This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 092 — Exam 2  This primrout should haw 32 questions.
Multiplochoioe qutstjono may cont{mm on
the next column or page _ ﬁnd all choices heron: answering.
This exam is ONLY for stndwrs in the
MW? 23pm soctionl! on: EM! points atom going from 1 Do 4 in order?
1.1,o,o,1 3” ~11 . l
2. —1,2, 1. o '
s. 1, —1,—1. 1 4.6,0,0,0 Q, 0, 1, —3 correct 6. M1, 0,1,9 Expiaxmh'on:
The formal charge is calcuiated by
FC in group #
—(# bonds + # unﬂw‘roti o“)
'l‘lmo Fostwwua‘owo
F‘sz6w>{2+4}m0
FCJm5~{l+0}ul
FC4m6~{i+6}—~:w1 902 EDI} poi'nm
Of the following combinations of comizounds
: names. the one which is incorrect. is
5.. 503 : sulfur trioxide 2. K28 : dipotassium sulﬁde correct 3. NagO : sodium amide Vomion 092 w Exam 2 ~ The C in CF; will have 4 RHED’s: three
single bonds to Cl ammo. and one lone pair.
Neither f" or G violate the octet rule. 000 10.0 points Choose the fomula for the compound man»
ganeaoﬂl) alumale. 1 Mam: 2. MnGlOg 3 MMCIDzle 4. Mn(C1{))p 5. MnC104 6. M58303): curred: 7. Mn{()]04)g 8. MnClOg
Ex Narration: 010 36.0 points
Balance the equation ?PBr3+Tl{gO ma?ligi’03+?HBr. using the smallest. possible integem. What
is the sum of the coefﬁcients in the balanceci
equation? 1.? 2.3 correct:
3. 10 4. 11 5.9 Explanation: A balanced equation has the. Alamo IIUJXI‘
bar of each kind of atom on both siéeﬁ of the
equation. We find the number of each kind
of atom using equation conﬁdentr. and com position stoichiometry. For example, we ﬁnti \ ‘_ {/(f ‘ Su2.cli.ﬂ‘e (53120) 1 4. MC] : lithium chloride
5. N205 : diniixogm pentoxide Explanation: i’rdlxos one}: as di, Lri, eLc. are used
what more than one compouno can be made
from the elements involved. This commonly
happen; when two or more nonmetal: come
together to £01m a compound. Hence pm
ftxes are used for sulfur trimdo and oinin jtoogen penloxidc. When metals bond with nonmetals, typically only one compound can
lie made bemoan thorn m pmﬁxrs are not
needed. This is the case with the compound
K25, which is properly named potassium sul
Side. 303 10.0 points
If the fdlowiog crysmliiae in the me type of
structure, wliixﬁa has the 10mm lalLlcc merry? Explanation:
Lowwt; charge liaisiﬂcs due to larger moo 00¢ 20.0 points
(formida urexeaction
MgQSiOr) + 4 1120(3) —v
2 M5(0H}2(aq} + Sil‘lﬂg) .
ﬂow mud: grams of silanc gas (Six!) is
formed if 25.0 g of Mgﬁii rm with mom
H90? 1.. 0.419 g
2. 0.095 g 3. 10.5 3 correct Sutcﬁffo w {53120) 3
there are 5 H alarm; on the reactant side:
2 ll
'7 m . m ‘
.llaborno ‘il‘lngngO {ill The balanced equation is
Why + 3&0 we» H3P03 + 31113] , “dmil’JBrﬁHandzommmemh
side.
7 sumooel'ﬁclonm: 14.34. 1 +3ﬂs or: room
Which of the following metal ion; has 1.1m momdnstam electron conﬁguralion {Arlads?
1. Po” curred:
2. Ca“
3. Mn”
4.. Cu+ 5. Ni” Explanation: Wriae the dmrtron conﬁgurations for the
ions llsboé to see thI one matches the outer
clearer» conﬁguration given. The Aufbau or»
do? of doctron ﬁlllmz is: is, 25, Tip, 3:, 3p, 63,
3a.. 49, 5s, 4d, 53;, 55, 4;, 5d, 51), etc. 5 orbitalr can hold 2 0105110115,? orbitals 6
electrons, and d orbimh 10 electrons. Non:
some Moeptlons do occur in the electron coo
ﬁmzration of atoms became of the stability of
oitlm a full or halffull outermost diorbltal.
so you my hood to 303mm. for this by ‘str‘
ﬂing’ an Flatiron Elam the (1': — 1):; 0):le
Wlim electrons are rcmnvod from orbitals in
a neutral atom to create a positive ion they
are taken in this order: outermost. (highest
value of n.) p, outermost. s, then. outermost d, 012 36.0 points
In the molecule N0. the antlbondlng moleeu.
lax orbitals have more 1. Nitrogenvlike charazber. correct; Version (392  Exam 2 — Subcllﬁ‘e — (53120) 2 4. 9.16.; Explanation:
mugs] “ 25.0 g Fin.2. we calculate the males Mags; present:
} mol MggSi
’26.? g MggSi
r: 0326 moi MggSi '! :Tml Mggsi 2 25.0 g M3231 x The balanced equation for the reaction Endl caba that 1 mol Siiid is producer! for read: mole of MggSi reacted. Reacting 0.326 mole; of Mggsi would produce 0.326 molar: Sin.
We amount from male; to grunts: 32.138ll‘14 7 gagi4 :: 0.325 mol 31H. x r morsm. w £0915 g Sill1 905 10.0 points
Which of the {allowing is a radical? 1. on;
2. BF;
3.0}1; 4. 13:0 unrrect Explanation:
Odd valence e“ coum. B 10.0 points
CHFg is (1 mo polar than CHI; because Qmore; the. Ge!“ bonds are more polar than
e C[inmda correct; 2.lv&s;t.l1e C—H bond in CH9; is a non—polar
bond. 3. km: tine Whahalral 1200mm dooream
the polarity of 0!“ bonds. 4. less; the ﬁlter: polar 0F bonds are gym.
metrical anti cancel the dlpolc moments. 5. more; the GI! bond in CHF; is 3 non
polarbond. Explanation: Since F is more clecuonegaijve than I, o
(l? bond will have a greater dill’ercnce in
éochonogaﬁviﬁes that: {I Cl bond. OD? EM} poinlo Whit}: of the following musics exhibits roar
(mace? 1. N03“ correct: '2. H20
3. None of the cholera has more than one
possible bonding scheme.
4. H95
Explanation:
:5: Jo'
l H
. In. ” . .N . M
£9.” \ 9: IQ.” ‘9:
:5:
l
N
. / .,
{9/ 39>. (308 10.0 points. Moor constructing the Lewis dot. formula for
CF“; , give its clochnic arrangement and
molecular olmpc. 1. trizooai bipyramldol, seesaw 2. trigonal planar, trigOnal planar 3. tetrahedral, tetrahedral 4. tetrahedral, angular 5. trigoml bipyramldal. 'llohaped 6. tetrahedral, trigoml pyramidal correct Exp] anaﬁ on: Version 092 w Exam ‘2 ~ Sutcliﬂo w (5312(3) 4 2. Oxygenlike diameter. Explanation: The bonding molecular orbitals reﬂect. le
more decimalnegative ﬁlament. 0 and the am
tibonding molmular orbitals ntﬂoci. the {m écclxonegaﬁvo element; N. 013 10.0 points
Which of the following is TRUE about: anti.
bowling orbitals ofthe molecular orbital the. cry? 1. Antibondiog orbitals are made from the
overlap ofulrpna and pi orbitals. 2. The strongman book between atoms have
no electrons in 311th and ing orbitals. 3. Alﬁmugh mjbomling orbitals may ac.
oopt. damom, the electrons never remain for
long. iboncl‘mg orbitals are higher in onorgy
than their currlxpmdlng bonding orbitals.
commit 5. Unslnrod eioclrons are 1215ch in auti
bonding orbitals. Explanation:
Anﬁhondﬁnﬁ orbitals are hiﬁm: in mvxgy
.than their corresponding bonding orbitals.
Molecules can most. with electrons in 3min
bonding orbimls, as loozas the molecule does
not place as many electrons in antibondlng
orbitalo as there are in bonding oriyltals. on 10.0 points
How many carbon atoms are in 1.00 X 1.072
[grams oftrlnltrotoluene, an explosive? 'll‘lniv
trotolueue is 07H; N305. 1. 4.22 X :0" atoms 2. 1.86 >< :0“ atoms curmet: 3. 1.75 x :0” atoms 4. 2.65 x 1523 atoms 5. 4.24 x 10” atoms Explanation:
manage. =z: r x {693 Each C7H5N309 molecule contains seven
carbon Moms. There are. Avogodro’s num
ber of QﬁsNaOs moleculw in one mole o£
CTH5N305 We need the molecule mass
of (”aHrNaOa so we can convert. grams of
CTHsNSOa to moioo CTHstOg: Molecular mass of ammo.
z 7(12.01 glmol) + 50 .01 33/1110!)
+ 304.0: gfmol) ‘5. 6(16 .00)
.—.— 22T.25 g/mol
We can use this molecular mass to convert
g 0;H5N305 to 11101 C7HsN305:
'1‘ mo] CrlisN305
1 mo] (77H5N305 z 100 N W
E G’H‘ 30“ x 227.15 a 073mm. =.: 0.4402 mol C’rHsNiiOo We can now we Avosgodro’s nmnlacr and
the ratio of C atoms to C7H5N305 molecules
to ﬁnd the number of carbon atoms: ? atoms 0 w 0.4402 mol Cyl'lﬁ‘v‘305
x 5.022 x 10“ CyHrNsOs 2 mol mama).
7 atoms C X 1 melee C7HrN305
— 1.556 >4 10M atoms 015 10.0 1:051:18
Whid: of the following is the correct Lewis
formula for hypoﬁuorous acid (HOF)? 1 = ilkom : Version 092 ~Exam 2 — 5. 6. 7. 3' H—0%F :
19 Hwéwl?
Explanation: The Lewls formula for laypoﬂuorous acid (HOF) is P] _0_p : 016 20.0 points
Which one hasthe groom number of atoms? 1. 3.05 moles of argon 2. 3.05 moles of CH4 curred;
3. 3.05 moles of helium 4. 3.05 mole; of water 5. All have the same number of atoms Explanation:
For 305 moles of water: (3.02 x i023 moloo '? atoms m 3953110ng0 x
lmo} 3 atoms lmolecule
225.51 K 10% atoms Pb) 3.05 melee of (3114:
. 2 33 .
r atoms _ 3.05 mol on. x 9M
lmol
5am
imolocule . 9.28». 10“ atoms Venr‘lon 092 —E)cam 2 —
3.4l mol C .= 1 me] C
3.4110
3.42 mo
W ... 1 mol 0
4.14 mu
T m 1.33molH To ﬁnd the lowest whole number ratio, do
termine the lowest number the above can be
multiplied by so that all will be whole mum
bets. in this caoe the number is 3: lmoleiimlimolC
lmoiOX3m3Jn010
1.33am] H x3rw3.99 mallI Therefore our lowert wholenumber ratio giver:
us His empirical formula angHqu. 021 10.0 palms
If a molecule has square plamr molecular ge~
ometry, what must be its hybridlwtlon? 13. war? «Aurod Explanation: A molecule with square planar molemic:
geometry has oclaboclrai electronic geometry
Whisk ooororpontio to mad? hybridization. 022 10.0 points
Whlch Mateorant is FALSE about 03? 1. lt is parmnaguetic. coxmoi: 2. 1:: would be expected to ham: 3 slawDer
bond length than C!" .
3. It 113m four electrons in wry, orbitals. 4. It has a bond order 01'2.
Explanation: SutcliEeu (53120) m For 3.05 molar of helimm 6.02 x :0” atoms ? atoms no 3.05 mol lie x
1 mo] $1.34 x 10?" atoms
For 3.5 moles of argon: 5.02 x i023 awnrs ? atoms —,—. 3.05 mol A: x
1 mol 2» 1.84 x 10““ atoms 017 10.0 points
Comidor the water molecule.
The i‘lOvH bond angle is " little less than the M‘trahodraé mg}.
 lone palm are more repulsive than 2. 90° bonanza: the electron dot structure is
drawn that way. 3. a little greater than the tetraheérﬁ :11:
glu became lone pairs are not as repulsive as
bonding pairs. 4. 180° because the twa pairs of bonding
electronsan to get. as far apart aspomlblu 5. aqua! to themMcdml Wedlovﬁ".
Explanation: 018 10.0 points
What is the percentage of carbon by weight
in table mgar {ngﬂnogll'e‘
1. Not enough information is given.
2. 3.5%
3. 48.5% 4. 411% coxred: 6. 51.5%
Explanah‘un:
.. part.
Percent Whol x 100%
Sutdil't‘e .. (53120) 7 023 10.6 prim
Consider the diatomic moieculm CD, which
hag a bond order of 3, and ()2, whirl: has a.
bond ardot of 2. One can state Elm 02 would
be (easier, hamtier) to break apart than C0.
and would also have (longer, more. Shorter)
bonds than CO. L easier; shorter 2. easier; longer correct
3. harder; more 4. harder; shorter 5. easier; more 6. border; longer Explanation:
The larger the bond order, the more Ruble the molmule. This corresponds to mmngor
and ehorter bonds. 021 10.0 points
The electronic geometry of SnClg' is I. triganal planar.
2. linear. 3. tetrahedral. 4. octahedral. E. trigonal bipyramidal. correct Explanation: 025 {(3.0 points
What is the shape of CS?“ 7 Lsoesaw 2. trigonal planar correct 3. T—shAped Version 032  Exam 2  In this case, the part would be the mass of C
in A certain mass 01' Gui1220“ {the whole).
The simplest amount of 0121122011 to use is
the mass of one mole ofcnl’an“. Eacbmnle CICuHmO” conLalru 12 mol at
C, 22 mol 0”], and 11 mol ofO. Welmowtlic
atomic mm of each of these elements from
the pmlodio table. Uﬁngtheso mile moses
we calculate the grams of each element in one
mole of Cul‘luOu: 12.010? p; (I f (I :.= 12
5; rom molC >< lrnol C == 144.1281; C ,  1.0mm g u
gftoml‘l —22molll x lmolll .. 22.1741 [5 H , and
15.999450 gliomOmllmole imolO n ”5.9933 0. To ﬁnd the mass of one mole oregano” we
add the masses of the component parts: molar mm 012322011
m 114.128 3'. + 22.1147 {:4 175,993 g
3 342.295 g Cial’lggOu
mul 01232202:
some percentage ofC‘ in (312113901; is 144423;;0 mum.  (53:20) a 020 10.0 point.“
A sample of vitamin C was analyzed and
found to contain 40.9% carbon. 4.58% hydro»
aw. and 55.5% oxygen. Whale the empirical
formula for vitamin C? I. (333403 correct
2. C2840:
3. CHO 4. Cgl‘lgo Explanation:
% C m 40.9%
% 0 m 54.5%
The empirical formﬂa ls theﬁmplest whole
number ratio ofatoms present in themolcculc.
The ratio of atumaln a compound is thesamc
as the ratio of moles. of atoms in a molecule
of the compound. The compound is 40.9%
C. 54.5% 0 and 4.58% H by mass. These
peroentagw are true no matter how large or
small our sample of the compound. Aluuming
we have a 100 5 mph of the compound, our
sample would contain 40.9 g C, 54.5 g 0, and
4.58 g H. Using; the atomic weight. of each
elcmmt, we ﬁnd the moles of each elmcm % K m 4.53% ‘1 Cam—m..— prwmtin 100 goftho sample:
95 342.293 a cannon " 100%
=1 42.106333. l mu] 0
? m . ..._...__
rnolC I509g0>< 12011 EC
019 10.0 points m 3.41 mol C
Which oompolmd in likely to haired low melh
in“ ml? 1 1 0
m“
?molO=’vl.5 0 ———~—
Lcnlciuo: chloride ° 3 x 15.999 3 o
m 3.42 mo] 0
2. None oftluzse molomicr. it: likely to have a
low melting point.
?moll~1 ”4.53;; H.. giggilili.
3. magnesium oxide ' g
z 4.54 mol 3
4' sull'ur moﬂde correct We now have a mole ratio, but wc want the
. lowest. wholeminim: ratio. gividing each of
5' b‘uhm ﬂuoride the above moles by the lowest number guar
Exptmmtinn: unites us at toast one whole number:
Version 092 — Exam 2  Sutciiffe .. (53120) 8
4. mahedral 5’ I?!“ f
5.Mgoml pyramidal H—(II—g EC ”H
Explanation: H b There are three rogo'om of electron [lonely
(Witlxno lune pairs) around the central atom: a. C
if \.91 028 10.0 palms
On the maul Poultry; scale ofeloctronegativl~
ties, the electronegativity o§ Selenium (Se) is
2.4, while that of chlorine (Cl) is 3.0. Barbed
on these values we should expect the bOndlng
between Se and CI to be i. ionic. 2. wvalmf.‘ and polar with the Cl and of the
bond ul'uzlnﬂy negative. eonact 3. covalent and nonpolar. 4. covalent auri pole: with the So end of the
bond slightly negative. 5. unstable in any eircmmtancm. Explanation: Since the (tlectrouegaﬁvity difference is
ﬁnal! (3.0 — 2.4 a 0.6), we tamed: and this
bond Will becovalmt and polar (because the
«dectwnegatiﬁty difference it: not eelo). Cl
has a larger declxunegalivily than So, so it
will draw electrons toward itself and away
from So. Ttu's will make the Cl and of this
bond slightly negative. 027 10.0 points Carbon ins a valence of four in nearly all ot'
iLs compounds and can form aiming and ring};
of C atoms. Consider the propync structure. What hybridizatious Would you expect for
the carbon atoms idmﬁﬁed by a. b, and c.
respectively? 1.529. 52>. or" 5/3359 2. 51):. 5?. s2 l/"g
3» £112. 5105'. 5323 ,3
4 ans. so. 3P3 ‘ ﬁ 893. Av. s1) curredc 0 3:73. 51>”. $293 7 spa. ml. or“
Explanation:
1i 3 5P
5?
EJ 3.
ENG —C EC —il
l 'E
H 5? 028 10.0 points Choose the species that is incorrectly matched
will: electronic geometry about the centml
atom. 1. H20 : tetrahedral 2. Fiﬁ ‘. pyramidal correct 3. CE: : [pushedral vi. NB: : tetrahedral 5. BeBm : linear Explanati on: 029 10.0 points Vmion 092 ' Exam 2 wSutdiffcﬁ (53120) 9 Um electronvdot notaticn to dm'lonstram
the formation of ionic compounds involving
the ﬁlaments Ca and l, L.Ca'+Ca'+Ca'+I'——+
. 3." Ca+ + Ca”? + (13+ +  ——> C331 2. Nona ol' {hm 3.614 él‘4nl:+I=+l:~—. {I : 4.: II : ——~»
Cazia 4.Ca'+12—> of + = = m» Ca]
5 Ca3+ Claw Ca=+~1 +1  m.
.. .74 _. a».
Ca“ +Cng++CaH+3 i : +2} = _.»
C3312 6. Ca=+'1=w~a “‘12 C3:+.’]‘:+.I:Mm, Ca“ 4 =1 = wwCaI Bxplmatiom
Ca: give; up two elecmms to form the (3'32" catiomand  I 53cquiwsanelmront9fmm the = I = anion, Two iodine sniom combine with one 03L
alum cation to form an electrically neural cfr++=1 : +=1 : "um$03.12 030 100 poinw
How many sigma and pi bonds are in the
molecule NHQC KOCH2? 1.85'tgzmaml3pi
2.655‘gmaand2pi
3. Seamaandﬂpi
4.8Wmd2pi correct
5.352311“: and2p': 6.6smaandtlpi Explanation: 031 20.0 point; The ion BeF' has a total of '3 shared
damom and Z nombomied daemons. 1.2;2
2.438
3. 2;4
4. 6:2 5.4;4 Version 092 M Erma; 2 w Sutcéiﬁu  (53120) B. 2; Seentact
7.4; 2
8.2; 3 9.6;3 3XpEanahun:
Beis one ofthe exocrptions to the octet xule.
The to»! umber ofvermce arcmm 'm Tm: Lewis Dot mucme 5:: [WW 5:] 032 10.0 points How many nanbonding Electron pairs are
amumi I in iF; ’? 1.4
2.0
3.2
4.1 5. 3 correct Explnmtiun: Iodine has seven valence. Plectrum. Flourine
has seven valence electrons. Each flourlne
atom wili form a covalent 50ml by sharing
one of iodine’s electrons. This leaves iodine
with 5 electrons 1)le the additional electron
that. glam; the molecule a not, negative énargc.
That makes.53): honbondim electrons around
iodine, or three W .. ,.._ .. wl—F: 10 ...
View
Full Document
 Fall '07
 Fakhreddine/Lyon

Click to edit the document details