8/11/2019 VSEPR 2

1/13

Atoms

and

Molecules

Clearly, there

are frxed

combining

capacitie.s

hat determine

the atomic

ratios

and limit

the variety

of

possible

molecules. It w'as

precisely

these

ratios

that

originally

gave

rise

to the

atomic theory

in

the early

19th

cen-

tury. The achievement

of

the

20th

century'r'i'as o explain the

phenomena

of valcnce

and chemical

combination

through the

quantum

theory.

2-+ . MOLECULES

Nuclei,

small and dense,

form

the skeleton of molecule5-1hg bricks.

The

electrons are the mortar

holding

the

bricks

toqether, the flesh

on the

bones. Lightwcighr

and di f fuse,

he-clcctrons

eteimine

molecular s ize,

shape, and reactivity.

What,

at this

point,

can

we

guess

about the electri-

cal and

geometric

structure

of molecules?

First there is

the

cuestion

ofhow

the

clectrons

are

distributed.

For

molecules

such

as

H,

b,, N,

and F,

thc

answer s dictated bv

the

most

basic

considerations

of symmetry.

Each structure (the singly bonded

halogen F-F, for example) remains the same whether rotated along the

bond

axis or flipped left

and right.

The two

nuclei, both

of the same

kind,

are

indistinguishable

and therefore incapable

of attracting the elec-

trons

unequally.

Since the elcctron

distribution

must respect

the

sym-

metry

of the nuclear

frameu,ork,

the

negative

charge spreads

out evenly

around and between

the

nuclei.

All

these diatomic

(having

two nuclei),

homonuclear

(of like kind)

molecules

are

nonpolar as a result, meaning

that there is no

skewing of

the elcctronic charge to either end. This re-

sult we demand

strictly

on the basis

of svmmetry,

requiring no

specific

theory of molecular

structure

for its

justification.

Go

no further

than the

arrangement

sketched

n Figure 2-12(a);nothing

elsewould mak e sense.

Heteronuclear

diatomic molecules,

containing tw.o unlikc nuclei,

make for

a

different

storv.

One of

the nuclei alwavs attracts the electrons

more

than the

other, u.ri ,.,

proportionatclv

more negative charge

accu-

mulates

on that

end. Ionic bonds

like

thosc

alreadv citcd in NaCl

are an

extreme

case of

unequal sharing.

Where the bonding

is

covalcnt, as u.ith

HCl,

the difference

in electronegativity

creates a skeweclbond and

a cor-

respondingly

polar

molecule.

The

chlorine side

gro\{rs

richer

in elec-

trons

than the hydrogen

side leaving

a

partiallv

negative

Cl and

partially

positive

H. This

separation

of charge

gives

rise to a dipole moment

(Figure

2-12b)

and u,e

use a

owercase

Greek

delta to represent

the

partial

charges:

D+-916-.

Triatomic

molecules

bring

another element of choice. Consider

water, HrO, where the singly boncledatoms HOH form a bent structure:

H,O.H

F

eilil{il,

q

lolot[r't

iulrlnn[tr|

Mm)lr*

lNlllliu

i(illLllrflii

mrlllLJq

l]lDlrlilU

8/11/2019 VSEPR 2

2/13

2-4. l4olecules

6 1

ta)

-

a

- c

he

tr

Frc;unr 2

-

12.

Symmetry anclstructure

in diatomic

molccules. (a) Homonuclcar

svstcms:

Since both

nuclei

are the samc,

no clistinctions exist

for

points

eithcr

ro-

tated around

thc

bond

axis or

flipped

end

to end.

Electrons are wrapped uniformly

around

thc axis, and thc

distribution

at

one end

is

alvuavs

he mirror imagc of the

other. Neither side accumulates more clcctrons.

The molcculc

is nonpolar. (b)

Heteronuclear systems: With differcnt nuclci,

the molecule now'

posscsses

wo

clearly rccognizablc cnds and

one of

them attracts

electrons at the expcnse

of its

neighbor. Thc structure has both

an

electron-rich

negative site (6-) and

an elec-

tron-poor

positive

site (6*): a dipole moment,

p.

Here each

of the O-H

bonds is

polar

(Od--11or; since oxygen

is more

clcctronegative than hydrogen,

and

the

bent configuration

gives

the mol-

cculc a net dipolc momcnt. The two

polar

bonds contribute

jointly

to a

combinecl

dipole

along the HOH

bisector:

ure

8/11/2019 VSEPR 2

3/13

o/ -

2.

Atoms

nd

Molecules

A

molecule

of

O:C:O,

however,

s

nonpolar

despite

its

polar

g6+:9E-

double

bonds.

The

structure

is

linear'

and

the

bond

dipoles

cancel

out

as

shown

below:

,o-:@-o'

One points

to the

right,

one

points

to

the

left'

and

their

combined

effect

is nil.

That

does

not

mea.,

he

electrons

are

evenly

distributedthroYghout

the

linear

structure,

which

remains

skewed

u'

6ti-:626+:Ob-'

There

s

iust

no

overall

dipole

moment

for the

species,

ven

hough

different

por-

iion,

of

the

mole^cule

isplay

nequal

oncentrations

[

charge'

We

should

wonder'*(y

tutbott

dioxide

is

linear

but

water

is

bent'

given

heir

similarities

n

oih",

ways.

Both

are

triatomic

molecules,

and

both contain two ,r.tll"i of one type bonded to

another

nucleus

in

the

center.

Where

they

differ,

certainiy,

s

in

the

particular

mechanism

of

bonding

(single.r"rrt,

double)'

'o

p"thup'

there

s some

elationship

e-

tween

electronic

structure

u"d

-ol"t"lar

geometry'

W"

need

to

un-

cover

t.

The

simplest

explanation

s to

attribute

the

shape

o

repulsive

orces

suffered

by th"

electrons

around

he

central

nucleus'

an

approach

now-n

as

rhe

valence-shell

electton-pair

repulsion

(

/SEPR)

-model

According

o

this picture,

h"

n"guiiu"ty

'tttu'g"d

clcclron

pairs

r'rill

*'tlTl':',,"l"ctrical

,"prririo.,,

by

mouiig

aw"i

*"^*1"":i

other-as

far

a*'av

as

possible'

hus

for

O:C:O,

the

two

,Jt'

of

doubly

bonded

pairs

on

carbon

ie

farthest

auart

when

the

C:O

bonds

stand

cliametricallv

oPPgt"q

to-form

a linear

;;il.;h.

In

H:O:H,

by

contrast,

here

arefour individuallyactingpairs

of

electrons

around

,h"'o"yg"t'

nucleus:

r'vo

bonding

pairs

and

tu'o

lonc

pairs.

All

lbur

pairs

*ou"

t'o-uuoid

ach

otheE

and

they

tat-t

do

so

most

cf-

fectively

by

pointing

toward

the

corners

of

a

tetrahedron

as

shown

in

Figure

-13.

VSEPR's

irst

prediction

for

\\"ater

s

almost

corrcct'

The

H-O-H

bond

angle

*'o.,ld

b"

109.5o

n the

forecasted

etraheclral

cometry,

a

p"r"-"rE,

reasonably

lose

o

the

exPerimentally

determined

value

of

iO+.S".

If

one

urrrrtt't".

urther

that

the

tlvo

lone

pairs

interact

more

,trongt,'.,'than

he

bonding

pairs,

hen

VSEPR

heory

predicts

a

comPres-

sion

of

the

H-O-fi

U"""a

ancl

a

smaller,

more

satiifactorv

angle'

The

presumeddifferencebetween one-pair and bonding-pairrepulsions

s

ascribed

to

their

respective

environments'

Thc

bonding

electrons'

shared

etween

nuclei'

take

up

less

space

nd

consequentlv

re

repelled

less

han

the

more

diffuse

one

pairs'

8/11/2019 VSEPR 2

4/13

polar

2-1. Molecules

63

FIcut

8/11/2019 VSEPR 2

5/13

2.

Atoms

and Molecules

should be

a trigonal

planar

plane.

and eachF-B-F

bond

molecule.

It is. The four

nuclei

lie in

one

angle

s 120o:

Th.

t - *

t l f 1 i

- t

_ :

a

: : -

- : : :

: t _ : :

_-

--- -

- -" " '

Note

also hat BF,

is

an

apparent

exception o the

octet rule,

since

the clcctron

configuration

placesonly six electrons

around the boron.

Suchdcviations

rom

the

simple

Lewis model

are

hardly

surprising,

and

r'r'e

shall resolve

he difficulty

evcntually by

using molecular

quantum

mechanics.Yct er.en n defcat hc concept of an octet hasa measureof

truth, for

the cleficient

structure

porr"r,r"t;

special

properties

precisely

becausef its "missing"

elcctrons.

The boron indeed

would

benefit

ener-

getically

bv acquiring

thosc

electrons,

and

hence

BF,

proves

to

be

a

strong

Lewis

acid:

a spccies

eady

to accept

a

pair

of electrons

rom

some

other sourcc.

Ammonia,

a

Lewis

base (an

electron-pairdonor),

is

just

such a

source. With

three

bonding

pairs

and one one

pair,

the molecule

adoptsa trigonal

pyramidal

geometry:

lL: i]-iiil

H : N :

H

IJ

8/11/2019 VSEPR 2

6/13

2-1.

14olecules

6 5

The

four

pairs of

elcctrons

are

rientccr

tetrahecrrally

ust

as n

r,r,ater,

ut

here

only

three

vertices

of

thc

tetrahetiron

are

actuaily

occupied

by

nu-

clei.

The

"1,.:g:"

sits

t

he

enter;

"u.t

orit

"

;i;;^/rilJ5"r.,",_,

"

orner;

and

the

lone

pair remains

unsharccr

at thc

fourlh";"Jil,

r.ertex

of

the

tetrahedron.

The

three

hl.clrogcns

all

lie

in

"".

pl;;,

,of,p"a

Uu

u

nitrogcn,bearing the nonbonded eiectron

pair and

ready

to'.,r-bi.r"

rvith

an

electron-deficient

Lew.is

acid.

BF,

and

NH,,

each

har.ing

rvhat

the

other

lacks,

thcn

come

together

to

form

the

united.o-pn.rnj

FH

l l

F-F-ry-H

l l

F H

in

rvhich

the

nitrogen

sharcs

ts

ronc

pair

to

give

boron

an

octet.

with

that

fourth

clectrrin

pair,

morcor".,

,h"

b,rrirr",

ubsequentll, .lopt. otetrahedralconfiguration

ust

as

VSEpRpredicts:

B'ron,and

nitrogcn

each

ha'e

fbur

bon

8/11/2019 VSEPR 2

7/13

(a )

66

2. Atoms

and Molecules

TRIGONAL

LANAR

( d )

t:.-.

TRIGONALIPYRAIVIDAL

OCTAHEDRAL

FtcuRE 2-14. VSEPR,

a summarv: orientation of electrons

about

a central

atom.

In

each configuration, thc

pairs

of electrons arc

f'arthest apart and thus suffer

minimal

repulsion. Bonding

and

onc

pairs

both

play

a

role, but sitescontaining one

pairs

arc not occupied bv atoms. (a) Two

pairs:

inear.

(b) Three: trigonal

planar.

(c )

Four:

tctrahedral. (d) Five:

trigonal

biptramidal. (c)

Six:

octahedral. To

predict

tle

molecular skeleton,

place

one atom at each

vertex

assigned o a

bonding

pair.

2-5 .

STOICHIOMETRY

Vierved at the most basic level, a chemical reaction is a microscopic en-

counter among individual

atoms and molecules. Some

integral number

of

particlcs

is transformed

to some other

number of particles, the event

summarized

concisclv by a balanced

chemical

equatlon such as

2H2+

02

*

2H2O

The

process

involves whole

species, each a separate

entity. Here tr.r,o

molecules of hvdrogen

and one molecule of oxygen

go

in; two molecules

of u,'ater come

out.

It is only a rearrangement,

with

particles

neither created nor de-

stroyed. Four hvdrogen atoms and two oxygen atoms go in; four hydro-

gen

atoms and tw-o oxygen atoms

come

out. More

precisely:

Four

hvdrogen

nuclei,

tu.o oxygen

nuclei,

and

twenty

electrons start out as

molccular hvdrogen

and oxygen; the

same

four hydrogen nuclei, two

(c)

TETBAHEDRAL

8/11/2019 VSEPR 2

8/13

(b) How many

calcium ons coexist

with 0.50'15

g

of chloride ons?

From

the

empirical

lbrmula

u.e

determinc first thc molcs of Cl-, then

the moles

of Ca'-, and inally

the

number of calcium ons:

35 .451 C l

x - - . - - - : i .

= r l . 9gc l

1 mol Cl

1 mol

Cl -

t 1 n ) I

I M O I L A

X - X

6.022

x

1023

ons

0.5045

Cl-

x

35.453g Cl- 2 mol Cl- mol

=

4.285

x

1021 a2+

ons

In doing

so,

we ignore

thc vcrl' slight

differences in mass bet'ivccn ions

and atoms.

A

fer,r,

more electrons,

give

or take, will have little effcct on

the calculation.

(c) Hon,mdn)i

Brams

of atomic

chlorine are contained

n

0.732

mol oJ

calcium

chloride? Notice by nou'hor,r'all

stoichiometry

problems

are

the

samc

grams

to

moles

to atoms

in all

possible

combinations:

2 mol Cl

R2.8 Review

nd Guide o Problems

0.732mol

CaCl,

x

I mol CaCl,

(d)

lVhat s

the

percentage

hlorineby

weight?

Onc

mole of CaCl,

contains one

mole

of calcium

(40.078

B)

and two

moles

of chlorine

(1O.906

g).

Their combined

ormula

u.'eight

s 110.984

g

mol-':

Weight

o/o

Cl

=

70.906

Cl

x

100%

63.888%

1 1 0 . 9 8 4 C a C l ,

Calcium's veight crccntages 36.112o/o.

ExR,nr,rprn-4. L Simple Molecule:

Electrons, Bonds,

and Geometry

Carbon and h-vdrogenbrm

covalent

bonds, not

ionic.

Compounds

con-

taining

ust

these

wo

elcments

are called

hydrocarbons,

any

of

*'hich

rve knou.

ascommon

fuels:

methane

700h

of natural

gas),

ethane,

ethvl-

ene,acetvlene,

ropane,

utane,

pentane,

exane, eptanc, ctane,an d

so

forth. They

exist as

ndcpendentmolecules.

PROsr-sru:A colorless,

dorless,

ighl)'f lammable

as

s fbund to

havea molecularu'eightof 16.04

g

mol-'

.

Analysis horvs

he material

o

be a

pure

hydrocarbon

75.0o/o

arbon

bv weight. (a)

What is the empiri-

cal

lbrmula? (b) What is

the

molecular

formula? (c)

Drarv

a

reasonable

Lcrvisstructurc or thc molecule.

(d) Proposea

geometric

structure.

8/11/2019 VSEPR 2

9/13

lons?

I

tncln

Sample

Problems

R2.9

SoLu.rtoN:

Thke

some

arbitrarl'

amount,

say

100

grams'

Since

cvery

100

grams

vi, i l l contain

75.0

grams

of

carbon

and

25.0

grams

of

hyclrogen,

i,'.

.un

immediately

comPute

moles

of

C and

moles

of

H:

l m o l C

T J . O s C X - = 6 . 2 4 m o l C

o

1 2 0 l 1 q C

. . " . 5

25 .0gHx

l m o l H

1.00794

H

=

24.8

mol

H

I O I 1 S

O I I

oJ

c

th e

con-

hich

r : thvl-

nd

o

o

(a) l4hat

is the

empiricalformula?

The

molar

ratio suggestt

C-u.z+Hzn.s,

o. CH,

,,

oo,hcn

educed

to lorvest

erms.

Exprcssed

as

ntcgers,

thc

empir-

ical

formula

is CHn.

(b7

l4lhat

is the

molecular

-formula?

The

formula

rveight

for

CHo

is

16.0+, exactly

equal

to the

moiccular

rveight

prcviouslv

given'

The

mol-

ecular

fbrmuia

therefore

is CH*

(and

the

gas

s

methane)'

(c) Suggest Lewis

structure.

First,

understand

.ivhat

a Leu'is

struc-

ture

is anciivhat

it is

not. A

dot

tliagram

is not a

flau,lcss,

complete,

or

e\rcn

unique

description

of a svstem's

bonds.

It

is a sketch'

a cartoon'

a

shorthani

\,va)/

o

pottruy

the

gross

distribution

of electrons.

It

sho'uvs

the

elcctronr

"ith"i

spread

betr,veen

nuclei

(in

pair

bonds)

or

localized

on

one

nucleus

(as

one

pairs).

Exactlv

how. man_v

electrons

arc

hcre

and

ho\r'

man)'

are

therc?

Why

clo

they

f"il *h"."

they

do?

What

are

their

energies?

Why

is

8 such

a

special

number-but

ivhrv, also,

are

there

so

many

excePtions

to

the

octet

rulc?

Expect.ro

onr.t"r,

from

a

Lervis

structure,

for such

diagrams

servc

onl) to

summarize'uvhat

rve

know-lrom

expcrimcnt.

The

ansrvers,

the reasons,

and

even

more

questions

u.ill

come,

soon

enough,

from

the

quantum

theory

ofbonding

(Chapters 4 through_7)'

Yet after

ali that,

thc

Liu,is

structure

u,ill still

give

us

the

bonding

in

a rcadily

appreciated,

almost

graphical

fashion.

It

survives

as

a useful,

p"rrrorir,.

..,ay

to reducc

a complex

system

to

a simple picture.

Here

is

u-hat

to

do, step

by step:

1

Ignoring

the

core, count

the

total

number

of

valence

electrons'

2. Arrange

the

atoms

according

to

thc

connectivitv

of

the

molecule.

Lay

th"em

out to

shorv

rvhat

ii connected

to

\vhat,

who

is

neighbor

to

\\'nom.

3.

Distribute

the

r.alence

electrons

around

the skeleton.

Start

bv

joining cach of the connected atoms r,vitha single bond, dravving

eithcr

a

pair

of

dots or,

equivalently'

a single

dash'

4.

Whcre

nccessar\.,

add

electrons

to

complete

the

valcncc

of

an-v

atom

boncled

to the

central

atom.

Hvdrogen

takes

tr'vo

electrons,

8/11/2019 VSEPR 2

10/13

R2.

10

Review

nd

Guide

o

Prcblems

and most

of the

elements

up through

argon ake

eight (but be

pre-

pared

or

exceptions).

5.

Usc the remaining

electrons,

f

any,

o ensure

an octet around

thc

central

atom.

One

option is

to

introduce

one

pairs;

another,

o

employ

multiple

bonds.

For

methane,

he bonding

s

comparatively

imple.

Carbon, a Group

IV

element,

contributes

our valence

electrons,

and eachof

the

four

hydro-

gen

atoms

supplies

one more

to

yield

a

total

of eight (step

1). Then,

knowing

that

H forms

one bond

and

C four,

we

place

hc

carbon at

the

center (step

2)

and draw

four

single

bonds (step

3):

H

I

H - C - H

t\

And, by sheer

good

luck,

rve

are

clone:

Each hydrogen

has its full

quota

of t'r'r'o

lectrons;

the

carbon

has

ts

octet; there

are

no

clectrons

to spare.

Wc have

nothing

more

to do.

(d) Propose

geometric

sftucturc.

A molecule of CHo must

accom-

modate

four

bonding

electron

pairs

about the central

carbon

atom. The

VSEPR

model

therefore

predicts

a

tetrahedral

geometrv.

rvith

H-C-H

bond

angles of 109.5o

a]l around.

Moreovcr,

since

nature

cannot

distin-

guish

among

the four

idcntical

hydrogcn

atoms, all

the C-H boncls

must

have

the same ength

and

polaritv

(skewed

slightlv

torvard the morc

elec-

troncgative

carbon).

But

there is

no

ovcrall molecular dipole

moment,

becausc

the four

polar

bonds

(orientcd

tetrahedrally) oppose each

other

to create a nonpolar moleculc.

The related

moleculc

CHrCl,

also

tetrahedral,

does

har.e

a molccu-

la r d ipo le

momcnt .

Why?

ExemprE

2-5.

More Molecules

PRoer-e^.,r:

uggest

el,vis

and

VSEPR

structures {br

the fbllou,ing

species:

a)

Sulfur tetrafluoride,

SF*.

(b,1

Acetvlcne,

CrHr. (c) Nitric

ox ide,NO.

(d)

The

n i t roson ium

on,

NO-.

Solu-rroN:

For each

system.

av

out the atomic skeleton

nd dis-

tribute

the valcnceelectrons n the most plausible ,vay. hen: C)rient he

clectron

pairs,

nonbonding

and bonding

alike, n the directions

specificd

h;

thc VSEPR

model .

(a) SuLfur

etrafluoride,

F

r.

Sulfur,

a Group

VI

atom, has

six

r.alence

electrons.Fluorine

has

seven.With

34 electrons o distribute,

rvc first

rt

t u

wuL

a@t

lilIr

T'rt

Tnu

nm

l

$ur

8/11/2019 VSEPR 2

11/13

I ) fC-

hc

tc)

V

len,

th e

{ ) I1 l -

T h e

-H

l l l L l s t

r ' lec-

; ther

] r ' cu -

ng

dis-

h e

t icd

Sample

Problems

R2. 11

connect each

F

to the central

S with

a single

bond:

The next step s to

complete

he

octet

on

eachF by adding

six

clectrons:

There

are t\l'o

electrons eft over

from the original

34,

evcn though

both

sulfur

and

all

four fluorines

alread,v

ave octets.

Still,

octet rule

notr,r'ith-

standing, he 33rd and

34th clectrons

hall

be assigned

o the

sulfur,n'hcre

the

fifth

pair

nor'r'contributes to

an

"expanded

octet"

of

10 clcctrons:

These five

pairs

of clectrons

(four bonding,

one

nonbonding)

rve

expect

to

point

toward the

vcrtices of a

trigonal

bipl'ramid.

Four

of the corncrs

r'vill contain

fluorinc nuclei,

while the

fifth lvill

rcmain

unoccupied.

Sulfur sits

at the center, as sho$'n

in the

drawing

belou':

F F

S

F F

r 'DCe

hrst

8/11/2019 VSEPR 2

12/13

R2. 12

Review nd

Guicle o Problems

Takc notc

as

$'ell:

Thc nonbonding

pair

I'alls specihcallv

in

the trigonal

plane

bctvreen

the uppcr

and

lorvcr

pvramids,

taking up an equatorial

po-

sition rather

than

an axial

position

on

top or bottom. Nominallv spaced

120o apart lrom

thc trvo

other equatorial

pairs

and 90o apart from thc

trvo

axial

pairs,

a

dilflse lonc

pair

in this

configuration suff-crs

he

rveak-

est repulsions possible. In gcneral, a ccntral atom surrounded b_v 10

elcctrons rvill

prcf'erentiallv

house

up

to three nonbonding

pairs

in equa,

torial sitcs.

(b) Acetylene,

C

rH

r.

After

drau'ing three single bonds to obtain thc

skeleton H-C-C-H,

u'e have

bur r,alence lectrons of the original 10

still to deplo-v.

Thev

go

automaticallr, to thc carbons, sincc the t'nl'ohy-

drogcns are

alrcaclv

satisfied; and

therc,

betu.een the carbons, thc lbur

clcctrons make

tn'o additional

pair

bonds: H-C=C-H. Thc triplc

boncl leaves

cach

carbon u.ith

an octet.

With no lone

pairs

in

the

molcculc,

VSEPR

prcdicts

a linear

arrangcmont

symmctric

about the HCCH axis. The tlr-o

C-H

bonds,

indistinguishable

and s-vmmetric,

must

har.e

he same

ength.

(c) ,\trric

oxirle,NO. Pooling

fir,e valcnce electrons from thc nitrogcn

and six from

thc oxygen,

NO has

a total

of 11 clcctrons. The

sum

vields

an odd number,

a sct impossiblc

tg arrangc

into tu,o

g.ctc$.

We can

givc

the

octet eithcr to oxvgen (:N:9)

or

to nitrogcn (N:O:), but not

to

both

at thc same

ime. Such

are thc failings of this too-simple model.

Each

of thcsc

tu'o

Leu-is

structurcs ofl'crs only a

partial

representa-

tion of thc bonding in

nitrogen

oxide.

For

a

more

complete

interpreta-

ticln, scc the description

of molecular

orltitals

in

Chapter 7.

(d) Ifie nitrosonium

on, NO+.

Thc

posit ively

charged on, dcficicnt

bv one

electron, has

a

r.alence

of 10. Allori.ed an evcn number of

elcc-

trons, u'e can satisfv

he octet rule

u,ith

thc structure

[:N:Q:]+.

ExR,uprE

2-6. Molecules

and Mass:

The Balancecl

Chemical

Equation

From

nuclei and

clcctrons to

atoms,

from atoms

to

molecules, from

moleculcs

to reactions--thc

lavr.

f

massconservation s at last

expressed

in

a

balanced

chcmical

equation.

A11 he

stoichiometric relationships

arc

embodiecl rvithin.

PRost.t..M

An

uniclentificd

compound

(relative

molecular r,r,cight

=

+6.07) rcacts

r.vith n-rolecular

oxvgcn

to

proclucc

carbon dioxidc,

r,t,atcr,and n'rolccular

nitrogen.

Analvsis shorvs hat

the

unknorvn mater-

ial

consistso{ 26.1%o

arbon, 13.1o/o vdrogen, and 50.8%onitrogen bv

mass. (a) What is its molecular lbrmula? (b) Brieflv describe a

possible

bonding

pattern.

(c) Write

a balanced equation {br the reaction.

Sorut-tor:

Gir,en he mass

perccntages

and

molecular

rveight, rve

knou.

hou, to determinc both

the

empirical

and molccular fbrmulas.

8/11/2019 VSEPR 2

13/13

ne

32.

33 .

3+.

35.

t o .

37.

38 .

Exercises

Ft2.21

31

Further

analvsis

hows

hat

the

compound

described

n

the

preced-

ing

exercise

as

a

molar

mass

of

:o.b

g

-or-f.-w;"1,

ir"*.r"."

lar

formula?

luggest

Levi'is

structures

or

:H6

(.thune)

and

CrHn

(et\lene).

Note that ethylenecontainso doubjebond.

Use

the

VSEPR

model

to

prerlict the

arrangemenr

of

hydrogen

atoms

around

each

carbon

n:

(a)

CrHu

"

(b)

C2H;

Dra-n'Lervis

tructures

or

HF,

HCl,

HBr,

and

HL

What

do

you

notice?

Show'the

erT'is

tructure-fbr

Hucl.

what

shape

oes

he vsEpR

theory

predict

for

the

molecule?

'

Draw'

Ler'vis

nd

vSEPR

structures

or

pclu

(phosphorusrichloride).

Drarv

Levvis

nd

VSEpR

structures

or

SFu

sulfur

hexafluoride).

A

molecule

of

ozone,

Or,

contains

hree

oxygen

atoms

n

a bent

configuration:

o

-o-o

Kno'wing

that

both

O-O

bonds

are

of

equal

length,

draw

two

equivalent

Lewis

structures

o represent

he

molecule.

All

thrce

N-O

bond

lengths

n

the

nitrate

ion,

NO'

are

equal.

Draw

thrce

equi'alcnt

Lewis

structures

consistent

with

this

obse.i.ation.

The

carbonate

on,

col-,

similar

to

the

nitrate

ion,

contains

hree

:x)gens

bound

to

a central

atom,

again

wit}

equai

bond

lengths.

Drar'r'

hree

equivalcnt

Lewis

structuies

o

represent

he

moleJular

carbonate

on.

Balance

he

following

equations:

( a ) _ N , + _ H , + _ \ H ;

( b )

_ H z * _ t r *

_ H I

(.)

-

CH4

+

_

02

------->

_

CO2

+

_

HrO

(,1)

-

coH,n

*

-

o:

------->

_

Co2

+

_

H2o

39.

+0.

+1