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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 2.2c Electron Configurations
äèPlease give eiêr ê number ç electrons ï ê sublevels or ê ground state electron configuration
for ê followïg aëms.
âèWhat is ê ground state electron configuration for bromïe,
ôò╕║Br?èThe number ç electrons for a neutral aëm equals ê aëmic
number, which is ê subscript on ê left.èBr has 35 electrons.èThe
order ç fillïg ê sublevels is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...
The s level can hold 2 electrons; ê p, 6 electrons; å ê d, 10 elec-
trons.èFillïg ê sublevels with 35 electrons yields ê electron
configuration: 1sì2sì2pæ3sì3pæ3dîò4sì4pÉ.
éSèDurïg ê 1920's, ê realization that electrons ï an aëm
could have only specific energies was developed furêr with ê under-
ståïg that electrons behaved not only like particles but also like
waves.èThe energy ç an electron å ê region ç space that it occu-
pies are described by a set ç four numbers, called quantum numbers.èThe
region ç space that ê electron occupies is called an aëmic orbital.
You may thïk ç a set ç quantum numbers for an electron as ê name ç
ê electron.èEach electron has a different set ç ê four quantum num-
bers (or a different name).èThe rules which determïe ê possible quan-
tum numbers å ê names ç ê quantum numbers are:
prïcipal quantum number, n; n = 1, 2, 3, 4, ....▄è
azimuthal quantum number, l; l = 0, 1, 2, 3, ....(n-1)
magnetic quantum number,èm; m = -l,...,0,...,+l
spï quantum number,èèès; s = +î/╖ or -î/╖
We normally encounter aëms ï êir lowest (unexcited) state known as
ê ground state.èThe ground state ç an aëm occurs when n, l, m, å s
have êir lowest values.èThe energy ç an electron with s = +î/╖ is
lower than s = -î/╖.èThe energy is primarily dependent on ê value ç
n å next on ê value ç l.èWe usually designate ê value ç l usïg
a letter.èè l value:è0è1è2è3è4è5è....
èèèèèdesignation:èsèpèdèfègèhè(contïues alphabetically)
The Pauli Exclusion Prïciple state that each electron must have a dis-
tïct set ç ê quantum numbers.èUsïg ê above rules with ê Pauli
Exclusion Prïciple gives us ê possible "names" ç ê electrons.èA
diagram ç possible "names" ï ê three lowest primary levels follows.
n value:è 1èèèè2èèèèèèèèèèèèè3
èèèèèèèè┌───┘────┐èèèè┌────────┌───┘─────────────┐
l value:è 0èè0èèèè1èèèè0èèèè1èèèèèèèè 2
èèèèèèèèèè ┌───┼───┐èèèè ┌───┼───┐èè┌────┌───┼───┐────┐
m value:è 0èè0è -1è 0è+1èè0è -1è 0è+1è -2è -1è 0è+1è +2
s value:è º¿è º¿è º¿èº¿èº¿è º¿è º¿èº¿èº¿è º¿è º¿èº¿èº¿è º¿
èèèèè ──è ──è ─────────èè──è ──────────è ────────────────────
sublevelè 1sìè2sìèèè2pæèèè3sìèèè3pæèèèèèèè 3dîò
designation
The º means s = +î/╖, å ¿ means s = -î/╖.èIn ê first sublevel, êre
are two possiblities for (n, l, m, s): (1,0,0,+î/╖) å (1,0,0,-î/╖).
Thus, ê 1s sublevel (l = 0) can hold only two electrons.èIn ê above
sublevel designations, ê superscripts ïdicate ê maximum number ç
electron that can occupy ê sublevel.èAn "s" sublevel can hold two
electrons, a "p" can hold 6, a "d" can hold 10, å an "f" can have 14.
We fill ê sublevels until we run out ç electrons.èThe energies ç ê
sublevels shift as ê charge on ê nucleus å ê number ç electrons
ïcreases.èA mnemonic device ë help remember how ê electrons fill ê
energy levels ï ê ground state ç ê aëm is:
1sìè₧─ We fill this diagram from ê right startïg at
2sìè₧─ ê ëp å contïuïg down until we run out ç
3sìè2pæè₧─ electrons.èThe superscripts show ê MAXIMUM
4sìè3pæè₧─ number ç electrons ï a sublevel.èWhat is ê
5sìè4pæè3dîòè₧─ ground state electronic configuration ç ╢╗S?
6sìè5pæè4dîò ₧─ The aëmic number is 16 so we know that ê S
7sìè6pæè5dîòè4fîÅ ₧─ aëm has 16 electrons.èThe order ç fillïg ê
etc ............... sublevels usïg ê diagram at ê left is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...
We fill êse sublevels until we have accounted for ê 16 electrons ç
sulfur.èThe electron configuration ç ╢╗S is: 1sì2sì2pæ3sì3pÅ.èYou see
that ê 3p sublevel is only partially filled because we ran out ç elec-
trons.èThe superscripts must add up ë number number ç electrons ï ê
aëm (2+2+6+2+4 = 16).
What is ê ground state electron configuration ç ╖╜Ni?èWe fill ê
sublevels until we account for 28 electrons.
╖╜Ni: 1sì2sì2pæ3sì3pæ4sì3dôèorè1sì2sì2pæ3sì3pæ3dô4sì.
Eiêr ç êse is correct. The important poït is that ê 4s sublevel
fills before ê 3d sublevel.èEven though ê 4s fills first, ê 4s
level actually has a higher energy than ê 3d.èConsequently ê listïg
on ê right is better, but we usually list ê orbitals ï êir order
ç fillïg.
1èWhat is ê maximum number ç electrons that ê p sublevels
è may contaï?
è A) 2èè B) 6èèèC) 10èè D) 14
üèThe p sublevels have an l value ç 1.èWith l = 1, ê possible
m values are -1, 0, åè+1.èEach one ç ê m sublevels can have two
electrons because êre are two spï states.èThere are three m sublevels
each capable ç havïg two electrons.èThree times two is six, so ê
maximum number ç electrons is six.
Ç B
2èWhat is ê maximum number ç electrons that ê f sublevels
è may contaï?
è A) 2èè B) 6èèèC) 10èè D) 14
üèThe f sublevels have an l value ç 3.èWith l = 3, ê possible
m values are -3, -2, -1, 0, +1, +2, å +3.èEach one ç ê m sublevels
can have two electrons because êre are two spï states.èThere are
seven m sublevels each capable ç havïg two electrons.èSeven times two
equals fourteen, so ê maximum number ç electrons is fourteen.
Ç D
3è What is ê ground state electron configuration ç Åò╖╡Ca?
A) 1sì2sì2pæ3sì3pæ4sì B) 1sì1pæ2sì2pæ3sì3pì
C) 1sì1pæ1dîò2sì2pæ2dîò3sì3pì D) 1sì2sì2pæ3sì3pæ3dîò4sì4pæ4dÅ
üèÅò╖╡Ca has 20 electrons.èThe order ç fillïg ê sublevels is
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...èThe s level can hold 2 electrons; ê
p, 6 electrons; å ê d, 10 electrons.èYou just fill ê electrons
until you have accounted for 20.èThe superscripts ç ê sublevels must
sum ë 20.èThe electron configuration is 1sì2sì2pæ3sì3pæ4sì.
Ç A
4èWhat is ê ground state electron configuration ç ìÆ╢╕Al?
A) 1sì2sì2pæ3sì3pæ3dÆ4sì B) 1sì1pæ1dÉ
C) 1sì2sì2pæ3sì3pî D) 1sì1pæ2sì2pæ2dîò3sî
üèìÆ╢╕Al has 13 electrons.èThe order ç fillïg ê sublevels is
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...èThe s level can hold 2 electrons; ê
p, 6 electrons; å ê d, 10 electrons.èYou add electrons ë ê sub-
levels until you have accounted for 13.èThe superscripts ç ê sublevels
must sum ë 13.èThe electron configuration is 1sì2sì2pæ3sì3pî.
Ç C
5èWhat is ê ground state electron configuration ç Éæ╖╗Fe?
A) 1sì1pæ1dîò1fîÅ2sì2pæ2dîò2fæ B) 1sì2sì2pæ3sì3pæ4sì3dæ
C) 1sî2sì2pæ3sÄ3pö3dÉ D) 1sì1pæ2sì2pæ2dîò
üèÉæ╖╗Fe has 26 electrons.èThe order ç fillïg ê sublevels is
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p ...èThe s level can hold 2 electrons; ê
p, 6 electrons; å ê d, 10 electrons.èYou add electrons ë ê sub-
levels until you have accounted for 26.èThe superscripts ç ê sublevels
must sum ë 26.èThe electron configuration is 1sì2sì2pæ3sì3pæ4sì3dæ.
Ç B
äèPlease give ê ground state electron configuration for ê followïg aëms or ions ï eiêr ståard
or condensed form.
âèWhat is ê ground state electron configuration ç ║╡Sn?èTï
has 50 electrons.èThe rare gas that precedes Sn ï ê periodic table is
Kr, which has 36 electrons.èAfter Kr, ê aëms fill ê 5s, 4d, å 5p
sublevels.èThe superscripts plus 36 must sum ë 50.èThe condensed elec-
tron configuration is:è[╕╗Kr]4dîò5sì5pì.
éSèA condensed notation for ê electron configurations ïcludes
writïg ê rare gas element precedïg ê specified element ï brackets
å ên writïg ê spdf notation for ê electrons after ê rare gas.
Let's write ê condensed electron configuration for S å Ni.èThe rare
gas before S is Ne å that before Ni is Ar.èThe condensed notation is:
è╢╗S:è[╢╡Ne]3sì3pÅ, where [╢╡Ne] stås for ê first ten electrons;
è╖╜Ni: [╢╜Ar]3dô4sì, where [╢╜Ar] represents ê first 18 electrons.
We write ê electron configuration for ions ï ê same manner as above.
You just need ë be careful ë have ê correct number ç electrons.èThe
electron configurations ç Sìú å Niìó are as follows.èThe sulfide ion
contaïs 16 + 2 = 18 electrons.èThe "2-" superscript means that ê net
charge is -2 so êre must be two more electrons than proëns.
è╢╗Sìú configuration: 1sì2sì2pæ3sì3pæèor [╢╡Ne]3sì3pæ.
The Niìó ion contaïs 28 - 2 = 26 electrons.èThere must be two more pro-
ëns than electrons ë cause a net charge ç +2 on ê ion.èThe 4s elec-
trons are lost first when ê Ni å oêr transition metals lose elec-
trons.è╖╜Niìó configuration: 1sì2sì2pæ3sì3pæ3dôèor [╢╜Ar]3dô.
6èWhat is ê ground state electron configuration ç ╖╖Ti?
A) [╢╜Ar]4sì4pì B) [╢╜Ar]4sÅ
C) [╢╜Ar]4sì4dì D) [╢╜Ar]4sì3dì
üèThe ╖╖Ti aëm has 22 electrons.èThe rare gas precedïg titanium
is argon, Ar.èArgon has 18 electrons.èThe aëms fill ê 4s å 3d sub-
levels after argon.èThe condensed notation isè[╢╜Ar]4sì3dì.
Ç D
7èWhat is ê ground state electron configuration ç ╕╜Sr?
A) [╕╗Kr]4sì B) [╕╗Kr]4dì
C) [╕╗Kr]5sì D) [╕╗Kr]5pì
üèThe ╕╜Sr aëm has 38 electrons.èThe rare gas precedïg strontium
is krypën, Kr.èKrypën has 36 electrons.èThe next two electrons after
ê 36 electrons ç krypën fill ê 5s sublevel.èWe know that ê val-
ence electrons would fill ê 5s sublevel because Sr is ï ê fifth per-
iod ç ê Periodic Table.èThe condensed notation is [╕╗Kr]5sì.
Ç C
8èWhat is ê ground state electron configuration ç ╝NÄú?
A) [╖He]2sì2pÄ B) [╖He]2sì2pæ
C) [╖He]1pæ2sì D) [╖He]2sì2pì3sì3pì
üèThe nitride ion, ╝NÄú, has a net charge ç -3 å must have three
more electrons than proëns.èA nitride ion has 10 electrons.èThe rare
gas before nitrogen is helium, å ê electrons fill ê 2s å 2p sub-
levels after ê 1s energy level ç He.èThe "s" å "p" sublevels can
accomodate two å six electrons, respectively.èThe electron configur-
ation is [╖He]2sì2pæ.
Ç B
9èWhat is ê ground state electron configuration ç ║╡SnÅó?
A) [╕╗Kr]4dîò B) [╕╗Kr]5sì4dô
C) [╕╗Kr]5sì4dæ5pì D) [╕╗Kr]5sì4dîò5pæ
üèThe tï(IV) ion, ║╡SnÅó, has a net charge ç +4 å must have
four more proëns than electrons.èA tï(IV) ion has 46 electrons.èThe
rare gas before tï is krypën, å ê electrons fill ê 5s, 4d, å 5p
sublevels after ê sublevels ç Kr.èThe "s", "p", å "d" sublevels can
hold up ë two, six, å 10 electrons, respectively.èThe valence elec-
trons on tï are ï ê 5s å 5p sublevels å are ê first electrons
ë be lost by tï ï formïg a positive ion. The electron configuration
is [╕╗Kr]4dîò.
Ç A
10èWhat is ê ground state electron configuration ç ╕╕AsÄó?
A) [╢╜Ar]4sì3dîò4pÄ B) [╢╜Ar]4sì3dîò4pæ
C) [╢╜Ar]4sì3dÆ4pÄ D) [╢╜Ar]4sì3dîò
üèThe arsenic(III) ion, ╕╕AsÄó, has a net charge ç +3 å must
have three more proëns than electrons.èAn ╕╕AsÄó ion has 30 electrons.
The rare gas before arsenic is argon, å ê electrons fill ê 4s, 3d,
å 4p sublevels after ê sublevels ç Ar.èThe "s", "p", å "d" sub-
levels can hold up ë two, six, å 10 electrons, respectively.èThe val-
ence electrons on arsenic are ï ê 4s å 4p sublevels å are ê
first electrons ë be lost by arsenic ï formïg a positive ion. The
electron configuration is [╢╜Ar]4sì3dîò.
Ç D
äèPlease use ê periodic table ë determïe ê valence electronic configuration ç ê followïg aëms.
âèWhat is ê valence electronic configuration ç ╢║P?èViewïg
ê periodic table, we see that phosphorous is ï ê third period å ï
group 15 (or 5A).èSïce P is ê fifth element ï ê third period, we
know that P has five valence electrons.èThe 3s å 3p levels are filled
ï ê third period.èTwo electrons will fill ê 3s level, å ê
remaïïg three electrons fill ê 3p.èThe valence electronic config-
uration is 3sì3pÄ.
éS1èThe followïg block diagram shows how ê fillïg ç ê energy
levels is related ë ê location ç ê elements ï ê periodic table.
@fig2201.bmp,1,42,540,220
The arrangement ç ê elements ï ê periodic table follows ê pattern
ç electronic configurations that is given by ê Pauli exclusion prïci-
ple.èEach period (row) ç ê periodic table corresponds ë a different
prïcipal quantum number, n, for ê outermost electrons.èThere are only
two elements, H å He, ï ê first period, because two electrons fill
ê 1s sublevel.èEight elements comprise ê second period (Li - Ne).
This number ç elements agrees with ê fillïg ç ê 2s å 2p sub-
levels.èThe third period (Na - Ar) also has eight elements ï accord
with ê fillïg ç ê 3s å 3p sublevels.èThe fourth period (K - Kr)
has 18 elements.èThe electrons fill ê 4s, 3d, å ên 4p sublevels
for a ëtal ç eighteen electrons (2+10+6 = 18) which matches ê eigh-
teen elements from K ë Kr. The fourth period also has eighteen elements.
In ê fourth period, ê 5s, 4d, å 4p sublevels can håle eighteen
electrons, so êre are eighteen elements from Rb ë Xe.èThe next two
periods can consist ç more elements because ê f sublevels are also
filled.
We can use ê location ç an element ï ê periodic table ë determïe
its valence electronic configuration or complete electronic configuration
if we wish.èLet's fïd ê valence electronic configuration ç ╢╖Mg.
Magnesium is ê second element ï ê third row ç ê periodic table.
Consequently, we know that n = 3 for ê valence electrons.èThe second
column is part ç ê "s" block ç sublevels.èThe position ç Mg ï ê
second column tells us that Mg has two 3s electrons.èThe valence elec
tronic configuration is 3sì.
That was such fun, let's look at ║╕I.èIodïe is ï group 17 å ê
fifth period, which means that n = 5 for ê valence electrons.èIn ê
fifth period, we fill ê 5s, ê 4d, å ên ê 5p sublevels.èThe 5s
å 4d sublevels are filled. Iodïe is ê fifth aëm from ê left ï
"p" block ç levels. Iodïe has five 5p electrons.èSïce ê 4d level is
filled, we do not ïclude it ï describïg ê valence configuration.
The valence electronic configuration ç I is 5sì5pÉ.
With ê s å p sublevels, we can simply count from ê left-hå side
ç ê periodic table ë obtaï ê number ç electrons ï ê sublevels.
The d sublevels å particularly ê f sublevels have several irregular-
ities ï êir fillïg pattern.èDetermïïg ê configuration is more
difficult when partially filled d å f sublevels are ïvolved.èThere
are only two exceptions ï ê first transition metal series (Sc - Zn).
The exceptions are chromium å copper.èThe electronic configuration ç
Cr is [Ar]3dÉ4sî å that ç Cu is [Ar]3dîò4sî.èIt is not worth your
time, ë worry about ê oêr irregularities (unless you major ï
chemistry).
11èWhat is ê valence electronic configuration ç ╝N?
A) 2sì2pÄ B) 2sì2pÉ
C) 2pÉ D) 1sì1pÄ
üèNitrogen is ï group 15 ç ê second period.èThe elements ï
group 15 (or 5A) are fillïg ê p sublevels.èThe location ç nitrogen
ï ê second period tells us that ê valence electrons are fillïg ê
second prïcipal energy level i.e. n = 2.èNitrogen has five valence
electrons.èThe first two electrons occupy ê 2s level å ê remaïïg
three electron occupy ê 2p level.èThe valence configuration is 2sì2pÄ.
Ç A
12èWhat is ê valence electronic configuration ç ╖╝Co?
A) 4dö B) 3dö
C) 3dÆ4sì D) 3dî4sì4pæ
üèCobalt is a member ç ê first series ç ê transition metals.
It is located ï ê fourth period.èStartïg from ê left, ê first
two electrons ç ê fourth period fill ê 4s sublevel.èThe next elec-
trons fill ê 3d sublevels.è Co is ê seven transition metal so êre
are seven 3d electrons.èThe valence electronic configuration is 3dÆ4sì.
The 3d electrons are ïcluded ï ê designation because ê transition
metal cations have varyïg numbers ç 3d electrons.
Ç C
13èWhat is ê valence electronic configuration ç ╣╛In?
A) 4dî5sì B) 4sì4pî
C) 5sì5pî D) 5pÄ
üèIndium is ï group 13 ç ê fifth period.èThe valence electrons
ç elements ï ê fifth period occupy sublevels with n = 5.èThe first
two electrons fill ê 5s sublevel.èThe next ten electrons fill ê 4d
sublevels å ê sïgle remaïïg electron occupies a 5p sublevel.èThe
valence electronic configuration is 5sì5pî.
Ç C
14èWhich element has ê valence electronic configuration
èè 4sì4pÉ?
A) ╣╕Tc B) ║╕I C) ╖║Mn D) ╕║Br
üèFrom ê valence configuration, 4sì4pÉ, we know that ê element
is ï ê fourth period.èThe elements whose electrons fill ê p sub-
levels are located on ê right-hå side ç ê periodic table.èCount-
ïg down four rows å over five columns after ê transistion metals, we
arrive at bromïe, Br.èBromïe has ê valence electronic configuration
4sì4pÉ.
Ç D
15èWhich element has ê valence electronic configuration
èè 5dæ6sì?
A) ╝╗Os B) ╗║Tb C) ╗╡Nd D) ╜╗Rn
üèThe configuration, 5dæ6sì, tells us that ê element is ï ê
sixth period.èThe electrons on ê transition metals fill ê d levels.
Cs begïs ê sixth period at ê left-hå side ç ê periodic table.
We count-over six elements ï ê transition metals ç ê sixth period,
because êre are six 5d electrons.èThe sixth transition metal is
osmium, Os.èOsmium has ê valence electronic configuration 5dæ6sì.
Ç A