Document 6594095

Transcription

Document 6594095
Chemistry l04A
Exam 2-1013012014
Valence Atomic Orbital Energies
tU" (15 points) MO Theory, UV-Vis Spectra
(or partially
Electrons can be promoted from full (or partially full) orbitals to vacant
in the
occuls
vacant) orbitals by photons of light. The lowest energy absorption usually
visible or ultraviotei (uv) regio-n of the spectrum, and results in the promotion of an
(LUMO)
electron from the highLst o..rli.a MO @OMO) to the lowest unoccupied MO
of the molecule or-ion. Using MO theory, explain the following observation: F2 is
virtually colorless, while Clz is yellow; Br2 is deep red, and 12 is purple.
fo" aW htl'o6orr+ U) , l'+0rlo 7 4, ",nA Lu440' fi,
tr'- 'b2o -V* &\/ LV"V<- W/4"- LLLlt4o A T)
4t :
9cr'-
ovbto
*
IL
,t
--rd
It
nbs W,,nu
^br*bg
bb^<---*d,t*,'t
Cy: abro'V*
0*;
- r^*L
Cen'il. Ko"t/u-lu*
*;" .-.
.!_
4d
r
pozoL o,l--l"f
zL-?,'\ 'ghu&
s*,o,/,( )",,r/- ilrl*t,*'
r>z&'"+
UI
A,@l)u
-/>--.-
/ 1\
f\-'l? a
>v : l*q)
boq/)- 5'latgPwealvn) e- r^zrl/r+
l^.p
t2]. (15 points) MO Theory, Electron
-'/
I
?Oor- a/%W
\erru,lrL
ftaao-cu* g^e
n1
L*
?-'
-n- r'-i,l
t-
Affinity
rl
-?_
oz (0.40 ev)
Explain why the electron affinity of cz is much larger (3.54 eV) than that of
orN2 (-2.2 eV).
4o
ftD Pi^'3''*' q"4)
/,-\ \ t-eA"_,fut/
;:--''-
r r,-_q,
I
?
'lb 7b'L
@@
uc
fr)
-4..4b..
,-T ly
-1t"
fuupts.cl+ aalX
e
cenn";'4",
lt4 e,
f+t*
d*{
{,,ilu
f^ unlret, 2>, 0- f,,/t h'04
v( l- W qrfii b,r/,1 4,{.o> (u# )
s'tryA L\a,o / q
^W7b
A- grnru,/'[,- AA,
giri^rtr
13] (30 points) Valence
Bond Theory
Experimentally, we know the axial and equatorial positions in PHs are non-equivalent,
with bond tengths such that raxiar ) requatoriat: Valence bond theory treatment of such
trigonal bipyrimidal molecules requires sp3d orbital hybridization. Determine the
wavefunctions for the 5 hybridizedorbitals use the following in-plane coordinate system,
and explain why rrgrl ) fequaroriar.
5 + cr % - c6tu +4"!;
?.r= r,t) -Lr?1 -Q(r 4c6J{
f,, = L\i
+ C7 fr + cq olr*
?., :.
c1
frvn 6u,,tof47. % ^
evtu
{
nurqil4''5,.
L(
E6 u',/'/e-ra
c,o*
)*
V)a,ur-l,,*t A
I
+c{=
/or*'
'b r* +f-a;
'(0.-, -i-fu *b"e;
?q, C, +i,-?g -,boo
\
?o,
Ofurf "lo8, ct c,t + gcy = o
7 6rbiW ! Ar-* + 5c? =
2ci ;5*:
ft)r*- fr,- z
=+ ,t=b B)
\2
'
l+C=
.I*r;_
C; .b
=
)-otx
- ,fo
' 6 fu
= *,
'op.
'?,
lg
tE=
I
i>
C.r =O.
An'oX
+;rn
f / ,o/.A
gSZ, s, 677" ?
br/+; toZ
&^M*1
bon&>"
0
brq*. g
.U
t/UlAt
Cl4^rrv+'r-'
a- dko*r-
t*ff
Vvld6e-
i^ eEuM.,/ b4
*w -t'r-^,
(
d''4,
"kn
L oUrn*+--^ )i,r nv',r^
bDruLg
Wr+ loryr--),
lq @A points) Molecular Orbital Theory
1.
(a). Start with a C2 cluster and sketch its MO diagram. Fill in the electrons, and find
the LUMO and HOMO. Sketch the shape of the LUMO and HOMO. What is the
CL4C
,,8,
/r\
Ir--^
,\
ttvfI u
'/
\
\
/t
l---"---j
-'loilol
ett.
''/'-T
-t/ [',R.
2ro
''4L'ry-'
e*q
lzt
-'\1,qil 'z
-..
.+lrfi'5
\1b
"
tq
\*o,h,,
lz
,1
G
c-- c
@
L.uw"
69
L-c
^7 €9
@
e,o-3-s/
f;*;TW
an4,'-ba"/,
T
(b). Now let's take a look at a CO molecule. Similarly, sketch its MO diagram. Fill in the
electrons, and find the LUMO and HOMO. Sketch the shape of the LUMO and HOMO.
What is the bond order of the CO molecule? Sketch the photoelectron spectrum for CO.
What would you look for in the PE spectrum to distinguish CO from Cz?
L
c.o
O
7@"'
-;
lz'
-1N,/:<-\\
t
l.
t\
\
t, '-'lL
^_
,4+Y 1*--
-11,*'v--
"
,-'\:
-15 X<-r/
'7u=^
lf ',
'1
)-
46'€e-/
'
lo
W4to:
Lo fEt '.u''*,Q h""v-
Af
orll;*'oar,l. , b*
0'@ oo
>U,l/t
:
I
zP
q -o c
@
@
UW
r1*rtv/
*&3,
) (rrr"l- ba"ah'g )
(9
kAL, ..^
(c).
Which species (C2 or CO) is more stable? Why?
D
br- wLo'rt/ gWe o\t*- *"
'U7
borr.& arok^. Orw cenW 4s, a/Xtac{-{4'"f C'Lo
v/o'4\il
,1o99e99e9
A-
hn'gh.--
No'{//) *rr(-
bwu
Lt4,t/z-t, ,a.akiV
tV"r-.1-;,rv. Lwk: CO cat, bc-- {r*&
q 79d,*.rX, A C",rr, rrt t )
i+
lU
Wor^t-,'
.
(d)"
Consider the following reactions,
Hr*C=C+HC=CH
BHr+CO+H3B-CO
Which MO is used for Cz and CO?
fr. Q'.
LU,//-()
/
o/c-,rla,,y>
tldl 4
Hu fta-U<:
lcz>taff o'*'^l'"6'
,LDV: WLuw
-/(- e
ff2"ue
I
L-a
C
fu-
CC', fro,qo
W
I
Ce"@
jh @..o@
14
vo{z'. Co
Lo<,trts vouloL Xi"'<-
L.z bol,*, du'oh 6{6,n>r?
vna,,4rrlr, ul^er^, art qu*4V'n,
Le*iz
tl
(,)
l4'ru12+*r\-
oo
C-
6
WW @
@
c
'
(e). C+ can be considered as the combination of two units of Cz. Let's consider two
possible geometries: linear Ca (butadiene form) and square planar Ca (cyclobutadiene
form). Sketch a partial MO diagram for each configuration considering only frontier MOs
and draw the orbital interactions. Is there any difference between the two configurations
in the bonds formed? What if we combined two Cz2' units (isoelectronic to CzHz)? How
would the bonding change?
Aar|h
Llne,,<'.
\ brt*
on
?X*&l
(
Naw/tu,tto, W,ilz//P./@, cuta/tu,a),
5r, y(ort.---,
(aL"'c-=2
D
I
f=
?
/22
v
-.
I
l,fuwt
tzr
W,
,
,1/
,o-
'"'1''#
-.1r
4H
'-__W_'Z
4t ,'.1\//--
-r3--;'J e->sltq ./rg*.1.-ra2
o
1/No,wt ffo,llo .z-
ban l<
o
No,r,tol /t4o,(//2
^4-b''A/
*Bf**
I
)
i
O
-La'"t
LU,IID ;Lu-nct
)
I
@e@
I
I
I
f* C- :
PaD l)/,,t/ttiy,**
l)w,-'.
O@oO
Q=Q--C=
@-6L@
{G\Lu4,L(,, t",i
9-=P
v
-o L7
" dW&il/), Aoq @ra.@
@42@
ioilud'brbo{4
,ray
z-ba,rl
,; (( ve{fl dz4'eaz/'t''7 21'1
er,L|e-r"<-?(r^ frl^i Jhz1e- -z-btil
"L{ot Cl,i. t)"Jln".^ 1a.,.- gcfuo- /n'
wiT yogil"}* {4/!u tur, cvr
Aq9,r/*.?
ru,di'bonfu'_V
olrlqt'tttt.)
,ttUt #"*
, r,*^!oV