UNIT 7 REVIEW PROBLEMS

NAME:~ ~

PERIOD:

1. Define each of the following intermolecular forces and give an example of the type of molecules theywould be involved with:

£> 0= ~'TI"EL"'" WEAl:.. FOR.(.ES Of:" ATTt.4C:f!O,J 77ff./T .aNOLVf3 IlTOfI1S 4",·i~a. London forces - "-C<..r' _ r-rfJ 1!:7rI Polll4/L 'f- . J)vu/J(, 7E/>1Pol.AeY l;)IPOLES .PJ oAlE A"'On-lc~. TJftE ,W\OLEC.uL~.s .DJ TM 1) Po - D/ SAf'A -A~.s 77-1E.'""SEDI PoLe=:J !-le.E 8tl..1G5L'" A-T77l.A c.7E1J I R.t: »eu.et: utJT7i.. E J U: eM"I=oe.(£.s ex 1ST eE'flvl=.1:::-iU A L.L /I10LECUL6.5 /3.uT 4~t: 7H€ (JAILY .n.I77::!-"""R..1'I1.0U:Wl.,q,(. r=oR.cr:

t3E1W8::-oJ NOJ...JPOU~~ ntOLECUU~..i. E'X: /.31J) CHlI) Cd.1-I1.(

b.Dipoleforces- T\4e- r-olt(£ OF 4TTflAc..T1oAJ f3cnvl:::-c>J POLAr< IVlOL.I::CULt=S. J::7"",ec--s.ULTJ t:-e.U/I1

PolAfl. n'\OLE"CuLt=S UN,,.JG uP so n+~T Jl.1IE"/le POSITlVt: 4"'0 ~f:FGA11vfE 1:::--;...)1)..5 4~1.:CLOSE TO O~E" 4I'JOTHEl'l. EX = AlO eo So

d I j ~

c. Hydrogen bonding - UlJV SVt\LL" SN.otJ6 l:>lrolt -t>IPou;o Fo~(.SS OF" arrencrro») 8r=rw(;.t7£A/Iv\OU;CuLES CoNTAI N IIIJ 6 J.+'1])t2..oc,r;~ dONUt:b DI/2Ecn:-( 10 Ii HI61-1LY ~L.~c..m..OtJt:GATI VI:

ATb"W\ Sv(...H 1\ S 0 J "-' I oil. F. l:X: I·ho ~ Ai H..,3

(8 points) 2. Using the prefix system, name each of the following compounds:

a. N203

nil..! 117(.0('&-1 TI2.IOXI/)Eb. XeF6

XeJoN liEXA~LvoR.lDt=

C. P40lO

TI::-II2APr-fOSPI-\OQ o.sbECo'IJDE

d. S4N4TfIIlA SULFuR...

Thlfl4 NI TR.I Dc

f. C-t;.O

D I CH wQ.I ,J E m.o"'O)<I})Ee.OF2

())('1~t=t.J blFLvO"R\OE"

g. N2F4

Oll\llrrlOGtp..J1FnA FLVoR. \J~

i. PF3

PliO SPHot2.. uS7R.1 FLUoRluE

j. S3F6

71(ISUL~Ut<... _HEXAf=LvOR..1 i)(!

1. eelsCAR..So,J

~4 CHL(fi. I iJt

k.NH3

NI1R.06eJ77C.1 1-/ 'I be I Df"

m, NZ05

Dt M I112.068'-1P~"'TO-i., llE

n. Si3N471<1Sill c.oAl

Terf2.,'J,N I ra IDfp. PBr5

Pl-\oSP !-IotlJ.! feJ1'flO((oll\. I DE

o. Ctz07

IJICHLORh.JiZ HEPrOxlcf

OVER

q. P4F5

~(.j P/fOSfJf/oit. u..JA:f.JrA F L.lJ of( I i)(f"

r. Si2Br6b I.II U COIJ If,,=-x P,BR.l7M.1 f)/E"

s. B2Si

D/8a20N ft1avOSIU (flUe

t. NZS3

biNITi?.O 6f=iJ71V SULFIDt

u.BZH6

»I t3uRo'"HEX4 f=.usoe. I De-

v. SbBr3

AAiT/ll1oI/Y 11!../ ~,e0t1/1r)f:

w.HI

H ....' DflOGt:M /YtONOrO 7:;)/ bt::

x.ceo2

tHUJe.ltJE" ))JO~ r D<!"

3. Write the chemical formula for each of the following compounds:

a. nitrogen trihydride b. diarsenic pentoxide

As~Osc. sulfur hexafluoride d. dichlorine monoxide

S t- (,

f. boron trihydride

8/13e. silicon tetrachloride

g. dichlorine heptasulfide h. antimony pentaiodide

ell. ~1

i. trisilicon diarsenide j. selenium hexabromide

Se 8r'f(J

k. iodine pentoxide 1.nitrogen trichloride

NCJ3m. carbon disulfide n. chlorine trifluoride

C S.1 C1. t,

o. dibromine monoxide p. antimony pentachloride

SbC~sq. hexaboron monosilicide

IS~ S.-<'r. antimony tribromide

s. sulfur dibromide t. dinitrogen monosulfide

u. tetrasulfur dinitride v. dicarbon tetrabromide

CO) 8'4w. tellurium tetrachloride x. iodine heptafluoride

Te CJIf ~F'7

.---1H

2.20

3 4 5 6 7' 8 9Li Be B C N 0 F

b.98 1.57 Pauling Electronegativity Values 2.04 2.55' 3.04 3.44 3.98

11 12 13 . 14 15 16 17Na Mg AI Si P S CI

0.93 1.31 1.61 1.90 2.19 2.58 3.16

19 20 21 22 . 23 24 25 26 27 28 29 30 31 32 33 34 35K Ca Sc Ti V Cr Mn .Fe Co Ni Cu Zn· .Ga Ge A.s Se Br

0.82 1.00 1.36 1.54 1.63 1.66 1.55 1.83 1.88 1.91 ·1.90 1.65 1.81 2.01 2.18 2.55 2.96

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52. 53Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag. Cd In Sn Sb .Te I

0.82 0.95 1.22 1..33 1.6 2.16 1.9," 2.2 2.28 2.20 1.93 1.69 1.78 1.96 2.05 2.1 2.66

55 56 57 n '. 73 74 75 '76 77· 78 79 80 81:' 82 83 84 85Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Ti Ipb Bi Po 'At

0.79 .0.89 1.1 1.3 1.5 2.36 1.9 2.2 2.20 2.28 2.54 2.00 1.62 2.33 2.02 2.0' 2;2

87 88Fr Ra0.7 0.9

How to Tell the Type of Bondthat will Form Between Two Elements

Type of BondDifference in

ElectronegativityNon-polar covalent less than 0.5Polar covalent between 0.5 and 2.1Ionic .~ greater than 2.1

Electronegativity Electronegativityof element of element

Difference ofelectronegativities Type of bond

CI,;:p6 CI, J.16 o non-polar covalent- electrons arcshared equally between hathchlorincs, Cl2

0, ;).44 0.89 polar covalent. oxygen.is marcelectronegative so the electrons arccloser to oxygen makingoxygen slightly negative andthe carbon slightly positive

CI,;j.16 2.23 ionic, chlorine "steals" one electronfrom sodium to make Cl- and Nai,NaCI

OVER

4. Using the table of electronegativity values classify each of the following as either ionic, polar covalentor nonpolar covalent:

a. C-F 8 P.J,SS 3.Q~

\.46

c. AI-Br fl.tJ I Z, l) Co~

e. Br-Br "-If

g. In-BrI. 7 ~ 2, qe"

~I, I~

i. Cl- Cl

p

5. Draw Lewis structures for:

a. HCI03 (Hint - put Cl in the center)

b. S-Cl PJ,S g 3,1ft,

~.58'

d.K-F.2''1- 3.18 I

~:3, I \0

f. Ti-Cl

~ f'I, c L

h. Ge-F~.O, 3.Qg

~1,17

p

j.N-Hs.ot( .:;.. .<.0

~I 6-l(

p

He • I ..

:0 CJ. 0;•• , .~

,0 .. ..~

ea'·

1-+-0-0-.-1

c. CH3CH2CH2OH 1+ I~ H3C= ,~ I I t • e

g H - C, - C - C - 0 - 118f-t s:I , \ ••

0:- ~ Ii H H- BltAJoJ. e -.:-~td t~ .: dl

L{

g. the three resonance structures for C032-

C,: •...I I. c1-g °0' o '.30 :- I C" I ••

(-) C"5: ~ 0 z: t: - 9':I 6 i

- 'i'A- '- t- ~l.

~ ,.-g,~

d. NOel

tJ=so ~ ~

Ct ~ 1'ip.L e - : 18

'B,oND e- -- 1{ 1.0_

KI , d.

e. CH3CNdC.;- 'g314: 3

tJ = 5

6. Name the following acids:

c. H2SH'I iROsuLFvR.rc At I i)

e. H3As04

~ AflSfNfC ncio

It Ir 0 •

HI

H' '. C -' C =- N ;!1--\

1-1\C:= .$

\

H

c •

o·

OVER

e O·. . '\ d-IIJ~\I .'

:o-c-o:I • f \c

b.HFH'f D12.0l-Lu0v2.l( (.It. Il)

f.HCIt-I YlJt?..o CI-I (,Oe (C /-I c. f 0

..: 0:, .,..:0- (.-== 0••,. .

7. Determine the shape and polarity of the following molecules:

b. SeC12

Se.. z: ~

J.C,Q ~ Ilf

vAL e s: aotach! t) e.: ;-.-:L

I~

T~;- ~30 = If'_.,fAt e :.d. t.f&JtJD e- ; 1,. ~

,it/to

d. OC12

o"it..2CO:IY-----VAL e,.- :. ~OMtJD e- : t.(

I~

e. CHC13

C. s: L-\1-1::- I

3d; J1-VAL e~"" ~ l,

C- z: L{

;< 0 ~/;;;"_

~le - _-1ft;~(hJDe.- -- A/. ~ 8

f9.N~

••; & or - A s - 13 r- :.. , ( .: 8....'.

c p~ .~ , 0

: eJ - Se- CJ :

00°• •(.

(! D 4 ,of)

: CQ - 0- CJl :• tI o· • D

t-I• .• , e"

• CJ - C - (J."•••• I' •

: Ci:••

• to tJ #

0.:::: C. -= 0

1/- 3 - I

171.IG()A/IJL PYt<AYtl/TJ4L

Pou~!l

,t.j- ;1- ;J.f;c/J TPOt.l11l

.0

·'O-I~ --O.~C I. •• ~ 7

\I

: 0 :3-.30

77U60N~L PLANt/if.

J..f-~-~dEN TPo LAil.

••~-;;lOLltJEAI(

AlON POL f.I~

.' fI •

= 0- Te =R•. I. 0 I. ..~