Chapter 4-Structure of Atom

8/8/2019 Chapter 4-Structure of Atom

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CHAPTER 4

STRUCTURE OF ATOM

NURUL AUNI BINTI ZAINAL ABIDIN

CHM 138

BASIC CHEMISTRY


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Properties of Waves

Wavelength, P (lambda)

- thedistancebetweenidentical points onsuccessivewaves.

Amplitude

- theverticaldistance fromthemidline ofawaveto the peak ortrough.

Frequency, R (nu)

- thenumberofwavesthat passthrougha particular pointin 1

second(Hz = 1 cycle/s).

The speed (u) of the wave = P xR


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Maxwell (1873), proposed that visible light consists ofelectromagnetic waves.

Electromagnetic radiation is the emission and transmission of

energy in the form of electromagnetic waves.

Speed of light (c) in vacuum = 3.00 x 108 m/s

ELECTROMAGNETIC RADIATION

All electromagnetic radiationP xR!c


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TYPES OF ELECTROMAGNETICRADIATION


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P x R = c

P = c/R

P = 3.00 x 108 m/s/ 6.0 x 104 Hz

P = 5.0 x 103 m

A photonhasa frequency of 6.0 x 104 Hz.Convertthis

frequencyinto wavelength(nm). Doesthis frequency fall

inthevisibleregion?

P = 5.0 x 1012 nm

P

R


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PLANCKS QUANTUM THEORY

Quantum thesmallestquantity ofenergythatcanbe

emitted(orabsorbed)inthe form ofelectromagnetic

radiation.

Whensolidsareheated, theyemitelectromagneticradiation overawiderange ofwavelengths.

Radiantenergyemittedbyan objectatacertain

temperaturedepends onitswavelength.

Atomsandmoleculescouldemit(orabsorb)energy(light)

indiscreteunits(quantum).


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E= h x R

Y!cP

E = hc

P

h = Plancksconstant

= 6.63 x 10-34 Js

R!frequency oftheradiation


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Lighthasboth:1. wavenature

2. particlenature

hR = KE + W

Photoelectric Effect

Photon isa particle oflight

KE = hR - W

hR

KE e-

W = is the work function and

depends how strongly electrons

are held in the metal

KE = kinetic energy of the ejected

electron


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E= h x R

E= 6.63 x 10-34 (Js) x 3.00 x 108 (m/s)/0.154 x 10-9 (m)

E= 1.29 x 10

-15

J

E= h x c/ P

EXAMPLE:

Whencopperisbombardedwithhigh-energyelectrons,X

raysareemitted. Calculatetheenergy(injoules)

associatedwiththe photonsifthewavelength oftheX

raysis0.154nm.


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LineEmissionSpectrum ofHydrogenAtoms

EMISSIONSPECTRA


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EMISSIONSPECTRAOF VARIOUSELEMENTS


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1. e- can onlyhavespecific

(quantized)energyvalues

2. lightisemittedase- moves from

oneenergylevelto alowerenergylevel

Bohrs Model oftheAtom(1913)

En = -RH ( )1

n2

n (principalquantumnumber) = 1,2,3,

RH (Rydbergconstant) = 2.18 x 10-18J


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Ephoton = (E = Ef- Ei

Ef = -RH ( )1

n2f

Ei = -RH ( )1

n2i

i f

(E = RH( )

1

n2

1

n2

nf = 1

ni = 2

nf = 1

ni = 3

nf = 2

ni = 3


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Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = (E = -1.55 x 10-19 J

P = 6.63 x 10-34 (Js) x 3.00 x 108 (m/s)/1.55 x 10-19J

P = 1280nm

Calculatethewavelength(innm) ofa photon

emittedbyahydrogenatomwhenitselectron

drops fromthen = 5stateto then = 3state.

Ephoton = h x c / P

P = h xc/ Ephoton

i f

(E = RH( )

1

n2

1

n2Ephoton =


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QUANTUM NUMBERS

Atomic orbital thewave function ofanelectroninanatom

3quantumnumbersrequiredto describethedistribution

ofelectronsinhydrogenand otheratoms.i)ThePrincipleQuantumNumber(n)

ii)TheAngular MomentumQuantumNumber(l)

iii)The MagneticQuantumNumber(ml)

Describesthebehaviorofaspecificelectron

i)TheSpinQuantumNumber(ms)

Quantum Numbers,

=

= (n, l, ml, ms)


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Principal Quantum Number (n)

n = 1, 2,3, 4, .

n=1 n=2 n=3

- distance ofe- fromthenucleus

- thelargern value, thegreateraveragedistance ofan

e- inthe orbital fromthenucleus, andthelargerthe

orbital


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Angular MomentumQuantumNumber(l)

foragivenvalue ofn,l= 0, 1, 2, 3, n-1

n = 1, l= 0

n = 2, l= 0or1

n = 3, l= 0, 1, or2

- Shape ofthe volume ofspacethatthee-

occupies/orbitals

l= 0 s orbital

l= 1 p orbital

l= 2 dorbital

l= 3 forbital


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Where 90% of the

e- density is foundfor the 1s orbital


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l= 0(s orbitals)

l= 1 (p orbitals)


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l= 2 (dorbitals)


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MagneticQuantumNumber(ml)

foragivenvalue ofl

ml= -l, ., 0, . +l

- orientation ofthe orbitalinspace

ifl= 1 (p orbital), ml= -1, 0, or1

ifl= 2 (d orbital), ml= -2, -1, 0, 1, or2

* Eachsubshell ofquantumnumber(l)

contains(2l + 1) orbitals.


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ml= -1, 0, or1

3 orientations is space


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ml= -2, -1, 0, 1, or2 5 orientationsisspace


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SpinQuantumNumber(ms)

ms = + or-

ms

= -ms

= +


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Quantum Numbers: (n, l, ml, ms)

Shell electronswiththesamevalue ofn

Subshell electronswiththesamevalues ofn andl

Orbital electronswiththesamevalues ofn,l, andml

Howmanyelectronscanan orbitalhold?

Ifn,l,andml are fixed, thenms = or-

]= (n, l, ml, ) or]= (n, l, ml, -)

An orbitalcanhold 2 electrons


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Howmany 2p orbitalsarethereinanatom?

2p

n=2

l= 1

Ifl= 1, thenml= -1, 0, or+1

3 orbitals

Howmanyelectronscanbe placedinthe3dsubshell?

3d

n=3

l= 2

Ifl= 2, thenml= -2, -1, 0, +1, or+2

5 orbitalswhichcanholdatotal of 10e-


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Existence(andenergy) ofelectroninatomisdescribed

byitsunique wave function].

Pauliexclusion principle - no two electronsinanatom

canhavethesame fourquantumnumbers.

SchrodingerWave Equation

quantum numbers: (n,l,ml

,ms)


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Energy of orbitals in a single electron atom

Energy onlydepends on principalquantumnumbern

En = -RH ( )1

n2

n=1

n=2

n=3


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Energy of orbitals in a multi-electron atom

Energydepends onn andl

n=1 l = 0

n=2 l = 0n=2 l = 1

n=3 l = 0n=3 l = 1

n=3 l = 2


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Fillupelectronsinlowestenergy orbitals

(Aufbauprinciple)

H 1 electronH 1s1

He 2 electrons

He 1s2

Li3electronsLi 1s22s1

Be4electronsBe 1s22s2

B5electronsB 1s22s22p1

? ?


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Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s


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C 6 electrons

Themoststablearrangement ofelectronsinsubshellsis

the onewiththegreatestnumber of parallelspins

(Hundsrule).

C 1s2

2s2

2p2

N 7 electronsN 1s22s22p3

O 8 electronsO 1s22s22p4

F 9 electrons

F 1s22s22p5

Ne 10 electrons

Ne 1s22s22p6


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ElectronicConfigurations ofSome

SecondRowElements


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Whatistheelectronconfiguration of Mg?

Mg 12 electrons

1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons

Abbreviated as [Ne]3s2 [Ne] 1s22s22p6

Whatarethe possiblequantumnumbers forthelast

(outermost)electroninCl?

Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3 l = 1 ml

= -1, 0, or +1 ms

= or-


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Outermost subshell being filled with electrons


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Paramagnetic

unpaired electrons

2p

Diamagneticall electrons paired

2p


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Example:

Howmanyelectronsin12Mgcanhavethe

followingquantumnumbers?a)s=+1/2

b)n=4

c)n = 1,s = -1/2

d)l =1


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1s2

Electronconfiguration:

1s2 2s2 2p6 3s2

Orbitaldiagram:

2s2 2p6 3s2

Answer:

a) 6 electronsb)0electron

c) 1 electron

d) 6 electrons

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Chapter 4-Structure of Atom