Chapter 1_Atomic Structure_PDF_GENERAL CHEMISTRY

Atomic Structure for Chemistry General

1. PhD. ĐặngVănHân
Office: 112B2 or 804H3 Building
Email: [email protected]
Chapter 1
Atomic Structure
Faculty of Chemical Engineering
Department of Inorganic Technology
General Chemistry

2. 2
Outline
2. Schrodinger wave equation
3. Quantum numbers for electron states in atoms
4. Atomic orbitals & shapes
5. Electron state in multi-electron atoms:
5.1. Shielding and Penetration effect
5.2. Electron distribution law
5.3. Electron configuration
1. Theory of atomic structure:
1.1. Classic
1.2. Bohr’s model
1.3. Quantum mechanical model

3. 3
John Dalton:
In1803,Daltonproposedtheatomictheorybasedontwolaws
ofconservationofmassandconstantcompositionasfollows:
1.Allelementsarecomposedoftinyindivisibleparticlescalledatoms.
Theycannotbecreatedordestroyedduringchemicalreactions.
2.Inchemicalreactions,atomsarecombined,separated,orrearranged
–butneverchangedintoatomsofanotherelement.
3.Compoundsareformedwhendifferentatomscombineindefinite
proportions;Ex:H
2O
Dalton
1983
Thompson
1904
Rutherford
1911
Bohr
1913
Schrodinger
1926
Dalton’s Atomic Theory

4. 4
Discovery of Electron
J.J, Thompson’s Exp.
Experiments:Whenthepowersupplyconnects2
electrodesinthevacuumtube:
1.Thecathoderayflowsfromthecathodetothe
anodeandhitsthefluorescentscreeninastraight
line.
2.Thecathoderayswillbedeflectedinthesame
directionasnegativelychargedparticlesinan
electricormagneticfield.
Dalton
1983
Thompson
1904
Rutherford
1911
Bohr
1913
Schrodinger
1926
The oil drop apparatus
Mass of the
electron is
9.11 x 10
-28
g
R. Millikan’s Exp.

5. 5
Thompson’s Atomic Theory
“Plum pudding” model:
1.Allatomsareelectricallyneutral.Totalpositive
charge(+)=Totalnegativecharge(-).
2.Thecharge(+)isevenlydistributedthroughout
thevolumeoftheatom,andelectronsmove
aroundthose(+)charges.
Dalton
1983
Thompson
1904
Rutherford
1911
Bohr
1913
Schrodinger
1926
Thompson’s Atomic Theory
Thomsonbelievedthattheelectronswerelike
plumsembeddedinapositivelycharged“pudding,”

6. 6
Alpha (α), beta (β) Particles, and gamma (??????)ray
-Alpha(α)Particle:twopositive
chargedelectronswiththesamemass
andenergyastheHenucleus.
-Beta(β)Particle:negativecharges
whichissimilartoanelectron.
-Gamma(γ)Ray:NOTaffectedby
electromagneticfield.It’sforminhigh
energyradiation.
Whenradiationinteractswithanelectrical
ormagneticfield,threetypesare
identified:
Rutherford and VilardExp.:

7. 7
Discovery of Proton & Neutron
Basedontheirepx.,E.GoldsteinandE.
Rutherfordclaimedthatthenucleushasasmall
volume,withpositivelychargedparticlescalled
“protons”locatedatthecenteroftheatom.Its
massis1.67x10
-24
g(relativemass=1)
1932–J.Chadwickconfirmedthe
existenceofthe“neutron”–aparticle
withnocharge,butamassnearly
equaltoaproton.

8. 8
Subatomic Particles
ParticlesCharge Mass (g) Location
Electron
(e
-
)
-1
(-1.6 x 10
-19
C)
9.11 x 10
-28
~5.5e-4 amu
Electron
cloud
Proton
(p
+
)
+1
(-1.6 x 10
-19
C)
1.67 x 10
-24
~1amu
Nucleus
Neutron
(n
o
)
0
1.67 x 10
-24
~1amu
Nucleus
Instead of grams, the unit we use is the Atomic Mass Unit(amu),
1 amu =
1
12
M
Carbon=1.67 ×10
-24
grams.

9. 9
Rutherford’s Atomic Model
Dalton
1803
Thompson
1904
Rutherford
1911
Bohr
1913
Schrodinger
1926
Based on his experimental evidence ⇢Atomic model:
Theatomismostlyemptyspace.
Allthepositivecharges,andalmostallthemassis
concentratedinasmallareainthecenter.Hecalledthisa
“nuclear”.
Thenucleusiscomposedofprotonsandneutronsbound
togetherbynuclearforces.
Theelectronsdistributedaroundthenucleus,andoccupy
mostofthevolume.
Theatomiselectricallyneutral.
Rutherford:

10. 1
0
Atomic Number
??????
??????
??????
X: element
Atomic number (Z)= n
e= n
p
Mass number (A) = Z + N
Pleaserememberthatspecific
atomsarecomposedofidentical:
PROTONS
NEUTRONS
ELECTRONS
The“atomicnumber(Z)”ofanelementisthenumberofprotonsinthenucleus
#protons=#electrons
Thesymboloftheelement:
Mass
number
Atomic
number

11. 11
Atomic Number
Example:
17
35
Cl
Z=n
e=n
p=17
A=35⇢n
n=35-17=18
Celllocationinperiodictable:17
Element:Chlorine
M
Cl= 35 amu

12. 12
Isotopes
Isotopesareatomsofthesameelementhavingdifferentmasses,
duetovaryingnumbersofneutrons.(differentneutronnumbers)
Frederick Soddy(1877-1956) proposed the idea of isotopes in 1912.
Naming Isotopes: Element-Mass number
Stable
Isotopes
Radioisotope
Cacbon-12Cacbon-13Cacbon-14
6
12
?????? 6
13
??????
6
14
??????
Stable
Isotopes
Radioisotope

13. 13
The Electromagnetic Wave (EW)
EWistheoscillationsoftheelectricandmagneticfieldsare
perpendiculartoeachotherandpropagateinavacuumwiththe
velocity=speedoflight(c~3x10
8
m.s
-1
)
Natural light has
electromagnetic wave
Specifications:
Wavelength (λ): distance between two adjacent peaks;
Frequency (ν): period of oscillation per unit of time:
ν = c/λ; (1 s
-1
= 1 Hz);
Energy (E): E = hν = hc/λ
Where: his Planck’s constant (6.626 10
-34
J.s)
and 1 eV = 1.6 x 10
-19
J

14. 14
The Electromagnetic Wave (EW)
A continuous spectrum of all wavelengths
Note that wavelength increases as frequency decreases

15. 15
The Electromagnetic Spectrum
Continuous Spectrum
Continuousspectrumistherainbowofcolors,containingallwavelengthsof
light,nodarkspotsinspectrum

16. 16
The Electromagnetic Spectrum
Line Spectrum
Linespectraisaphenomenonwhichoccurs
whenexcitedatomsemitlightofcertain
wavelengthswhichcorrespondtodifferentcolors.

17. 17
Line Spectrum
Atomic emission: an emission spectrum, is a line spectrum.
This technique is applied to qualify chemical elements

18. 18
Bohr’s Model
Bohr’s Model
In1913,Bohrproposedanewatomictheorybasedonthemerger
ofRutherford'satomicmodelandPlanck'squantitytheoryoflight,
whichincludedthreemaintopics:
1.Electronsrevolvearoundthenucleuswithcertain
concentricorbitsarecalledstationaryorbits;
2. When rotating in orbits, electronsare quantified and do
not emit electricity;
3. Energyis absorbed and emitted only when the electron
movesfrom one stationary orbit to another orbit.
E = ??????
�−??????
�=ℎ??????
Dalton
1803
Thompson
1904
Rutherford
1911
Bohr
1913
Schrodinger
1926

19. 19
Limitations of Bohr’s Model
1.Cannotexplainwhyelectronscanonlylocatepositionswhenmovingin
orbits
2.Canonlyexplainthelinespectrumofhydrogenorone-electronions
(He
+
,Li
2+
)adequately.
4.It’snotsuitabletocalculateelectronenergyformulti-electronatoms.
3.Cannotdetermineclearlytheline-spectrumintensityofspecific
elements.

20. SModern Atomic Theory
Based on Quantum
Mechanics
1.Themovementofmicroscopicparticle:
Wave-particledualityproperties;
Theuncertaintyprinciple;
Schrodingerwaveequation.
4. Electron State in Multi-electron Atoms
3. Electron State in Single-electron Atoms
2. Four quantum numbers

21. 21
Wave-Particle Duality Properties
Louis de Broglie
(1892 –1987)
Example:
-Electron has m
e= 9.1 x 10
-28
g, v ~6 x 10
5
m/s → λ~100 nm → Wave properties
-A ball has m = 200 g, v = 30 m/s → λ≪(~10
-39
nm) → No wave properties.
UsingEinstein’sandPlanck’sequations,deBroglieshowedthatall
mattershavingmass(m),velocity(v)willpropagatewithwavelength
(λ)bytheformula:

m
h

Particle Property
Wave Property
Microscopic objects such as electronsexhibit simultaneously
wave and particle properties.

22. 22
The Uncertainty Principle
Werner Heisenberg
(1901 –1976)
Δ??????.Δ??????�
??????≥
ℎ
4??????
Example: Electron has m
e= 9.1 x 10
-28
g, Δx ~10
-10
m→ Δv
x~6.6 x 10
6
m/s
Conclusion:Cannotdeterminetheexactpositionofanelectron,only
knowtheprobabilityofitspresenceatagivenpointinspace
Itisimpossibletodetermineaccuratelyboththemomentumand
thepositionofanelectron(oranyotherverysmallparticle)
simultaneously.
Where: x is the uncertain position and mv
xis the uncertainty in momentum

23. 23
Quantum Mechanics
Electronsinatomarecloudssurroundingtheirnucleusandhas
wave-particleproperties.
So,whycanweuseSchrodingerEquationtodescribeatomic
structure?
1.SchrödingerEq.containsbothwaveandparticleterms.
2.Solvinghisequationleadstowavefunctions:
Thewavefunctiongivestheshapeoftheelectronicorbital.
Thesquareofthewavefunction(
2
)givestheprobabilityoffindingthe
electron.Thatis,givestheelectrondensityfortheatom.

24. 24
The Schrodinger Wave Equation 0
8
2
2
2
2
2
2
2
2











VE
h
m
zyx
Where:
:wavefunctioncorresponding
to3-dimensionalwaveamplitude.
V:potentialenergyofparticle.
x,y,z:coordinatesofparticle.
Describe the motion of microparticlesin a stationary state0
8
2
2
2
2
2
2
2
2
2




















r
e
E
h
m
zyx
ForHydrowithV=-e
2
/r,then:

2
:alwayspositivetoshowtheprobabilitypresenceofe.

2
unit:probabilitydensityofeinavolumeunitdV.

25. 25
The Schrodinger Wave Equation),().(),,(
,,,,


 mnmn
YrRr 
Solve thisEq. for Hydrogen
AO ShapesAO Size
ThemotionofelectroninspaceoftheHydrogenatom
determined3quantumnumbers:
n = 1, 2, ...;
ℓ = 0,1,..(n-1);
m
ℓ= -ℓ,…,0,.,+ ℓ

26. 26
The Atomic Orbital (AO)
1.Electron State:
Electronscanbepresentinanywherewithvaryingprobabilityforminga
regionofspacesurroundingthenucleusknownasanelectroncloud.
Probability of finding an
electron in a hydrogen atom
in its ground state
2. Atomic Orbital (AO):
Theregionofspacearoundthenucleusinwhichthe
probabilityofanelectron’spresenceislager90%,
characterizedby3quantumnumbers:n,l,m
l
AOdescribessize,shapeandorientationinspace
throughthreequantumnumbers(n,l,m
s).

27. 27
The Atomic Orbital (AO)
Choosethecorrectstatement(s).Anatomicorbital(AO)is:
1)Theregioninwhichthereismaximumprobability(>90%)offindinganelectron.
2)Thesurfacewithequalelectrondensitycloud.
3)Theorbitalmotionofelectronsinanatom.
4)Theenergystateofanelectroninanatom.
5)AnAOisawavefunctionthatdescribesthestateofanelectroninanatomdefinedby4
quantumnumbers:n,ℓ,m
ℓ,m
s.
A.1and5 B.only1 C.1,2and3 D.1,2,3,4and5.
Review:

28. 28
Three Quantum Numbers
1.Principal Quantum Number,n
(lượngtửchính)
2. Azimuthal (Angular) Quantum Number,l
(lượngtửphụ)
3. Magnetic Quantum Number,m
l
(lượngtửtừ)
Schrödinger’sequationrequires3quantumnumbers:

29. 29
1. Principal Quantum Number, n
ThesameasBohr’sn.
n,describesthemainenergylevel,orshell,electronoccupies.
Asnbecomeslarger,theatom&energybecomeslarger
nispositiveinteger=1,2,3…
n 1 2 3 4 5 6 7
Electron
Shell
K L M N O P Q

30. 30
2. Azimuthal Quantum Number, l
n 1 2 3 4 5 6
l 0 1 2 3 4 5
Subshells p d f g h
Note: Electronswith the same n and l values ​​will form a quantum subshell
Ex: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, ….
Maximum number of electrons in subshell = 2(2l+1)
Dependsonthevalueofn(l<nandl=0,1,2,…,n-1)
l,describesthename&shapeoftheorbitals.
Weusuallygivealetternotationtoeachvalueofl:

31. 31
3. Magnetic Quantum Number, m
l
Dependsonthevalueofl
Hasintegralvaluesfrom–lto+l(l=-l…0…+l)
m
l,describesthe3Dorientationofeachorbitalinspace.
Example:
ℓ = 0 m
l = 0 1s-orbital
ℓ = 1 m
l= -1,0,+1 3 p orbital
ℓ = 2 m
l= -2, -1, 0, +1, +25orbital d
-10+1
-2-10+1+2

32. 3. Magnetic Quantum Number, m
l
In the n
th
shell
Number of AO = n
2 Max e
-
number = 2n
2
In the l
th
subshell
Number of AO = (2l+1) Max e
-
number = 2(2l+1)
Example.Determinethemaximumnumberofelectronsandtheprincipalquantum
numbernofLandN-shells:
a)L-shell:18e,n=3;N-shell:32e,n=4b)L-shell:8e,n=2;N-shell:32e,n=4
c)L-shell:8e,n=2;N-shell:18e,n=3.d)L-shell:18e,n=3;N-shell:32e,n=5

33. 33
4. Spin Quantum Number, m
s
Apartfrom3quantumnumbers(n,l,m
s),Usingm
stodescribetherotation
aroundelectronaxisbyitself.
Convention:
m
s= +1/2: Clockwise rotation; m
s= -1/2: Counter clockwise rotation of electron
↑ ↓
Referstothespinofanelectronandtheorientation
ofthemagneticfieldproducedbythisspin.
NotaffecttheAOmotion;
Foreverysetofn,landm
lvalues,m
s=+½or-½.

34. 34
Short Summary
The4quantumnumbers(n,l,m
landm
s)completelydeterminethestateofan
electroninanatomincludingsize,energy,shapeandmotion.
3p
6
e
-
no. in this
orbital
SubshellShell
Convention:
Note:
n = positive integer 1, 2, 3, …
l < n and l = 0, 1, 2, …, n-1
m
l< n and m
l= -l….0….+l
m
s= +1/2 or -1/2
Number of Orbitals
In Shell = n
2
2n
2
In Subshell = 2l+1 2(2l+1)
Max. Electrons

35. 35
Some Examples
Ex.1:Whichsetsofthethreequantumnumbersareacceptable?
1)n=4,ℓ=3,m
ℓ=-3 2)n=4,ℓ=2,m
ℓ=+3
3)n=4,ℓ=1,m
ℓ=0 4)n=4,ℓ=0,m
ℓ=0
a)1,3,4 b)1,4 c)2,3,4 d)3,4
Ex.2:Namesoftheorbitalscorrespondton=5,ℓ=2;n=4,ℓ=3;n=3,ℓ=0:
a)5d,4f,3s b)5p,4d,3s c)5s,4d,3p d)5d,4p,3s
Ex. 3: Orbital 3p
xis defined by the following quantum numbers:
a) only n, ℓ, and m
ℓ b) only nand m
ℓ
c) only ℓand m
ℓ d) n, ℓ, m
ℓ, m
s

36. 36
Atomic Orbital Shapes
ℓ = 0 s-orbital: sphere
ℓ = 3 f-orbital: complex
s: sharp
p: principal
d: diffuse
f:fundamental
ℓ = 1 p-orbital: dumbbell
ℓ = 2 d-orbital: clover

37. 37
The s-Orbitals
Alls-orbitalsarespherical.
Asnincreases,thes-orbitalsgetlarger.
Asnincreases,thenumberofnodesincrease.
Anodeisaregioninspacewherethe
probabilityoffindinganelectroniszero.
Atanode,
2
=0
l = 0 → m
l= 0 → No. AO = (2l+1) = 1

38. 38
The d-Orbitals
l = 1 → m
l= 0, ±1; → No. AO = (2l+1) = 3
There are three p-orbitals p
x, p
y,and p
z.
The orbitals are dumbbell shaped.
The three p-orbitals lie along the x-, y-and z-axes of a Cartesian system.
As nincreases, the p-orbitals get larger.
All p-orbitals have a node at the nucleus.
Electron
distribution of a
2p orbital.

39. 39
The d-Orbitals
There are five d-orbitals.
l = 2 → m
l= 0, ±1; ±2 → No. AO = (2l+1) = 5

40. 40
The d-Orbitals
Two of the d-orbitalslie in a plane aligned along the x-, y-and z-axes.
Three of the d-orbitalslie in a plane bisecting the x-, y-and z-axes.

41. The f-Orbitals
41
l = 3 → m
l= 0, ±1; ±2; ±3 → No. AO = (2l+1) = 7

42. 42
Electron State in Multi-electron Atoms
For Multi-electron Atoms:
Appeartheattraction&repulsionbetween
electronsandnucleus.
→Formationofshieldingandpenetration
effectsintheatom;
→Theenergystateofelectronsdependson
bothquantumnumbern&l.
SimilarwithH,thee
-
stateisalsodefinedby4quantumnumber:n,l,m
l,m
s.
AOshapesaresimilarwiththatofHydrogen

43. 43
Shielding (S) and Penetration (P) Effect
Theshieldingeffect(S)describesthe
repulsiveinteractionofthee
-
layerscreatesa
shieldingscreen→thedecreaseinattraction
betweenanelectronandthenucleus.
Z
Thepenetrationeffect(P):Incontrastto(S)-thepenetrationability
ofelectronfromtheouterlayertothenucleus.
As(n+l)↑,theshieldingeffect↑andpenetrationeffect↓Note:
Electron-nuclearattraction↓ E
n,l↑ AufbauPrincipleE
n,l~(n+ℓ)

44. 44
Aufbau Diagram
Orbitals and Their Energies
1. The largest energy gap is between
the 1s and 2s orbital
2. The gap between np and (n+1)s is
fairly large.
3. The gap between (n-1)d and ns is
quite small.
4. The gap between (n-2)f and ns is
even smaller.

45. 45
Electron Distribution Law
Threerules:
Aufbauprinciple(Germanaufbauen,"tobuildup")
Pauli’sExclusionsPrinciple
Hund’srule.
Electronconfigurationstellsusinwhichorbitalstheelectrons
foranelementarelocated.
4 quantum number: n, l, m
l, m
s
Electron
Arrangement

46. 46
Aufbau Principle
StabilizationPrinciple:themoststablestateofelectronsinan
atomhasthelowestenergy.
Electronsfillorbitalsstartingwithlowestnandmovingupwards
Electronsareassignedtoorbitalsinorderofincreasingvalueof(n+l).
Twogeneralruleshelpustopredictelectronconfigurations:
Forsubshellswiththesamevalueof(n+l),electronsareassigned
firsttothesubshellwithlowern.

47. 47
Aufbau Principle (KleshkovskiPrinciple)
Subshell:1s2s2p3s3p4s3d4p5s4d5p6s4f5d6p7s5f6d
(n+ℓ) 12 3 4 5 6 7 8
E
Example:
Oxygen(O)hasZ=8
⇢1s
2
2s
2
2p
4
Titanium(Ti)hasZ=22
⇢1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
2

48. 48
Pauli Exclusion Principle
Pauli’sExclusionsPrinciple:notwoelectronscanhavethesame
setof4quantumnumbers.
Paired Electron Un-paired Electron

Therefore,twoelectronsinthesameorbitalmusthaveoppositespins
3s
2

Example: n = 3
ℓ = 0
m
ℓ= 0
m
s= +1/2
n = 3
ℓ = 0
m
ℓ= 0
m
s= -1/2

49. 49
Hund’s Rule
Hund’sRule:electronsinthesamesubshellmustbe
distributedsothattheabsolutevalueofthetotalspinorunpair
electronsismaximized
Electronsfilleachorbitalsinglybeforeanyorbitalgetsasecondelectron
6
12
C
→ Z = 6 → 1s
2
2s
2
2p
2
→↑↓
1s
2
↑↓
2s
2
↑↓
2p
2
↑↑
-10+1
-10+1
Example:
[Trong phân lớp, điền spin dương (↑) đủ các obitals sau đó điền spin âm (↓)]
Nitrogen(N)hasZ=7➡
1s2s2p

50. 50
Hund’s Rule
Example: The Orbital diagram for the ground state of the Oxygen
atom (Z=8 )

51. 51
Determination of Electron Configuration
Step 1: Write Electron Distribution based on Energy (Aufbau Principle);
Step 2: ⇢Write Electron Configuration with ascending n↑;
Step 3: Fill electrons in AOs based on the rules of electron arrangement
→ Z = 7 → E
n: 1s
2
2s
2
2p
3
→
7
14
??????1. ↑↓
1s
2
↑↓
2s
2
↑↑↑
-10+1
2p
3
Example:
→ Z = 22 → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
2
22
48
????????????2.
↑↓
4s
2
↑ ↑
3d
2
⇢E
-
: 1s
2
2s
2
2p
6
3s
2
3p
6
3d
2
4s
2

52. 52
Unstable-Electron Configuration
Semi-saturated Configuration: ns
2
(n-1)d
4
→ ns
1
(n-1)d
5
Full-saturated Configuration:ns
2
(n-1)d
9
→ ns
1
(n-1)d
10
24
52
????????????1. → Z = 24 → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
4
(unstable)
→ E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
5
(stable)
E
-
: 1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
4s
1
29
64
??????&#3627408482;2. → Z = 29 → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
9
(unstable)
→ E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
10
(stable)
E
-
: 1s
2
2s
2
2p
6
3s
2
3p
6
3d
10
4s
1

53. 53
Electron Configuration of Ions
Outermost shell: shell with maximum n
(lớpngoàicùng)
3d
2
4s
2
Outermost
shell: 4s
Last shell: 3d
Final shell: shell with the highest energy
(lớpcuốicùng)
CationM
n+
:M
lose&#3627408475;(e−)
M
n+
AnionX
m-
:X
add&#3627408474;(e−)
X
m-
1.Fe (Z = 26) → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
6
Fe
2+
(24) → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
3d
6
Example:
Fe
3+
(23) → E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
2.
Cl
1-
(18)→ E
n: 1s
2
2s
2
2p
6
3s
2
3p
6
Cl (Z = 17) → E
n: 1s
2
2s
2
2p
6
3s
2
3p
5

54. 54
Valence Electron (VE)
Definition: are electrons in the outer shell associated with an atom
In Group A (ns
a
np
b
): VE = ∑e
-
in n
maxshell = a + b
Example:
15P: 1s
2
2s
2
2p
6
3s
2
3p
3
→ 5 VEs (3s
2
3p
3
)
In transition metals ns
e
(n-1)d
f
:
VE = ∑e
-
in (ns) subshell+ ∑e
-
in (n-1)d subshell =e+f
Example:
25Mn: 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
5
→ 7 VEs (4s
2
3d
5
)

55. 55
Practices
Problem1:Whichofthefollowingatomoriondoestheoutermost
electronconfiguration3s
2
3p
6
represent:
a) X (Z = 17) b) X (Z = 19)c) X
-
(Z = 17)d) X
+
(Z = 20)
Problem 2: The electron configuration of Cu
2+
ion (Z = 29) in the ground
state is:
a) 1s
2
2s
2
2p
6
3s
2
3p
6
3d
9
4s
0
b) 1s
2
2s
2
2p
6
3s
2
3p
6
3d
7
4s
2
c) 1s
2
2
s2
2p
6
3s
2
3p
6
3d
8
4s
1
d) 1s
2
2s
2
2p
6
3s
2
3p
6
3d
10
4s
0

56. 56
Practices
Problem 3: The valence electron configuration of Fe
3+
ion (Z= 26) in
the ground state is:
a) 3d
4
4s
1
b) 3d
3
4s
2
c) 3d
6
d) 3d
5
Problem4:ThevalenceelectronconfigurationofCo
3+
ion(Z=27)in
thegroundstateis:
a)3d
6
(nosingleelectrons)b)3d
4
4s
2
(existsingleelectrons)
c)3d
6
(existsingleelectrons)d)3d
4
4s
2
(nosingleelectron)

57. 57
Practices
Problem5:Determine4quantumnumberfromelectron
configurationofatomsandions:
1. 2p
5
Shell (n)Orbital (l)
electrons
↑↓↑↓↑
-10+1
2p
5
Last
electron
m
l= 0
m
s= -1/2
(n,l,m
l,m
s) = (2,1,0,-1/2)
2. 3d
4
→
↑↑↑
-2-10
3d
4
↑
+1+2
↑↓↑↓↑
-2-10
3d
7
↑↑
+1+2
(n,l,m
l,m
s) = (3,2,-1,-1/2)
(n,l,m
l,m
s) = (3,2,+1,+1/2)
3. 3d
7
→
L=2
OB=2l+1=5