Chapter Nine Coordination Compounds Coordination Compound: a compound in which a central metal ion is attached to a group of surrounding molecules or.
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Chapter Nine
Coordination Compounds
Coordination Compound:
a compound in which a central metal ion
is attached to a group of surrounding
molecules or ions by coordinate covalent
bonds.
Anemia( 贫血症 )
CH2OH
CHSH
CH2SH
CH2OH
CHS
CH2S
Hg+ Hg
•Anti-tumour (肿瘤) agent
Coordination Compounds
• 9-1 Basic Concepts
• 9-2 The Chemical Bond Theory
9-2.1 Valence Bond Theory
9-2.2 Crystal Field Theory
• 9-3 Coordination Equilibrium
• 9-4 Chelates
9-1 Basic Concepts An introduction to complex ions with an explanation
of what ligands are and how they bond to the central metal ion.
central metal ion transition metals (but not all) complex ion ligands (anions or polar molecules)
Transition Metals (T.M.)
• This gives rise to the following properties:– Distinctive color – paramagnetic compounds – catalytic activity– great tendency to form complex ions
•Zn, Cd, Hg are not considered T.M.
Ligands and Donor atom
• Ligands: ions or molecules that is bound
directly to the metal atom. e.g. NH3, CN-,
H2O, Cl-, I-
• Donor atom: the atom in a ligand that is
bound directly to the metal atom , has lone
electron pairs.
e.g. C, N, O, S, F, Cl, Br, I
Ligands
• Depending on the number of donor atoms present in the molecule or ion, ligands can be classified as :
monodentate : (H2O: :NH3 )
bi dentate :( H2N-CH2-CH2-NH2 )
polydentate : (EDTA)
also called chelating agents
Coordination number• Coordination number: the number of donor
atoms surrounding the central metal atom in a complex ion. Commonly, it is 2, 4 ,5 or 6
For monodentate : [Cu(NH3)4] SO4 , [Fe(CN)6]4-
Coordination number = ligand number
For bidentate or polydentate : Coordination number ≠ ligand number
e.g. [Cu(en)2]SO4 ( en = H2N-CH2-CH2-NH2)
Coordination number = 4 ≠ 2
Charges of coordination ion:
Charges of coordination ion = the sum of charges of central ion and ligands
e.g. K3[Fe(CN)6] Fe3+
[Fe(H2O)6]Cl3 Fe3+
K4[Fe(CN)6] Fe2+
• Contains a complicated ion - coordination ion
[Cu(NH3)4]2+ , [Ag(NH3)2]+, [Fe(CN)6]4-
• Metal ion bonded with other ion or molecule by coordination bond
• Has definite stability:
KCl•MgCl2 • 6H2O : K+, Cl-, Mg2+
KAl(SO4) • 12H2O : K+, Al3+, SO42+
• The features of coordination ion
Simple ion and complex ion
CuSO4 BaCl2 BaSO4
CuSO4
CuSO4
NaOH Cu(OH)2
NH3•H2O Cu(OH)2 BlueNH3•H2O
Cu2+ + 4NH3 == [Cu(NH3)4]2+ Complex ion
AgNO3NaCl AgCl NH3•H2O [Ag(NH3)2]+
[Cu(NH3)4] SO4
[Ag(NH3)2]Cl
The composition of coordination compound
• the coordination sphere:
1. The central metal and the ligands bound
to it constitute.
2. square brackets to set off the groups
within the coordination sphere from
other parts of the compound.
for example: [Co(NH3)6]Cl3
[PtCl6]2+
[Cu(NH3)4]2+ SO42-
Inner spherecoordination sphere Outer sphere
Central ion
or atom
Ligand
Coordination number
Charges of coordination ion
coordination atom or
donor atom
bidentate
What are the oxidation numbers of the central metal in the complexes below?
• K3[FeF6]
• Na2[Ni(CN)4]
中文命名原则 :1. 内界和外界 :
同一般简单化合物 - 某化某 ; 某酸某 CuCl2 ; BaSO4
[Pt(en)2]Cl2 ; [Cu(NH3)4]SO4 ; K4[Fe(CN)6]
2. 内界 : 按下列次序 :
[Cu(NH3)4]2+: 四 氨 合 铜 ( )Ⅱ 离子 [Fe(CN)6]4- : 六 氰 合 铁 ( )Ⅱ 酸根离子
配体数 - 配体名称 -“ 合” - 中心离子名称( 氧化数 )
3. 当有多种配体时 : 先无机后有机 ; 先阴性离子后中性分子 ; 先简单后复杂 ; 先常见后不常见 ;同类按配位原子字母顺序 ; 如先 NH3 后 H2O
[Co(H2O)(NH3)3Cl2]Cl: 氯化二氯三氨一水合钴 ( )Ⅲ[Pt(NH3)4(NO2)Cl]CO3 : 碳酸一氯一硝基四氨合铂 ( )Ⅳ
[Cu(NH3)4](OH)2 : 氢氧化四氨合铜 ( )Ⅱ K[Pt(NH3)Cl5] : 五氯一氨合铂 ( )Ⅲ 酸钾 Fe[Fe(CN)6] : 六氰合铁 ( )Ⅲ 酸铁 Fe2[Fe(CN)6] : 六氰合铁 ( )Ⅱ 酸亚铁 [Pt(NH3)2Cl2] : 二氯二氨合铂 ( )Ⅱ [Ni(CO)4] : 四羰基合镍
Naming of Coordination Compounds • the International Union of Pure and Applied
Chemistry (IUPAC) • 1. The cation is named before the anion.
NaCl: sodium choride• 2. Within a complex ion the ligands are named
first, in alphabetical order, and the metal ion is
named last. • 3. To name the ligands:
anionic ligands end in -o neutral ligands usually called the name of the molecule
LIGAND Name of Ligand in Coord. Cpd.Bromide, Br- BromoChloride, Cl- ChloroCyanide, CN- CyanoHydorxide, OH- HydroxoOixde, O2- Oxo
Carbonate, CO32- Carbanato
Nitrite, NO3- Nitro
Oxolate, C2O42- Oxolato
Ammonia, NH3 AmmineCarbon monoxide, CO Carbonyl
Water, H2O AquoEthylenediamine(en) Ethylenediamine
Nam
ing
Coo
rdin
atio
n
Com
pou
nd
s
4. When several ligands of a particular kind are present, we use the Greek prefixes to name them.• Di (2)• tri (3) • tetra (4)• Penta (5)• Hexa (6)
[Co(NH3)4Cl2]+are
tetraamminedichloro
If the ligand itself contains a Greek prefix, we use the prefixes bis, tris, tetrakis to indicate the number of ligands present. e.g. [Cu(en)2]2+ bis(ethylenediamine)
5. The oxidation number of the metal is written in Roman numerals following the name of the metal.
[Cr(NH3)4Cl2]+, which is called
tetraamminedichlorochromium(Ⅲ) ion.
6. If the complex ion is an anion, its name ends in -
ate. K4[Fe(CN)6] the anion [Fe(CN)6]4 - is
called hexacyanoferrate(II) ion.
Metal in anion complex
Aluminum AluminateChromium ChromateCobalt CobaltateCopper CuprateGold AurateIron FerrateLead Plumbate
Metal in anion complex
ManganeseManganate
Nickel NickelateSilver ArgentateTin StannateTungsten TungstateZinc Zincate
• Example 9-1 :
(a) Ni(CO)4, (b) [Co(NH3)4Cl2]Cl,
(c) K3[Fe(CN)6], (d) [Cr(en)3]Cl3.
• Solution:
(a) tetracarbonylnickel(0)
(b) tetraamminedichlorocobalt( )Ⅲ chloride
(c) potassium hexacyanoferrate( ). Ⅲ potassium ferricyanide
(d) tris(ethylenediamine) chromium( ) Ⅲ chloride.
. tetraamminedichlorochromium(Ⅲ) ion.
hexacyanoferrate(II) ion.
the cation [Cr(NH3)4Cl2]+
the anion [Fe(CN)6]4 -
Give the formula for the following coordination compounds.
tetracarbonylnickel(0)
tetraammineaquochlorocobalt(III) chloride
[Co(NH3)4H2OCl]Cl2
Ni(CO)4,
How did I know there had to be two chlorides at the end?
9-2 The Chemical Bond Theory • Several different bonding theories have
been applied to transition-metal coordination compounds. We shall consider two of these.
• the valence-bond theory: being covalent and examines the hybridization of orbitals on the metal.
• the crystal-field theory: from an ionic point of view and focuses on( 集中 ) the effect of the surrounding ligands on the energies of the metal d orbitals
9-2.1 Valence Bond Theory
• Magnetism • Isomerism ( 异构体 )
• stability
9-2.2 Crystal Field Theory
• The Splitting (分裂) of the d orbitals in
octahedral Field
• High-spin and Low-Spin
Coordination Compounds • The color of coordination compounds
9-2.1 the valence-bond theory
• a ligand orbital containing two electrons overlaps an unoccupied orbital on the metal atom.
• donate a pair of electrons into a suitable empty hybrid orbital on the metal,
1. Central ion bonds with ligands by
coordination bond.
2. The empty orbitals of central ion must
hybridize to increase bonding ability.
3. There are two types of coordination compounds:
_ outer-orbital coordination compounds
_ inner-orbital coordination compounds
the valence-bond theory Outline :
Example [Ag(NH3)2]+
• 47Ag [Kr]4d105s1
• Central ion: Ag+ 4d105s05p0
hybrid
sp hybridization( outer orbital )
2 : NH3
[Ag(NH3)2]+ ( linear )
2 electron clouds
Ag+:
Ni(NH3)42+ 28Ni 3d84s2
• Central ion: Ni2+ 3d8 4s0 4p0
hybrid
sp3 hybridization( outer orbital )
4 :NH3
[Ni(NH3)4 ]2+ (tetrahedral)
Ni
[Ni
[Ni(CN)4 ]2-
• Central ion: Ni2+ 3d8 4s0 4p0
hybrid
dsp2 hybridization inner orbital
[Ni(CN)4 ]2-( square planar)
4 :CN-
realignment
[Co(NH3)6] 3+
• Co atom
• Co3+ ion
• [Co(NH3)6]3+
3d74s2
3d6
d2sp3 hybridization
6 :NH3
Co(NH3)6 3+ (octahedral)(inorbital complex)
CoF63+
• Co atom 3d74s2
• Co3+ ion
• CoF63+
3d 4s 4p
4d
sp3d2 hybridization
6 :F-
(octahedral)
3d6
Magnetism
• Paramagnetism: substances containing unpaired electrons are paramagnetic.
• diamagnetic: substances without unpaired electrons are diamagnetic.
• magnetic moment: μ μ=√n (n+2) • n : is the number of unpaired electrons
• Example : For [Fe(H2O)6]SO4
μ=5.26B.M according to above equation, n = 4 ,so there are 4
unpaired electrons in this coordination compound.
• Example : For K4[Fe(CN)6]
μ=0 B.M according to above equation, n = 0 ,so there are 0
unpaired electrons in this coordination compound.
• Example : For [Fe(H2O)6]SO4 , μ=5.26B.M according to above equation, n = 4 ,so there are 4
unpaired electrons in this coordination compound.
3d 4s 4p 4d
26Fe: 3d64s24p04d0
outer orbital
• • • • • • • • • • • •Sp3d2 hybridization
Fe2+: 3d6 4s04p04d0
• Example : For K4[Fe(CN)6] , μ=0 B.M according to above equation, n = 0 ,so there are 0
unpaired electrons in this coordination compound.Fe2+: 3d64s04p04d0
d2sp3 hybridization, inner orbital
3d 4s 4p
• • • • • •• • • • • •
Table 9-4: Some common types of hybridization and geometries.
Isomerism( 异构体 ) • Isomers : Two or more compounds that have
the same formula but a different structure (that is, the same collection of atoms but arranged in different ways) are called isomers.
• Isomers Geometric isomers (Stereo 立体 isomers)
Optical isomers
Structural isomers
• Structural isomerism• Structural isomers are that differ in how the atoms
are joined together.
• Example: [Co(NH3)5(SO4)]Br red
[Co(NH3)5 Br] SO4 violet
CH3-CH2-CH2-COOH
CH3-CH-COOH
CH3
硝基四氨合钴(Ⅲ) 亚硝酸四氨合钴(Ⅲ)
• Geometric isomerism (Stereoisomerism,
cis-trans isomerism) • are isomers that have the same chemical
bonds but different special arrangements.
• the isomer with like groups close together is
called the cis-isomer.
• whereas the one with like groups far apart is
called the trans-isomer.
Geometric Isomerism (stereoisomerism)
• Pt(NH3)2Cl2
Cl NH3
cis Pt
Cl NH3
NH3 Cl
trans Pt
Cl NH3
cis-diamminedichloroplatinum(II)
both coordination compounds are named: diamminedichloroplatinum(II)
cis trans
Optical Isomerism
“ 反应停”的手性结构分子
化学名称叫做酞胺呱啶酮,是一种常用于安定精神和抑制妊娠恶心的镇静药和催眠药。
被反应停夺去胳膊的孩子们
“ 海豹”式畸形,不是手足全无,就是缺胳膊少腿,甚至手掌直接长在肩膀上;或者出现心脏等内脏器官畸形、脑障碍、听力或视力丧失、发生自痹症和癫痫症等
9-2.2 Crystal Field Theory (中文 P217 ) 1. the ligands in a transition-metal complex are
treated as point charges. Thus, a ligand anion becomes simply a point of negative charge. metal ion becomes simply a point of positive charged.
2. For the formation of a complex ion or molecule is the electrostatic attraction
3. this theory explains both the paramagnetism and color observed in certain complexes.
中心离子 d 轨道角度分布图
The Splitting of the d Orbitals in Octahedral Field
eg (dγ)orbitals
t2g (dε)orbitals
Δ = eg - t2g
0)(3)(22
gg te dEdE
DqdEdEgg te 10)()( 02
0
0
4.0)(
6.0)(
2
g
g
t
e
dE
dE
:solutionenergy) (increase
energy) (decrease
Δ 在数值上等于一个电子由 t2g 轨道跃迁到 eg 轨道所需的激发能。
crystal field splitting energy, Δ
High-Spin and Low-Spin Coordination Compounds
• Fe(H2O)62+ Fe 3d64s2 Fe2+ 3d6
Figure: Occupation of the 3d orbitals in complexes of Fe2+. (a) Low spin. (b) High spin.
pairing energy P
• pairing energy P: the energy required to put two electrons into the same orbital.
• 1. P > Δ : the fourth electron will go into one of the higher d orbitals.
a high-spin complex 2. P <Δ : an electron in one of the lower
energy orbitals. a low spin complex
影响分裂能 () 的因素
中心离子和配体构型一定,值与配体有关:① 配体:
配体固定时,中心离子电荷越高,值越大② 中心离子
① 配体: spectrochemical series,
Weak-bonding ligands Strong-bonding ligands I – < Br – < Cl– < SCN–< F –< OH – < H2O <NCS–
< edta < NH3 < en < NO2–< CN –< CO
Increasing Δ →
[Co(H2O)6]3+ [Co(NH3)6]3+ [Co(CN)6]3-
o /cm-1 13000 22900 34000
which is a list of ligands arranged in order of their abilities to split the d orbitals
强场: o > P : low-spin complex
弱场: o < P: high-spin complex
CN- is astrong-field
ligand
F- is a weak-field ligand
● ② 中心离子 M 对的影响:
[CrCl6]3- [MoCl6]3-
o /cm-1 13600 19200
主量子数 n 增大, o 增大 :
[Cr (H2O)6]3+ [Cr (H2O)6]2 +
o /cm-1 17600 14000
电荷 Z 增大, o 增大 :
例:下列四种络合物中, d-d 跃迁能量 ( 分裂 能 ) 最低的是 ( )
a. [Fe(H2O)6] 2+ b. [Fe(H2O)6]
3+
c. [FeF6] 4- d. [FeF6]
3-
八面体场中电子在 t 2g 和 eg 轨道中的分布
八面体场中电子在 t 2g 和 eg 轨道中的分布
d4 d5 d6 d7
4 5 4 3
2 1 0 1
高自
旋低
自旋
d4 d5 d6 d7
4 5 4 3
2 1 0 1
高自
旋低
自旋
1 2 3 2 1
d1 d2 d3 d8 d9
只有
一种
排列
1 2 3 2 1
d1 d2 d3 d8 d9
只有
一种
排列
1. Crystal Field Stabilization Energy(CFSE)
• A measure of the net energy of stabilization gained by a metal ion's nonbonding d electrons as a result of complex formation.
● 定义 : 晶体场稳定能 (Crystal field stabilization energy,
CFSE) 是指电子占据分裂的 d 轨道后而产生的高于平 均能量的额外稳定能 。
晶体场理论的应用
• ligand field stabilization energy:
a measure of the increased stability of a complex showing ligand field splitting.
In general,
CFSE = (# electrons in t2g) × (-0.4 Δ 0 )
+ (#electrons in eg) × (0.6 Δ 0 )
CFSE= 6 × (-0.4 Δ 0 )
+ 0 × (0.6 Δ 0 ) = -2.4 Δ 0
CFSE= 4 × (-0.4 Δ 0 )
+ 2 × (0.6 Δ 0 )
= -0.4 Δ 0
▲ 所吸收光子的频率与分裂能 大小有关
▲ 颜色的深浅与跃迁电子数目 有关
2. 解释配合物离子的颜色 过渡金属配合物的颜色产生于 d 电子在 d 轨道之间的跃迁( d-d 跃迁)。
ΔO
hνhν = Δ
The larger the crystal-field splitiing ( Δ 大) , the higher will be the frequency of light absorbed most strongly ( ν 大) , and the shorter its wavelength( λ 短) .
The color of coordination compound
• Many of the colors of octahedral transition-metal compounds arise from the excitation of an electron from an occupied lower energy orbital to an empty higher energy orbital.
• The frequency (ν) of light that is capable of inducing such a transition is related to the energy difference between the two states, which is the crystal-field splitting energy.
hν = Δ
Δ 大ν 大λ 短
Δ 小ν 小λ 长
hν = Δ
• As we have noted earlier, strong-field ligands cause a large split in the energies of the d orbitals of the central metal atom.
• Transition metal coordination compounds with these ligands are yellow, orange, or red since they absorb higher-energy violet or blue light.
•On the other hand, coordination compounds of transition metals with weak-field ligands are blue-green, blue, or indigo since they absorb lower-energy yellow, orange, or red light.
[Co(NH3)5Cl]2+ Purple compound
absorbs yellow green region
( wavelength of 530 nm)
[Co(NH3)6]3+ orange compound
absorbs violet region
(wavelength of 410nm)
d10 (Zn2+, Ag+ complexes) is colorless
Ti3+ : 3d1 , [Ti(H2O)6]3+
吸收可见光中蓝绿色的光, 使溶液呈红色。
= 492.7nm ( 蓝绿色 ) ,E=h = 242.79 kJ·mol-1 ,波数 1/=20300cm-1
(1cm-1=11.96J·mol-1) ,恰好等于该配离子的分裂能
Δo=20 300 cm-1
例:已知 [Fe(CN)6]3- 和 [FeF6]
3- 的磁矩分别为 1.7B 和 5.9
B , (1) 计算这两种络离子中心离子未成对电子数。 (2) 写出中心离子 d 轨道上电子排布。 (3) 它们是强场还是弱场络合物,说明原因。解:
(1)[Fe(CN)6]3-中,由√ n × ( n + 2 ) =1.7B.M, 得 n=1
[FeF6]3中,由√ n× ( n + 2 ) = 5.9B.M, 得 n=5
(2) [Fe(CN)6]3-中, d 轨道上电子排布为 t2g
5eg0,
[FeF6]3中, d 轨道上电子排布为 t2g
3eg2,
(3) [Fe(CN)6]3-是强场络合物,因为强场低自旋;
[FeF6]3 -是弱场络合物,因为弱场高自旋。
9-3 Coordination equilibrium coordination
Cu2+(aq) + 4NH3(aq) [Cu(NH3)4]2+ (aq)
ionization
[Cu(NH3)4]2+ [Cu2+][NH3]
4
• Ks = -------------------- Kis= -------------------
[Cu2+][NH3]4 [ Cu(NH3)4]
2 +
Stability constantor formation constant(Kf)
Instability constant
Cu2+ + NH3 Cu(NH3)2+
K1 =——————————
[Cu(NH3)2+]
[Cu2+]×[NH3
]Cu(NH3)2+ + NH3 Cu(NH3)2
2+K2 =————————————
[Cu(NH3)22+
][Cu(NH3)2+][NH3]Cu(NH3)2
2+ + NH3 Cu(NH3)3
2+K3 =————————————
[Cu(NH3)32+
][Cu(NH3)22+]
[NH3]Cu(NH3)3
2+ + NH3 Cu(NH3)4
2+K4 =————————————
[Cu(NH3)42+
][Cu(NH3)32+]
[NH3]
βn --Accumulate stability constant
Cu2+(aq) + NH3(aq) [Cu(NH3)]2+ (aq)
K1=1.4×104
Cu2+(aq) + 2NH3(aq) [Cu(NH3)2]2+ (aq)
K2=3.17×103
Cu2+(aq) + 3NH3(aq) [Cu(NH3)3]2+ (aq)
K3=7.76×102
Cu2+(aq) + 4NH3(aq) [Cu(NH3)]2+ (aq)
K4=1.39×102
sKKKKK 43214
β3= K1 • K2 • K3
β2= K1 • K2
lgβ4 = lgK1 + lgK2 + lgK3 + lgK4
• Generally, for the same type complex ions:
the larger value of Ks(Kf), the great
stability of the complex ion in solution and
accounts for the very low concentration of
metal ions at equilibrium.
• For the different type complex ions: you need compare their stability by calculation. (见中文 p224 【例 9-1 】)
1. Influence of acidity on coordination
equilibrium • [Fe(C2O4)3]3- Fe3+ + 3C2O4
2-
+ Equilibrium shift 6H+
3H2C2O4
酸效应 : 当溶液酸度增大时, H+ 离子可与配体结合生成弱酸,使配离子向解离方向移动,配离子的稳定性降低。这种因溶液酸度增大而使配合物稳定性降低的现象 .水解效应 : 因金属离子与溶液中 OH -结合而使配离子离解的作用叫做金属离子的水解效应。
2. Influence of precipitation on coordination equilibrium
[Ag(NH3)2]+ Ag+ + 2NH3
+
Equilibrium shift Br-
AgBr
• AgBr(s) Ag++Br-
+
Equilibrium shift 2S2O32-
[Ag(S2O3)2]3-
配位平衡与沉淀平衡的转化,取决于沉淀剂与配位剂争夺金属离子能力的大小及其浓度。实验证明一些阴离子或分子争夺Ag+ 离子的能力: Cl -< NH3 < Br -< S2O3
2 -< I -< CN -<S2 -
这个顺序与形成难溶盐的溶度积 KSP 和形成配离子的 KS 大小有关
[ 例 9-2] 向 [Ag(CN)2] -和 CN -的浓度均为 0.10mol·L-1 溶液中加入 NaCl ,能否产生 AgCl 沉淀? 若改加 Na2S ,能否产生 Ag2S
沉淀? 解:溶液中有下列配位平衡: Ag+ + CN - [Ag(CN)2] - KS = 1.0×10 21
由平衡式得: (mol·L-1)
生成 AgCl 沉淀的条件是 [Ag+][ Cl - ] > KSP = 1.77×10 - 10
假设加入 NaCl 后溶液体积无变化,要生成 AgCl 沉淀,则 :
[Cl - ] > (mol·L-1)
显然仅靠加入 NaCl 是无法使 [Cl - ] > 1.77×1010mol·L-1 。 所以向上述溶液加 NaCl 不可能产生 AgCl 沉淀。
202212
2 100.1)10.0(100.1
1.0
][
])([][
CNK
CNAgAg
s
1020
1010
1077.1100.1
1077.1
][
1077.1
Ag
加入 Na2S 产生 Ag2S 沉淀的条件是 :
[Ag+]2 [ S2-] > KSP = 6.69×10-50
[ S2-] >
假设加入 Na2S 后溶液体积无明显变化,则有
[S2 - ] > (mol·L-1)
加入 Na2S很容易使溶液中 [ S2-] > 6.69×10-10 mol·L-1 ,
因此向上述溶液加 Na2S 时可产生 Ag2S 沉淀。
10220
50
1069.6)100.1(
1069.6
][ Ag
K sp
• 解:在溶液中有两个平衡: KSP = 1.8×10-10
KS = 1.6×107
将两式相加得总的平衡式:
])[Ag(NH NH2Ag 233
[ 例 9-3] 25℃ ,在 1L 氨水中使 0.1molAgCl固体溶解,氨水的浓 度至少为多少 mol·L-1 ?
[Ag(NH3)2]+ + Cl-AgCl(s) + 2NH3
][][
]][][)([
][
]][)([2
3
232
3
23
AgNH
AgClNHAg
NH
ClNHAgK
此反应的平衡常数式为:
-ClAg (s) AgCl
SSP KKK
• 设溶解了的 Ag+ 全部转化为 [Ag(NH3)2]+ 配离子,则[Ag(NH3)2
+] = [Cl-] = 0.1mol·L-1
• 由平衡常数表示式解出平衡时 NH3 的浓度为:
[NH3]= (mol·L-1)
• 氨水的总浓度 至少应为 :
86.1108.1106.1
1.01.0]][)([107
23
sps KK
ClNHAg
3NHc
3NHc
= 1.86 + 2×0.1 = 2.06 ( mol·L-1 )
SSP KKK
计算说明氨水能否溶解 AgI 沉淀 ?
Influence of redox on coordination equilibrium
Fe3+ + I- Fe2++ 1/2I2
+
6F- Equilibrium shift
[FeF6]3-
[FeCl4]- Fe3+ + 4Cl-
+
Equilibrium I-
Shift
Fe2+ + 1/2I2
根据标准电极电势表得知 > ,当加入 NaF 晶体时, Fe3+ 与 F -结合生成 [FeF6]3 -配离子,使 Fe3+ 离子浓度降低, Fe3+/ Fe2+ 电对的电极电势也随之降低,当Fe3+/ Fe2+ 的电极电势小于 I2/ I -的电极电势时,则氧化还原反应方向发生改变 .
23 / FeFe
II /2
另一方面,配位平衡也可以转化为氧化还原平衡,使配离子离解。如 I -可使[FeCl4] -配离子中 Fe3+ 还原
成 Fe2+ ,从而使 [FeCl4] -
配离子离解 .
电极电势改变 Cu+/ Cu Ag+/Ag Au3+/Au
Φ0/v 0.52 0.799 1.50
[Cu(CN)2]-/Cu [Ag(CN)2]-/Ag [Au(CN)2]+/Au
-0.43 -0.31 -0.58
Influence of on coordination equilibrium
• [Ag(NH3)2]+ +2CN- [Ag(CN)2]- + 2NH3
• [Mn(en)3]2+ + Ni2+ [Ni(en)3]2+ + Mn2+
利用 KS 值可以判断配位平衡转化的方向和程度。
[ 例 9-4] 在 [HgCl4]2- 配离子的溶液中加入 KI 溶液, 能否生成 [HgI4]2- 配离子?
解:溶液中存在下列平衡:
查表得知 [HgI4]2 - KS = 6.8×1029; [HgCl4]2 - KS = 1.17×1015
代入上式得:
K值很大,说明由 [HgCl4]2 -转化为 [HgI4]2 -的反应完全可以实现
[HgI4]2- + 4Cl-[HgCl4]2- + 4I-
24
24
][
][
2424
2424
424
424
][]][[
][]][[
]][[
]][[
HgCl
HgI
K
K
HgIHgCl
HgClHgI
IHgCl
ClHgIK
1415
29
108.51017.1
108.6
K
9-4 Chelates
• polydentate ligands have two or more
donor atoms situated so that they can
simultaneously coordinate to a metal ion.
• chelating agents appear to grasp the
metal between two or more donor atoms.
• claw ∶ NH2-CH2-CH2-H2N∶
polydentate ligand is the ethylenediaminetetraacetate ion:
• This ion, abbreviated EDTA4 -, has six donor atoms. It can wrap around a metal ion using all six of these donor atoms as shown in figure (9-9).
• Figure 9-9: • The CoEDTA- ion
showing how the ethylenediamine-
tetraacetate ion is able to wrap around a metal ion, occupying six positions in the coordination-sphere.
• In general, chelating agents form coordination compounds containing rings, these coordination compounds are called chelates.
• The chelates are more stable than related monodentate ligands, which is called chelating effect.
Ni2+(aq) + 6NH3(aq) =[Ni(NH3)6]2+(aq) K = 4 ×108
Ni2+(aq) + 3en (aq) = [Ni(en)3]2+(aq) K = 2 ×1018
• Although the donor atom is nitrogen (N) in both
instances, [Ni(en)3]2+ has a stability constant
nearly 1010 times larger than [Ni(NH3)6]2+.
• The stability of chelate depends on the
number and size of chelate ring.
• The larger number of chelate ring, the
more stable chelate is.
• When the chelate ring is formed by 5
or 6 members, the chelate is most
stable.
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