d10 is supposed to be more stable than d9. So Cu(I) with d10 configuration should be more stable but on the contrary, Cu(II) with d9 is more stable. Please explain.
This a quick answer, and I will have to come back and revise this later today. With ligands that are nitrogen donors etc Cu(I) complexes tend to have Td symmetry and Cu(II) complexes form 5 coordinate square planar complexes. The d10 Cu(I) complexes have electrons which will interact repulsively with Lewis bases, and depending on the ligands ie pi acceptors that can pull electron density away from the metal center impacts the ligand exchange kinetics. One has to deal with electron-electron repulsion in the metal centered orbitals. Cu(II) complexes with a d9 configuration have reduced electron repulsion as an electron has been removed from an anti-bonding orbital, and a fifth ligand increases the LFSE in the Cu(II) complex. I realize that the explanation is not complete, and I will return to this response later today.
The primitive answer:
It depends on the ligand. CuCl2 is more stable than CuCl, but with iodide...CuI is more stable than CuI2. The quick answer is originated from Pearsons's hard-soft acid-base theory: Cu+ is a soft "acid", Cu2+ is a hard "acid", Cl- is hard and I- is a soft "base". Soft acids love the soft bases...and hard acids loves the hard bases.
The stability of 3d9 is mainly related to the ligand field stabilization, since in the typical square-planar geometry the dx2-y2 orbital has low energy. The 3d10 configuration does not have such a stabilization
In aqueous solution (that we suppose you are referring to), Cu+ spontaneously disproportionates into Cu2+ and Cu0 as is expressed by the favorable (positive) redox potential (therefore negative free energy) of the corresponding equilibrium. However, this order does not hold for all solvents as the affinity of the solvating ion for Cu+ vs Cu2+ will condition it (for example, Cu+ will supersede Cu2+ in acetonitrile). In the same way, in a given solvent into which a ligand is added, the predominance of either Cu+ or Cu2+ will depend on the stability of the complexes that each of these will form with this ligand. In water for example, NH3 (monodentate ligand) will favour Cu+ whereas ethylenediamine (bidentate) will favour Cu2+, respective stability constants modifying the redox potential of the Cu+ / Cu2+ couple accordingly.
Mr. Swu,
Answering to your question we should take into consideration the following two factors:
1. Intra-ionic factor
Please find as attachment NBO analysis of CuI and CuII ions in continuum, where there are shown the energies of d-orbitals. In CuI - ion there is found a full degeneracy (E = -0.27002) to all d-orbitals (D = 0). By contrast, CuII-ion where a removing of the degeneracy (JT effect) to all orbitals with D values within [0.1815-0.0428] is obtained.
In other words, there is a removing of the d-orbital degeneracy even in structureless continuum in the case of CuII-ion (d9), contrary, to CuI - ion (d10).
2. Environmental factors
Please find as attachment the thermodynamics data, too. In polar structureless continuum where is taken only macroscopic e-value of the medium, there is obtained D(DG) = -292.11 kcal.mol-1, showing CuII - ion as more stable ion. The main contribution to this stability is due to polar solute-continuum and unpolar solute-continuum interactions. Please bear in mind that DG involvs, thermal motion, cavitation and repulsion-dispersion terms. The cavitation and the repulsion-dispersion ones are equat to both ions (Ecav = 3.35 kcal.mol-1, Erep = Edis = 0 kcal.mol-1).
Solute polarization energies (ESP) of both ions have small absolute quantities, but ESP(CuI) > ESP (CuII). Taking into consideration that those quantities, as values, are difference to gas-phase unpolarized total energies, the value 0.05 kcal.mol-1 in fact indicates a very close to the gas-phase data of the CuII-ion in polar continuum on the one hand. On the other hand the solute polarization of the electron density effects the electronic dispersion properties like spin population, dipole moment, charge distribution and the fourth. As energy term it augments the electrostatic part of the solvation energy, so that the higher polarization energy of CuI additionally increases in the electrostatic quantity, contrary to he close to zero (0.05 kcal.mol-1) value for the system CuII/continuum.
Please take into consideration as well that the conclusion stated about the stability of CuII is valid both using computations in polar and non-polar solvents (attachment). There are used the e data for CH3OH and C6H12, respectively. In spite of the continuum's polarity (polar or non-polar one), CuII is more stable system than CuI system. Given that we cannot assign the higher stability of CuII - ion in polar medium only with its 2+ charge, because of this ion is more stable even in low polar medium like C6H12. Furthermore we cannot assign the higher stability of CuII in CH3OH continuum with any kind of specific solute - solvent interactions yielding of any kind complexes of CuII with CH3OH, because of first of all they are not taken into consideration in PCM continuum approach as well as they absent in the CuII/continuum-C6H12 system. Or we have only solute - to - continuum non - specific interactions causing the higher stability of CuII - ion.
The difference is only that the solute polarization energy in non-polar continuum is 0.0 kcal.mol-1 for both the systems computed (attachment).
NB! Detail description about the terms, used, in point 2 and the PCM approaches you may find in [Ref.1].
[Ref.1] J. Tomasi, B. Mennucci, R. Cammi, Quantum Mechanical Continuum Solvation Models, Chem. Rev. 2005, 105, 2999-3093
Dear Researcher
Good answers have been given already. Some important correlations can be made to highlight the beauty of transition metal chemistry.
The generalization that copper (I) as a d10 system with completely filled orbitals should be more stable than copper(II) with d9 configuration which do not have completely filled orbitals. However this is not always so, this brings first correlation that stability is not absolute but relative, as copper(I) in aqueous solution disproportionate to copper(II) and metallic copper(0) meaning when we consider aqueous solution copper(I) is unusual oxidation state and unstable relative to copper(II) which is stable and usual oxidation state in aqueous solution. While as in a non-aqueous solvent like Acetonitrile copper(I) is way more stable to copper(II).
The stability of an oxidation state in a coordination compound depends on its hard soft nature. Copper(I) is a soft metal ion and hence prefers soft donor sites like sulfur and Iodide and not so much preference for borderline or harder donor sites. e.g the most common complexes of copper(I) are with sulfur donor ligands like thiourea. The soft -soft interaction of copper(I) and thiourea is so dominating that when thiourea is added to the solution of copper(II) it oxidizes thiourea to reduce copper(II) to copper(I) (see attached ic PDF and journal of Analytical chemistry). Copper(II) is borderline hence forms more stable complexes with borderline donor like nitrogen e.g the famous dark blue [Cu(NH3)4]2+ complex.
The stability of an oxidation state also depends on the preferred geometry. Copper(I) being d10 system can SP3 hybridize and hence prefer tetrahedral geometry while copper(II) being d9 system prefers distorted octahedral geometry on account of Jahn Teller Stabilization energy. Hence Copper(I) in tetrahedral systems is more stable than copper(II) while as copper(II) in distorted octahedral systems shall be way more stable than copper(I). This can illustrated from the binding constants of copper(II) and copper(I) with neocuproine ligand (see attached pdf 2 which highlights how complexation changes the stability of copper(II)/copper(I) and influences the redox potential of Cu2+/Cu+ redox couple)
The stability of an oxidation state in a given complex depends on the type of bonding. Synergistic bonding (ligand donation and back acceptance) brings in electrical neutrality in complex which brings additional stability in the complex. Copper(I) on account of its lower charge and completely filled d orbitals favors a good degree of back acceptance to ligand compared to copper(II) hence copper(I) forms more stable complexes with pi acceptor type ligands while as the stability of copper(II) complexes with pi acceptor ligands is less as compared to copper (I) due to low degree of back donation from copper(II).
copper(II) is a typical case of Jahn Teller effect and hence has a stability in distorted octahedral geometry where it enjoys Jahn Teller Stabilization e.g in complexation of copper(II) with ethylenediamine(en) in aqueous solvent only two en ligands get attached to copper while as the binding of third en is restricted as it will make the copper to lose the distorted octahedral configuration and hence Jahn Teller stabilization.(see Inorganic chemistry by James Huheey 4th edn page:454-455) No such distortion/ stabilization is seen in case of copper(I)
In summary we cannot generalize whether copper(II) is more stable to copper(I) or copper(I) is more stable to copper(II). It can only be decided as per the system under consideration. Copper(II) and copper (I) have same atomic number same nucleus but just one difference of electron which makes the huge difference in the stability of copper(II) and copper(I) compounds make one to appreciate the diverse and beautiful chemistry of transition metal systems which has little or no parallel in carbon or main group chemistry.
Mr.Rizvi,
Towards your posting I have remarkds in the following context:
1. The reaction of disproportionation in CuI is a result of the higher stability of CuII towards CuI and formation of colloidal Cu0. Because of the thermodynamics of Cu0 both in CH3OH and C6H12 environment, like computations of CuI and CuII, shown in my previous posting indicate "instability" (Attachment, page 1). Both DGs are positive and a full degeneracy of d-orbitals along with same 4s energies in both continuum models have occurred.
2. Towards your comments about the stability of "CuI" in acetonitrile I have, particularly, remarks, due to the following reasons:
2.1. There is not stability of a ion (CuI or CuII) as isolated particle, because of those ions form complexes with solvents. Some of those complexes are stable (with CH3CN, too) and can be observed in solution, gas-phase and solid-phase too, even in the presence of competitive ligands, even chelating agents. Please pay attention to our mass spectrometric data on AgI- and ZnII-complexes with organics and organometallics showing stabilization of [Ag(CH3CN)x]y+ complexes in gas-phase [Refs. 1 and 2] or [Zn(H2O]6]2+ containing complex salt in crystalline state [Ref.3] (Attachment, page 2). Please pay attention to ref. [4] where other authors have observed same phenomena, as well. Even chelating effect can be neglected, towards monodentate coordination of solvent molecules to the inner coordination sphere of the metal ion.
Those complex species, with different solvent stechiometry, have different individual thermodynamics and behaviour depending on environmental factors. A theoretical modelling of such as systems can be done, using mixed solvation approach, where given complex model is studied in continuum, too (next data).
For example, data about model complexes [CuI(CH3CN)3]+ and [CuII(CH3CN)3]2+ show in the first case DG = -35.54 kcal.mol-1, while in the second one, a value DG = -130.54 kcal.mol-1, respectively. In other words, CuII - acetonitrile complex [CuII(CH3CN)3]2+ is about 4 times more stable in acetonitrile continuum than [CuI(CH3CN)3]+ complex CuI - containing analogous in the same medium.
2.2. Particularly CH3CN as solvent is among less suitable medium to compare stability of ions, due to reasons pointed out in 2.1; as well as, this solvent has ability to form organometallics, bonding metal ions to CH3-CN and CH3-CN atoms. Particularly our data on AgI-complexes have often shown formation of colloidal Ag0 in CH3CN [Ref. 2,3]. You can pay attention to works [5 and 6], too, where other authors have reported crystallographic data about covalently bonded CH3CN to various metal ions, via manners mentioned above.
To have however in a polar solvent like CH3CN (e = 37.5) a more stable 1+ charged species, rather than +2 charged ones, is less probable. So that if there are such observations, most probably in CH3CN are stimulated redox processes with participation of CH3CN and may be O2 and/or H2O yielding to intermediates, having higher stability where there are CuI-organometallics; or there is a formation of neutral CuI-acetonitrile mixed-ligand complexes. An example of complex [CuI-CH2-CH=NH(O)]2- in CH3CN environment shows DG = -261.20. The corresponding analogue with CuII - ion, [CuII-CH2-CH=NH(O)]-, has DG only -82.90 kcal.mol-1. Or those are examples where complex of CuI-ion has higher stability, than CuII - derivative.
We however cannot talk about stability of CuI-ion, we have determined that complex ensemble [CuI-CH2-CH=NH(O)]2-, involving CuI-ion as a more stable one than the same ensemble [CuII-CH2-CH=NH(O)]-, involving CuII-ion.
3. In an isolates state CuII is more stable than CuI as shown in both polar and non-polar environment. You have pointed out that the oxidation state of the ions depens on the geometry of the chromophore, but in continuum you have not geometry of the chromophore. The PCM model of the solvent environment is based on the Onsager's understanding about the structureless environment. Please pay attention to the theory in [Ref. 1] to my past posting. The stability of CuII is due to intra-ionic factors, observed even in continuum, which are not typical for CuI or Cu0 as shown in the attachments here and to my previous posting.
Or, there is no dough that CuII is more stable than CuI. It is more stable! But in the presence of suitable ligands and set of environmental factors there is possible to change the stability of a complex ensemble, including CuII- (or CuI) ion.
The computations carried out (both attachments), are by Gaussian (http://www.gaussian.com/). The PCM model is involved in, so that the data are fully reproducible. You have not effect of the ligands in the classical concept of the ligand-field-theory and the effect of the geometry, you have, in fact, ion-to-structureless continuum interaction.
[Ref. 1] B.Ivanova, M. Spiteller, MOLECULAR STRUCTURE AND PROPERTIES OF SILVER(I/II/III)-CONTAINING ORGANOMETALLICS, GRIN Verlag, Muenchen, 2015, pp. 1 - 129; ISBN:978-3-6680-423-9-1
[Ref. 2] Ivanova, B.; Spiteller, M. (2011), AgI and ZnII complexes with possible application as NLO materials - Crystal structures and properties, Polyhedron, 30, 241-245.
[Ref. 3] Lamshoeft, M.; Storp, J.; Ivanova, B.; Spiteller, M. (2011) Gas-phase CT-stabilized Ag(I) and Zn(II) metal-organic complexes-experimental vs. theoretical study, Polyhedron, 30, 2564.
[Ref. 4] A. Li, Q. Luo, S. Park, R. Cooks, Synthesis and Catalytic Reactions of Nanoparticles formed by Electrospray Ionization of Coinage Metals, Angew. Chem. Int. Ed. 53 (2014) 3147-3150
[Ref. 5] A. Oertel, V. Ritleng, M. Chetcuti, L. Veiros, C-H Activation of Acetonitrile at Nickel: Ligand Flip and Conversion of N-Bound Acetonitrile into a C-Bound Cyanomethyl Ligand J. Am. Chem. Soc. 2010, 132, 13588-13589
[Ref. 6] E. Suzuki, T. Komuro, M. Okazaki, H. Tobita, Three- and Five-Membered W/C/N Metallacycles Formed by Incorporation of Acetonitrile Molecules into Silyltungsten Intermediates, Organometallics 2007, 26, 4379-4382
Madam Ivanova
Thanks for the comments, I have limited access to Computational Chemistry Literature and your information has added to my knowledge. I think Dr. Swu wanted a broader answer as could be seen from his response to Prof.Thompson and Prof Szabo,answer to which he has responded and i quote " thank you so much for your enlightening answers."
In my post also i have started with the line good answers have been givenalready, as many answers had come before my post. However as an Inorganic chemist i thought to elaborate on the answers which have been already given and more importantly to highlight the beauty of the transition metal chemistry wherein the two copper oxidation states are just one electron different , yet they have such a large difference in the stability of their compounds.I would repeat the precise answer to this question ," We cannot generalize whether copper(II) is more stable to copper(I) or copper(I) is more stable to copper(II). It can only be decided as per the system under consideration."
Mr. Rizvi,
I have tried to provide evidences that there is unambiguous answer which of both CuII or CuI is more stable ion. The answer to this question is that this is CuII, not CuI.
If you have payed attention to my postings (attachments), in CuI you have full degeneracy of 3d AOs. According the JT theorem such as system is not stable and it aims to achieve stability removing the degeneracy. The theorem focuses on the orbital degeneracy. Please pay attention to the original paper of Jahn and Teller: http://rspa.royalsocietypublishing.org/content/161/905/220
In Cu II-you have removed degeneracy in a structureless continuum. So that those results confirm, in fact, JT theorem and answer to the question why CuII is more stable than CuI - due to intra ionic factors, causing for a removing of the degeneracy of 3d orbitals in Cu II.
In addition to the above nice answers, I would like to add the following comments:
1- Cu(I) disproportionates to Cu(II) and Cu(0) in aqueous solution because the E0 of the reaction is 0.35. The E0 of this reaction is even higher in the presence of chelating ligands such as ethylenediamine.
2- The Cu(II) oxidation state is more stable than Cu(I) for complexes with nitrogen or oxygen electron donating ligands because of the CFSE. The d9 of Cu(II) configuration has more CFSE than d10 of Cu(I) which is zero.
3- As mentioned in the above answers, Cu(I) is stable in some organic solvents and with ions such as iodide which forms insoluble CuI.
I think that there are two reasons for this. First, the constants for the corresponding Cu(I) complexes are smaller due to the smaller charge and a larger ionic radius of that ion. Second, the univalent ions Cu(I) disproportionate in water, and as a result they only exist as insoluble compounds or complexes. The equilibrium constant for the disproportionation of Cu(I) in solution is high.
Dear All, how can I and which aliovalent I will doped in Cu2O for its enhancement electronic property so that I can use in gas sensor. Because Cu2O changes CuO after heating/ in air ?
Thank you
when we work with Cu(I) complex in aquaus solvent under N2 atmosphere, How convert Cu(I) to Cu(II)? thank you
Dear Saeideh Jajarmi,
Why not directly use Cu(II). However, if your reaction is such that you need Cu(I) as a starter and then Cu(II) in the later stage, then you can use oxidizing agents to convert Cu(I) to Cu(II).
TBHP works very well as an oxidizing agent for Cu(I).
All the best.
Dear researcher
If you question to copper complexes, the CuII more than stable CuI .This due to most of the copper complexes are square planar geometry (dsp2 hybridization)
Dear researcher,
For Cu (II), there is Jahn Teller effect in d9 configuration. But in the d10 configuration there isn’t the Jahn Teller effect. The d9 configuration is gained by tetragonal distortion extra Stabilization energy. Hence Cu(II) with d9 is more stable. The Jahn Teller effect for Cu (II) can also be seen in the Irving-Willliams series.The Irving-Williams series: Mn(II)
What do you mean by this question yogesh? Do you want to see which is more stable CuO or Cu? If that is the question, I should say that you can not compare the stability of a metal with its oxide. To give you more information, Cu does not react with nonoxidizing acids , such as HCl, whereas CuO reacts with all acids to give Cu(II) salts. On the the other hand, if you heat Cu metal it will be converted to CuO but CuO does not change on heating.
The stability of each one depends on the type of anion. Pearson Principle See " Soft base with Soft Acid and Hard base with Hard Acid"
Please have a look at these useful links.
https://www.meritnation.com/ask-answer/question/cu2-is-more-stable-than-cu-even-former-have-3d9-configurat/structure-of-atom/4962935
Cu(II) has a higher charge than Cu(I). Thus, Cu(II) has a higher attraction to water than Cu(I), and has a more exothermic hydration energy. In fact, it has the highest hydration energy of all the first row transitional metals since TM ions get smaller along a period, and Cu is at the end of the first TM period.
So, because of that highly exothermic hydration energy, Cu(II) is stable.
Additionally, because Cu(II) is d9 octahedral, if you draw out the octahedral splitting diagram for it, there is an option of whether the ninth highest energy electron can go into the d(z2) or d(x2-y2) orbital. If it hops into the d(z2) orbital, it repels the ligands which are located exactly along the z axis. This brings the z ligands further away on average from the metal d(z2) electrons (increased bond length), reducing repulsion on average. This stabilises the ninth electron by repulsion. Same argument if it was d(x2-y2) orbital. This leads to a Jahn-Teller effect that stabilises the ninth electron in Cu(II), ----relative--- to it being in Cu(I). This also stabilises Cu(II) relative to Cu(I).
The hydration energy, coupled with the Jahn-Teller effect, are strong stabilising effects. They out-weigh the endothermic penalties of the ionization energy needed to convert Cu(I) into Cu(II), in converting the stable, closed shell d10 Cu(I) to the open shell d9 Cu(II).
Just want to make an amendment to my original answer as I got it wrong. If the electron hops into d(z2) and pairs with the electron already in there, that leads to additional shielding of the positive metal from the attraction of the negative ligand (in this theory we assume the metal to be positively charged and ligands to be negative). That shielding, relative to the shielding provided by the single electron in d(x2-y2), leads to the z ligands being pushed further from the metal. That means the z ligands are further from the z electrons and repel them less. That stabilises the z electron, hence Jahn Teller effect. Same arg for d(x2-y2).
The important thing is that Cu(II) is d9, Cu(I) is d10. Since there is no Jahn Teller effect in Cu(I), no stabilisation occurs, so Cu(II) is more stable.
The Cu(II) is d9 where as Cu(I) is d10.There is no Jahn Teller effect in Cu(I), no stabilization hence Cu(II) is more stable.
I would not say that Cu(II) is more stable than Cu(I) because its stability depends on the redox status of the environment. Cu(II) is stable in relatively oxidizing conditions and Cu(I) in reducing conditions.
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