CHEMISTRY 161 Chapter 7 Quantum Theory and Electronic Structure of the Atom
CHEMISTRY 161 Chapter 4
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Transcript of CHEMISTRY 161 Chapter 4
CHEMISTRY 161
Chapter 4
REVISION
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
(Arrhenius)
HCl(aq) + NH3(aq) → NH4Cl(aq)
H+(aq) + NH3(aq) → NH4+(aq)
(Bronsted)
KOH(aq) + HF(aq) → ???
EXAMPLE 1
1. KOH(s) → K+(aq) + OH-(aq)
2. HF(g) → H+(aq) + F-(aq)
1.+2. K+(aq) + OH-(aq) + H+(aq) + F-(aq) →
reactants
H2O(l) + K+(aq) + F-(aq)
KOH(aq) + HF(aq) → H2O(l) + KF(aq)
Mg(OH)2(aq) + HCl(aq) → ???
EXAMPLE 2
1. Mg(OH)2(s) → Mg2+(aq) + 2 OH-(aq)
2. 2 HCl(g) → 2 H+(aq) + 2 Cl-(aq)
1.+2. Mg2+(aq) + 2 OH-(aq) + 2 H+(aq) + 2 Cl-(aq) →
reactants
2 H2O(l) + Mg2+(aq) + 2 Cl-(aq)
Mg(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + MgCl2(aq)
Mg(OH)2(aq) + 2 HCl(aq) → ???
Ba(OH)2(aq) + H2SO4(aq) → ???
EXAMPLE 3
1. Ba(OH)2(s) → Ba2+(aq) + 2 OH-(aq)
2. H2SO4(l) → 2 H+(aq) + SO42-(aq)
1.+2. Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) →
reactants
2 H2O(l) + Ba2+(aq) + SO42-(aq)
Ba(OH)2(aq) + H2SO4(aq) → 2 H2O(l) + BaSO4(s)
CHEMICAL REACTIONS
a) precipitation reactions
b) acid-base reactions (proton transfer)
c)redox reactions (electron transfer)
H → H+ + e-
(1) K → K+ + e-
(2) S + 2 e- → S-2
/ × 2
/ × 1
(1) 2 K → 2 K+ + 2 e- (2) S + 2 e- → S-2
2K(s) + S(s) → K2S(s)
REDOX REACTIONS
reactions involving transfer of electrons
(solid state reaction of potassium with sulfur)
oxidation
reduction
(1)+(2) 2K + S + 2e-→ 2 K+ + S2- +2e-
(1)+(2) 2K + S → 2 K+ + S2-
1. oxidation
KEY CONCEPTS
loss/donation of electrons (minus)
2. reduction gain/acceptance of electrons (plus)
mo r p
m o p r
3. balance electrons
4. add oxidation half reaction and reduction half reaction
(1) Ca → Ca2+ + 2e-
(2) ½ O2 + 2 e- → O2-
Ca(s) + ½ O2(g) → CaO(s)
EXAMPLE
reaction of calcium with molecular oxygen
oxidation
reduction
(1)+(2) Ca + ½ O2 + 2e-→ Ca2+ + O2- +2e-
(1)+(2) Ca + ½ O2 → Ca2+ + O2-
2 Ca(s) + O2(g) → 2 CaO(s)
/ ×2
OXIDATION NUMBER
ionic compounds ↔ molecular compounds
NaCl HF, H2
Na+Cl- ?electrons are fully transferred covalent bond
‘oxidation number’
charges an atom would have if electrons are transferred completely
HF H+ + F-
molecular compound ionic compound
F- oxidation state -1
H+ oxidation state +1
EXAMPLE 1
H2O
molecular compound ionic compound
2 H+ + O2-
H+ oxidation state +1
O2- oxidation state -2
EXAMPLE 2
H2
molecular compound ionic compound
H+ + H-
EXAMPLE 3
OXIDATION NUMBER OF FREE ELEMENTS IS ZERO
RULE 1
OXIDATION NUMBER OF FREE ELEMENTS IS ZERO
H2, O2, F2, Cl2, K, Ca
RULE 2
monoatomic ions
oxidation number equals the charge of the ion
group I M+
group II M2+
group III M3+ (Tl: also +1)
group VII F-
RULE 3
oxidation number of hydrogen
+1 in most compounds
(H2O, HF, HCl, NH3)
-1 binary compounds with metals (hydrides)
(LiH, NaH, CaH2, AlH3)
RULE 4
oxidation number of oxygen
-2 in most compounds
(H2O, MgO, Al2O3)
-1 in peroxide ion (O22-) (H2O2, K2O2, CaO2)
-1/2 in superoxide ion (O2-) (LiO2)
+1 in FO
RULE 5
oxidation numbers of halogens
F: -1 (KF)
Cl, Br, I: -1 (halides) (NaCl, KBr)
Cl, Br, I: positive oxidation numbers if combined with oxygen (ClO4
-)
RULE 6
neutral molecule: sum of oxidation numbers must be zero
NH3 H: +1 N: -3
3 × (+1) + 1 × (-3) = 0
polyatomic ion: sum of oxidation numbers equals charge of ion
NH4+ H: +1 N: -3
4 × (+1) + 1 × (-3) = +1
RULE 7
charges of polyatomic molecules must be integers
(NO3-, SO4
2-)
oxidation numbers do not have to be integers
-1/2 in superoxide ion (O2-)
MENUE
1.oxidation states of group I – III metals
2.oxidation state of hydrogen (+1, -1)
3. oxidation states of oxygen (-2, -1, -1/2, +1)
4.oxidation state of halogens
5.remaining atoms
NONO2
NO+
NO-NO2-
NO3-
PO43- SO4
2-
SO3
SO2
KO2
K2O
BrO-
SUMMARY
1.redox reactions
2. oxidation versus reduction
3. oxidation numbers versus charges
4. calculation of oxidation numbers
Homework
Chapter 4, p. 116-121problems