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1、Chapter 10 ligand compoundsChapter 10 ligand compoundsNo water CrCl3 and ammonia can form two complexes, the equivalent of CrCl3? Six nh3 and CrCl3? 5 nh3. Can join AgNO3 solution from the first complexes in aqueous solution, almost all of the chloride precipitation for AgCl, but from the second kin
2、d of complexes in aqueous solution, only can precipitate out the equivalent of composition of chlorinity two-thirds of AgCl, when heated both solution without adding NaOH and NH3. The internal boundary and the outside world are calculated from the form of the complexes, and indicate the charge numbe
3、r of the ions, the oxidation number of the central ions and the names of the complexes.Answer: the first one: Cr (NH3) 6 the charge of the Cl3 ion is 3 + :The Cr (NH3) 6 3 +, Cr (), three six ammonia chloride and chromium ().The second: CrCl (NH3) 5 2 +, Cr (), a chlorine chloride? Five ammonia chro
4、mium ().Name the following complexes, and indicate the central and oxidation Numbers, ligand and number.(1) Co (NH3) 6 Cl2 (2), K2 PtCl6 (3) Na2 SiF6(4) CoCl (NH3) 5 Cl2 (5) Co en 3 Cl3 (6) CoCl (NO2) +Answer: namingCenter ion oxidation numberligandCoordination number(1) six ammonia and cobalt chlor
5、ide ()+ 2NH36(2) six chlorine platinum () acid potassium+ 4Cl -6(3) six fluorine and silicon () acid sodium+ 4F -6(4) chloride? Five ammonia or cobalt ()+ 3Cl -, NH36(5) three chloride (ethylenediamine) and cobalt ()+ 3En6(6) a chlorine? The nitro? Four ammonia or cobalt () complex ion+ 3Cl -, NO2 -
6、, NH36Write the formula of the following coordinates(1) disulfide generation or silver () sodium sulfate (2) three nitro ammonia or cobalt ()(3) a hydrated cobalt dichloride three ammonia chloride () (4) dichloride dihydroxy diamine and platinum ()(5) a chloride ammonium sulfate 2 (ethylenediamine)
7、and chromium ()(6) dichloride a root oxalate (ethylenediamine) and iron () ionsAnswer: (1) Na3 Ag (S2O3) 2 (2) Co (NO2) 3 (NH3) 3 (3) CoCl2 (NH3) 3 (HO2) Cl(4) PtCl2 (NH3) 2 (OH) 2 (5) CrCl (NH3) en 2 SO4 (6) FeCl2 (C2O4) -According to the theory of valence bond, the bonding and spatial configuratio
8、ns of the following ions are indicated(1) Fe (CN) 6 3 - (2) FeF6 3 - (3) c 2 + (4) 2 -Answer: (1) d2sp3 hybridized orbital form, octahedron. (2) sp3d2 hybridized orbital bonds, octahedral.(3) sp3d2 hybridized orbital bonds, octahedron. (4) the dsp2 hybridized orbital, the plane square.According to t
9、he effective magnetic moment measured by the experiment, the test is determined that the following complexes are the internal or outer rails, which are the reason, and are expressed in their electronic layer structure(1) Mn (SCN) 6 4 - mu = 6.1 B.M. (2) Mn (CN) 4 - mu = 1.8 B.M.(3) Co (NO2) 6 3 - mu
10、 = 0 B.M. (4).(5) K3 FeF6 mu = 5.9 B.M.A: (1) there are five forms of single electron and external rail.Mn (SCN) 4-3d5 4s 4dUp the arrow up the arrow? ? ? ? ? ? Sp3d2 mixed into a key(2) there is a single electron, in-orbit type.Mn (CN) 4-3d 4s 4p? ? Write? ? ? ? ? ? D2sp3 mixed into a key(3) no sin
11、gle electron, inner track type.Co (NO2) 6 3-3d 4s 4p? ? ? ? ? ? ? ? ? D2sp3 mixed into a key(4) there are three single electrons, and the outer rail.Co (SCN) 4 2-3d 4p 4d? ? Up the arrow? ? ? ? ? ? Sp3d2 mixed into a key(5) there are five single-electron, external, and orbital complexes.FeF6 3d5 4s
12、4p 4dUp the arrow up the arrow? ? ? ? ? ? Sp3d2 mixed into a key(6) there is a single electron, the inner track type.K3 Fe (CN) 6 3d 4s 4p? ? Write? ? ? ? ? ? D2sp3 mixed into a key6. Some known platinum complexes, such as cis - PtCl4 (NH3) 2, cis - PtCl2 (NH3) 2 and cis - PtCl2 (en) can be used as
13、active anticancer agent (no anticancer activities all trans-isomer), the experimental measured they are inverse magnetic materials. So the theory of valence bond is to draw the hybridized orbitals of these complexes, are they internal or external? What type of hybrid orbital bonding?Answer: platinum
14、 Pt 5d96s1, Pt4 + 5d6Cis - PtCl4 (NH3) 2 5d 6s 6p? ? ? ? ? ? ? ? ? D2sp3 mixed into a keyNo single electron, inverse magnetic, inner rail.Cis - PtCl2 (NH3) 2 and cis - PtCl2 (en)Pt2 + 5d8 6s 6p? ? ? ? ? ? ? ? The dsp2 is the hybridised bond,No single electron, inverse magnetic, inner rail.7. PtCl4 2
15、 - known for planar square structure, HgI4 2 - to the tetrahedron structure, draw their electronic distribution and points out they use what kind of hybrid orbital bonding?Answer: PtCl4 2 - dsp25d8 6s 6p? ? ? ? ? ? ? ? Inner rail type complexesHgI4 2 - sp35d10 6s 6p? ? ? ? ? ? ? ? ? Outer railWhat a
16、re the basic points of crystal field theory? What is its advantage over the valence theory?A: the basic points of the crystal field theory:(1) crystal field theory, in the complexes, which center is positively charged metal ions, ion ligand is often anion or polar molecules, the interactions between
17、 them can be regarded as is, the interaction between anion ion crystal, center due to electrostatic attraction between ion and ligand and release energy, reduce the system energy.(2) the center of transition metal ions around five degenerate d orbitals are non spherical symmetric ligand negative ele
18、ctric field effect of ligand negative d orbit electrons repel each other, not only makes the orbital electron energy is generally higher, and different d electronic affected by different orbit, higher the orbital energy values are different, resulting in d orbital energy level splitting.(3) the d el
19、ectron is rearranged because d orbitals are divided. D electrons are never divided before the d orbitals go into the dissected d orbital, causing the total energy of the system to decrease, and it gives extra stability to the complexes.Crystal field theory to overcome the shortcomings of valence bon
20、d theory, it not only consider the center between ion and ligand electrostatic effect, but also consider the covalent properties between them, its better explains the magnetism of the complexes, can better explain the visible spectrum and ultraviolet spectrum, to a certain extent, can also be quanti
21、tatively explain the stability of the complexes.Use crystal field theory to explain why the octahedral ion CoF6 3 - is high spin? And NH3 6 3 + is low spin? And judge their stability?A: octahedral crystal field theory: CoF6 3 - (high spin)Arrow arrow? Arrow up arrow (f-weak field, high spin)? Up arr
22、ow d epsilonCFSE = (-4) x 4 + (+ 6) x 2 = -4dqOctahedral crystal field theory: Co (NH3) 6 3 + (low spin)Dr? Arrow up arrow (NH3, low spin)? ? ? D epsilon.CFSE = (-4) x 6 = -24dqFrom the calculation of CFSE, Co (NH3) 6 3 + ratio CoF6 3 - stability.The experiment measured (NH3) 6) 3 + ion is reverse m
23、agnetic.(1) what is its space configuration? Based on the valence bond theory Co3 + ion what hybridization orbital and the ligand NH3 molecule are used to form a ligand?(2) according to the crystal field theory to map out the complex ion two possible d electron configuration, the discretion of the s
24、hows that they spin and magnetic conditions, to complex ion which is correct?(3) when Co (NH3) 6 that is 3 + to Co (NH3) 6 2 +, magnetic moment is about 4 5 Bohr magneton, draw its possible d electron configurations, explain their magnetic condition.A: (1) Co3 + 3d6 4s 4pNH3 6 3 +? ? ? ? ? ? ? ? ? D
25、2sp3 mixed into a keyOctahedron, inner track.(2) Co3 +Arrow arrowD6? Arrow up arrow (weak field, high spin)? Up arrow d epsilonDr? Arrow up arrow (strong field, low spin)? ? ? D epsilon.Because NH3 is a strong ligand, its a low spin coordinate, and mu = 0 is inverse magnetic.(3) Co2 plus 3d7 4s 4dNH
26、3 6 2 +? ? Up the arrow? ? ? ? ? ?Sp3d2 is the bondWith three unpaired electrons, it has a paramagnetic octahedral ion.11. Have sensing Co () in aqueous solution to form a with three unpaired electrons, match octahedron with paramagnetic ions. Which of the following statements is consistent with the
27、se conclusions and explains why?(1) 6 Co (H2O) 2 + the crystal field splitting can () is greater than the electron pairs can be (P)(2) when the d orbital is divided, the electron population is (3) the d orbital is divided, and the electron is filled inAnswer: (2) agree with the above conclusion.Co2
28、+ d7 arrow up Dr? ? Write write write? ? Write d epsilon.Only in this way can there be three unpaired electrons, which have a paramagnetic octahedral ion.Cr3 +, Cr2 +, Mn2 +, Fe2 + ions in the octahedron and octahedral weak fields each have a number of unpaired electrons, plot notes and the number o
29、f electrons in orbit.A: Cr3 + DrD3 write write write 0 P strong fieldUp arrowAll three unpaired electrons,There is only one arrangement for d3.Cr2 + arrow DrD4 write write write write weak field, the four into a single electronUp arrowDrUp the arrow up the top of the arrow, two electrons? Up arrow d
30、 epsilonMn2 + arrow up DrThe d5 arrow up the arrow up the arrow up to fiveUp arrowDrUp the arrow up the arrow up a single electron? ? Write d epsilon.Fe2 + arrow arrowD6? The arrow is up to the top of the arrow and up to four? Up arrow d epsilonDr.? The arrow is up and up to nowhere? ? ? D epsilon.I
31、t is known that AlF6 3 - the level-by-grade stability constant is 6.13, 5.02, 3.85, 2.74, 1.63, and 0.47, and try to find its K stable and K unstable.A:To predict the stability of the two groups formed by the following groups, and give a brief explanation:(1) Al3 + with F - or Cl - coordinates (2) P
32、d2 + with RSH or ROH(3) Cu2 + and NH2CH2COOH or CH3COOHAnswer: (1) both are hard and hard formed with ions, and the f-ratio Cl is smaller, so AlF6 3 - stability is greater than AlCl6 3. (2) Pd (RSH) 4 2 + stability is greater than Pd 4 2 + because of the electronegative sulfur. (3) the combination o
33、f Cu2 + and NH2CH2COOH is more stable than the combination of Cu2 + and CH3COOH. Because the former forms a chelate.Indicate which of the following compounds may serve as an effective chelating agent.(1) H2O (2) hydrogen peroxide(3) H2N - CH2 - CH2 - CH2 - NH2 (4) (CH3) 2N - NH2A: only (3) can be us
34、ed as an effective chelating agent, and only ethylenediamine can form a ring chelate, and the other cannot form a five-membered ring.16.0.10 mol? L - 1 AgNO3 solution 50 ml, join the relative density of 0.932 containing NH3 30 ml, 18.24% of the ammonia water to 100 ml, seek the solution Ag +, Ag (NH
35、3) 2 +, and NH3?We know that Ag (NH3) 2 + K = 1.7 x 107.A: Ag + + 2NH3? Ag (NH3) 2 +So we start with (moles? L-1) = 0.05 = 3.0Equilibrium (mol? L - 1) x 3.0-0.05 x 2 = 2.9 0.05 - x is approximately 0.05X = Ag + = 3.5 x 10-10.Ag (NH3) 2 + = 0.05 (mol? L - 1) NH3 = 2.9 (mol? L - 1).The calculation contains 0.10 moles. CuSO4 and 1.8 L - 1 mol? L-1 ammonia solution, Cu2 + ion concentration?(Cu (NH3) 4 2 + K unstable = 5 x 10-16)Answer: Cu (NH3) 4 2 +? Cu2 plus 4 nh3So were going to start (mol? L minus 1) 0.101.8 minus 4 x 0.10 is 1.4Equilibrium (mol? L minus 1) 0.1 minus x is approximately 0.