a)
(c) Nz*
5 + 5 -1= 9 valence gives paramagnetic with BO = 1.5
An extra electron will INCREASE the bond order.
Ne22*
with B0 = 1
An extra electron will DECREASE the bond order (enters antibonding MO).
(d)
8 + 8 -2 = 14 valence gives diamagnetic
Bond Order (BO; = (# BONDING e- minus # ANTIBONDING e-) l2
o" 2p
A
o* 2p
I
rr* 2p
o2p
I
n* 2p
I
I
rt 2p
I
Ir2p
I
o" 2s
I
o2p
I
o" 2s
I
o2s
I
o2s
I
Use for Li. Ee. B. C. and N species
Use for O. F. and Ne species
ENERGY