SLATER’S RULE.pdf

This is the way of finding the effective nuclear charge.credit goes to the professor

1. PRESENTER:
SEEMASAINI
ASSOCIATEPROF.INCHEMISTRY
GOVT.COLLEGE,RUPNAGAR
SLATER’SRULE

2. Slater’sRule
INTRODUCTION:
In1930,ascientist proposedasetof
empiricalrulestounderstandtheconceptofEffective
NuclearChargeandtocalculatetheScreening
ConstantorShieldingConstant.
HeproposedaformulaforcalculationofEffective
NuclearCharge
Zeff=Z–S
whereSistheSlater’sscreeningconstant,
ZistheNuclearcharge

3. PriortoexplainingSlater’srules,certaintermslikeNuclear
Charge,ShieldingEffectandEffectiveNuclearChargehave
tobeUnderstood.
WhatisNuclearCharge?
Itisthechargeonthenucleuswithwhichitattractsthe
electronsoftheatom.Basically,theNuclearChargeissaid
tobeequaltotheAtomicNumber(i.e,.theNumberof
protons)inanatom.ItisdenotedbythesymbolZ.
WhatisShieldingEffect?
IncaseofMultielectronatoms,astheorbitalsarefilled
up,theelectronsintheInnerorbitalsshieldtheelectronsin
theOuterorbitalsfromtheNucleus.

4. So,theelectronsintheOuterorbitalsdonotfeelthefullforce
orchargeofthenucleus.Thus,thereductionofnuclear
chargeontheOutermostelectronsiscalledShielding.Effect
orScreeningEffect.ShieldingEffectisdefinedasameasure
oftheextenttowhichtheinterveningelectronsshieldthe
outerelectronsfromthenuclearcharge.Itisdenotedbythe
symbolS
WhatisEffectiveNuclearCharge?
EffectiveNuclearChargeistheactualchargefeltbythe
outerelectronsaftertakingintoaccountshieldingofthe
electrons.ItisdenotedbythesymbolZ*orZeff

5. Slater'sRules:
1)Writetheelectronconfigurationfortheatomusing
thefollowingdesign;
(1)(2,2)(3,3)(3)(4,4)(4)(4)(5,5)(5d)
(5f)(6s,6p)…….etc.
2)Anyelectronstotherightoftheelectronofinterest
contributesnothingtowardsshielding.
3)Allotherelectronsinthesamegroupasthe
electronofinterestshieldtoanextentof0.35
nuclearchargeunitsirrespectiveofwhetherthe
electronsareins,p,d,orforbitals.

6. 4)Incaseof1selectronshieldinganother1selectronthe
screeningconstantvalueistakentobe0.30.
5)Iftheelectronofinterestisanorelectron:Allelectrons
withonelessvaluei.e.(n-1)valueoftheprincipal
quantumnumbershieldtoanextentof0.85unitsofnuclear
charge.Allelectronswithtwoormorelessvaluesi.e.
(n–2,n–3,n–4etc.)valuesoftheprincipalquantum
numbershieldtoanextentof1.00units.
6)Iftheelectronofinterestisanorelectron:Allelectronsto
theleftshieldtoanextentof1.00unitsofnuclearcharge.
7)Sumtheshieldingamountsfromsteps2through5and
subtractfromthenuclearchargevaluetoobtaintheeffective
nuclearchargevalue.

7. CalculateZ
*
foravalenceelectronin
fluorine(Z=9).
Electronicconfigurationoffluorineis1
2
,2
2
,2
5
Groupingitacc.toslater’srule:(1
2
)(2
2
,2
5
)
Rule2doesnotapply;
Now,oneelectronoutofthe7valenceelectrons
becomestheelectronofinterest.Theotherremaining
6valenceelectronswillcontribute0.35eachtowards
shielding.
Theelectronsin(n–1)orbitalsi.e.1sorbitalwill
contribute0.85eachtowardsshielding.

8. S=0.35x(No.ofelectronsinthesameshelli.e.
norbital)+0.85x(No.ofelectronsinthe
(n–1)shell)
S=0.35x6+0.85x2=3.8
Z
*
=Z–S=9–3.8=5.2foravalenceelectron.
CalculateZ
*
fora6electroninPlatinum
(Z=78)

9. TheelectronicconfigurationofPlatinum(Z=78)is
1s
2
,2s
2
,2p
6
,3s
2
,3p
6
,4s
2
,3d
10
,4p
6
,5s
2
,4d
10
,5p
6
,
6s
2
,4f
14
,5d
8
,
.
Groupingitacc.toSlater’srule:
(1
2
)(2
2
,2
6
)(3
2
,3
6
)(3
10
)(4
2
,4
6
)(4
10
)(4
14
)
(5
2
,5
6
)(5
8
)(6
2
)
Rule2doesnotapply;
Now,oneelectronoutofthetwovalenceelectrons
becomestheelectronofinterest.Theother
remainingvalenceelectronswillcontribute0.35
eachtowardsshielding.

10. Theelectronsin(n–1)orbitalsi.e.(6–1)=5
th
orbitalwill
contribute0.85eachtowardsshielding.
Theelectronsin(n–2),(n–3),(n–4)….etc.orbitals
{i.e.(6–2),(6–3),(6–4)……etc.}orbitalswill
contribute1.00eachtowardsshielding.
S=0.35x(No.ofelectronsinthesameshelli.e.
norbital)+0.85x(No.ofelectronsinthe
(n–1)shell)+1.00x(No.ofelectronsinthe(n–2),
(n–3)….etc.orbitals)
S=0.35x1+0.85x16+60x1.00=73.95
Z
*
=Z–S=78–73.95=4.15foravalenceelectron.

11. CalculatetheEffectiveNuclearChargeforone
ofthe4felectronsofCerium(Z=58)
TheelectronicconfigurationofCeriumis:
1s
2
2s
2
2p
6
3s
2
3p
6
3d
10
4s
2
4p
6
4d
10
4f
2
5s
2
5p
6
6s
2
Groupingtheorbitalsacc.toSlater’s:
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
10
)(4s
2
4p
6
)(4d
10
)(4f
2
)
(5s
2
5p
6
)(6s
2
)
Aswehavetocalculatetheeffectivenuclearcharge
of4felectronsofCerium,theelectronslyingafterthe
4felectronwillnotcontributetoshielding

12. Now,
S=0.35x(No.ofelectronsinthesameorbital+
1.00x(alltheelectronsinthelowerorbitals)
S=0.35x1+1.00x46=46.35
Zeff=Z–S=58–46.35=11.65
CalculatetheEffectiveNuclearChargein
theperipheryofNitrogen(Z=7)
InordertocalculatetheEffectiveNuclearChargeinthe
peripheryofanatomorion,theshieldingofnuclear
chargebyalltheelectronspresentintheelectronic
configurationoftheatomorion.

13. TheelectronicconfigurationofNitrogenis1s
2
2s
2
2p
6
Groupingacc.toSlater’srule:(1s
2
)(2s
2
2p
6
)
S=0.35x(No.ofelectronsinthesameorbital)+
0.85x(No.ofelectronsinthe(n–1)orbital
S=0.35x5+0.85x2=3.45
ZeffintheperipheryofN-atom=Z–S=7–3.45
=3.55

14. APPLICATIONSOFSLATER’SRULE
Itprovidesaquantitativejustificationforthe
sequenceoforbitalsintheenergyleveldiagram.
Ithelpstoexplainthefillingofns-orbital(4s,5s,6s
etc.–orbitals)priortothefillingof(n-1)dorbital(3d,
4d,5d…etc.).
LetusconsiderthecaseofPotassium(Z=19),in
whichthelastelectronisaddedto4sorbital

15. TheconfigurationofPotassiumacc.toSlateris
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(4s
1
)
Astheeffectivenuclearchargeonelectronin4s
orbitalhastobecalculated,theelectronsinthe
sameorbitali.e.norbitalwillcontribute0.35each,
theelectronsin(n–1)orbitali.e.3sand3p
orbitalswillcontributeS=0.85eachandallthe
electronsin(n–2,n–3….Etc)orbitalsi.e.(2s,
2p,1s)orbitalswillcontributeS=1.00each.

16. So,
=0x0.35+8x0.85+10x1.00
=16.80
Therefore,EffectiveNuclearCharge
Z*=Z–S=19–16.80=2.20
Now,Letusassumethatthelastelectronenters
the3dorbitalratherthan4sorbital,
Thentheconfigurationacc.toSlateris
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
1
)

17. HerethedelectronisunderInterest,sotheelectrons
inthesameorbitali.e.3dorbitalwillcontributeS=
0.35each,whereastheelectronsinalltheother
orbitalsi.e.(3s,3p,2s,2p,1s)willallcontributeS=
1.00each.
So,
S=0x0.35+18x1.00=18.00
ThereforeEffectiveNuclearCharge
Z*=Z–S=19.00–18.00=1.00
OncomparingtheEffectiveNuclearChargeofboth4s
and3dorbitals,weseethatthe4selectronisunderthe
influenceofgreaterEffectiveNuclearcharge(Zeff=2.20)
ascomparedto3delectron(Zeff=1.00)inPotassium
atom.

18. So,theelectronin4sorbitalwillbemoreattractedbythe
nucleusandwillhavelowerenergythanthe3delectron.
Thus,thelastelectronwillenterinthe4sorbital,rather
thanthe3dorbitalincaseofPotassiumatom.
Slater’sruleexplainwhy4selectronsarelostprior
to3delectronsduringcationformationincaseof
Transitionelements.
Letusconsiderthecaseof (Z=23)
TheelectronicconfigurationofVanadiumis
1s
2
2s
2
2p
6
3s
2
3p
6
3d
3
4s
2

19. Afterlosing2electrons,theelectronicconfigurationof
V
2+
is
1s
2
2s
2
2p
6
3s
2
3p
6
3d
1
4s
2
Andnot
1s
2
2s
2
2p
6
3s
2
3p
6
3d
3

TheaboveElectronicConfigurationcanbeexplainedby
Slater’srules
TheEffectiveNuclearChargefor4selectronis
calculatedas:
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
3
)(4s
2
)
Now,oneoftheelectronsofthe4sorbitalbecomes
electronofinterest.Thesecondelectronhowever,will
contributetowardsshieldingeffect

20. S=1x0.35+11x0.85+10x1.00=19.70
So,Zeff=Z–S=23–19.70=3.30
Now,letscalculatetheEffectiveNuclearCharge
fora3delectron
AccordingtoSlater’srule
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
3
)(4s
2
)
As4sorbitalliesaftertheelectronunderinterestitwill
contributesnothingtowardsshielding.
Oneelectronof3dorbitalbecomeselectronofinterest.The
othertwo3delectronswillcontributetowardsshielding
effect.
S=2x0.35+18x1.00=18.70
Zeff=Z–S=23–18.70=4.30

21. ComparingtheEffectiveNuclearChargeofboththe
3dandthe4selectron,itisseenthatEffectiveNuclear
Chargeon3delectronis4.30whereas4selectronhas
3.30.
Theforceofattractionexperiencedbythe3d
electronsismoreascomparedto4selectrons.
The3delectronsaremoretightlyheldtothenucleus
than4selectrons.
Thus,the4selectronsareremovedinpreferenceto3d
electrons.

22. Theforceofattractionexperiencedbythe3d
electronsismoreascomparedto4selectrons.
The3delectronsaremoretightlyheldtothenucleus
than
4selectrons.
Thus,the4selectronsareremovedinpreferenceto3d
electrons.
Ithelpstoexplainwhysizeofacationis
alwayssmallerthanitsneutralatom.
Let’staketheexampleofLithiumatomandLithiumion
TheElectronicconfigurationofLithiumatomis1s
2
2s
1
WriteitaccordingtoSlater’s(1s
2
)(2s
1
)
As2sorbitalhasonlyoneelectron,itbecomesthe
electronofinterest.

23. Onlythe1selectronswillcontributetowardsshielding.
S=2x0.85=1.70
Zeff=Z–S=3–1.70=1.30
IncaseofLithium(Li
+
)Ion
TheElectronicConfigurationofLi
+
is1s
2
Groupingacc.toSlater:(1s
2
)
InCaseofLi
+
,oneofthe1selectronsbecomeselectron
ofinterestandtheother1selectroncontributestowards
shielding.
S=1x0.30=0.30
Zeff=Z–S=3–0.30=2.70

24. ComparisonoftheEffectiveNuclearChargeofLiatom
(Zeff=1.30)andLi
+
ion(Zeff=2.70),showsthat
EffectivenuclearchargeofLi
+
ionismorethanLiatom.
So,thesizeofLi
+
ionissmallerthanLiatom.
Itexplainswhyaanionisalwayslargerthanitsneutral
atom
TakingtheexampleofChlorineatomandChlorineion.
IncaseofChlorineatom(Z=17),theelectronic
configurationis1s
2
2s
2
2p
6
3s
2
3p
5
Groupingitacc.toSlater’srule
(1s
2
)(2s
2
2p
6
)(3s
2
3p
5
)
S=6x0.35+8x0.85+2x1.00=10.90
Zeff=Z–S=17–10.90=6.10

25. IncaseofChlorineionCl
-
Theelectronicconfigurationis1s
2
2s
2
2p
6
3s
2
3p
6
Groupingitacc.toSlater’srule
(1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)
S=7x0.35+8x0.85+2x1.00=11.25
Zeff=Z–S=17–11.25=5.75
ComparisonoftheEffectiveNuclearChargeofClatom
(Zeff=6.10)andCl
-
ion(Zeff=5.75),showsthatEffective
nuclearchargeonClatomismorethanCl
-
ion.
So,thesizeofCl
-
ionislargerthanClatom.

26. Slatergroupedbothsandporbitalstogetherfor
calculatingeffectivenuclearcharge,whichis
incorrect.Thisisbecauseradialprobability
distributioncurvesshowthatsorbitalsaremore
penetratingthanporbitals.So,thesorbitals
shouldshieldtoagreaterextentascomparedtop
orbital.
AccordingtoSlater,allthes,p,dandfelectrons
presentinshellorenergylevellowerthan(n–1)
shellwillshieldtheouternelectronswithequal
contributionofS=1.00each.Thisisnotjustified
asenergeticallydifferentorbitalsshouldnot
contributeequally.
Slater’srulesarelessreliableforheavierelements
LIMITATIONSOFSLATER’SRULE

27. THANKYOU
*THEEND