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You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\... 6 Theory guides and experiment decides (I. M. Kolthoff). Sometimes thermodynamics predicts a reaction but it does not occur very fast because its kinetics is slow. If you dissolve iron (III) chloride in pure denionized water, assuming it is from a fresh bottle, it will take hours to form a precipitate of iron hydroxides. Therefore, in order to perform this ... 5 As said by Oscar Lanzi, they are called "ternary metal chalcogenides". However, the name depends on the type of metal present in the compound. From here: Systems to be studied include the layered intercalates$\ce{AMX2}$(A=alkali metal;$\ce{X=S, Se, Te}$) which are potential ionic conductors suitable for use in solid state batteries: Group 2 ... 4 Actually it is diamagnetic but due to a phenomenon called charge transfer spectrum (CTS), oxygen transfers one electron to manganese and$\ce{KMnO4}$as a whole becomes paramagnetic. This is also the reason why it shows a characteristic colour while diamagnetic complexes are colourless due to absence of d-d transitions. 4 Because you are comparing apples with donuts. If I eat enough apples they will add more calories than that donut I gobbled for breakfast. To compare food calories properly you need to match serving sizes. Similarly, to compare reaction energies fairly you need to use a common basis for the extent of reaction. For oxidation of metals the reactivity of the ... 4 Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer. Iron is fairly early in the transition metal series, so when pushed beyond the$+3$oxidation state it has few$d$electrons in the central core. As explained in the referenced answer, this makes the iron strongly ... 4 You likely had either or both of the following: Copper had some oxide on it, and this oxidized the alcohol leaving the metal behind. If there was air in your system, especially on heating copper could have reacted with it to generate the oxide (a relatively fast reaction with heating) and you're back to (1). The metal would then, of course, be available ... 4 When heated in air, a piece of copper metal gets oxidized into black copper oxide$\ce{CuO}$according to $$\ce{2 Cu + O2 -> 2 CuO}$$ When hot copper oxide$\ce{CuO}$is in contact with some ethanol vapor$\ce{C2H5OH}$, it reacts according to : $$\ce{CuO + C2H5OH -> Cu + CH3CHO + H2O}$$ And the copper metal is regenerated. So if a copper plate is ... 4 Metallic osmium is harmless but it reacts with oxygen at room temperature, forming volatile osmium tetroxide. Any other osmium compounds are also converted to the tetroxide if oxygen is present. This makes osmium tetroxide the main source of contact with the environment and human body. Exposure to osmium tetroxide cause harm to the skin, eyes, and ... 4 Aqua regia doesn't actually dissolve lead metal. It converts to insoluble lead salt. From here: When we are working with an acidic chloride solution, like HCl (with an oxidizer) or aqua regia, most of the lead will become an insoluble lead chloride which if it is left to settle can be removed by careful decanting of the solution and filtering, what lead ... 3 There are two ways to approach this: simple crystal field theory or more complex molecular orbital theory. Both can describe and predict Jahn-Teller distortion well. In simple crystal field theory, one always starts with a set of five degenerate d orbitals on the metal. The ligands are considered as a point-sized negative charge. (This consideration, while ... 3 Before you consider J-T distortion, take a look at the d orbitals of an octahedral complex. Note that the orbitals that are on the axes where the ligands are are higher in energy. This is because these "d orbitals" are actually anti-bonding orbitals comprised primarily of the metal d orbitals. Because they are anti-bonding, they get higher in ... 3 As you note, many hydride phases have the hydrogen occupying interstitial sites within a metal phase, not unlike many carbide/nitride/oxide phases as well. In theory this could range from no hydrogen at all up to full occupancy (and perhaps even double occupancy). Since you mentioned lanthanum hydride, I pulled up a Calphad analysis for the H-La-Ni system, ... 3 Wikipedia renders multiple stages of decomposition as the common heptahydrate is heated. Based on the data summary at the beginning of the article and the quoted passage below, the anhydrous salt is obtainable between 300 and 680°C. A lower temperature, starting at 60-64°C, gives the monohydrate. An oxygen-free atmosphere is necessary to prevent iron ... 3 "f-hole is critical to describe DoS of IrO2 correctly." [...] This implies, that f-electrons are somewhat involved into chemistry of Ir. Perhaps naïvely, but I would dispute that the second statement follows from the first. While f-electrons are generally not critical for the chemical reactivity of molecules or materials, their inclusion indeed ... 3 Dry ferrous sulfate heptahydrate is green(ish) and is expected to make a green(ish) aqueous solution. Anhydrous ferrous chloride is described as green to yellow, and the dehydrate and tetra hydrate are green to blue-green (CRC Handbook). Now color is broad and continuous (see the visible spectrum), not digital, like how many fingers am I holding up. This ... 3 The oxidation state of$\ce{Zn}$as calculated by you is incorrect. The correct oxidation state of$\ce{Zn}$is$+2$as it is associated with two mono negatively charged$\ce{NO3-}$ions. Thus the unbalanced reaction with correct oxidation states would be:$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(... 3 TL;DR: Decomposition of nitrites always afford$\ce{NO2}$if you apply the correct reaction condition to it. Long answer To generalize the decomposition of metal nitrates (I concluded from the article above, and hereby using a generic monovalent metal M for the equations), we can say this: The first stage of decomposition is always the decomposition to ... 2 Chemical Book describes nickel(II) thiocyanate (CAS #: 13689-92-4) as "Nickel Thiocyanate Greenpowder" (see Freddy's answer elsewhere), but according to the Ref.1, it is yellow in color: [...] This chloroform solution was green in contrast to the usual intense blue of nickel amino complexes. When the chloroform was allowed to evaporate, a viscous ... 2 This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth. We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was ... 2 First and foremost, d-block elements and transition elements are two different entities. These terms become not exchangeable as you start using current IUPAC definition of transition element (see M. Farooq's answer). Older textbooks included entire d-block, including Group 12 elements, into transition elements cohort because historically the name "... 2 I decided to do some more research for this question as I had faced it before. Turns out the proper name of this ligand is diethylenetriaminepentaacetate ($\ce{DTPA^5-}$). It's an expanded version of EDTA and is primarily used in MRI scanning where it forms complexes with the Gadolinium ion. It forms complexes with a coordination number of six and seven with ... 2 I would be hesitant to apply the term transition metal chalcogenide to$\ce{CrSbX3}$– and for the others, it depends on what the actual structure on a molecular level is. A term like transition metal chalcogenide (to me) invokes the idea of the transition metal and the chalcogen being in close contact. Basic examples would include iron sulphide (where there ... 2 For the record, after M. Farook's agreement I am making my comment an answer. Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate. Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water. 2 I think, the answer to number 1 will remove the need to answer the subsequent questions: 1) Which method gives the true d-electron count? Neither! You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible ... 2 It seems that you have missed quite a few key points while explaining your question. Firstly, Hg is a d-block element which is not considered to be a transition element since it has complete d-orbital. The same holds for other members of the group like Zn and Cd Also, the statement that complexes containing fully filled d-orbitals are not coloured is ... 2 Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one [1]: The$\ce{Cu(II)}$atom has square-planar geometry with two bidentate oxalate ligands. Each$\ce{K+}$cation is coordinated by eight oxygen atoms from one bidentate and two ... 2 Your premise Wikipedia's statement about transition metals seems to be flawed. (1) It isn't possible to have a transition metal that has the electronic structure$\mathrm{(n-1)d^{10} ns^2np^6}$. Such an electronic structure would imply a noble gas. (2) Gold$\mathrm{[Xe]4f^{14} 5d^{10} 6s^1}$and platinum$\mathrm{[Xe]4f^{14} 5d^{9} 6s^1}$are the most noble ... 2 Research notes in this ACS article Kinetics of Hydrogen Reduction of Manganese Dioxide that even heating$\ce{MnO2}$in an atmosphere of hydrogen gas only results in$\ce{MnO}\$. The source further notes that very high temperatures (some 1,600 K) are required for any appreciable formation of the metal. As such, a more facile approach, simply dissolve the ...