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1. M‘ 112w W '. KEY
(12 points)
(i) Use VSEPR theory to draw the best structures, including resonance forms (if any);
(ii) Indicate any formal charges other than zero
(iii) List any distortion by giving approximate bond angles.
a. XeOZF4
Xe is least electronegative and in the center . . (f . a
. FI} \F.~
Xe 1x8e' 8e‘ ‘ "'Xew‘
FXe 1x4e‘ 4e' .. V ‘ ..
W .,F.. H .. F,
Octahedral Geometry 12e‘ Q .
The two oxygens, both having double bonds will orient themselves to have the most room, therefore
opposing themselves in the structure. There are no formal charges other than zero in this structure, and
since the effect of each oxygen is cancelled by the other, no distortions are seen and FXeF and OXe—F
bond angles are all 90°. O—XeO bond angle is 180°.
b. BrIzCl;
l is least electronegative and in the center . "I. +
I 1x7e‘ 7e' {{ _‘j
ll 1x1e‘ 1e' 6' w‘“
lBr 1x1e‘ 1e' \ 0.
lCl 2x1e' 2e' I BE
Cation 1x—1e’ 1e' ‘Ql‘
TBP Geometry we
The unconstrained lone pair will occupy the equatorial position followed by the largest and least
electronegative substituents  iodine and bromine. The central iodine will have a +1 formal charge. The
lone pair will force all atoms away from it, causing distortions that will result in the IlBr bond angle to be
less than 120°, and the Clll and CllBr bond angles to be less than 90°. The CllCl bond angle will be
less than 180°. The ﬁnal shape of the molecule will be a distorted see saw.
c. IO; . s e ‘ _. .. ‘ .. .. ‘ .. ..
l is least (r C“) O (I?
electronegative and /I 4—» / <—> Ill <—> a l
in the center " / \ . " / \9. ’ 4 \ “ / \ .
‘Q // \Q‘ ‘0 // O: 'O /e\0‘~ Q // \o
l 1x7e' 7e' _
lO 4x0e' Oe' 5
Anion 1x1e‘ 1e' .0
Tetrahedral Geometry 8e' :I
5
. . . ./. \ 5
There exrsts a formal charge of 1 on one oxygen (0 :5 more electronegative than 0' / O
I). Since there are four resonance structures resulting in the charge to be evenly Q'
spread among all the oxygens and each lO to have a bond order of 1.75, the 3'
resulting structure will be a perfect tetrahedron with all bond angles being 109.5°.
d. For 10,; What is the highest proper rotation axis Cu? C3
For 104‘ what is the highest improper rotation axis Sn? 84 nmllll M ,‘lSAGE3 NAMEL
/ 2. (16 points) V a. The start of a molecular orbital diagram for the CN‘1 ion is given below, assuming no s—p mixing. Label
C, N and all atomic and molecular orbitals. Indicate the atomic and molecular electron distribution and
use dotted lines to show which atomic orbitals are used to form which molecular orbitals. gr
,1 ’ _ \
7'? 4—ll—" _‘,_~’ 4¢'Prmm7(_\\(‘1r7‘\
5H \
\ x ’ r I (39‘ \ \\\
\ I I
, ’ \ ‘ ~ \ Iq— +— 1—— 7.?
/ / \ " ~(
15 ‘H” \ \“’V 4l7“'"nﬂ/<
\ \ "\ / ‘
\ \ \4b— 6’01" I \\
‘ v
‘ \ _ a. «H'— ’L3
. \ l b ((16)'
\5 ll
l
c 4*" s
V Based on your M.O. diagram in part a: b. Calculate the bond order for CN‘l. , Ls . 2»; rah ~ 3 w
c. If CN'lis oxidized to CNO, how will this affect the C—N bond length? _ CM" ’9 (No his all: la in anbom‘fg Orbllhl d. Is CN'1 paramagnetic or diamagnetic? 6‘0' 3°“ b 3'5" 55°40! all pairCal : (liAchne‘l'a'c 1 I‘.‘ Hal ('8)
e. Which are the HOMO and LUMO orbitals in CN'I; Homo: 04:5 Luna: 1! 1,? f. Cyanide, CN'I, is a deadly poison because it acts as a Lewis base and binds strongly to iron in
hemoglobin. Based on your M.O. dia , which end C or N will bind to iron? Why?
LEI’is bow. 5 aloudM3 (Jae ran Paw
HOMO = (is 1 an +r badly arbil‘ql MON. like C) +51; M55 ﬁlec‘l'lbnsaHw demi g. If extensive s—p mixing occurred how would this affect the MD. diagram for CNl? 071') orbi'lul would bum is 'lqlar “zeal and increase ,‘n 3”” Y
(50 “Pi n M O)
V 606) orbi‘lnl would buomc $bl Luci qndl lowir In My (30 down in MO) i/PAGE 4 NAME: ! E 1
/ 3. (15 points) v a. Which compound or element do you expect to have the highest melting point? Brieﬂy explain your
reasoning. (2+) (2 5 Egg. M H “Y
 l 0 ‘g L E z u n,
(1) CaBr2 ' y n, CW Mn (ii) BCl3 R‘Wv'ZuLll'll‘y: LN"; Lou3n: ‘t' mrL Fblwzmlola. .. ngv Aisfwslo, {>ch (i'ms ‘ A\ Tr A: MLI'H‘S Paln" Dun994A Dun +91, A‘ Kan, M!
0.0 c. M p (huh ,m‘n‘d/ﬁ m sue ("2* "A'MPJN' grim
4&1th 64‘] I Va'evm 9’) b. Which compound has the smallest bond angle? Briefly explain your reasoning. w' PC13 or BBr3 Br 85(3. :3 +P£30M\ Planar Bo—A é; (ZOO
V '{9 15%“ mo“; 1 w PU:
/ ”‘4 AVG / \3 PF3 l” we may . . .
F ‘12; F c: It? I 5r r bwc F3 smalle “A loss SWIM lam! (ll) 13' 0.0 “ML 9‘" Paw“; '*' *6 5M‘h’ “as“, E M w
I. (a "sour/ac? ... If wham I
l
c. What two lanthanides are most likely to have a +2 oxidation state? For each, write out the ground state
electronic structure (noble gas core + valance electrons). Which one is larger? Europlom (E03 —‘9 C13. XXL] 931:.”7 i:
Y‘H'ulolom (no) A (3.5. [x4 4,51%” V EU is ‘aaat/ AVL +0 \W‘Stf ZtrQ‘ LC. IQHMHI¥1 COH+ML+:OA
/ ﬁAGE 5
/ / 4. (8 points) v a. Identify the two lattice types on the right. b. How many nearest neighbors (in 3 dimensions) does each atom have in diagram 3? c. For diagram b mark an “X" to show an octahedral site. Page 6 KEY
5. (15 points) a. Draw the angular density function and the radical distribution function (1A1!2 vs. r) for a 5d,2
orbital. Indicate all angular and radial nodes. Determine the point group. Z y X i
' ~—‘—.v—~7— yea—#wi in, W” m
2 nodes The angular density function has two angular conal nodes that are located at the origin opening in the +2 and —2
directions. The radial distribution function has two nodes where rzllJ2 reaches zero (note the graph should
approach but not reach zero as r—aw.) A picture of the two functions combined is shown on the right. This orbital belongs to the point group Dnh. b. What combinations of atomic orbitals can produce a molecular orbital with two nodes?
State the atomic and molecular orbitals involved and draw a picture of the molecular orbital.
Does g, u, or no symbol apply? 2 + 8 ’ ' b There are several correct answers to this problem, three of which are shown here. In
2p 2p 1t 9 each case, the resulting molecular orbitals
have inversion symmetry and are y y centrosymmetric, resulting in a g designation.
as + as — z
y y c. For SF.s state all the symmetry elements (i.e. rotation axes >C1, mirror planes, inversion
centers and rotationreﬂection axes, Sn, n>2) and give the point group. This molecule belongs to the octahedral point group Oh. It is highly symmetric and contains the following
symmetry elements: An example of each is shown below. E (identity)
3 F 3 C3 rotational axes (4)
f . «‘ '° 02 rotational axes (9)
. F//,, I \\\F  C4 rotational axes (3)
" S“ an inversion center i f \ o. 84 improper rotation axes (3)
l , F , Se improper rotation axes (4)
. . ' ‘ horizontal mirror plane oh (3)  F . dihedral mirror planes od (6) /9AGE77 V NAME: 6 (17 points) Structures: a. Tungsten adopts a BCC structure. If the W—W bond length 18 2.0 A what 15 the length of the unit cell
edge? How many tungsten atoms are in the unit cell? ‘ E
“2 .:_2 ¢ A
BC c
o
ZDCleS/Ce,” 0L . 82 B4 A
b. LiCl adopts a rock salt structure. If the ionic radii of Li"” 18 0.9 A and Cl is 1. 7 A what 1s the length of
the unit cell edge? How many fonnula units are in the unit cell of LiCl'?
a + Vat/2 \ V 90L: 5 2—
4 Mi on UMi‘S
WZ‘I/
c. MgSe adopts a cubic structure. If the 1onic radii of Mg+21s 0.7 A and Se'2 is 1.8 A, predict the likely
structure and determine the length of the unit cell edge How many long are in the unit cell? 34 II; 325/:
T/Y—:¢>3;9:¢5_3ﬁ __y CM: Li ‘7:
" aﬁﬁA
2  1+ f
FCC Se wlrlM IVE M '/Z “£1413; ZMC 5‘01; Q.
m a" a gtons
7. (17 points) Gold has a facecentered cubic structure with a ’AuKu bon d1 ce 0 1.7 2A. 2.
:1. Calculate the density of gold. Show all your work. 4157121 703A)
4
mass: 1616.13,. 4w. ,3 ,o W ‘8 11/2 24 , L4A
not 4 no owl Cn—U '2‘ o
‘— 1' a, =/ 20" A '_——~
\  ' \ v 3
bowi3: HINDI}. , “gm“ = afﬁx“); x 4140*: 7443.. A onrl’ 05“ b. Explain why gold is the most malleable element kn FCC ‘f séls 0F alWPaCOMEWSPWe’A c. Perovskite 15 a mineral containing calcium, titanium and oxygen. The following diagram shows the unit
cell. What IS the chemical formula for Perovskite? Cat, maﬁa: 4
O Aéxyz’ g
'77—» A. Q :03 ...
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 Fall '08
 Kaner

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