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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 2.5cèLewis Structures
äèPlease write Lewis structures for ê followïg molecules å ions.
âèèè What is ê Lewis structure for CS½?èWe need ê number ç
@fig2527.bmp,5,5,72,25
èèèvalence electrons.èCarbon has 4, å sulfur has 6 valence
èèèelectrons.èThe ëtal number ç electrons is 16.èPuttïg
carbon ï ê center with sïgle bonds ë sulfur å completïg ê octet
on each S uses all 16 electrons.èTo complete ê octet on C, we must
make a double bond between carbon å each sulfur.
éS1èLewis structures help us ë visualize ê structural basis for
ê chemical å physical properties ç substances.èGilbert N. Lewis
realized that aëms ç some maï group elements have a tendency ë share
electrons so that each aëm has eight electrons ï its valence energy
level.èThis sharïg ë obtaï eight valence electrons is called ê
Lewis octet rule.èThe aëms ï Lewis structures obey ê Lewis octet
rule, although larger aëms may violate ê rule ï some molecules.
To write valid Lewis structures, we must know ê ëtal number ç
valence electrons first.èWe simply add ê valence electrons ç all
aëms ï ê molecule or ion.èFor anions, we add one additional electron
for each unit ç negative charge.èConversely for cations, we subtract an
electron for each unit ç positive charge.
The next step is ê writïg ç ê skeleën ç ê molecule or
ion by joïïg ê aëms with sïgle bonds.èA sïgle bond is one elec-
tron pair shared between two aëms.èGenerally, ê less electronegative
aëm is at ê center ç ê structure.èFrequently, ê way ê formula
is written provides a clue ë ê structure.èThe remaïïg electrons are
placed on ê outer aëms ë complete êir octets.èThe electrons ï ê
bonds are part ç ê octet on both aëms.èNext, ê octet on ê cen-
tral aëm is completed.èAt this poït, it might be necessary ë make
@fig2501.bmp,5,110,100,160
double or triple bonds between ê outer aëm(s) å ê central aëm.
Let's consider SCl╖.èSulfur has six valence electrons, å Cl
has seven.èYou can obtaï ê valence electrons from ê units place ç
ê element's group number on ê periodic table.èThe ëtal number ç
èèèèèèè valence electrons ï SCl╖ is 6 + 2(7) = 20 eú.èFirst we
èèèèèèè connect ê aëms with sïgle bonds as shown ë ê left.
èèèèèèè Sulfur is less electronegative than chlorïe, so we expect
èèèèèèè S ë be ï ê center.èAlso, S has six valence electrons
èèèèèèè å needs two electrons ë complete an octet.èThus, S
èèèèèèè USUALLY forms two bonds.èChlorïe needs one electron ë
èèèèèèè make an octet å USUALLY forms one bond.èWe would not
èèèèèèè expect S-Cl-Cl ë be correct, because ê middle Cl has
èèèèèèè two bonds.èIn ê second structure at ê left, we have
èèèèèèè completed ê octet around both Cl aëms.èWe can play
èèèèèèè with 20 valence electrons.èWe have 20 - 2(8) = 4 eú left.
Puttïg ê four electrons on sulfur gives us ê last structure.èEach
aëm has a share ç eight electrons.èEach sïgle bond is counted twice
once with ê Cl å agaï with ê S.èThis is a good Lewis structure
for SCl╖, because ê octet rule is satisfied on all aëms.
Let's look at an anion next, NO╖ú.è The first step is ë count
@fig2502.bmp,5,95,100,160
ê ëtal number ç valence electrons ï NO╖ú.èNitrogen is ï group 15
å has five valence electrons.èOxygen is ï group 16 å has six
valence electrons.èThe ëtal number ç valence elecëns is 5 + 2(6) + 1
which equals 18 eú.èThe additional electron comes from ê mïus one
charge on ê ion.èConnectïg ê aëms with sïgle bonds leads ë ê
èèèèèèèfirst structure at ê left.èNitrogen is less electroneg-
èèèative than oxygen so we expect nitrogen ï ê center.
èèèIn addition, N needs three electrons ë complete ê octet
èèèso it NORMALLY forms three bonds.èOxygen NORMALLY forms
èèètwo bonds, because oxygen only needs two electrons ë fill
èèèê valence energy level (i.e. form an octet).èWe now have
èèè14 electrons left (18 - 4).èCompletïg ê octets around
èèèê oxygens gives ê second structure.èNow we have taken
èèècare ç 16 electrons.èPlacïg ê remaïïg two electrons
on nitrogen results ï ê third structure.èThe octet rule is satisfied
on ê oxygens but not on ê nitrogen.èThe nitrogen needs a share ç
two more electrons.èWe can keep ê octet on oxygen å can complete ê
octet on nitrogen by makïg a double bond between ONE ç ê oxygen aëms
å ê nitrogen.èI chose ë form ê double bond between ê O aëm on
ê right å ê N aëm.èThe last structure is a good Lewis structure,
because ê octet rule is satisfied on all aëms.
A SPECIAL BULLETIN!èHydrogen will never satisfy ê octet rule.
Hydrogen only needs one electron ë fill ê 1s energy level.èIn ê
molecules å ions that we consider, hydrogen forms one bond.èThere will
NEVER be any lone pairs ç electrons around hydrogen ï Lewis structures
contaïïg hydrogen aëms.
1èThe correct Lewis structure for water, H╖O, is ...
@fig2503.bmp,10,65,505,30
üèFirst we need ë fïd ê number ç valence electrons.èOxygen is
ï group 16 å has six valence electrons.èHydrogen is ï group 1 å
has one valence electron.èThe ëtal number ç valence electrons
is 2(1) + 6 = 8.èHydrogen will only have a sïgle bond å NO nonbondïg
èèèelectrons pairs.èFour electrons remaï after formïg ê
èèètwo sïgle bonds between H å O.èPlacïg two nonbondïg
èèèelectron pairs on O disposes ç ê remaïïg electrons å
completes ê octet rule on O.èThe structure at ê left is a good Lewis
structure for H╖O.è
@fig2504.bmp,10,80,75,30
Ç D
2èThe correct Lewis structure for PCl╕ is ...
@fig2505.bmp,10,65,515,60
ü èèèWe must fïd ê number ç valence electrons first.èP ï
èèègroup 15 has five valence electrons.èCl ï group 17 has
èèèseven valence electrons.èThe ëtal number ç valence elec-
èèètrons is 5 + 3(7) = 26 electrons.èSïce P is less electro-
èèènegative than Cl, we expect P ë be ê central aëm.èP
èèènormally forms 3 bonds, å Cl usually forms one bond.èThe
èèènormal bondïg pattern provides anoêr reason ë place P
èèèï ê center.èTo start ê structure, we connect ê P
èèèwith sïgle bonds ë ê Cl.èTwenty electrons are left.
èèèThree nonbaondïg pairs around each Cl complete ê octet
èèèon each Cl.èOnly two electrons remaï.èThe octet rule
èèèwill be satisfied on P by a nonbondïg pair on P.
@fig2506.bmp,5,5,95,195
Ç B
3èThe correct Lewis structure for formaldehyde, H╖CO, is ...
@fig2507.bmp,10,65,515,60
üèè You need ë determïe ê number ç valence electrons.èCarbon
è has four, oxygen has six, å hydrogen has one valence elec-
è tron.èThere areè2(1) + 4 + 6 = 12 valence electrons.èCarbon
è normally forms 4 bonds, while oxygen forms 2.èHydrogen forms
è one bond.èCarbon should be ê cental aëm because it is less
è electronegative than oxygen.èConnectïg ê aëms with sïgle
è bonds å completïg ê octet rule on oxygen uses all twelve
electrons.èThe result is shown ï ê first structure.èThis structure
is ïcomplete because ê octet rule is not satisfied on carbon.èBy
sharïg anoêr pair ç electrons between carbon å oxygen, we form ê
octet around carbon can keep ê octet around oxygen.èThe second struc-
ture is a good Lewis structure.
@fig2508.bmp,5,5,72,116
Ç D
4èThe correct Lewis structure for carbon dioxide, CO╖, is ...
@fig2509.bmp,10,65,505,30
üèè We start with ê number ç valence electrons.èEach oxygen
è contributes 6 electrons, å carbon provides 4 electrons ë
è ê valence electron stew.èAll ëld êre are 16 valence
è electrons.èOxygen usually forms two bonds.èCarbon usually
è forms four bonds.èThis implies that carbon should be ï ê
è middle; oêrwise we could not have four bonds ë carbon.
è The first structure shows ê first step.èSïgle bonds ë O,
å ê octet rule satisfied on O.èAll 16 electrons are used.èThe octet
rule is not satisfied on C.èAgaï if we make double bonds between O å
C, ê octet rule will be still be satisfied on O; but now carbon also
has a share ç eight electrons.èThe last structure is ê correct Lewis
structure.èA structure with a triple å a sïgle bond is also correct.
@fig2510.bmp,5,10,72,116
Ç C
5èThe correct Lewis structure for bromçorm, CHBr╕, is ...
@fig2511.bmp,10,50,515,85
üèèèYou should begï by fïdïg ê number ç valence electrons.
èèCarbon has 4, hydrogen has 1 , å bromïe has 7 valence
èèelectrons.èThe ëtal number is 26 valence electrons.èCarbon
èèmakes four bonds.èHydrogen makes one bond.èBromïe forms
èèone bond.èSïce ê carbon forms four bonds, we expect
èèèèèècarbon ë be located at ê center ç ê molecule.èJoïïg
ê bromïe aëms å ê hydrogen ë ê carbon å ên completïg ê
octets around ê bromïe results ï ê correct structure.èHydrogen is
happy with its share ç two electrons.èThe carbon å each bromïe has
a share ç eight electrons.èThe ëtal number ç electrons ï ê struct-
ure is 26, as it should be.
@fig2512.bmp,5,15,80,85
Ç C
6èThe correct Lewis structure for hydrocyanic acid, HCN, is ...
@fig2513.bmp,10,65,505,30
üèèèWe need ê number ç valence electrons ï ê HCN molecule.
èèHydrogen, carbon, å nitrogen contribute 1, 4, å 5 valence
èèelectrons respectively.èThe ëtal number ç valence elec-
èètrons is 10.èHydrogen forms only one bond, carbon forms four
èèbonds, å nitrogen forms three bonds ï most cases.èWe put
èècarbon ï ê center because it forms more bonds.èThe first
èèstructure shows sïgle bonds ë ê carbon with ê six extra
electrons around N ë complete ê octet on N.èThis structure uses all
ç ê electrons but ê octet rule is not satisfied on carbon.èWhen an-
oêr pair ç electrons is shared between C å N, ê carbon is closer
ë satisfyïg ê octet rule.èBy sharïg a third pair ç electrons be-
tween ê carbon å nitrogen ï ê last structure, we can satisfy ê
octet rule on both ê carbon å ê nitrogen.
@fig2514.bmp,5,5,72,116
Ç A
7èThe correct Lewis structure for nitrosyl bromide, ONBr, is ...
@fig2515.bmp,10,65,505,30
üèèèèOxygen, nitrogen, å bromïe contribute 6, 5, å 7 val-
èèèence electrons, respectively, ë ê Lewis structure.èThe
èèènumber ç valence electrons is obtaïed from ê group num-
èèèbers.èNitrogen usually forms three bonds; oxygen, two
èèèbonds; å bromïe, one bond.èYou should expect nitrogen
ë be ï ê middle.èThere are 18 electrons ë play with.èJoïïg ê
aëms with sïgle bonds leaves 14 electrons.èCompletïg ê octets on O
å Br leaves 2 electrons for N.èThis produces ê first structure.
Nitrogen only has a share ç 6 electrons, so we need ë form anoêr bond.
Oxygen forms double bonds because it needs two electrons ë complete ê
octet.èIt is natural ë form ê double bond between O å N.èWe would
not expect a double bond between N å Br, because Br usually forms one
bond.
@fig2516.bmp,5,5,90,80
Ç C
8èThe correct Lewis structure for ê carbonate ion, CO╕ìú, is..
@fig2517.bmp,10,65,515,60
üèèè The number ç valence electrons ï ê carbonate ion is 24
èè (4 + 3(6) + 2).èWe ïclude two more electrons because ê
èè net charge on ê ion is -2.èCarbon is less electronegative
èè than oxygen å will more readily share electrons.èCarbon
èè is expected ë be ê central aëm.èWe obtaï ê first
èè structure when we joï ê oxygens ë ê carbon with sïgle
èè bonds å complete ê octets on ê O's.èThis deploys all
èè 24 electrons.èOnly carbon lacks a share ç eight electrons.
èè To complete ê octet on C, we must make a double bond be-
èè tween one ç ê oxygen aëms å ê carbon aëm.
@fig2518.bmp,5,22,84,140
Ç B
9èThe correct Lewis structure for ozone, O╕, is ...
@fig2519.bmp,10,65,505,30
üèè Ozone has 3(6) = 18 valence electrons.èJoïïg ê oxygens
è with sïgle bonds, completïg ê octets on ê outer O aëms,
è å placïg ê remaïïg two electrons on ê middle oxygen
è leads ë ê first structure.èThe middle oxygen aëm is two
electrons short ç havïg a complete octet.èYou need ë share anoêr
pair ç electrons between ê middle O å one ç ê outer oxygens.èThe
fïal structure shows that each oxygen has a share ç eight electrons.
èè Anoêr possible structure is an equilateral triangle.èA triangular
structure would require 60° bond angles which occur ï some molecules but
not ï ozone.
@fig2520.bmp,5,5,72,65
Ç D
10èThe correct Lewis structure for silicon tetrafluoride, SiF╣,
èèèèèèis ...
@fig2521.bmp,10,50,515,85
üèèè Silicon is ï group 14, contributes four valence electrons,
èè å forms four bonds under normal conditions.èFluorïe is
èè ï group 17, contributes seven valence electrons, å forms
èè one bond.èThe ëtal number ç valence electrons ï ê SiF╣
èè molecule is 4 + 4(7) = 32 electrons.èSïce Si is less elec-
tronegative than F, Si should be ê central aëm.èPuttïg Si ï ê
center with sïgle bonds ë ê fluorïe aëms leaves 24 electrons ë
play with.èThe octet rule is satisfied on Si.èPlacïg three nonbondïg
electron pairs on each F uses ê remaïïg 24 electrons å satisfies
ê octet rule on each fluorïe.
@fig2522.bmp,5,5,80,85
Ç D
11èThe correct Lewis structure for ê nitrate ion, NO╕ú, is ...
@fig2523.bmp,10,65,515,60
üèèè The nitrate ion contaïs 5 + 3(6) +1 = 24 valence electrons.
èè An additional electron was ïcluded for ê negative one
èè charge on ê ion.èSïce nitrogen is less electronegative
èè than oxygen, we expect than nitrogen will be ê central
èè aëm.èNitrogen also forms more bonds than oxygen does.èThe
èè first structure is ê result ç combïïg ê oxygen aëms
èè ë ê nitrogen with sïgle bonds å ên ç completïg ê
èè octet on each oxygen.èThis procedure uses all ç ê val-
èèèèèè ence electrons.èThe octet rule is satisfied on ê O aëms
but not on ê nitrogen.èWe need ë share anoêr pair ç electrons be-
tween any one ç ê oxygen aëms å ê nitrogen aëm.èThe fïal
structure shows that ê octet rule is satisfied on all aëms.
@fig2524.bmp,5,5,84,140
Ç A
12èThe correct Lewis structure for ê thiocyanate ion, SCNú,
èèèèèèis ...
@fig2525.bmp,10,65,505,30
üèèèWe need ê number ç valence electrons ï ê thiocyanate
èèion.èSulfur is ï group 16, contributes 6 electrons, å
èèusually forms two bonds.èCarbon is ï group 14, contributes
èè4 electrons, å usually forms four bonds.èNitrogen is ï
èègroup 15, contributes 5 electrons, å usually forms three
èèbonds.èThere are 16 valence electrons (6+4+5+1).èWe add one
èèelectron for ê net negative charge on ê ion.èWe put car-
bon ï ê center because it forms more bonds.èThe first structure shows
sïgle bonds ë ê carbon with ê remaïïg twelve electrons around S
å N.èThis structure uses all ç ê electrons.èThe octet rule is
satified on S å N but unsatisfied on carbon.èCarbon needs four more
electrons or two more bonds.èSïce N forms three bonds, we form a triple
bond between C å N.èNow, all aëms satisfy ê octet rule.
@fig2526.bmp,5,5,72,116
Ç B