New answers tagged inorganic-chemistry
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After washing by water, get rid of the residual water with methanol or acetone.
The residue of solvent will finally evaporate, leaving much less water than without this step. Similar procedure is used in preparation chemistry to remove water from crystaline products.
Ethanol is usable too, but there is usually some 4-5% of water.
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Color is somewhat subjective, that is why people study wavelengths with atomic spectra rather than color. For elementary classes, this is fine. Certainly A is a wrong choice because I checked the transmission characteristics of cobalt glass,* the blue glass transmits both the violet and the deep red doublets of potassium. So, the flame color appears like ...
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You first focus on the groups that undergo oxidation or reduction. The sulfate does neither, so you can set it aside, just like the potassium cations.
The half reactions you wrote, however, need some work. The oxidation state of manganese in permanganate is +7, so there is a transfer of 5 electrons. For bromine to bromide, the oxidation state changes by one, ...
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According to PubChem, the IUPAC name of nickel bis(glycinate) is carboxymethylazanide;nickel(2+), which has a given structure of:
However, according to suggestions in Ref.1, the correct IUPAC name should be bis[glycinato(2-)-$\kappa^2$N,O]nickel.
Since the complex is simply given as $\ce{[Ni(gly)2]}$, it is no need to go for the compllicated name since I ...
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In the Handbook of Chemistry and Physics, the solubility of calcium citrate in water is $8.5$ g/L at room temperature. This correspond to $1.79$ g Calcium per liter, or $1790$ ppm $\ce{Ca}$
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Due to two contradictory factors, one is the effective nuclear charge which is greater in $\mathrm{5d}$ making it of lesser energy and the second one is a nodal factor. Nodal factor depicts localization of electrons, which indicates of higher energy. Hence to balance these two effects out, for Ce the second electron goes to $\mathrm{4f}$ and its electronic ...
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The picture you provided shows half the story.
In the diagram above you can see that copper has a charge of +2 i.e. Cu$^{2+}$ which leads to a 3d$^9$ configuration. Since H$_2$O will here act as a weak field ligand, no pairing of electrons will be done in d-orbitals which will lead to sp$^3$d$^2$ hybridization.
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Rules for construction of Lewis dot structures
Calculate the number of total valence electrons, including charges (ions) ($V$).
Calculate the number of total electrons required to complete octet/duplet ($T$) = $(2 × \ number \ of \ H \ atoms) + (8 × \ number \ of \ other \ atoms)$
Calculate the number of bonding electrons $B = (T-V)$
Calculate the number ...
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Semiconductors conduct because some electrons are easily bumped into the conduction band
The basic definition of a semiconductor is a substance where there is an empty bundle of orbitals at an energy just a little higher than the valence band where the bonding electrons reside.
This gap is close enough to the valence band in energy that even thermal energy ...
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Hydrogen and chlorine atoms of $\ce{HCl}$ in gaseous state are covalently bound and is termed hydrogen chloride. When this gas is bubbled into water, it ionizes completely to give $\ce{H3O+}$ (free proton + water molecule) and $\ce{Cl-}$ ions and becomes an acid solution which is termed hydrochloric acid. Even in gaseous $\ce{HCl}$, the charge is not ...
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Dissolution enthalpy is kind of scale balancing as being combination of lattice energy ( positive solid matrix breaking enthalpy) and hydration enthalpy(usually negative), following the formal process
$$\ce{CaCl2(s) -> Ca^2+(g) + 2 Cl-(g) ->[H2O] Ca^2+(aq) + 2 Cl-(aq)}$$
The hydration enthalpy is very negative for anhydrous $\ce{CaCl2}$, but just ...
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No bond is completely ionic or completely covalent, Fajan's rule. Also $HCl$ is predominantly a covalent compound rather than ionic.
Shouldn't, H+Cl- be a salt since hydrogen is positive and Cl is negative?
Sure, $\ce{HCl}$ does dissociate into $\ce{H+}$ and $\ce{Cl-}$ in a polar solvent, but that doesn't mean $\ce{HCl}$ is simply $\ce{H+}$ and $\ce{Cl-}$. ...
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First of all, charged species are in general less stable than neutral species. So $\ce{O_2}$ will be the most stable species among them.
In the next step compare their bond orders. The species with more bond order is more stable (in general). So next place goes to $\ce{O_2^+}$.
Finally, in comparison of $\ce{C_2^+}$ and $\ce{O_2^-}$ both will have the same ...
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A picture, specifically from WP user Dnn87, beats a thousand words. The gray background is ideal for bringing out the true golden color of caesium metal.
The color comes from plasma oscillations, in which electrons in a conducting medium spontaneously oscillate due to inherent instabilities. These may be traced to the electromagnetic interactions as ...
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$\ce{SO2}$ does react with $\ce{H2O}$ according to your second reaction (which is a half-equation) : $$\ce{SO2 + 2 H2O -> SO4^{2-} + 4 H+ + 2 e^- }$$ But $\ce{SO2}$ cannot react according to the first equation, which produces two non-combined (nascent) $\ce{H}$ atoms. Usually the electrons produced in this half-equation do not react with $\ce{H+}$ to ...
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In fact the nonbonding electron pairs on the end atoms do affect ozone's polarity, and because the end atoms have more nonbonding electrons they are actually the ones at the negative end of the dipole. The central atom with only a single nonbonding pair (and a positive formal charge) is at the positive end. See also this answer.
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And then don't forget we have nitroprusside, still used as a short-term, if fairly rough way to reduce peripheral hypertension in a hypertensive crisis (and on the WHO list of essential medications). I'm not sure of its Fe oxidation state, but I'm sure it's one of the above. A remarkable drug, still in use, if even for dramatic (and short term) purposes.
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It is not common to find the ferrous ferrocyanide because it oxidizes rapidly. As you can see the remaining salts or pigments are pretty stable in air. This one is white. I quote from a very old Analytical Chemistry book by Treadwell, pg. 155, this should address your query. It states:
Potassium Ferrocyanide, $\ce{K4[Fe(CN)6]}$, produces in solutions of ...
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Calcium sulfate dihydrate has a solubility in water of 0.26 g/100ml at 25 °C, so it dissolves slowly - you already know this. No gases will be evolved on dissolution so there is nothing to worry about there.
With respect to the resulting solution this is from the Wikipedia page on Calcium Sulfate:
The calcium sulfate hydrates are used as a coagulant in ...
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Chelate complexes are more stable IF EVERYTHING ELSE IS THE SAME.
In your case, the anionic nature(cf. ethylenediamine is neutral) of cyanide makes cyanide a better sigma-donor than is ethylenediamine, while the pi-accepting nature(cf. ethylenediamine is pi-neutral, if not (hyperconjugative) pi-donating) of cyanide makes cyanide a better pi-acceptor than is ...
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Sodium carbide isn't hypothetical. It does exist and can be synthesized. In @Oscar's answer, there is a patent which shows the synthesis reaction of sodium carbonate and carbon at temperature 1050 °C-1200 °C and contacting the resulting gaseous mixture of carbon monoxide and sodium vapor on steel surface.
There is another patent which shows a different ...
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As the comments imply, sodium carbide certainly does exist. However, it is difficult to get because it is only metastable. Therefore it cannot be made from direct combination of the elements in thermodynamically stable form, a procedure often used (in some cases indirectly) with stable binary compounds.
Sodium carbide can be made by using the sodium/CO ...
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The disproportionation rection of $\ce{HIO}$ cannot produce $\ce{HIO3}$ and $\ce{HI}$ (or $\ce{I^-}$ ions) in acidic solution. The reason is that $\ce{I^-}$ and $\ce{IO3^-}$ ions do react with one another, if $\ce{H^+}$ ions are present, to give : $$\ce{IO3^- + 5 I^- + 6 H^+ -> 3 I2 + 3 H2O}$$ If no ion $\ce{H^+}$ is present, the previous reaction does ...
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Thanks for all the feedback ;) Now I have found out the answer on the Internet.
For the diluted ( or very diluted, for nitric acid) form of these acids, the anions are quite stable as there are sufficient water molecules that form bonds with them. Moreover, there's the concentration effect-- hydrogen ions are more abundant as they come fro two sources-- ...
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Pentahalides are more covalent than trihalides of the same element with the same halogen. For instance $\ce{TaCl5}$ is more covalent than $\ce{TaCl3}$ (the respective WP articles give their melting points as 216°C for the pentachloride and 440°C with decomposition for the trichloride); but we cannot make such a simple comparison of $\ce{TaCl5}$ with either $\...
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The problem is, we want to see a trend in everything. There are various factors that govern melting point and boiling point which is the reason of perceived anomalies. "The equation which has more number of variables, is harder to solve."
For melting point, few of them are: Crystal system, size of atom, atom-atom distance, distance between two ...
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Definitely, you should boil the solution of silica and sodium hydroxide few hours. I did this experiment many years ago. And, as I remember this, it is extremely dangerous and very boring experiment. So boring that I still remember it approximately after 40 years.
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Have you checked Wikipedia?
Typically the reaction to form Grignard reagents involves the use of magnesium ribbon. All magnesium is coated with a passivating layer of magnesium oxide, which inhibits reactions with the organic halide. Many methods have been developed to weaken this passivating layer, thereby exposing highly reactive magnesium to the organic ...
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The earliest discovered molecular form of the elemental phosphorus is white phosphorus $(\ce{P_\mathrm{W}})$, which was isolated in 1669 by a German alchemist, Hennig Brand (Ref.1). Later in 1914, Percy Bridgman has introduced the phase transition of white phosphorus to black phosphorus $(\ce{P_\mathrm{B}})$ under high pressure (Ref.2). The irreversible ...
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The strange property of $\ce{Al2O3}$ is due to the position of aluminium in the periodic table. As it is not obvious, I have to start a rather long explanation. Sorry !
Let's start with the definition of acids used in the beginning of the $20$th century. At this time an acid was a substance containing a Hydrogen atom and able to produce $\ce{H^+}$ (or $\ce{...
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First of all, let's be clear on the the formula for chalcopyrite. It is $\ce{CuFeS2}$. The other copper-iron sulfide ore is bornite, $\ce{Cu5FeS4}$. Now, let's talk about copper extraction. What is written in your question is not correct and I am quite sure that's not how copper is extracted. If you heat the ore (roast) in reverberatory furnaces, it happens ...
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Chalcopyrite $(\ce{CuFeS2})$ is the most common copper bearing mineral on earth (Approximately 70% of the world’s copper reserves are contained in the mineral chalcopyrite), but also the most stable minerals because of its structural configuration, face-centered tetragonal lattice (Ref.1). At present, there are basically two main methods employed worldwide ...
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