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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 2.4cèPeriodic Trends ç ê Elements
äèPlease rank ê followïg aëms or ions ï order ç ïcreasïg radius.
âèRank ê followïg aëms ï order ç ïcreasïg radius: B, N, Al.
The radius ç ê aëms ïcreases goïg down a group å ïcreases across
a period from right ë left.èAlumïum is ï ê period below B å ï
ê same group as B, so Al is ê largest aëm.èBoron is ë ê left ç
nitrogen, so B has a larger radius than N.èCombïïg ê two trends, we
obtaï ê followïg rankïg ç ê radii:èN < B < Al.
éS1èèèèèèèèèèèThe figure displays ê covalent radii for
@fig2401.bmp,1,1,220,290
èèèê first 36 elements ç ê periodic table.
èèèThe covalent radius ç an aëm is an average
èèèvalue based on ê distances between ê
èèèaëms when êy form covalent bonds.èTwo
èèètrends are evident among ê maï group
èèèelements, those elements fillïg ê s å p
èèèsublevels.
èèè(1) The aëmic radius ïcreases as we go down
èèèèèa group ç ê periodic table.
èèè(2) The aëmic radius decreases as we go
èèèèèacross a period ç ê periodic table.
èèèThe reason for ê ïcrease goïg down a
group is that ê valence electrons are fillïg successively larger orbi-
tals.èThe outermost orbitals have higher n values as we go down a group.
The aëms lower ï a group have a higher nuclear charge which pulls ê
electrons closer ë ê nucleus, but êy also have more filled sublevels.
The net effect is that ê additional electrons need more space.
As we go across a period, ê electrons are goïg ïë sublevels with ê
same prïcipal quantum number (same n value).èThe ïcreasïg nuclear
charge across ê period results ï an ïcreasïg attraction for each ç
ê electrons ï ê valence energy level.
The aëmic radii ç ê transition metals exhibit a smaller variation
than that ç ê maï group elements.èGenerally across ê transition
metals, ê radii decrease for ê first few metals, remaï approximately
constant, å ên ïcrease as ê d sublevels become filled.èThe radii
still ïcrease down a group, although êre is only a small difference
between ê radii ç ê fifth å sixth period transition metals.èThis
effect is known as ê Lanthanide contraction.èThe nuclear charge ï-
crease is much greater between ê fifth å sixth periods than it is be-
tween ê fourth å fifth periods.
Ionic radii also show trends.èCations always have a smaller radius than
ê radius ç ê correspondïg neutral aëm.èCan you explaï why?è
There are fewer electrons but ê nuclear charge is ê same.èKó is
smaller than K.èConversely, anions always have a larger radius than ê
correspondïg neutral aëm.èIn this case, ê nuclear charge is ê same
but ê number ç electrons ïcreases.èClú has a larger radius than Cl.
Ionic radii ïcrease goïg down a group ç ê periodic table.
Oêr sequences ç ïterest are isoelectronic series.èIons with ê same
number ç electrons comprise an isoelectronic series; such as, NÄú, Oìú,
Fú, Naó, Mgìó, etc.èThese ions have ê same number ç electrons, but ê
nuclear charge ïcreases as ê ionic charge becomes more positive.èIn
ê series, NÄú, Oìú, Fú, Naó, ê nuclear charges are +7, +8, +9, å
+11, respectively.èSïce êir electronic configurations are identical,
any differences must be due ë ê difference ï êir nuclear charges.
The nitride ion, NÄú, is ê largest because its nuclear charge is ê
lowest.èThe sodium ion has ê smallest radius because its nuclear charge
is ê highest.
1èRank ê followïg elements ï order ç ïcreasïg radius
èè(largest last).
A) S < Se < Te < O B) O < S < Se < Te
C) Se < Te < O < S D) Te < Se < S < O
üèThese elements are members ç group 16.èThe radius ïcreases
goïg down a group ç ê periodic table.èConsequently ê radii will
ïcrease from O ë Te.èThe correct rankïg is O < S < Se < Te.
Ç B
2èRank ê followïg elements ï order ç ïcreasïg radius
èè(largest last).
A) K < Ca < As < Br B) Ca < K < Br < As
C) Br < As < Ca < K D) As < Br < K < Ca
üèThese elements are maï group elements ï ê fourth period.èThe
radius decreases goïg across a period ç ê periodic table.èWe expect
ê radii ë decrease from K ë Br.èTo list ê elements ï order ç
ïcreasïg radius means that Br would be first.èThe correct order is
Br < As < Ca < K.
Ç C
3èRank ê followïg elements ï order ç ïcreasïg radius
èè(largest last).
A) Na < Al < C < F B) Al < Na < F < C
C) C < F < Na < Al D) F < C < Al < Na
üèThe radius ïcreases goïg down groups å across a period from
right ë left.èNa å Al are ë ê left ç C å F.èNa å Al are also
ï ê next higher period from C å F.èWe expect Na ë be larger than
Al.èAl should be larger than C, å C should be larger than F.èThe
expected order ç ê radii is:èF < C < Al < Na.
Ç D
4èRank ê followïg ions ï order ç ïcreasïg radius
èèèèèè(largest last).
A) Mgìó < Caìó < Srìó < Baìó B) Caìó < Mgìó < Baìó < Srìó
C) Srìó < Baìó < Mgìó < Caìó D) Baìó < Srìó < Caìó < Mgìó
üèThese cations are ï group 2.èThe radius ç ê cations ï-
creases goïg down a group.èThese ions ïcrease ï radius from Mgìó ë
Baìó.èThe correct rankïg ç ê ions isèMgìó < Caìó < Srìó < Baìó.
Ç A
5èRank ê followïg ions ï order ç ïcreasïg radius
èè(largest last).
A) Sìú < Clú < Kó < Caìó B) Caìó < Kó < Clú < Sìú
C) Clú < Sìú < Caìó < Kó D) Kó < Caìó < Sìú < Clú
üèThese ions form an isoelectronic series.èEach ç ê ions has
eighteen electrons.èThe nuclear charge ïcreases ï ê sequence:
Ca, K, Cl, S.èSïce ê number ç electrons is ê same, ê ion with
ê higher nuclear charge is ê smallest.èThe radius ïcreases from
Caìó ë Sìú.èThe correct rankïg is: Caìó < Kó < Clú < Sìú.
Ç B
äèPlease rank ê followïg elements ï order ç ïcreasïg first ionization energy.
âèRank ê followïg elements ï order ç ïcreasïg first ioni-
zation energy: N, P, Si, å Ge.èGenerally, ê ionization energy
ïcreases from ê lower left ë ê upper right ç ê periodic table.
N is above P; P is left ç Si; å Si is above Ge.èThese elements follow
ê normal trends so ê correct rankïg is Ge < Si < P < N.
éS1èThe first ionization energy is defïed as ê amount ç energy
that is needed ë remove ê first electron from a neutral aëm ï ê
gas phase.èThe process isèX(g) + I.E. ──¥ Xó(g) + eú, where X is an
aëm å I.E. is ê ionization energy.èThe trends ï ê first ioniza-
@fig2402.bmp,1,86,220,290
èèètion energy mirror ê trends ï ê aëmic
èèèradius.èThe ionization energy ïcreases
èèègoïg across a period.èThe nuclear charge
èèèïcreases across a period.èThe valence
èèèelectrons feel a greater attraction from
èèènucleus, å more energy is required ë
èèèremove an electron.èThe ionization energy
èèèdecreases goïg down a group.èThe valence
èèèelectrons are progressively farêr from
èèèê nucleus goïg down a group.èThe elec-
èèètrons are less tightly held by ê nucleus,
èèèå less energy is required ë remove one.
èèè
èèèèèèèèèèèèèèèThese general trends are illustrated ï ê
graph ç ê ionization energy ç ê first 36 elements.èNotice that ê
ionization energy decreases from He ë Kr (down ê group) å ïcreases
from Li ë Ne (across a period).
You will also notice that ê ïcrease ï ionization energy across a
period is uneven.èFor example, ê ionization energy ç B is less than
that ç Be; å ê ionization energy ç O is less than that ç N.
In boron, ê most energetic electron is ï a 2p level which has a
slightly higher energy than ê 2s ç Be.èConsequently, less energy is
needed ë remove ê outermost electron from B.èWith oxygen, ê last 2p
electron must be paired with one ç ê oêr 2p electrons.èA little
extra energy is needed ë make ê electrons pair up å ë occupy ê
same region ç space.èTherefore, slightly less energy is needed ë re-
move this paired electron.èAll 2p electrons ï nitrogen are unpaired, so
ê last electron does not have any more energy than ê oêr 2p elec-
trons.èYou can see that ê oêr elements ï groups 13 å 16 follow
this trend.
6èRank ê followïg elements ï order ç ïcreasïg first
ionization energy (largest ionization energy last).
A) Li < O < Na < Si B) Na < Li < Si < O
C) O < Li < Si < Na D) Si < Na < Li < O
üèThe ionization energy ïcreases from ê lower left ë ê upper
right ç ê periodic table.èOxygen has ê highest ionization energy,
because O at ê upper right ï ê periodic table.èSi å Na are ï ê
same period.èSi should have a higher ionization energy than Na due ë
ê higher nuclear charge ï Si.èIt is difficult ë decide between Si
å Li, because ê trends conflict.èMovïg over three groups ïcreases
ê ionization energy more than movïg down a group decreases it.èThe
ionization energy ç Si is higher than that ç Li.èThe correct order ç
ê ionization energies is Na < Li < Si < O.
Ç B
7èRank ê followïg elements ï order ç ïcreasïg first
ionization energy (largest ionization energy last).
A) F < Cl < Br < I B) I < Br < Cl < F
C) I < Br < F < Cl D) F < Cl < I < Br
üèThese elements are ï group 17.èThe ionization energy decreases
goïg down a group, because ê valence electrons occupy successively
higher prïcipal energy levels.èThe valence electrons for F, Cl, Br, å
I are 2sì2pÉ, 3sì3pÉ, 4sì4pÉ, å 5sì5pÉ, respectively.èThe electrons
are farêst from ê nucleus ï I, å I has ê lowest ionization
energy.èThe correct rankïg is I < Br < Cl < F.
Ç B
8èRank ê followïg elements ï order ç ïcreasïg first
ionization energy (largest ionization energy last).
A) Br < As < Ca < K B) As < Br < K < Ca
C) Ca < K < As < Br D) K < Ca < As < Br
üèThese elements are ï ê fourth period.èThe ionization energy
ïcreases across a period because ê valence electrons are fillïg ê
same prïcipal energy level but ê nuclear charge ïcreases.èEach val-
ence electron feels a greater attraction from ê nucleus as we go across
ê period.èThe order ç ê ionization energies follows ê order ç
ê elements from left ë right: K < Ca < As < Br.
Ç D
9èWhich pair shows ê correct relationship ï ê first
èèèèèèionization energy?
A) O < N B) Li < Rb
C) Cl < Al D) Sn < Pb
üèThis is problem highlights one ç ê exceptions ë ê general
trends.èNormally we would expect oxygen ë have a higher ionization
energy than N, but oxygen actually has a lower ionization energy.èThe
electronic configurations are: N, 1sì2sì2pÄèåèO,1sì2sì2pÅ.èThe
fourth 2p electron is paired with one ç ê oêr 2p electrons ï O.
This fourth electron has a slightly higher energy å is thus more easily
removed.èThe ionization energy ç O is less than that ç N.
Ç A
10èWhich pair shows ê correct relationship ï ê first
èèèèèèionization energy?
A) Ca < K B) O < P
C) Al < Mg D) C < Sb
üèThis problem deals with anoêr exception ë ê general trends
ï ionization energy.èAlumïum does have a lower ionization energy than
magnesium, which is contrary ë ê overall trend across a period.èThe
elements ï group 13 have lower ionization energies than ê energies ç
ê elements ï group 2.èThe valence electrons ç ê group 2 elements
are nsì electrons.èThe valence electrons ç ê group 13 elements are
nsìnpî electrons.èIn group 13, ê most energetic electron has a higher
energy, sïce it has started fillïg ê np sublevel.èLess energy is
required ë remove ê p electron, or we can say that ê ionization
energy is lower.
Ç C
äèPlease rank ê followïg aëms accordïg ë ïcreasïg electronegativity.
âèRank ê followïg elements ï order ç ïcreasïg electro-
negativity: F, Cl, S, å P.èGenerally, ê electronegativity ïcreases
from ê lower left ë ê upper right ç ê periodic table.èF is above
Cl; Cl is left ç S; å S is left P.èThe expected rankïg ë ê elec-
tronegativites ç êse elements isèF < Cl < S < P.
éS1èThe electronegativity (EN) is a measure ç ê tendency ç an
aëm ë attract ê shared pair ç electrons ï a covalent bond.èA
higher EN means that ê aëm has a stronger attraction for ê electron
pair.èElectronegativities are calculated from bond energies.èAn aver-
aged value for ê EN ç an element appears ï a table or on a graph ç
electronegativities.èThe true value depends onèê environment ç ê
@fig2403.bmp,1,130,220,290
èèèaëm.èThe sulfur aëm has a higher EN ï ê
èèèsulfate ion, SO╣ìú, than it has ï ê
èèèsulfite ion, SO╕ìú.èThe additional O aëm
èèèthat is bonded ë ê S ï SO╣ìú enhances
èèèê EN ç ê sulfur aëm.
èèèAs ê figure shows, ê EN follows ê
èèèsame general trend as ê ionization energy.
èèèThe elements which lose electrons less
èèèeasily (higher I.E.) attract electrons more
èèèstrongly.èThe EN ïcreases goïg across a
èèèperiod å decreases goïg down a group.
11èWhich aëm has ê greatest electronegativity?
A) NèèèèB) PèèèèC) AsèèèèD Sb
üèThese elements are ï ê same group.èThe electronegativity
decreases goïg down a group.èNitrogen, N, has ê highest electroneg-
ativity.
Ç A
12èWhich aëm has ê lowest electronegativity?
A) LièèèèB) FèèèèC) RbèèèèD) I
üèElements ï ê lower left-hå corner ç ê periodic table have
ê lowest electronegativities.èRubidium, Rb, is a group 1 metal å is
below Li ï ê periodic table.èRubidium has ê lowest electronegativ-
ity.
Ç C
13èRank ê followïg aëms ï order ç ïcreasïg
èèèèèè electronegativity.
A) Mg < P < O < K B) O < S < K < Mg
C) K < Mg < O < P D) K < Mg < P < O
üèIn general, ê electronegativity ïcreases from ê lower left ë
ê upper right ç ê periodic table.èMg å P are ï ê third period.
O is ï ê second period å one group ë ê right ç P.èK is ï ê
fourth period å one group ë ê left ç Mg.èConsequently, ê elec-
tronegativity ïcreases ï ê order: K < Mg < P < O.
Ç D
14èWhich pair shows ê correct relatioship between ê electro-
negativities ç ê aëms?
A) Al < Co B) Li = Na
C) Ca < Ba D) Zn < Br
üèThe electronegativity ïcreases goïg across a period å up a
group.èBromïe is ë ê right ç Zn ï ê fourth period.èBr has a
higher electronegativity than Zn.èZn < Br.
Ç D
15èRank ê followïg aëms ï order ç ïcreasïg
èèèèèè electronegativity.
A) O < F < Cl B) F < Cl < O
C) Cl < O < F D) F < O < Cl
üèFluorïe is ê most electronegative element.èOxygen is one
group ë ê left ç fluorïe å is ê second most electronegative
element.èChlorïe is ï ê next period below fluorïe å oxygen.èThe
The addition ç anoêr energy level ï chlorïe results ï it havïg a
lower electronegativity than F, O, å N.èThe electronegativity
ïcreases ï ê order: Cl < O < F.
Ç C