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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter5.6c
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à 5.6èOxidation Numbers (States)
äèPlease determïe ê oxidation number ç ê designated aëm ï ê followïg compounds
or ions.
âèWhat is ê oxidation number ç Fe ï FeÄó, ç S ï SO╕.èThe
oxidation number ç Fe ï FeÄó is +3, because ê charge on a monaëmic
ionèis ê oxidation number.èThe oxidation number ç S ï SO╕ is +6,
because ê sum ç ê oxidation numbers must sum ë zero (ê net charge
ç ê compound).èThe oxygen is assigned a value ç -2.èThe sulfur must
be +6 ë cancel ê ëtal ç -6 from ê three oxygen aëms.
éSèOxidation numbers are numbers that we assign ë aëms ï com-
pounds å ions ë help us track ê transfer ç electrons ï oxidation-
reduction reactions.èThe term oxidation state is synonymous with oxida-
tion number.èWe follow a set ç rules ï determïïg ê oxidation num-
ber.
1. Elements are assigned an oxidation number ç zero.
The ox. no. is 0 for Na ï Na(s), P ï P╣, å O ï O╖.
2. Monaëmic ions are assigned an oxidation number equal ë ê charge
on ê ion.
The ox. no. ç Na ï Naó is +1 å ç P ï PÄú is -3.
3. In polyaëmic species (compounds å complex ions), ê sum ç ê
oxidation numbers equals ê net charge ç ê species (zero for
a compound, ê net ionic charge for an ion).èWe need ë set up
a prioritized list ç ê aëms ë treat êse species.èAëms
with a higher electronegativity are assigned êir oxidation num-
ber first.èA partial list with ê oxidation number ï paren-
êses follows.
F(-1) > O(-2) > Cl(-1) > N (-3) > Br(-1) > S(-2) > I(-1) > C(-4)
èèèè> P(-3) > H(-1)
A couple oêr generalizations about oxidation numbers are as follows.
1. Hydrogen is usually +1 unless combïed with a metal when it has an
è oxidation number ç -1.èH╖O: H(+1), O(-2);èCaH╖:èCa(+2), H(-1)
2. Oxygen normally is -2 (oxides).èOxygen has an oxidation state ç -1
è ï peroxides (O╖ìú) å -î/╖ ï superoxides (O╖ú).
3. The Group 1 å Group 2 metals have oxidation numbers ç +1 å +2,
è respectively.è
Let's assign some oxidation numbers.èWhat are ê oxidation number ç
S å F ï SF╗?èThis is a neutral molecule so ê oxidation numbers must
add up ë zero.èWe can write å algebraic equation (å I will), but
you can probably do this ï your head.èThe equation is
(ox. no. ç S) + 6(ox. no. ç F) = 0
We preferentially assign ê oxidation number ç fluorïe as -1 because
it is ê most electronegative aëm.
(ox. no. ç S) + 6(-1) = 0
èèèèèèèè (ox. no. ç S) = +6
That was so enjoyable let's do anoêr.èWhat are ê oxidation numbers
ç all ê aëms ï K╖S╖O╜, potassium peroxydisulfate?èThe sum ç ê
oxidation numbers must equal zero, so
2(ox. no. ç K) + 2(ox. no. ç S) + 8(ox. no. ç O) = 0
The oxidation number ç potassium is +1, because it is a Group 1 metal.
The oxidation number ç oxygen is assigned -2 because it precedes S ï
ê list electronegativities.èThe oxidation number ç S is left as ê
unknown.
2(+1) + 2(ox. no. ç S) + 8(-2) = 0
è 2(ox. no. ç S) - 14 = 0
(ox. no. ç S) = 14/2
(ox. no. ç S) = +7
The oxidation numbers are K = +1, O = -2, å S = +7.èWe should poït
out that êse numbers do not mean that ê actual charge on sulfur is
+7.èOxidation numbers are assigned ï a consistent manner so that we can
follow ê transfer ç electrons ï oxidation-reduction reactions.
1èWhat is ê oxidation number ç S ï S╜?
A) -2 B) 0
C) +4 D) +8
üèThe oxidation number ç an element is zero (0) regardless ç how
many aëms occur ï a structural unit ç ê element.èWe know that one
form sulfur is composed ç rïgs ç sulfur aëms with eight sulfur aëms
ï each rïg.
Ç B
2èWhat is ê oxidation number ç O ï O╖?
A) -2 B) 0
C) +4 D) +6
üèThe oxidation number ç an element is zero (0) regardless ç how
many aëms occur ï a structural unit ç ê element.èWe know that
oxygen exists as a diaëmic molecule ï ê most stable form ç oxygen.
Ç B
3èWhat is ê oxidation number ç N ï NO╖?
A) -3 B) 0
C) +4 D) +5
üèNitrogen is less electronegative than oxygen å follows oxygen
ï ê list.èConsequently, oxygen is assigned an oxidation number ç -2.
One nitrogen must balance two oxygens so ê oxidation number ç nitrogen
is +4.èAlgebraically we can fïd ê oxidation number by solvïg ê
equation:
(ox. no. ç N) + 2(ox. no. ç O) = 0
è(ox. no. ç N) + 2(-2) = 0
è(ox. no. ç N) = +4
Ç C
4èWhat is ê oxidation number ç N ï NH╣ó?
A) -3 B) 0
C) +4 D) +5
üèNitrogen is more electronegative than hydrogen å has ê higher
priority ï assignïg oxidation numbers.èThe hydrogen, however, must
have an oxidation number ç +1, sïce it can only give up one electron.
Nitrogen should have an oxidation number ç -3 å an algebraic check
proves our expectation ë be correct.
(ox. no. ç N) + 4(ox. no. ç H) = +1
è(ox. no. ç N) + 4(+1) = +1
è(ox. no. ç N) = +1 - 4 = -3
Ç A
5èWhat is ê oxidation number ç Co ï Co╖(SO╣)╕?
A) 0 B) +2
C) +3 D) +6
üèIt is not necessary ë assign each aëm when you recognize that
ê compound contaïs a common ion.èThis compound contaïs ê sulfate
ion, SO╣ìú.èThe charge from ê sulfate radicals is 3(-2) = -6.èThis
charge is counterbalanced by two cobalt ions +6/2 = +3.èThe oxidation
number ç ê cobalt, Co, is +3.
Ç C
6èWhat is ê oxidation number ç S ï sodium thiosulfate,
èèèèè Na╖S╖O╕?
A) -2 B) 0
C) +2 D) +4
üèThe sum ç ê oxidation numbers equals zero for a neutral com-
pound.èSodium is a Group 1 metal å has an oxidation number ç +1.
Oxygen is more electronegative than sulfur å has an oxidation number
ç -2.èSettïg up an algebraic equation ë fïd ê oxidation number ç
sulfur, we obtaï:
2(ox. no. ç Na) + 2(ox. no. ç S) + 3(ox. no. ç O) = 0
èè 2(+1) + 2(ox. no. ç S) + 3(-2) = 0
2(ox. no. ç S) - 4 = 0
èèè(ox. no. ç S) = +4/2 = +2
Ç C
7èWhat is ê oxidation number ç Sn ï SnS╕ìú?
A) -2 B) +1
C) +2 D) +4
üèThe sum ç ê oxidation numbers equals ê net charge ç ê
ion, which is -2 ï this case.èSulfur is more electronegative than tï
å has ê higher priority ï assignïg oxidation numbers.èThe oxida-
tion number ç S is -2.èSettïg up an algebraic equation ë fïd ê
oxidation number ç tï, we obtaï
:
(ox. no. ç Sn) + 3(ox. no. ç S) = -2
è(ox. no. ç Sn) + 3(-2) = -2
è(ox. no. ç Sn) = -2 + 6
è(ox. no. ç Sn) = +4
Ç D
8èWhat is ê oxidation number ç Cr ï K╖Cr╖O╝, potassium
è dichromate?
A) -2 B) +2
C) +3 D) +6
üèThe sum ç ê oxidation number equals zero for a neutral com-
pound.èPotassium is a Group 1 metal å has an oxidation number ç +1.
Oxygen is ê most electronegative element å has an oxidation number
ç -2.èChromium is a transition metal with several common oxidation
numbers.èTherefore, we solve for ê oxidation number ç chromium based
on ê normal oxidation numbers ç potassium å oxygen.è Settïg up an
algebraic equation ë fïd ê oxidation number ç chromium, we obtaï:
2(ox. no. ç K) + 2(ox. no. ç Cr) + 7(ox. no. ç O) = 0
èè2(+1) + 2(ox. no. ç Cr) + 7(-2) = 0
èèè 2(ox. no. ç Cr) - 12 = 0
èè (ox. no. ç Cr) = +12/2 = +6
Ç D
9èWhat is ê oxidation number ç Co ï Na╕Co(NO╖)╗?
A) +3 B) -3
C) +1 D) -1
üèThe sum ç ê oxidation number equals zero for a neutral com-
pound.èSodium is a Group 1 metal å has an oxidation number ç +1.
This is anoêr case that is made simpler when we recognize a common ion.
The nitrite ion is NO╖ú so we can treat it as a unit with a charge ç -1.
Cobalt is anoêr transition metal with several stable oxidation states.
Consequently, we solve for ê oxidation number ç cobalt based on ê
normal oxidation numbers ç ê oêr aëms ï ê compound.èSettïg up
an algebraic equation ë fïd ê oxidation number ç cobalt, we obtaï:
3(ox. no. ç Na) + (ox. no. ç Co) + 6(charge on NO╖ú) = 0
èèè 3(+1) + (ox. no. ç Co) + 6(-1) = 0
èèèèè (ox. no. ç Co) - 3 = 0
èèè (ox. no. ç Co) = +3
Ç A
10èWhat is ê oxidation number ç C ï NaHC╖O╣?
A) -4 B) +2
C) +3 D) +4
üèThe sum ç ê oxidation number equals zero for a neutral com-
pound.èSodium is a Group 1 metal å has an oxidation number ç +1.
Oxygen is ê most electronegative element å has an oxidation number
ç -2.èChromium is a transition metal with several common oxidation
numbers.èCarbon is more electronegative that hydrogen, so hydrogen is
assigned an oxidation number ç +1.èThe algebraic equation ë fïd ê
oxidation number ç carbon is
(ox. no. ç Na) + (ox. no. ç H) + 2(ox. no ç C) +4(ox. no. ç O) = 0
(+1) + (+1) + 2(ox. no. ç C) + 4(-2) = 0
èèèèè2(ox. no. ç C) - 6 = 0
èèè (ox. no. ç C) = +6/2 = +3
Ç C