A solution of K2MnO4 has a deep purple colour with a strong band in its UV/V is spectrum Explain the origin?
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Solutions of some chemicals are coloured because the molecules in the solution selectively absorb certain colours from the light passing through. For example, if the chemical absorbed red light then the solution would appear blue-green. In the case of potassium permanganate almost all colours are absorbed, leaving only the blue/indigo/violet range (to use the popular names for the rainbow colours) to get through and cause the purple colour.\n\nNot sure by "band" whether you mean absorption band or transmittance band but the mechanism to produce one would be almost the same. An absorption band is caused by a strong removal of a certain colour (or range of wavelengths) in the area of the band and a transmittance band by the strong removal of colours on each side of the band.\n\nColours are absorbed because the energy the photons having that particular wavelength contain is exactly enough to excite an electron within a molecule. This excitation raises the electron to a higher energy level and the photon is absorbed. Somewhat later, the energy is returned to the system as the electron returns to its former energy level (often via other intermediate levels). The energy is not usually returned as light again however, but more often as a form of heat (a vibration of the crystal lattice, actually). In those cases where the energy does reappear as light the phenomenon is called luminescence, or in some circumstances laser action.\n\nElectrons associated with atoms in molecules and/or complex crystal lattices can have their energy levels changed by the close association with other atoms and so make it more likely they will absorb visible light. This is one of the reasons there are many more coloured compounds than elements. In particular, certain minerals can develop intensely absorbing locations within their structure called "colour centres" which is what produces the deep colours of precious stones. Diamond, on the other hand, is transparent because its highly ordered and stable structure means that a photon would need to have quite a high energy for it to be absorbed and so all visible light passes through unchanged.\n\nHope this is what you were after.
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Potassium Permanganate is an oxidant which will accept electrons depending on the pH at the time of the reaction. In neutral to alkaline solutions, it will form insolubile manganese oxide (MnO2), accepting three electrons, in acid solutions it will form Mn2+ ions.\n\nOxalate is an organic acid rest with a C-C bond and the sum formula (CO2)2 2-. It reacts as a reductive agent and will donate two electrons and accept negative charged two Oxygen atoms (from the water or the Permanganate), forming two carbonate ions.\n\nStrictly speaking, not carbon dioxide but dissolved sodium and  potassium carbonate will be the product of the reaction.
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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.
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Usually the salts of Cu2+ - like CuSO4 - forms blue colored solutions. But it is not the only. Some cobalt salts also form blue colored solutions.
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K2CrO4 +Pb(NO3)2 ---> PbCrO4(s)  + 2KNO3\n\nNice yellow ppt.
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Group I elements almost always have an oxidation number of +1, so the K (potassium) has a +1.  Unless they're in peroxide compounds, which is not the case here, oxygen usually has an oxidation number of -2.  The hardest one is the Mn.  We know that the whole compound has to have an oxidation number of 0, so let's do some math.  There is one +1, and there are 4 oxygens, each with -2.  \n+1 + ? + 4*(-2)=0\nThe answer is +7.  Therefore, Mn must have an oxidation number of +7. \nK=+1, Mn=+7, O=-2
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It is a silvery white metal, it looks shiny.\nAnd i can tell you it doesn't glow in green :D\n\nNote: the color shown by Plutonium solutions depend on both the oxidation state and the nature of the acid anion, which influences the degree of complexing of the plutonium species by the acid anion.
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Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
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no, its extreme dark blue, that has a hue of purple, but its more blue then violet...but dark
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It is not due to the chemistry of copper or gold.\n\nIt is due to the electronic band structure. The two metals unusually have a band structure that causes them to absorb light at certain frequencies rather than reflecting it.
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Coz chlorophyll interacts with the iodine soln and the purple colour will not be clearly visible Instead u will end up in black colour.
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This solution probably has fluorescence at UV (ultra violet) wave length. try looking at it in a room with red light, if there is no light using the lenses then it's confirmed. but you can't simply identify it this way. you have to know the exact wave length that it has fluorescence at. even then you might not be able to identify it. try using UV-visible spectroscopy and when you have its spectra, you can identify it. for more information see: principles of instrumental analysis by Skoog
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Some of the matals have multiple oxidation states and these must be indicated in the names:\nAmericurium (III) fluoride\nIndium (III) fluoride\nBarium permanganate\nNeodymium (III) iodide\nCadmium sulfite\nSilicon tetrathiocyanate\nVanadium (III) fluoride\nIron (III) dichromate\nGold (I) cyanide\nBarium bromate\nMagnesium molybdate\nPlutonium (III) bromide\nSamarium (III) sulfate\nTungsten (II) iodide\nPotassium borate
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try Vogel's "Qualitative Inorganic Analysis", I'm pretty sure it has all the answers 4 u.\n\ndont care editions, but for 7th ed, ISBN is 0582218667\n\n****\nokey then, from Vogels\n\n1) when heated, lead nitrate decomposes to white solid lead oxide (PbO), brown gas (NO2) and oxygen gas. Brown gas decolorizes when contact with water. \n\n2) ammonia odur. ammonium compounds are all volatile towards heat and decomposes to ammonia and an acid but quickly recombine if the acid, in this case, hydrochloric acid, is a strong acid.\n\n3) anhydrate copper sulphate is white, hydrate copper sulphate is blue. I have no idea which one you actually heated. If it's anhydrate, no change; if hydrate, blue ---> white.\n\n4) blue precipitate forms initally, but resolve upon excess reagent, clear blue soln results due to the formation of copper complex [blue precipitate: Cu(OH)2*CuSO4, complex: [Cu(NH3)4]^2+]\n\n5) brown precipitates froms [Cu(OH)2], no change with excess reagents.\n\n6) metal-gray stipe (magnesium stipe) gains a redish brown color due to formation of copper metal. Possible decolorization of the soln (depends on how much magnesium u actually put in)\n\n7) volatile reaction, solid metal solves,ignitable gas is formed. (if phenolphthalein is added to water initally, soln becomes pink)
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It's not polar; It's ionic.\n\nBaO consists of Barium in the +2 oxidation state and Oxygen in the -2 oxidation state.
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NaHSO4 itself is acidic, so the solution's color should be yellow due to the bromcresol green being in the presence of acid
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The sodium vapor is absorbing the electromagnetic energy in that specific band.  It is actually two lines, close together, in the yellow-orange part of the visibile spectrum.\n\nThe energy of that yellow-orange light contains precisely the quanta of energy to excite electrons into higher energy states, so the sodium vapors abosrb very strongly at those frequencies.  The other frequencies are at the wrong energy level, and are "ignored".