Study Problem Solutions - Experiment #6 研究问题解答 - 实验6

Q1 问题1

Coordination complexes or complex ions 配位化合物或配合物离子

Q2 问题2

Six 六

Q3 问题3

Six (oxalate is a bidentate ligand) 六（草酸根是双齿配体）

Q4 问题4

+3 +3

Q5 问题5

+3 +3

Q6 问题6

6: Octahedral; 4: Tetrahedral or square planar 6：八面体；4：四面体或平面正方形

Q7 问题7

Are oriented directly towards the incoming ligand electron pairs 直接朝向进入的配体电子对

Q8 问题8

[Zn(en)3]2+\left[\mathrm{Zn}(\mathrm{en})_{3}\right]^{2+} will be colorless because zinc has a complete d subshell and is not a transition metal. [Zn(en)3]2+\left[\mathrm{Zn}(\mathrm{en})_{3}\right]^{2+}将是无色的，因为锌具有完整的d亚层，并且不是过渡金属。

Q9 问题9

Because transition metals form bonds to species that donate lone pairs of electrons, transition metals are seen as Lewis acids (electron-pair acceptors). The Lewis bases in coordination compounds are the ligands, which have an lone pair of electrons to donate. The coordinate covalent bond between the ligand and the transition metal just indicates that both electrons in the bond originally came from one of the atoms in the bond. Here, the electrons in the bond come from the ligand. 因为过渡金属与提供孤对电子的物种形成键，所以过渡金属被视为路易斯酸（电子对受体）。配位化合物中的路易斯碱是配体，它们具有孤对电子可以捐赠。配体和过渡金属之间的配位共价键仅表示键中的两个电子最初来自键中的一个原子。在这里，键中的电子来自配体。

Q10 问题10

Increasing wavelength means lower energy and therefore a smaller crystal field splitting energy. From shortest to longest absorbed wavelength (based on ligand strength): 波长增加意味着能量降低，因此晶体场分裂能更小。从最短到最长的吸收波长（基于配体强度）：

[Co(CN)6]3−<[Co(en)3]3+<[Co(OH2)6]3+<[Co(l)6]3−\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}<\left[\mathrm{Co}(\mathrm{l})_{6}\right]^{3-}

Q11 问题11

The purple complex ion (Tube 1) absorbs yellow light ( λ≈570 nm\lambda \approx 570 \mathrm{~nm} ) and is [Cr(OH2)6]3+\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}. The yellow complex ion (Tube 2) absorbs purple light ( λ≈415 nm\lambda \approx 415 \mathrm{~nm} ) and is [Cr(NH3)6]3+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}. The green complex ion (Tube 3) absorbs red light ( λ≈650\lambda \approx 650 nm)\mathrm{nm}) and is [Cr(OH2)4Cl2]+\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{4} \mathrm{Cl}_{2}\right]^{+}. 紫色配合物离子（试管1）吸收黄光（λ≈570 nm\lambda \approx 570 \mathrm{~nm}），是[Cr(OH2)6]3+\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}。黄色配合物离子（试管2）吸收紫光（λ≈415 nm\lambda \approx 415 \mathrm{~nm}），是[Cr(NH3)6]3+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}。绿色配合物离子（试管3）吸收红光（λ≈650\lambda \approx 650 nm\mathrm{nm}），是[Cr(OH2)4Cl2]+\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{4} \mathrm{Cl}_{2}\right]^{+}。

In Beran lab manual: 在Beran实验手册中：

Prelaboratory Assignment from Experiment #36: all Transition Metal Complexes 实验36的预实验作业：所有过渡金属配合物

Q1 问题1

a) The ligands in the coordination compound, [Co(NH3)4(OH2)Cl]SO4\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl}\right] \mathrm{SO}_{4}, are Cl−\mathrm{Cl}^{-}, NH3\mathrm{NH}_{3}, and H2O\mathrm{H}_{2} \mathrm{O}. a) 配位化合物[Co(NH3)4(OH2)Cl]SO4\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl}\right] \mathrm{SO}_{4}中的配体是Cl−\mathrm{Cl}^{-}、NH3\mathrm{NH}_{3}和H2O\mathrm{H}_{2} \mathrm{O}。 b) The complex of the coordination compound is [Co(NH3)4(OH2)Cl]2+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl}\right]^{2+}. b) 配位化合物的配合物是[Co(NH3)4(OH2)Cl]2+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl}\right]^{2+}。 c) The coordination sphere is " Co(NH3)4(OH2)Cl\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl} ". c) 配位层是" Co(NH3)4(OH2)Cl\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right) \mathrm{Cl} "。 d) The coordination number of cobalt (III) in this complex is six (each ligand is monodentate). d) 该配合物中钴(III)的配位数是六（每个配体都是单齿）。

Q2 问题2

a) [PtCl4]2−\left[\mathrm{PtCl}_{4}\right]^{2-} a) [PtCl4]2−\left[\mathrm{PtCl}_{4}\right]^{2-} b) [Cr(en)3]3+\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+} b) [Cr(en)3]3+\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+} c) [Zn(H2Y)];H2Y\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{Y}\right)\right] ; \mathrm{H}_{2} \mathrm{Y} is a tetradentate ligand c) [Zn(H2Y)]\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{Y}\right)\right]；H2Y\mathrm{H}_{2} \mathrm{Y}是四齿配体

Q3 问题3

Part A: concentrated HCl A部分：浓HCl

Parts B, C, D: concentrated NH3\mathrm{NH}_{3}, ethylenediamine B、C、D部分：浓NH3\mathrm{NH}_{3}，乙二胺

Q4 问题4

Ethylenediamine and the oxalate ion are both bidentate ligands. 乙二胺和草酸根离子都是双齿配体。

Q5 问题5

NaOH(aq)\mathrm{NaOH}(\mathrm{aq}) is used to test the stability of the complex ions. The OH−\mathrm{OH}^{-}competes with the ligand for bonding to the metal ion. NaOH(aq)\mathrm{NaOH}(\mathrm{aq})用于测试配合物离子的稳定性。OH−\mathrm{OH}^{-}与配体竞争与金属离子的键合。

Q6 问题6

[Co(Cl)6]3−(aq)+6CN−(aq)→[Co(CN)6]3−(aq)+6Cl−(aq)\quad\left[\mathrm{Co}(\mathrm{Cl})_{6}\right]^{3-}(\mathrm{aq})+6 \mathrm{CN}^{-}(\mathrm{aq}) \rightarrow\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}(\mathrm{aq})+6 \mathrm{Cl}^{-}(\mathrm{aq}) [Co(NH3)6]3+(aq)+Cl−(aq)→\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow No reaction [Co(CN)6]3−(aq)+NH3(aq)→\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightarrow No reaction

Q7 问题7

5.9 g[Cu(NH3)4]SO4⋅H2O5.9 \mathrm{~g}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} 5.9 g[Cu(NH3)4]SO4⋅H2O5.9 \mathrm{~g}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}

Q8 问题8

a) Cr(OH)3( s)\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{~s}) a) Cr(OH)3( s)\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{~s}) b) [Cr(OH)4]−\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-} b) [Cr(OH)4]−\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}

Q9 问题9

Monodentate ligands are bound through only one atom. With the addition of hydroxide ion to a complex with monodentate ligands, it is likely that the hydroxide ion will displace the monodentate ligands from the complex. However, in the case of polydentate ligands in a complex, given the multiple ligating atoms to the metal center, it is less likely for the hydroxide ion to displace polydentate ligands from the complex. 单齿配体仅通过一个原子键合。向含有单齿配体的配合物中加入氢氧根离子时，氢氧根离子很可能会取代配合物中的单齿配体。然而，在含有多齿配体的配合物中，由于有多个配位原子与金属中心键合，氢氧根离子不太可能取代配合物中的多齿配体。

Laboratory Questions from Experiment #36: 1-5, 7

Q1 问题1

Water. Only in a highly concentrated Cl−\mathrm{Cl}^{-}solution does the chloro complex form. But upon addition of more water, the aqua complex re-forms. Water displaces the chloride ion from the coordination sphere and is therefore a stronger ligand than the chloride ion. 水。只有在高浓度的Cl−\mathrm{Cl}^{-}溶液中才会形成氯配合物。但加入更多水后，水合配合物会重新形成。水将氯离子从配位层中取代出来，因此是比氯离子更强的配体。

Q2 问题2

Ammonia. With the addition of ammonia to the solution containing Cu2+\mathrm{Cu}^{2+}, the ammonia complex immediately forms, displacing water from the coordination sphere to form the deep-blue [Cu(NH3)4(OH2)2]2+\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+} complex. Ammonia is the stronger ligand. 氨。向含有Cu2+\mathrm{Cu}^{2+}的溶液中加入氨后，氨配合物立即形成，将水从配位层中取代出来，形成深蓝色的[Cu(NH3)4(OH2)2]2+\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+}配合物。氨是更强的配体。

Q3 问题3

Ethylenediamine. Given that the [Cu(NH3)4(OH2)2]2+\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+} complex is a royal blue color, whereas the [Cu(en)2(OH2)2]2+\left[\mathrm{Cu}(\mathrm{en})_{2}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+} complex is dark blue/purple, this indicates that the ethylenediamine complex is absorbing higher energy light (yellow as opposed to orange, respectively). Therefore, ethylenediamine is the stronger ligand. 乙二胺。由于[Cu(NH3)4(OH2)2]2+\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+}配合物是宝蓝色，而[Cu(en)2(OH2)2]2+\left[\mathrm{Cu}(\mathrm{en})_{2}\left(\mathrm{OH}_{2}\right)_{2}\right]^{2+}配合物是深蓝/紫色，这表明乙二胺配合物吸收了更高能量的光（分别是黄色而不是橙色）。因此，乙二胺是更强的配体。

Q4 问题4

Ethylenediamine appears to be the strongest ligand. Its complexes do not form precipitates upon the addition of hydroxide ion. The chloride ligand is the weakest because it functions only as a ligand in a very concentrated HCl solution. 乙二胺似乎是最强的配体。它的配合物在加入氢氧根离子后不会形成沉淀。氯配体是最弱的，因为它只在非常浓的HCl溶液中作为配体起作用。

Q5 问题5

From Co2+\mathrm{Co}^{2+} to Ni2+\mathrm{Ni}^{2+} to Cu2+\mathrm{Cu}^{2+}, the experimental data appears to indicate a slight trend toward increased stability of the complexes. 从Co2+\mathrm{Co}^{2+}到Ni2+\mathrm{Ni}^{2+}再到Cu2+\mathrm{Cu}^{2+}，实验数据似乎表明配合物的稳定性有轻微的增加趋势。

Q7 问题7

Even though we did not complete Part E, this is an important question pertaining to the addition of aqueous ammonia to any transition metal. Because ammonia is a Bronsted-Lowry base in water, thereby producing OH−(aq)\mathrm{OH}^{-}(\mathrm{aq}) in solution, the initial precipitate that forms is Cu(OH)2( s)\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{~s}). 尽管我们没有完成E部分，但这是一个重要的问题，涉及向任何过渡金属中加入氨水。由于氨在水中是布朗斯特-劳里碱，因此在溶液中产生OH−(aq)\mathrm{OH}^{-}(\mathrm{aq})，最初形成的沉淀是Cu(OH)2( s)\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{~s})。