Upload
dn-zack
View
235
Download
0
Embed Size (px)
Citation preview
8/8/2019 Chapter 4-Structure of Atom
1/49
CHAPTER 4
STRUCTURE OF ATOM
NURUL AUNI BINTI ZAINAL ABIDIN
CHM 138
BASIC CHEMISTRY
8/8/2019 Chapter 4-Structure of Atom
2/49
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
8/8/2019 Chapter 4-Structure of Atom
3/49
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
8/8/2019 Chapter 4-Structure of Atom
4/49
TYPES OF ELECTROMAGNETICRADIATION
8/8/2019 Chapter 4-Structure of Atom
5/49
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
8/8/2019 Chapter 4-Structure of Atom
6/49
8/8/2019 Chapter 4-Structure of Atom
7/49
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).
8/8/2019 Chapter 4-Structure of Atom
8/49
E= h x R
Y!cP
E = hc
P
h = Plancksconstant
= 6.63 x 10-34 Js
R!frequency oftheradiation
8/8/2019 Chapter 4-Structure of Atom
9/49
8/8/2019 Chapter 4-Structure of Atom
10/49
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
8/8/2019 Chapter 4-Structure of Atom
11/49
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.
8/8/2019 Chapter 4-Structure of Atom
12/49
8/8/2019 Chapter 4-Structure of Atom
13/49
LineEmissionSpectrum ofHydrogenAtoms
EMISSIONSPECTRA
8/8/2019 Chapter 4-Structure of Atom
14/49
EMISSIONSPECTRAOF VARIOUSELEMENTS
8/8/2019 Chapter 4-Structure of Atom
15/49
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
8/8/2019 Chapter 4-Structure of Atom
16/49
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
8/8/2019 Chapter 4-Structure of Atom
17/49
8/8/2019 Chapter 4-Structure of Atom
18/49
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 =
8/8/2019 Chapter 4-Structure of Atom
19/49
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)
8/8/2019 Chapter 4-Structure of Atom
20/49
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
8/8/2019 Chapter 4-Structure of Atom
21/49
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
8/8/2019 Chapter 4-Structure of Atom
22/49
Where 90% of the
e- density is foundfor the 1s orbital
8/8/2019 Chapter 4-Structure of Atom
23/49
l= 0(s orbitals)
l= 1 (p orbitals)
8/8/2019 Chapter 4-Structure of Atom
24/49
l= 2 (dorbitals)
8/8/2019 Chapter 4-Structure of Atom
25/49
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.
8/8/2019 Chapter 4-Structure of Atom
26/49
ml= -1, 0, or1
3 orientations is space
8/8/2019 Chapter 4-Structure of Atom
27/49
ml= -2, -1, 0, 1, or2 5 orientationsisspace
8/8/2019 Chapter 4-Structure of Atom
28/49
SpinQuantumNumber(ms)
ms = + or-
ms
= -ms
= +
8/8/2019 Chapter 4-Structure of Atom
29/49
8/8/2019 Chapter 4-Structure of Atom
30/49
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
8/8/2019 Chapter 4-Structure of Atom
31/49
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-
8/8/2019 Chapter 4-Structure of Atom
32/49
Existence(andenergy) ofelectroninatomisdescribed
byitsunique wave function].
Pauliexclusion principle - no two electronsinanatom
canhavethesame fourquantumnumbers.
SchrodingerWave Equation
quantum numbers: (n,l,ml
,ms)
8/8/2019 Chapter 4-Structure of Atom
33/49
Energy of orbitals in a single electron atom
Energy onlydepends on principalquantumnumbern
En = -RH ( )1
n2
n=1
n=2
n=3
8/8/2019 Chapter 4-Structure of Atom
34/49
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
8/8/2019 Chapter 4-Structure of Atom
35/49
Fillupelectronsinlowestenergy orbitals
(Aufbauprinciple)
H 1 electronH 1s1
He 2 electrons
He 1s2
Li3electronsLi 1s22s1
Be4electronsBe 1s22s2
B5electronsB 1s22s22p1
? ?
8/8/2019 Chapter 4-Structure of Atom
36/49
Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
8/8/2019 Chapter 4-Structure of Atom
37/49
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
8/8/2019 Chapter 4-Structure of Atom
38/49
38
8/8/2019 Chapter 4-Structure of Atom
39/49
8/8/2019 Chapter 4-Structure of Atom
40/49
40
8/8/2019 Chapter 4-Structure of Atom
41/49
8/8/2019 Chapter 4-Structure of Atom
42/49
ElectronicConfigurations ofSome
SecondRowElements
8/8/2019 Chapter 4-Structure of Atom
43/49
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-
8/8/2019 Chapter 4-Structure of Atom
44/49
Outermost subshell being filled with electrons
8/8/2019 Chapter 4-Structure of Atom
45/49
8/8/2019 Chapter 4-Structure of Atom
46/49
Paramagnetic
unpaired electrons
2p
Diamagneticall electrons paired
2p
8/8/2019 Chapter 4-Structure of Atom
47/49
47
8/8/2019 Chapter 4-Structure of Atom
48/49
Example:
Howmanyelectronsin12Mgcanhavethe
followingquantumnumbers?a)s=+1/2
b)n=4
c)n = 1,s = -1/2
d)l =1
8/8/2019 Chapter 4-Structure of Atom
49/49
1s2
Electronconfiguration:
1s2 2s2 2p6 3s2
Orbitaldiagram:
2s2 2p6 3s2
Answer:
a) 6 electronsb)0electron
c) 1 electron
d) 6 electrons