why copper complexes almost always show z-out jahn teller distortion?
102316
Jahn-Teller distortions are when a complex changes itself geometry to gain additional stability.  Without the enlongation of the z-axis, there would be two degenerate half filled orbitals.  After the distortion, which can be described as a linear combination of the two degenerate orbitals, the orbitals are no longer at the same energy.  One is much lower in energy and has both of the electrons.  The other is higher in energy and unfilled.  The molecule is more stable.  Since copper II has nine d electrons, the higher energy orbital in the Jahn-Teller distortion would be half filled, but the complex would still be more stable than in the case of the undistorted molecule.
837624
Cassandra C has given you a very good answer.\nYou are interested in a single type of reaction: single displacement. It is a redox reaction so you have to think which compound is more happy having extra electrons (more electronegative).\n\nIn the case of A+BX-> AX + B, B(+) in BX is a cation and is missing an electron. If B is a more electronegative negative element than A, then it will steal away the electron from A and the reaction will occur. If A is more electronegative then B is not "strong" enough to steal the electron and the reaction will mot occur.\n\nIn the case of Y+BX -> BY+X you have the opposite. X(-) in BX has already an extra electron. So if X is more electronegative than Y it will not give away the electron and the reaction will not occur.\n\nThis is an oversimplified reasoning. The proper one is that you have to look at the standard reduction potentials (E0) which give the reactivity series. The half-reaction with the highest reduction potential will be the one favoured or more correctly the reaction will occur if the redox potential of the whole reaction is positive.\n\nFor instance you can find a detailed table at\nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\n\nLet's see your reaction\nCu+FeSO4 -> CuSO4+ Fe \nIn this case the half reactions would be\nCu - 2e -> Cu(+2)\nFe(+2) +2e-> Fe\n\nFrom the website we get that for\nCu(+2) + 2e->Cu  E01=+0.342 Volt\nFe(+2) +2e-> Fe E02=-0.447 Volt\n\nE01 is more positive than E02 thus Cu(+2) tends to get electrons from Fe and not the other way around so the reaction will not occur \nOR\nYou find the E of the reaction like this.\nThe second half-reaction is in the form we want it. The first is in the reverse. So we just reverse the reaction and the sign for the potential\n\nCu -> Cu(+2) +2e E01'=-0.342\nFe(+2) +2e-> Fe E02=-0.447 and add\n\nCu + Fe(+2)+2e -> Cu(+2)+ Fe+ 2e with E=E01'+E02= -0.342+(-0.447) \n\nTherefore Cu+Fe(+2) -> Cu(+2) + Fe E= -0.789<0 and so the reaction will not occur.
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
725081
The answer is Cu + Cu(NO3)2 -> Cu + Cu(NO3)2\n\nThere is no reaction.
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
596804
3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O
1086836
Zinc is a much more reactive metal than Copper.  Zinc is much higher on the activity series than Copper.\n\nWhen Zinc metal is added to a Copper +2 solution, the Zinc will displace the Copper in solution, precipiating out Copper metal while Zinc ions go into solution.\n\nZn (s) + Cu+2 (aq) --> Zn+2 (aq) + Cu (s)\n\nSince Zinc (II) Sulfate is soluble in water, the Sulfate (SO4-2) ion is a spectator ion in the reaction, it is in solution on both sides of the reaction.
453985
Because they effectively bind to but also dissociate from the metal center.  They protect the metal from reaction with undesired things until something which is a better ligand comes along and knocks it off.\nIt also serves to generate low-valent or even zero-valent forms of the metal (such as palladium tetrakis).
55435
jtkeel is mostly right, except the structure of copper is not crystaline.  Al3K is right about the structure of copper, which is the basic structure of all metals and is what gives them their physical characteristics, like their malleability, conductivity and so forth.
760435
You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
188677
Figure 21: Heteropolymolybdate structures. (A) The structure of [Mo< ...  \n  \nchemical compound\n\nCrystal field, ligand field, and molecular orbital theories \n\nConsiderable success in understanding certain coordination compounds also has been achieved by treating them as examples of simple ionic or electrostatic bonding. The German theoretical physicist Walther Kossel's ionic model of 1916 was revitalized and developed by the American physicists Hans Bethe and John H. Van Vleck into the crystal field theory (CFT) of coordination, used by physicists as early as the 1930s but not generally accepted by chemists until the 1950s. This view attributes the bonding in coordination compounds to electrostatic forces between the positively charged metal ions and negatively charged ligands--or, in the case of neutral ligands (e.g., water and ammonia), to charge separations (dipoles) that appear within the molecules. Although this approach meets with considerable success for complexes of metal ions with small electronegative ligands, such as fluoride or chloride ions or water molecules, it breaks down for ligands of low polarity (charge separation), such as carbon monoxide. It also requires modification to explain why the spectral (light-absorption) and magnetic properties of coordinated metal ions generally differ from those of the free ions and why, for a given metal ion, these properties depend on the nature of the ligands. Thus, the complex ion hexafluoroferrate(III) (FeF63-) has magnetic properties to be expected from a substance with five unpaired electrons, as does the free iron(III) ion (Fe3+), whereas the magnetic properties of the closely related hexacyanoferrate(III) ([Fe(CN)6]3-) correspond to only one unpaired electron.\n\nSince 1950 it has been apparent that a more complete theory, which incorporates contributions from both ionic and covalent bonding, is necessary to give an adequate account of the properties of coordination compounds. Such a theory is the so-called ligand field theory (LFT), which has its origin in the more general, but more complicated, theory of chemical bonding called the molecular orbital (MO) theory. (Molecular orbitals describe the spatial distributions of electrons in molecules, just as atomic orbitals describe the distributions in atoms.) This theory accounts with remarkable success for most properties of coordination compounds.\n\nAn important conclusion from LFT is that two types of bonds, called sigma () bonds and pi () bonds, occur in coordination compounds just as they do in ordinary covalent (organic) compounds. The more usual of the two are  bonds, and they are symmetrical about the axis of the bond;  bonds, which are less common, are unsymmetrical with regard to the bond axis. In coordination compounds,  bonding may result from donation of electrons from ligands, such as fluorine or oxygen atoms, to empty d orbitals of the metal atoms (the designation d merely being a way of indicating the particular orbitals involved). An example of this type of bonding is the chromate ion, (CrO4)2-, in which the oxygen atoms donate electrons to the central chromium ion (Cr6+). Alternatively, electrons from d orbitals of the metal atom may be donated to empty orbitals of the ligand. This is the case in the compound tetracarbonylnickel, Ni(CO)4, in which empty  orbitals in the carbon monoxide molecules accept d-orbital electrons from the nickel atom.
889558
copper metal is listed as "very unreactive".  This means it is relatively stable as a metal, and relatively unstable as an ion.\n\nMagnesium metal "reacts with acids" so it is more stable in ionic form than copper,and less stable as pure metal.\n\nThe goal of nature is to go from an net unstable form to a more stable form.  (This is called reducing the free energy, or delta G of a system).\n\nIn order to use the activity series:\n\nany metal below a given metal is less stable in ionic form and more stable as a pure metal.\n\nThis means that a pure metal can reduce the ions of any metals below it.\n\nFor example, aluminum metal can reduce manganese ions, zinc ions, and so on because they are below it.\n\nAluminum metal cannot reduce magnesium ions, calcium ions, and other metals above it.\n\nI hope you can get it now.  Ok I am sure what I typed is correct now.  Good luck!\n\nhttp://en.wikipedia.org/wiki/Activity_series
414105
I presume you asking for the N15 shift since there is no proton or carbon.  There is variability of the shift depending on it's environment (is it aliphatic or aromatic?)
273309
wat r u ppl talking about....u ppl r absolutely wrong.....its alkynes n i am 100% sure about\n\nthey r more reactive becoz of there triple bonds...\n\n\n2 ppl in a row a row giving a wrong answer can confuse a person completely
79044
Here is a centralized page of information, with several links along the left side for more information:\nhttp://www.webelements.com/webelements/elements/text/Cu/key.html
332147
Those ions do follow Hund's rule, though. The d orbital looks like this:\n\n\n_________ ___________\n\n__**_______ ____*____ ___*____\n\nThe electrons are only paired when absolutely necessary.
848068
ICl2 + should be bent like water since I will have two lone pairs and two Cl's on it.\n\nI3- will be linear\n\nSbCl6- will be octahedral