What is the chemical structure / hybridization of sulfuric acid? I was looking at the chemical structure of $\ce{H2SO4}$.

Intuitively, I would have expected this molecule to be square planar in accordance with $p^2d^2$ or $sp^2d$ hybridization, but instead it is shown to be in a tetrahedral geometry consistent with $sp^3$ hybridization. Why is this?

I guess that an alternative way of asking this question is: What is the hybridization of the sulfur atom in $\ce{H2SO4}$?

It is much easier to explain it on simpler example – $\ce{O3}$ molecule. It has structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, central atom has hybridization state $sp^2$. One bond here is normal covalent bond and another bond is dative: an electron pair is donated onto vacant orbital of $\ce{O}$ atom with all electrons paired. in $\ce{H2SO4}$ molecule to bonds are simple covalent ($\ce{S-OH}$ ones) and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved brocken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.

A little more interesting example is $\ce{XeF2}$ molecule, where three-atom four-electron bond $\ce{F-Xe-F}$ are formed, that can be think of as reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$

Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take course of quantum chemistry, it should be enough. However, I'll recommend to search for "MO LCAO model": it is quite simple and very useful. It is also often used in advanced chemistry books and articles.


If Electronegativity of Cl is greater than H2 then why is the bond angle of Cl2O greater than H2O? If Electronegativity of Cl is greater than H2 then why is the bond angle of Cl2O greater than H2O?
Cl2 has more EN than H2 and size of Cl2 is more than H2 but if we go according to EN then more angle should be of H2O as electrons in H2O would more nearer to O2 than the electrons in Cl2O near to O2?

Bond angle is largely governed by the location of electron cloud with regard to atoms in a molecule.
In dichlorine monoxide, the electron cloud is more spread out between oxygen and chlorine while in the water molecule it’s much closer to the oxygen atom. Therefore, the net bond paid-bond pair repulsion is higher in dichlorine monoxide than in water, while the lone pair-lone pair repulsion is almost the same.

Also, chlorine atoms are much bigger than hydrogen atoms, because of which they also undergo steric repulsion causing the bond angle to further increase.
That’s why the bond angle in dichlorine monoxide is greater than in the water molecule.


Determine minimum C-H bond length in ethane derivatives Which of the following compound has minimum C-H bond length?

  ethane
  ethene
  1,2-dibromoethene
  1,2-dibromoethane

How can we know the comparison of $\ce{C-H}$ bond length (not $\ce{C-C}$)? Also, there is not any reagent given, so that we would do dehydrohalogenation reaction at it. How to compare it simply?

One way to compare $\ce{C-H}$ bond lengths is to compare their stretching frequencies using IR spectroscopy. However, since ethane and ethylene are gases at ambient conditions, we are going to have some difficulties in accuracy. Yet, it is worth comparing. Since we are comparing $\ce{C-H}$ bond in each case, we can suggest that the stronger the stretching frequency of $\ce{C-H}$ bond is, the shorter the bond should be.

1,2-Dibromoethane is a liquid at room temperature and $\pu{760 torr}$ with boiling point in the range of $\pu{131-132 ^\circ C}$. 1,2-Dibromoethene is from Aldrich is a mixture of cis- and trans-isomers, which boils at $\pu{110 ^\circ C}$ under $\pu{754 torr}$ pressure. IR spectra of both are extracted from NIST.gov webbook and illustrated as follows:

1,2-Dibromoethene: Range of $\ce{C-H}$ stretching is around $\pu{3150-3050 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3100 cm^{-1}}$):

1,2-Dibromoethane: Range of $\ce{C-H}$ stretching is around $\pu{3010-2950 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3000 cm^{-1}}$):

Accordingly, we can easily suggest that $\ce{C-H}$ bond in 1,2-dibromoethane is longer than that in 1,2-dibromoethene. However, this is hardly a surprise since we are comparing a $\mathrm{sp^3-s}$ bond versus a $\mathrm{sp^2-s}$ bond. Nonetheless, this is a good exercise because it confirm the validity of the test, in a way. Yet, the $\ce{C-H}$ stretching at around $\pu{3000 cm^{-1}}$ is undoubtedly suggested that  $\ce{C-H}$ bond in 1,2-dibromoethane is shorter than a $\ce{C-H}$ bond in a regular alkane.

Now, let's see two good liquid alkane and alkene examples to represent gasious ethane and ethene, respectively.

2-Methylpent-1-ene (a terminal alkene) is a liquid at room temperature and $\pu{760 torr}$ with boiling point of $\pu{62 ^\circ C}$. n-Pentane (the shortest n-alkene at ambient condition) is a liquid at room temperature and $\pu{760 torr}$ with its boiling point around the range of $\pu{35-36 ^\circ C}$.

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{3100-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3075 cm^{-1}}$):

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{2990-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{2900 cm^{-1}}$):

Thus, we can confidently suggest that the $\ce{C-H}$ bond lengths of given compounds are in following order:

ethane > 1,2-dibromoethane >ethene > 1,2-dibromoethene


Does this C17H28 isomer exist? Can anyone tell me what this $\ce{C17H28}$ molecule is? Has it been synthesized?  Does it have a name?

I was just interested in the molecule because it looks interesting.

This compound (formally called "hexadecahydronaphtho[1,8-de]naphthalene", but in common parlance also "[6.6.6.6]fenestrane" - thanks Mithoron!) has never been actually synthesised before, but some of its properties have been studied computationally.

It does not particularly matter, but it is worth pointing out that this compound possesses stereochemistry which was not indicated in your original drawing. Hence your question is essentially asking about a collection of stereoisomers.

Gund and Gund1 investigated the strain energies of a few $\ce{C17H28}$ isomers, shown above, using empirical force field methods. The compound you are interested in is 4b in the scheme (the various 4a are constitutional - as opposed to configurational isomers - of 4b). The numbers next to the names are the strain energies in kcal/mol. The higher the energy, the more strained; hence it seems that 4b, if synthesised, would be generally thermodynamically unstable with respect to 4a (kinetic stability is of course entirely separate). The letters 'c' (chair), 't' (twist boat), and 'b' (boat) refer to the conformations of each six-membered ring in the structure. The labels cis and trans refer to the relationship between each neighbouring pair of fused six-membered rings.

The potential for systems containing fused six-membered rings to undergo "ring flips" was also investigated computationally (using DFT) in a recent paper by Alder et al.2 In particular, the authors were interested in the all-cis isomer of your compound, i.e. ccct/4b6 in the screenshot above - which incidentally has a rare $S_4$ symmetry. They found that inversion has a whopping barrier of +88.9 kJ/mol (B3LYP-gCP-D3/6-31G*//B3LYP/6-
31G*) - almost double that of cyclohexane (~45 kJ/mol). The authors conclude that this isomer of [6.6.6.6]fenestrane "probably has the highest inversion barrier for any nonpolymeric multi-ring structure built from cyclohexane rings".

Although the compound has never been made, the authors propose that at high temperature the inversion process (which interconverts equatorial and axial hydrogens) may be studied by NMR spectroscopy. However, due to the significantly slower rate of inversion relative to cyclohexane, temperatures of 100–200 °C would be needed (as compared to cyclohexane which requires low temperatures for NMR studies).

References

Gund, P.; Gund, T. M. How many rings can share a quaternary atom? J. Am. Chem. Soc. 1981, 103 (15), 4458–4465. DOI: 10.1021/ja00405a028.
Alder, R. W.; Butts, C. P.; Sessions, R. B. Perhydrohelicenes and other diamond-lattice based hydrocarbons: the choreography of inversion. Chem. Sci. 2017, 8 (9), 6389–6399. DOI: 10.1039/C7SC01759F.


What is the correct nomenclature for diphenyl methane? Can $\ce{Ph-CH2-Ph}$ be also be named as  as (phenyl methyl) benzene instead of diphenyl methane?

I came to this thinking that $\ce{CH2-Ph}$ can be considered as side chain at side chain.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): “1,1'-methylenedibenzene” (preferred IUPAC name; ring preferred to chain)

Method (2): “diphenylmethane” (treats phenyl groups alike as substituents)

(This example is is explicitly mentioned in the Blue Book.)


How to generate fully coloured electrostatic potential maps in Avogadro? I want to generate electrostatic potential maps of molecules using Avogadro, that show the colour scheme ranging from blue through green and yellow to red.

In this Youtube video, at 1:32, the video author does just that.

However, when I try the same thing on my Avogadro (version 1.2.0), it shows this:

How do i display the colourful electrostatic potential map on my computer? My computer has a 4 GB RAM and an NVIDIA GeForce 210 Graphics Card.

Thank you!

The answer is that the color scheme changed, due to user (and developer) feedback. More people preferred the red to white to blue scheme for electrostatics.

In Avogadro 2, we hope to have a full-featured gradient editor, allowing you to set the colors you wish.


Reaction (or lack thereof) between oxygen and halogen gas I understand that oxygen and halogens are both very electronegative elements and it would makes sense that they don't want to associate with each other at all. However, I know that oxides of halogens ($\ce{Cl2O, ClO2, ClO, Cl2O2}$ etc.) do exist. When reading about halogens, it says that chlorine gas doesn't react with oxygen gas. I tried to find the intuitive reasoning behind this phenomenon online, but to no avail.
I know that chlorine and fluorine gas can react to produce chlorine trifluoride (fluorination of chlorine), proving that two diatomic, covalent and non-polar gases of electronegative elements can react.
$$\ce{Cl2 + 3F2 -> 2ClF3}$$
Also, drawing comparison with the reaction between nitrogen and oxygen gas:
$$\ce{N2 + O2 -> NO}$$
$$\ce{N2 + 2O2 -> NO2}$$
Nitrogen's triple bond is loads stronger than the covalent bonds from oxygen or chlorine, and it can still react with oxygen gas, albeit only under extreme conditions (lightning). Is this the case with chlorine? Even so, it only makes sense that it requires a lower temperature, as the energy needed to atomize them is lower (I haven't done the calculations with enthalpy and entropy yet). I can't find any reference to a reaction such as this online, though.
Most oxides of chlorine can only be made from compounds like chlorites and chlorates or by other novel methods. Is there a reason oxygen and chlorine (or other halogens, but chlorine in particular) can't react with oxygen gas? Is there something stopping them from reacting that I don't know about? Is it due to undesirable entropy changes? Or are the reactions I mentioned special, allowing two relatively nonreactive gases to react?

Your on the right track: in order to form a new compound, old bonds must be broken. This requires energy, even though the reaction may, overall , be exothermic. This is called the activation energy.
Think of it as pushing a car a short distance up one side of a hill before it reaches the peak and then rolls a long distance down the other side.
For example, in the dark, at room temperature, $\ce{Cl2}$ and $\ce{H2}$ coexist quietly in a flask... until light activates the chlorine (wait for the bang).
On the other hand, add enough energy, and even endothermic reactions take place, and, if the product is comparatively stable, or metastable, some can be left afterwards. For example $\ce{O2}$ is fairly stable, but given some energy input, e.g. from an electric discharge or from shortwave ultraviolet radiation, $\ce{O3}$, ozone, is produced, though that reaction is not thermodynamically favorable. The ozone is waiting to decompose... once it's activation energy is achieved.


Why isn't neptunium used in nuclear reactors in nuclear power plants? Why isn't neptunium used in nuclear reactors in nuclear power plants?  Uranium is, and plutonium is. But neptunium isn't and it is in the middle of them.  Is it like it is too hard to make it do fission?  Can someone please kindly explain it to me?

There are several reasons that neptunium may not be used.

The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium.

Next would be the likelihood of fission when struck with a neutron. If it is the same or less than uranium, I would go with uranium. It would have to be substantially more than uranium to offset the cost of refinement.

Also, there is the energy output per gram. I don't think it would be much greater than uranium, probably less.

The isotopes of neptunium that have long half-lives are $\ce{{}^{236}Np}$ and $\ce{{}^{237}Np}$. These are not produced in great abundance from $\ce{{}^{235}U}$ or $\ce{{}^{238}U}$.

To make plutonium from uranium, the uranium must absorb neutrons without fission. This is much more likely to happen then to make $\ce{{}^{236}Np}$ or $\ce{{}^{237}Np}$


Reference request on potassium cyanide production method I kindly ask for a reference for the following claim about potassium cyanide, found at potassium.atomistry.com:

  It is also obtained by the action of a mixture of steam and nitrogen on potassium carbonate.

I'd be very interested to know the temperatures required and yields obtainable, and of course if this works at all.

I did some literature research and I wasn't able to find any methods involving steam (as water in the gas phase) for the synthesis.
This in fact makes sense as alkali metal cyanides are prone to hydrolysis and tend to decompose to formate $\ce{K(HCOO)}$ over time or in hot water:

$$\ce{KCN(aq) + 2 H2O(l) -> K(HCOO)(aq) + NH3(g)}$$

which further decomposes to potash $\ce{K2CO3}$ upon further heating, so presence of water, especially at elevated temperatures, appears to be highly undesirable for the synthesis of cyanides.

I suspect the aforementioned article contains a typo.
Probably it should read "stream of nitrogen" — then it would make more sense as there are several industrial processes utilizing atmospheric nitrogen fixation for making cyanides with coal proposed in late XIX – beginning of XX centuries.

First, there is a German patent by Caro issued around 1900 (I couldn't find the full text or the exact date) mentioning synthesis of the cyanides of group 1 and 2 elements from carbonates using nitrogen and coal [1].

Second, there is Bucher process [2], which, I think, http://potassium.atomistry.com might refer to.
There is a comprehensive review of Bucher process [3]:

  The Bucher process, as known, consists in heating a mixture of sodium carbonate, charcoal, and iron powder (catalyst) in a stream of nitrogen at a temperature above 900°. The reaction is written as follows:

  $$\ce{Na2CO3 + 4 C + N2 -> 2 NaCN + 3 CO}$$

  […] A stream of nitrogen was passed  through the tube while a certain temperature between 900° and 1000° was maintained for several hours.
  The reaction  product was rapidly cooled by lifting the tube out of the furnace without interrupting the flow of nitrogen and was then examined microscopically.
  In all experiments sodium cyanide, together with some of the ingredients of the original mixture, proved to be present, which confirms the reaction given by Bucher.

There is also a brief overview of industrial methods used for manufacturing alkali metals cyanides throughout the past 170 years in Ullmann’s Encyclopedia of Industrial Chemistry [4]:

1850s–1900s: semi-industrial process (aka "the blood process") of fusion of potassium carbonate with potassium hexacyanoferrate(II) (obtained from animal blood and waste) initially proposed by F. and E. Rodgers [5].

1900s–1960s: Castner–Keller process; primarily employed at Degussa AG [6]:

$$\ce{Na + C + NH3 ->[\pu{600 °C}] NaCN + 3/2 H2}$$

1960s–present sodium and potassium cyanides are produced exclusively by the neutralization process, between a solution of sodium or potassium hydroxide with liquid or gaseous hydrogen cyanide:

$$\ce{MOH + HCN -> MCN + H2O}$$

  Gaseous or liquid $\ce{HCN}$ is first introduced continuously into concentrated solutions of $\ce{NaOH}$ or $\ce{KOH}$ at elevated temperature.
  The heat of neutralization is used for the subsequent vaporization of water. The use of special mixing devices and the addition of $0.2–3\%$ excess $\ce{NaOH}$ prevents polymerization.
  Simultaneously or in a second step, water is vaporized under reduced pressure and the alkali metal cyanide is precipitated below $\pu{100 °C}.$
  The shorter the residence time of the cyanide in solution, the cleaner the product and the lower the content of hydrolysis products and iron impurities.

  Figure 6. Flow chart of the main production steps for cyanide
  production ($\ce{NaCN}$ as example)

References

Caro, Nikodem. German pat. anmeld. C15812 (8-4-09). Verfahren zur Darstellung von Cyaniden aus Alkali oder Erdalkalioxyden, bezw. Carbonaten, Kohle und Stickstoff.
Bucher, J. E. The Fixation of Nitrogen. J. Ind. Eng. Chem. 1917, 9 (3), 233–253. DOI: 10.1021/ie50087a011.
Posnjak, E.; Merwin, H. E. Note on the Bucher Cyanide Process for the Fixation of Nitrogen. Journal of the Washington Academy of Sciences 1919, 9 (2), 28–30. https://www.jstor.org/stable/24521552
Gail, E.; Gos, S.; Kulzer, R.; Lorösch, J.; Rubo, A.; Sauer, M.; Kellens, R.; Reddy, J.; Steier, N.; Hasenpusch, W. Cyano Compounds, Inorganic. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley, 2011. DOI: 10.1002/14356007.a08_159.pub3.
F. and E. Rodgers; Ueber Darstellung und Anwendung des Cyankaliums. Ann. Chem. Pharm. 1842, 41 (3), 285–294. DOI: 10.1002/jlac.18420410303.
Degussa, DE 124 977, 1900; DE 126 241, 1900; DE 148 045, 1901; DE 148 046, 1901; DE 149 678, 1901; GB 265 639, 1926 (H. Freudenberg); DE 851
056, 1943 (W. Stephan).


Reaction of chloride of 4th A group Why CCl4 does not react with water,while SiCl4 do react,as both central atom are surrounded by 4 chlorine atom?

Reaction of a compound with water is called hydrolysis.During hydrolysis, Si atoms of the chlorides make up of their empty d-orbitals in their valence shells to form coordinates bonds with water molecules.
However, C is in period 2 does not have d-orbitals in its valence shell to form coordinate bond with water and is not hydrolysed by water.

extra reading https://www.chemguide.co.uk/inorganic/group4/chlorides.html


What is the mechanism of formation of Grignard's Reagent? What is the mechanism of formation of Grignard reagents? I couldn't find much in my book or on the Internet. This page says something which I am unable to understand. Why are the $2$ electrons in the $\ce{Mg}$ separated? Shouldn't they be together?

Looking at the mechanism:

We see that these are not double headed arrows, but single headed arrows. This indicates the movement of single electrons, rather than electron pairs as is common in most mechanisms.

Mechanisms that involve movements of single electrons tend to form free-radicals.

The lewis structure of Mg has the electrons seperated in order to make it clearer to read.

In step 1, the shared electron pair in $\ce{R-X}$ has been split to form two radicals: $\ce{R}{\small{•}}$ and $\ce{X}{\small{•}}$. However the $\ce{X}{\small{•}}$ radical is further attacked by an electron, this time coming from the $\ce{Mg}$. Thus forming a free-radical magnesium halide: $\ce{X^-Mg^+}{\small{•}}$.

After this, the two free-radicals undergo a termination step to form the Grignard reagent.


How does aluminium react with bases to form aluminates? An example of reaction:

$$\ce{2Al + 2KOH + 6H2O->2 K[Al(OH)4] + 3H2 ^}$$

Aluminium is not ionic, how then does it attract the $\ce{OH-}$ groups to bond with them into the complex ion $\ce{[Al(OH)4]-}$?

I started re-reading non-organic chemistry at a slightly deeper level recently after going through the basics of organic chemistry. This reaction stumped me. I've re-read some stuff about aluminium and complex ions, but can't seem to understand it yet.

There's no positive charge on $\ce{Al}$, yet it reacts with hydroxide groups floating around. Where will the electron from a non-ionic aluminium atom go?

I cannot 'imagine' the reaction, visualize its mechanism, and this makes it hard to remember it.

P.S. Aluminium hydroxide seems to be mysterious in its structure. I quote Chemguide:

  The chemistry textbooks that I have to hand aren't too clear about the structure of aluminium hydroxide as far as the degree of covalent character is concerened, and a web search (until I got totally bored with it!) didn't throw up any reliable chemistry sites which discussed it. Several geology sites describe the mineral gibbsite (a naturally occurring form of aluminium hydroxide) in terms of ions, but whether it actually contains ions or whether this is just a simplification as a convenient way of talking about and drawing a complicated structure, I don't know.

There's no positive charge on Al, yet it reacts with hydroxide groups floating around.

Your observation is correct. In the beginning, aluminium in its elemental state has the oxidation state 0. Apparently, that changes in the course of the reaction:
$$\ce{Al <=> Al^3+ + 3e-}$$

  Where will the electron from a non-ionic aluminium atom go?

Again, your observation is correct. If one species is oxidized, another one must get reduced. In this case, protons from the water are reduced to form hydrogen gas:

$$\ce{ 2H2O + 2e- <=> 2OH- + H2 ^}$$

If you combine both half reactions and balance the stoichiometry, you end up with the initial equation for the whole redox reaction.


How to dilute a stock solution when the resulting solution has a smaller volume? Describe how you would dilute $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution to make $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.

I know the $$c_1 V_1 = c_2 V_2$$ formula and that magnesium hydroxide's molar mass is $58.32~\mathrm{g/mol}$. However, I am confused because $V_2$ is lower than $V_1$.

Maybe the question was posed this way to make you think practically.

Approach this problem as if you were in a lab and you actually had a bottle of $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution, and you wanted to make a new solution of $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.

You do not need to dilute the entire stock, but just a portion of the stock solution.  Describe what volume of stock solution you would take out, and how you would dilute it, using the $$c_1 V_1 = c_2 V_2$$ formula, to reach $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ of magnesium hydroxide.


What is the value for the reverse reaction? My question deals with the reaction $\ce{2NO(g) + O2(g) -> 2NO2(g)}$ with $K_c = 1752$.

This is what I have so far; $K_c$= products over reactants, so is the reverse reaction reactants over products to find the answer
Such as
$$1752 = \frac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$
So then does the answer become
$$1752 = \frac{[x]^2}{[x]^2[x]}$$

${K_c}$ of reverse reaction should just be the reciprocal of forward ${K_c}$. So, ${K_c}$ of reverse reaction is $\frac{1}{1752}$.

Now, I am not sure, why you used 2nd equation. Suppose you have 1 unit conc of $\ce{NO}$ and $\ce{O_2}$, then after time $t$, if the conversion is $x$, then your equation should be
$1752 = \frac{[2x]^2}{([1-x])^3}$.

Can you please check your 2nd equation again?


Artificial foods — Could you make them? And would they compare to their "organic" counterpart? This is partly just a hypothetical, but basically.
If you were to take all the chemicals in an apple and put them together, could you make an apple? (Probably closer to apple sauce though.)
If this is possible, How would my almost-apple paste compare to a normal apple when it comes to nutrition?
I imagine it would not be nearly as healthy as a normal apple. As from my research, supplements to a vitamin do not have the effect that the vitamin would have if consumed naturally.

I know apples contain Alpha-linolenic acid, Ferulic acid, Isoquercetin, Lutein, Neoxanthin, Niacin, Thiamine, Riboflavin, Pantothenic acid, Vitamin B6, Folate, Vitamin C, Vitamin K, Carbohydrates, Sugar, Calcium, Iron, Magnesium, Manganese, Phosphorus, Potassium, Zinc, Asparagine, D-Categin, Farnese, P-Coumaric-Acid, Avicularin, Quercitin, Rutin, Ursolic-Acid, and Protocatechuic-Acid.

This is presumably not a comprehensive list by any means. I also only really used apples as an example. If there's another food you feel would be easier to "build" then please use that.

Thank you for giving this a read! And I apologize if I posted in the wrong place, I'm new to stackexchange.

No, it is not currently possible to create whole foods (like an apple) (as opposed to processed foods, like sugar or vinegar*) from their component chemicals.  The reason is simple:  All whole foods are composed of tissues, which are in turn made of cells, and we aren't close to being able to synthesize cells, especially eukaryotic cells, from their component chemicals. Their structures are far too complex. [https://en.wikipedia.org/wiki/Artificial_cell]
*In this sense I would also consider honey a processed food, though it is processed by bees instead of humans.
The closest we can come to creating artificial tissues is to clone existing cells; but that wasn't your question.
So then the question becomes whether we could fully characterize all the components of a whole food (again, let's use the apple as an example), and create a mixture containing all those components—and whether, in so doing, we would have something nutritionally equivalent.
The answer to that is not known, and I would say the burden of proof lies with those who wish to claim it would be nutritionally equivalent.  Even if we could characterize every possible micronutrient in an apple, we still wouldn't necessarily know, much less be able to recreate, the states in which they exist (in particular, their states of association with other compounds) within the apple.
Certainly there is a difference between a micronutrient that is complexed with other micronutrients, and then contained within a structure and then contained within another structure, and one that is mixed within a paste and thus complexed with compunds with which it might not be naturally associated in the intact apple. Of course, we do "paste" the apple ourselves when we chew it, but that process likely does not lead directly to the type of paste you would have if you tried simply mixing all the apple's constituent compounds together.
And it's not just how we directly would react to the difference.  There's also how the organisms of our microbiome would react.  It's certainly possible that consuming intact apples would favor one species of microorganism, while consuming synthetic apple paste would favor another, thus leading to changes in the balance of our microbiome.
Then we have to combine all the above uncertainty with the enormous added uncertainty that exists surrounding human physiology and nutrition.


Two liquids that react to produce a gas less dense than air at room temperature Does anybody know two liquids (or a solid and a liquid) that can react to produce a gas less dense than air (or similar) at room temperature?

This question is intended to make a chemical induced reaction when a lever is pulled to produce gas and decrease an object's mass.

If the other products aren't significant, I'll recommend the all famous alkali metals' reaction with water:
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
Hydrogen gas is less dense than air, both at RTP and at STP, because the main components of air (by mole percent) are $\ce{N2}~(78\%)$, $\ce{O2}~(20.95\%)$ and $\ce{Ar}~(0.934\%)$ - all of which have a molar mass larger than hydrogen.


Are photons ejected from atomic nuclei? I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct? Can atomic nuclei eject photons?

I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct?

Not exactly.  The atom as a whole emits or absorbs the photon.  There is no reason to single out the electron versus the nucleus in such transitions.

  Can atomic nuclei eject photons?

Yes there are two ways a nucleus in particular (as opposed to an atom taken as a whole) can emit photons.

First, nuclei can exist in excited states. When an excited nucleus transitions to the ground state, a photon in the gamma or X-ray range of the electromagnetic spectrum is emitted.  Electrons do not have an analogous mechanism for emitting photons, because electrons are fundamental particles, while nuclei are not.

Second, if a nucleus has a nuclear spin (S), then in a magnetic field there will exist  2S + 1 quantized energy states of the nuclei.  Transitions of nuclei between these states emit or absorb photons, the energy of which depends upon the strength of the field, for example in the radio frequency range.  This is the basis of NMR and MRI.  The electron has an analogous mechanism for emitting photons, known as ESR or EPR.


Why does menthol (e.g. peppermint) feel cool to the tongue? Especially when drinking water after the fact, mint can give a sharp cold sensation inside one's mouth.  What process causes the sensation to occur?

Menthol it self gives a cold feeling in the mouth because it is active at the same receptor (an ion channel) on the tongue that cold temperature triggers. Interestingly, although they act at the same receptor, they act at different sites, so that provides the intensified response when eating a mint and then drinking water. This reference gives an excellent detailed answer, with references to the original papers, which I'll  summarize here.

Menthol acts at the TRPM8 protein which forms an ion channel that allows $\ce{Na+}$ and $\ce{Ca^2+}$ ions to flow into cells and this sends a signal saying "cool" to the brain. (As an aside, this protein monitors temperature across the body and not just on the tongue.) Cold temperatures actually change the confirmation of this protein, which allows the ions to flow more freely, and sends the signal to the brain. Menthol, on the other hand, stabilizes the open channel (allowing ions to flow even more freely) and also...

  “menthol shifts the voltage dependence of channel activation to more negative values by slowing channel deactivation”.  This is very significant to my question because it supports a claim made by the first web page I visited which stated that menthol acts on the receptors, leaving them sensitized for when the second stimulus is applied (i.e. cold water) resulting in the enhanced sensation.  This mechanism of binding is very clearly different from the mechanism of cold affecting the TRP channels.  This is why the sensation is increased when both stimuli are applied, yet is not affected after addition stimulation from the same stimuli (i.e. eating another mint).

All in all, pretty cool.


Covalent Bond/Hydrogen Bond During the formation of a molecule of water ($\ce{H2O}$), what kind of bond occurs between the elements? I'm studying this subject on my own and from some sources I saw that there is a covalent bond and others showed an hydrogen bond, responsible for the non-evaporation of water at $-90\ ^{\circ} \rm{C}$. Can you guys explain to me exactly how this bonds behave? Both of them occurs to the molecules of water?

Covalent bonds are the bonds between the atoms within the same water molecule. Hydrogen bonds are the bonds between two water molecules. All molecules have covalent bonds, but only some molecules have hydrogen bonds. As an example, water has hydrogen bonds, but carbon dioxide does not.

One of the requirements for hydrogen bonding is that the molecule must be polar. Water molecules are polar because of two effects. The first is that oxygen is more electronegative than hydrogen, so each of the OH bonds will be polar. The electrons shared in each of those bonds will spend more time near the oxygen than near the hydrogen. The other effect is the geometry of the molecule: the oxygen atom has two free electron pairs in addition to the two bonds, so it assumes a tetrahedral arrangement. The $\ce{O-H}$ bonds are two corners of that tetrahedron, and the free electron pairs are the other two corners. This arrangement means that, if you only look at the atoms, the molecule has a bent geometry. Because of that, the polarity of the two bonds partially adds together. By contrast, $\ce{CO2}$ has a linear geometry, and the two polar bonds cancel each other out, leaving the molecule non-polar.

So the water molecule is polar, with a lot of excess positive charge around the hydrogen atoms, and a lot of negative charge on the side of the oxygen atom away from the hydrogen atoms. Hydrogen bond is the name given to the electrostatic interaction between the positive charge on a hydrogen atom and the negative charge on the oxygen atom of a neighboring molecule. The covalent bond is the electrostatic interaction between two atoms in the same molecule. Covalent bonds are much stronger than hydrogen bonds: the $\ce{O-H}$ has a strength of 467 kJ/mol, while the hydrogen bond is usually between 4 to 40 kJ/mol.


What is the name of this ester? Can you tell me the name of this ester? It is two $\ce{C8}$ fatty acids ester-linked with a molecule of 1,3-butanediol in the middle. I would be tempted to say (R)-butanediol-1,3-diyl dihexanoate, but I am really not sure.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are usually formed by placing the alcoholic component in front of the name as a separate word. The particular subsections concerning esters from a single alcoholic component and multiple acid components read as follows:

  P-65.6.3.3.3 Polyesters formed from a single ‘alcoholic’ component

  Esters derived from a single ‘polyhydroxylic’ component are named by placing the name of the ‘polyhydroxylic’ component denoted by a multivalent organyl group (alkyl, aryl, etc.) in front of the name(s) of the acid component denoted by the anion names derived from the appropriate acid(s) (…).

  P-65.6.3.3.3.1 When anions are identical, functional class multiplicative nomenclature is used. Names are formed by citing the multivalent group, the multiplicative prefix, and the multiplied anionic component name. Multiplicative prefixes ‘di’, ‘tri’, etc. are used when anions are unsubstituted; when substituted, prefixes ‘bis’, ‘tris’, etc. are used.

In this case, the polyhydroxylic component is denoted by the multivalent organyl group ‘butane-1,3-diyl’. The acid component is denoted by the anion name ‘hexanoate’, which is derived from the corresponding acid ‘hexanoic acid’. Since both acid components are identical and unsubstituted, the multiplicative prefix ‘di’ is used in functional class multiplicative nomenclature. Therefore, the preferred IUPAC name for the compound given in the question is (3R)-butane-1,3-diyl dihexanoate.


Why Does Electrical Work Cause Changes in Internal Energy of the System? If we look at system at constant temperature and volume which is galvanic cell, first law of thermodynamics states: $$ dU = dQ + dW'$$
Where W' is electrical work done by galvanic cell and Q is heat exchanged with surroundings.
As far as I know electrical work is work done by electric field when charge moves between certain potential difference. In galvanic cell, potential difference exists between two electrodes.
Internal energy is sum of total microscopic kinetic energy of all molecules and atoms and their potential energy of interaction (energy stored in intermolecular and chemical bonds).
What does work done by electric field have to do with internal energy of the system? Why does electrical work cause changes in internal energy of the system?
In galvanic cells, electrical work is negative as work is done by the system and from the first law it tends to decrease internal energy of the system.
I can't really see the connection between work done by electric field in the galvanic cell and sum of microscopic kinetic energy and microscopic potential energy of interaction.

If water is electrolyzed producing $\ce{H2}$ and $\ce{O2}$ gases, a great amount of energy has to be added to this water. The obtained substances are not hotter than the original water. But they possess a much higher potential energy. Their internal energy (and enthalpy, and free enthalpy) is much higher than the same atoms in water. This internal energy may be liberated into heat if these substances are sent in a flame, or into electric energy in a combustion cell.
It is no use thinking about the work done by an electric field on an electron in the atoms. The electric field inside an atom is difficult to calculate, because of all other electrons interfering with the field of the nucleus inside the atom.


Is HCl a lewis base? Cl in HCl has three lone pairs of electrons. So it can donate a lone pair and thus become a lewis base. Right? Just like water is called a lewis base because oxygen has 2 lone pairs and can donate a lone pair.

The strength of Lewis acids relative to your Lewis base has been commented on. A fuller answer to your question needs to include the Lewis acid you are considering.
The original Bronsted-Lowry theory of comparing acidity and basicity, or acids and bases, involved $\ce{H^+}$ and $\ce{OH^-}$ and the transfer of a proton.
The broader Lewis theory allowed consideration of other cations besides the proton as acids, and other nucleophiles than hydroxide ion to be bases. So water can be considered a Lewis acid relative to chloride ion because $\ce{H2O}$ hydrates chloride ions by hydrogen bonding. And water is a Lewis base relative to metal cations because water hydrates the metal ions through electron donation from oxygen.
So the answer to the question is yes and no. Yes, $\ce{HCl}$ can be considered a Lewis base - relative to a very strong acid, as mentioned in the comments - but in the ordinary world, I would not first think of $\ce{HCl}$ as an alkali. Stretching the word "base" by adding "Lewis" in front of it needs more, like adding the reference acid, to make the comment clear rather than confusing.


UV-Vis emission from charge-separated states? Are there examples of UV-Vis emission from charge-separated states? The cases I have encountered have only absorption. Is the recombination process of a nonradiative kind by necessity?

There are at least cases in which charge recombination results in an excited species that deactivates via fluorescence.

Imagine the ionization of a hydrocarbon ($\ce{RH}$) to yield a radical cation and a solvated electron.
\[\ce{RH ->[\mathrm{e-beam}] RH^{+\cdot} + e-}\]

Trap the electron by reducing an aromatic hydrocarbon ($\ce{ArH}$) to the coresponding radical anion.

\[ \ce{ArH +e- -> ArH^{-\cdot}}\]

Charge recombination between the radical cation and the radical anion can yield $\ce{ArH}$ in an excited state.

\[\ce{RH^{+\cdot} + ArH^{-\cdot} -> RH + ArH^{\ast}} \]

Back in the days, Alexander D. Trifunac did a lot of work on that at ANL.


IUPAC name of H[AuCl4] Is $\ce{H[AuCl4]}$ tetrachloroauric [III] acid or hydrogen tetrachloroaurate[III]? As per the rules I have studied the latter name is correct. Then why am I getting the former name mentioned everywhere on the internet as the IUPAC name, although Wikipedia does mention the other name (hydrogen tetrachloroaurate[III]) though not as the IUPAC name?

The authoritative source Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (Red Book) lists $\ce{H[AuCl4]}$ as an example of a salt in the subsection IR-4.4.3.4 Generalized salt formulae [1, pp. 61–62].
Further, introduction to the section IR-8 Inorganic Acids and Derivatives underlines that IUPAC nomenclature is established from composition and structure, not properties, hence the word "acid" should not be used in systematic names [1, p. 124]:

Certain inorganic and simple carbon-containing compounds are commonly given non-
systematic or semi-systematic names containing the word ‘acid’. Examples are boric acid or
orthoboric acid, metaboric acid, phosphoric acid, diphosphoric acid, cyclo-triphosphoric
acid, catena-triphosphoric acid, dithionous acid, peroxodisulfuric acid or peroxydisulfuric acid, etc. These names are unique in modern nomenclature in that, interpreted literally, they describe a particular chemical property of the compounds in question. Systematic names are otherwise based solely on composition and structure.
All such acids may also be given structure-based systematic names using principles
already described in preceding chapters on substitutive and additive nomenclature, so in that respect the ‘acid’-containing names are superfluous. Furthermore, many species which
would be classified as acids based on their chemical properties are never named as such,
e.g. aqua ions such as hexaaquaaluminium(3+), and hydrides and derivatives such as
ammonium, hydrogen sulfide (sulfane), etc. The term ‘acid’ is thus not used consistently.
Based on these considerations, the use of the word ‘acid’ in any new name in inorganic
nomenclature is discouraged. However, a number of the existing ‘acid’ names are so
commonly used (sulfuric acid, perchloric acid, etc.) that it would be unrealistic to suggest replacing them altogether by systematic alternatives.

According to Wikipedia, $\ce{H[AuCl4]}$ exists in aqueous solutions as square planar $\ce{AuCl4-}$ anion, which is also retained upon crystallization as tetrahydrate, but there is a rapid
exchange of all protons in the unit cell [2]. For compounds of this class where the structural aspects of protons are either not essential or cannot be precisely established, a hydrogen name can be used. From the section IR-8.4 HYDROGEN NAMES [1, pp. 134–136]:

An alternative nomenclature for hydrogen-containing compounds and ions is described here. The word ‘hydrogen’, with a multiplicative preﬁx if relevant, is joined (with no space) to an anion name formed by additive nomenclature and placed within appropriate enclosing marks (see Section IR-2.2). This construction is followed (again with no space) by a charge number indicating the total charge of the species or structural unit being named (except for neutral species/units).
Hydrogen names are useful when the connectivity (the positions of attachment of the
hydrons) in a hydron-containing compound or ion is unknown or not speciﬁed (i.e. when
which of two or more tautomers is not specified, or when one does not wish to specify
a complex connectivity, such as in network compounds).
[…]
Examples:
[…]

$\ce{H4[Fe(CN)6]}$
tetrahydrogen(hexacyanidoferrate)
$\ce{H2[PtCl6]·2H2O}$
dihydrogen(hexachloridoplatinate)—water (1/2)

[…]
The strict deﬁnition of hydrogen names proposed here is meant to eliminate such
confusion by imposing the requirements:
(i) that ‘hydrogen’ be attached to the rest of the name,
(ii) that the number of hydrogens must be speciﬁed by a multiplicative preﬁx,
(iii) that the anionic part be placed in enclosing marks, and
(iv) that the charge of the total structure being named is speciﬁed.
Hydrogen names constructed in this way cannot be mistaken for other types of name.

To sum it all up, I think the IUPAC name for $\ce{H[AuCl4]}$ is hydrogen(tetrachloridoaurate).
Note that new recommendations also suggest using chlorido instead of chloro for anionic ligands. Also, in names formal oxidation states are enclosed in parentheses, not square brackets.
Wikipedia might not always mention IUPAC names, and even those which are listed as such should not be automatically trusted. Also, acceptance of systematic names in chemical community is inertial; not every systematic name is immediately replacing the traditional one which everyone is familiar with. I suspect this is the case as to why chloroauric acid still gets a pass.
References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.
O’Reilly, D. E.; Peterson, E. M.; Scheie, C. E.; Williams, J. M. Nuclear Magnetic Resonance of the Aquated Proton. II. Chloroauric Acid Tetrahydrate. Phase Transitions and Molecular Motion. J. Chem. Phys. 1971, 55 (12), 5629–5635. DOI: 10.1063/1.1675731.


Having trouble naming hydrocarbons with the IUPAC nomenclature rules First of all, thanks a lot for taking the time. I'm homeschooling and don't currently have any proper teachers, so I thought this would be a good place to ask. Its an easy question I think, but it has me very confused.
In my chemistry book, they showed this hydrocarbon
Translated from Spanish, it would be something like 4-ethyl -1,2 dimethylcyclohexane. Now, my question is, according to the rules, when there is more than one different type of ramification, you number the carbons giving the radical with alphabetical preference the lowest value, in this case, ethyl is first in the alphabet than methyl, so shouldn't it be 1-ethyl- 3,4 dimethylcyclohexane? They would all still be between the first 4 carbons, so the rule of lowest value wouldn't really change. Unless, the di, indicating the 2 methyl radicals counts and gives it alphabetical preference, but to my understanding, those prefixes aren't considered when giving alphabetical preference. Thanks for your time and have a great day!

The first name has the correct alphabetical order of groups: ethyl before dimethyl. The "d" of dimethyl is not involved in alphabetizing. The numbering is correct because the one-to-one comparisons of locants is {1,2,4 < 1,3,4}. The correct name is 4-ethyl-1,2-dimethylcyclohexane. For a brief set of IUPAC rules applicable to this question see, https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm


What is the logic behind complex splitting? Like, on a superficial level I know that due to different coupling constant value complex splitting follows a different rule than when splitting involves same hydrogen neighbors. But I don’t understand why and how complex splitting occurs at a intuitive level. Like in complex splitting, is the magnetic moment from the two different neighboring hydrogen adding or subtracting together to affect nmr peak of one of the peak? Is that what’s happening?

Spin coupled nuclei are affected by the spin states of their coupled partners. That is to say, they can detect whether their spin coupled partner is spin up or spin down (for spin 1/2 nuclei), and the energy level transitions are different, depending on whether that other spin is up or down. If a nucleus is coupled to more than one spin, it is affected by the spin state of each of those nuclei. 'Complex' coupling or splitting is a qualitative term used to describe first order couplings that involve more than two coupled nuclei. They can easily be described using standard splitting diagrams.

Consider the example below for the spin system AMX

In the absence of any spin coupling, spin A will appear as a singlet - it has only one energy level transtion available. If it is coupled to spin M, the energy levels of spin A will depend on whether spin M is up or down, and so there are two non-equivalent energy level transitons possible for spin, different in energy determined by the magnitude of the coupling constant. Now, if Spin A is coupled to M AND X, then each of the combinations of the AM transitions will also depend on the spin state of X, and so each of the AM transitions will also have two possible transtions - a total of 4 lines, or what we would call a doublet of doublets.

A useful reference for interpreting complicated, but first order, splitting patterns is by Hoye:

http://pubs.acs.org/doi/abs/10.1021/jo001139v


Galvanic Cell: Temperature vs Voltage $$\ce{Cu^2+ + Ni -> Cu + Ni^2+}$$

The standard potential of this redox reaction is 0.57 volts (which was obtained by 0.23 + 0.34)

$$E = E_0 - \frac{RT}{nF}\ln Q$$

I used 0.1 M nickel sulphate and 0.5 M copper sulphate in my galvanic cell.

$$E_{25} = 0.57 - \frac{8.31 \cdot 298}{2 \cdot 96485}\ln(0.2)=0.59065\ldots$$

$$E_{45} = 0.57 - \frac{8.31 \cdot 318}{2 \cdot 96485}\ln(0.2)=0.59204\ldots$$

Shouldn't the two be inversely proportional? Is the effect of temperature really this minuscule?

Indeed, it should be inversely proportional. In case you divide the copper concentration by the nickel concentration, the sign of the RT/nF part should be plus and not minus. If you divide the product of the reduction by the precursor concentration, you have to use minus.

The temperature has a great effect on the current, especially in the "Cottrell-determined" area. The Nernst equation, which you use in this case, is true for a near-equilibrium environment with fast electrode kinetics. If you drive the potential too high, mass transport can become so slow (in comparison to electron transfer) that an increase in potential will not yield a higher current flow as predicted by Butler-Volmer kinetics.


The boiling points of Alkali metal group As I know, the melting points of Alkali metals decrease down to the group due to decreasing of the energy of the metallic bond.

But according to my understanding, after it has become to liquid the metallic bonds are not affecting furthermore because of the lattice has destroyed, so the boiling point affects only the atomic mass. Then according to my logic the boiling points should be increase down to the group. But the above graph shows the opposite of that.

What is the wrong in my logic. Please help me.

You can explain the trend in terms of increased charge density, and ensuing charge repulsion between nuclei and between electrons as you go down the group.
An alternative explanation is provided in a libretext, which draws information from a popular educational chemistry site:

When any of the Group 1 metals is melted, the metallic bond is weakened enough for the atoms to move more freely, and is broken completely when the boiling point is reached. The decrease in melting and boiling points reflects the decrease in the strength of each metallic bond.
The atoms in a metal are held together by the attraction of the nuclei to electrons which are delocalized over the whole metal mass. As the atoms increase in size, the distance between the nuclei and these delocalized electrons increases; therefore, attractions fall. The atoms are more easily pulled apart to form a liquid, and then a gas. As previously discussed, each atom exhibits a net pull from the nuclei of +1. The increased charge on the nucleus down the group is offset by additional levels of screening electrons. As before, the trend is determined by the distance between the nucleus and the bonding electrons.

Although the two perspectives (my opening explanation and that in the libretext) would seem to be equivalent, they aren't identical, since a description in terms of Coulomb's law is more fundamental and does not require invoking concepts such as polarizability or moments. I would argue that appealing to nuclear screening to explain bonding is not the best way to understand the trend.


Why does ozone have an electrostatic potential map similar to that of SO2? This is the electrostatic potential for ozone.

This is the electrostatic potential for sulfur dioxide.

(Click to enlarge)

They look almost exactly the same except one is all oxygen atoms and the other has sulfur in the center instead of oxygen.

Why does ozone have an MEP (molecular electrostatic potential) similar to that of SO2?

Resonance structures of $\ce{SO2}$ and $\ce{O3}$.

As you can see from the diagram $\ce{O3}$ and $\ce{SO2}$ are isoelectric and have the same shape and this accounts for their very similar electrostatic potential maps. Both molecules are polar although $\ce{SO2}$, with a dipole moment of 1.62D, is more polar than $\ce{O3}$ (0.53D). As the diagram shows, the polarity of both molecules is due to the their trigonal planar shape with a formal positive charge on the central atom.


Names of NAD+ and NADH As part of the biochemical process of glycolysis, NAD+ is reduced to NADH. In other words, nicotinamide adenine dinucleotide is reduced to... nicotinamide adenine dinucleotide?

Everywhere I look tells me that NAD+ and NADH refer to the same particle with the same name, just in an oxidized or reduced form. However, NADH has one more hydrogen atom then NAD+, so shouldn't these particles have different names?

The H stands for hydride, but most everyone just shortens nicotinamide adenine dinucleotide hydride to NADH.


When is it necessary to check wavefunction stability in density functional theory? In Gaussian, there is a stable keyword that checks the stability of the wavefunction. Using stable=opt reoptimizes the wavefunction until a stable one is found if there is an instability.
My question is - when do you use this stability check? Presumably one must have some hunch in advance that the wavefunction may be unstable in order to use this keyword, but what insight would lead one to this conclusion?

To add some context, this question came to mind after following instructions on how to model a system with antiferromagnetic coupling (e.g. as shown here). When I tried this procedure on cupric acetate hydrate, which has a dinuclear structure of copper(II) atoms and is known to be antiferromagnetic, I had to use the stable=opt keyword to stabilize the wavefunction resulting from the "fragment guess job" otherwise it erroneously appeared that the antiferromagnetic singlet was higher in energy than the triplet. This made me wonder when I should be using stable=opt in general (perhaps outside the context of antiferromagnetic coupling calculations) and also what it is specifically doing.

The very short answer is: whenever you are in doubt.

The stability check is probably unnecessary when you are dealing with closed shell molecules, where you can be pretty sure that a restricted methodology is reasonable. For example, when you calculate water at the RHF level of theory, you won't really need to look for a more stable wave function, because it would not be physically sensible. An UHF approach in principle will still lead to a lower energy since it has more flexibility.

Whenever you are calculating molecules in which you would assume some open shell character you should be considering checking wave function stability. It usually doesn't hurt to perform one check too many. Discovering instabilities later can be a lot more frustrating. That does not necessary mean that you are gunning for the correct ground state, as you are still restricted on spin. I'll include a small, but well-known example below: $\ce{O2}$. I am using Gaussian 09 Rev. D.01 for the calculation, but it should work with anything more recent. (It won't work the same way with anything older.) Also note, that I am not conducting geometry optimisations, but use $\mathbf{d}(\ce{O-O})=\pu{120pm}$ to keep it simple.

First let's run an ordinary, restricted calculation. If you are not using the SCF(xqc) keyword, you should immediately notice, that something is wrong.

%chk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit             ! Use density fitting to spead up calculation
gfinput            ! Prints basis set in a form suitable for use as general basis set input
gfoldprint         ! Prints the basis set information in the Gaussian format
iop(6/7=3)         ! Switches on printing of all MO
symmetry(loose)    ! Use correct symmetry
scf(xqc,conver=10) ! Quadratic convergence after conventional (for tough cases)
INT(SuperFineGrid) ! Largest grid, most accurate

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

You'll find a summary of the energies at the end, if you want to check and run your calculations.

Next up, we'll check and optimise the wave function.

%chk=bp86svp.rhf.sing.stab.opt.chk
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10) INT(SuperFineGrid)
stable=opt

title card unused

0 1

You should find the following in your output:

 The wavefunction has an RHF -> UHF instability.

Followed by another SCF run. Pay close attention to the following:

 SCF Done:  E(UB-P86) =  -150.201660211     a.u. after    6 cycles
            Convg  =    0.3033D-11                    52 Fock formations.
              S**2 =  1.0016                  -V/T =  2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

You are still finding a singlet state, but now as a UHF wave function. You should later find:

 The wavefunction is stable under the perturbations considered.

What you have found is a broken symmetry singlet. We do know that the ground state should be a triplet. Therefore we'll perform a restricted open shell calculation. We already know this one will have and RHF -> UHF instability, and since we cannot use the quadratic convergence, we cannot use stable either. But it'll provide us with a good starting point for the UHF calculation, or at least an energy to compare others to.

%chk=bp86svp.rohf.trip.chk
%oldchk=bp86svp.rhf.sing.chk
#p ROBP86/Def2SVP/W06
DenFit
geom=check guess=alpha
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(conver=10) ! Quadratic convergence not available for RO
INT(SuperFineGrid)

title card unused

0 3

The important part here is, that it generates a pure spin state wave sunction:

 = 0.0000 = 0.0000 = 1.0000 = 2.0000 S= 1.0000

Another calculation is using the the mixing of HOMO and LUMO to generate a broken symmetry ansatz wave function directly. You'll find the same result as with the approach above.

%chk=bp86svp.uhf.sing.chk
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
geom=check guess=(read,mix)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

title card unused

0 1

You'll therefore find again:

 SCF Done:  E(UB-P86) =  -150.201660211     A.U. after   11 cycles
            NFock= 11  Conv=0.88D-12     -V/T= 2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 KE= 1.494731832724D+02 PE=-4.114759744301D+02 EE= 8.357834648903D+01
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

The wavefunction is stable under the perturbations considered.

Now for the last part, run a real UHF calculation:

%chk=bp86svp.uhf.trip.stab.opt.chk
%nproc=8
%mem=8000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

title card unused

0 3

You'll find as expected

 Annihilation of the first spin contaminant:
 S**2 before annihilation     2.0031,   after     2.0000

 The wavefunction is stable under the perturbations considered.
 [...]
 The wavefunction is already stable.

For giggles, why not consider a quintet?

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155030787 (      4 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201660211 (      6 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213771982 (     10 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201660211 (     11 )
bp86svp.uhf.trip.stab.opt E(UB-P86)       =       -150.215679340 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651916836 (      9 )

In conclusion, whenever you suspect some open shell character, check for stability. If you have well-behaved molecules (singlet), like most organics are, you can probably skip it.
Be aware that a stability optimisation will not necessarily give you the ground state.


Psychoactive ingredients in this soap? Could any ingredients from this product be psychoactive in some people, if the are sensitive to them?
https://www.lush.co.uk/products/honey-i-washed-kids

I have found the product to be relaxing as advertised, but more so that I would expect from a nice scent alone.

None of the ingredients listed would be classified as a psychotropic substance.  However, many of the compounds, when studied in mice, do have effects on brain activity and behavior characteristics.  For example,

rapeseed oil (in a mixture with other vegetable oils): "the rapeseed oil group exhibited much higher locomotor activity than that expected from the alpha-linolenate/linoleate ratio. Additionally, the rapeseed oil group exhibited unusual behavior patterns, including higher ambulation and rearing activities, faster acquisition of the water maze task and slower habituation behavior as compared with the control group. Susceptibility to pentobarbital anesthesia tended to be higher in the rapeseed oil group." abstract link
bergamot: "produces dose-dependent increases in locomotor and exploratory activity that correlate with a predominant increase in the energy in the faster frequency bands of the EEG spectrum" abstract link
linalool: "Inhaled linalool showed anxiolytic properties in the light/dark test, increased social interaction and decreased aggressive behavior" abstract link

So I am not surprised to read that when used in combination they have a noticeable affect on you.


Why don't calcium chloride and sodium nitrate react? $$\ce{CaCl2 + 2NaNO3 -> 2NaCl + Ca(NO3)2}$$

Why doesn't the above reaction occur? Does it have to do with the metal activity series or solubility rules?

It's because all the species are soluble in water and there is no driving force such as precipitation or gas evolution. You just have a mixture of ions.


IUPAC name of trisubstituted benzene ring What is the IUPAC name for the compound shown below? I named it as 6-Bromo-3-hydroxy benzonitrile according to the numbering scheme in 1b, since the priority order is $\ce{-CN} > \ce{-OH} > \ce{-Br}$.

However, the answer given in the book is 2-Bromo-5-hydroxy benzonitrile, which implies the numbering scheme in 1a. Please help me to identify whether my answer or the book's answer is correct.

As -CN has the highest priority, root-word is benzonitrile. Lowest locant is 2 for -Br.
So, the name given is:
2-bromo-5-hydroxybenzonitrile

As the functional group is -CN, with highest priority, no need to look for priority of -OH and -Br; just look for lowest locant.


Reason for the greater energy of the eclipsed conformer of alkanes? My textbook states:

  In the staggered form of ethane, the electron clouds of C-H bonds are as far apart as possible, thus there are minimum repulsive forces, minimum energy and maximum stability. On the other hand, when the staggered form changes into the eclipsed form, the C-H bond clouds come closer to each other resulting in increased electron cloud repulsion. To check the increased repulsive forces, molecule will have to possess more energy. And therefore it is less stable

My question pertains to the statement in bold text. Am I correct in thinking that in the eclipsed conformer the electron clouds repel more, this repulsion would (sort of) want to break up the molecule so to fight this breaking tendency, the molecule has to possess greater energy to control these new repulsive forces?

If I am wrong, what does this statement mean?

Just take two electrons. If you want to push them closer together you need to input some energy and the system overall will have higher energy. There is a preference for the two electrons to be farther apart.

This situation is not any different if we look at pairs of electrons in bonds instead of individual electrons.


Schrödinger equation and wave function What is the Schrödinger wave equation?
What are radial probability functions, nodes-angular and radial and electron cloud?

I am really confused with all these terms and all my book seems to offer me is a set of very complex words and definitions. I am relatively new to atomic structure and related things, so a brief answer in simple words would really be helpful!

I also would like to know more about Hamiltonian and Laplacian operators and related subjects.

There are a lot of questions in this one post which is probably why there has not yet been an attempt to answer this post.

1.) What is the Schrödinger wave equation?

The wave equation (i.e. wave function) is a function which contains all the information about a chemical system.  With an appropriate operator, one can extract this information and gain insight into the system.  For example, the Hamiltonian operator will extract the energy of a system from the wave function.  The wave function is an eigenfunction while the energy (in this example) is the eigenvalue of the eigenfunction.  For example, in the Schrödinger equation

$$ \mathcal{H}\Psi = E\Psi$$

$\mathcal{H}$ is the Hamiltonian operator, $\Psi$ is the wave function (eigenfunction) and $E$ is the energy (eigenvalue).  Notice that by operating on $\Psi$ with $\mathcal{H}$ returns $E$ AND $\Psi$, which by definition, makes this an eigenfunction-eigenvalue equation.

2.) What are radial probability functions?

These are probability distributions for the probability of an electron residing some distance from the nucleus of an atom such as the hydrogen atom.  Consider this plot of various radial probability distribution functions of the hydrogen atom.

Here we see that the probability of an electron (in the hydrogen atom) in a 1s orbital is zero at the nucleus (at 0,0) and a maximum a hair shy of $2 a_0$.  Interestingly an electron in a 2p orbital has a small probability of residing close to the nucleus ($< 1a_0$) but is more likely to reside further away from the nucleus than an electron in a 1s orbital.  This makes sense given that a 2p orbital is higher in energy because 'it' is further away from the nucleus than a 1s orbital.

3.) What are nodes in radial and angular functions?

A node is where the function is zero at a particular point.  This simply means in a physical sense that at this point (at the node), the probability of an electron existing at that point is ZERO.  Nodes are a result of the separation of variables technique when solving the appropriate equation(s). When any one component (or variable) is zero then the resulting function will be zero at that point.

4.) What is electron cloud?

Electron 'cloud' is just a nicer away of saying electron density.  Electrons are not strictly particles.  Electrons do not just 'orbit' a nucleus.  Electrons also exhibit wavelike behavior.  (see: double-slit experiment - wave-like behavior and J. J. Thompsons cathode ray tube experiments - particle-like behavior).  The 'cloud' is obtained simply by 'squaring' the wave function.  More technically, its taking the product of the wave function and its complex conjugate.  The 'cloud' is just a probability distribution of where an electron is likely to be (usually given in 3-dimensions).

5.) What is the Hamiltonian operator?

The Hamiltonian operator is the energy operator to the Schrodinger equation.  Because it is Hermitian, it will return a 'real' result (as opposed to a complex result).  There are different versions of the Hamiltonian depending upon what approximations you want to make.  The full Hamiltonian operator can be written as (in terms of atomic units)

$$\mathcal{H} = -\sum\limits_{i=1}^{N}\dfrac{1}{2}\nabla_i^2
              -\sum\limits_{A=1}^{M}\dfrac{1}{2M_A}\nabla_A^2
              -\sum\limits_{i=1}^{N}\sum\limits_{A=1}^{M}\dfrac{Z_A}{r_{iA}}
              +\sum\limits_{i=1}^{N}\sum\limits_{j>i}^{N}\dfrac{1}{r_{ij}}
              +\sum\limits_{A=1}^{M}\sum\limits_{B>A}^{M}\dfrac{Z_A Z_B}{R_{AB}}
$$

Each term on the right hand side of the equation (going from left to right) is defined as

the kinetic energy of the electrons
the kinetic energy of the nuclei
the potential energy of the electron-nuclear attractions
the potential energy of the electron-electron repulsions
the potential energy of the nuclear-nuclear repulsions

We can write the Hamiltonian using a bit more compact notation such as

$$
\mathcal{H} = \hat{T}_{e} + \hat{T}_{N} + \hat{V}_{eN} + \hat{V}_{ee} + \hat{V}_{NN}
$$

where negative signs are assumed for each $\hat{T}$ operator.  The terms denoted here correspond to the respective terms in the equation before this one.

If we take into account the Born-Oppenheimer approximation (the B.O. approximation assumes that the electron motion is orders of magnitude larger than the nuclei they belong to... so much faster that the nuclei are, relatively, standing still with respect to the electrons), we can make two changes to the full Hamiltonian giving us the electronic Hamiltonian.

$$
\mathcal{H} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee} +\ \mathrm{constant}
$$

The nuclei are treated as frozen in space when the B.O. approximation is applied for reasons explained above.  Therefore the kinetic energy of the nuclei goes to zero.  The nuclear-nuclear repulsion is now a constant because the nuclei are fixed.  The repulsion will never change.

6.) What is a Laplace operator?

The Laplace operater (or Laplacian) is simply a differential operator of the form

$$
\nabla^2 = \dfrac{\partial^2}{\partial x^2} +  \dfrac{\partial^2}{\partial y^2} +  \dfrac{\partial^2}{\partial z^2}
$$

if you are working with Cartesian coordinates. This is simply a set of second-order derivative operators within the kinetic energy operator.

While I may have addressed each question, there is a lot more depth to these answers that lie well beyond the scope of this post.  I encourage you to find more detailed and insightful answers in well-established texts.  I highly recommend Szabo & Ostlund for a solid introduction to quantum mechanics.  You can snag a paperback version of this excellent work for about $10.


Which pH approximation to choose when? When I learned about pH approximations in school, I was told that each approximation fits to a certain acidity strength/pKa range. Those ranges are grouped in common categories (pKa borders are only approximately, as they change from book to book):

very strong acids: $\mathrm{pK}_a < 0$
strong acids: $0 \le \mathrm{pK}_a < 4$
medium strong acids: $4 \le \mathrm{pK}_a < 8$
weak acids: $8 \le \mathrm{pK}_a < 14$
very weak acids: $\mathrm{pK}_a \ge 14$

From this point on, the arguments were made and approximations for certain strengths of acids were evaluated. But is it possible to choose the "right" approximation only based on acidity strength? Do those equations really only apply to it's categories?

This question originates in my wish to implement a program on my calculator that computes the pH for a given concentration and pKa value. I thought it might be wrong to simply ask for the pKa value and choose one of the approximations solely based on that.

Preface

It is being assumed throughout this post that the concentration c equals the activity a. This is of course not true for concentrations above $\approx 10^{-3} \ldots 10^{-2}~\mathrm{mol\,L^{-1}}$ but it is common practice in schools and universities as long as the exercise is not very specifically about activities.
$$\mathrm{pH} = -\lg a(\ce{H+}) \xrightarrow{a=c} \mathrm{pH} = -\lg [\ce{H+}]$$
The other general assumption is that temperature is $22~^\circ \mathrm{C}$ and therefore $k_\text{W} = 10^{-14}~\mathrm{mol^2\,L^{-2}}$.

Given error contours are not "real measurements vs. calculations" but absolute differences between approximation n and the cubic equation that is not more simplified than using concentrations instead of activities.

General

If you have a calculator that can solve cubic equations, you don't need to simplify. In addition you only would have to remember one equation, no matter which pKa or concentration occurs (see above for real world restrictions).

Solve the following equation for x, with $x=[\ce{H+}]$:
$$x = \underbrace{\frac{k_\mathrm{W}}{x}}_{[\ce{OH-}]} + \underbrace{\frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}}_{[\ce{A-}]}$$
or as a single equation (that is harder to remember):
$$x^3+x^2 k_\mathrm{a}-x([\ce{HA}]_0 k_\mathrm{a}+k_\mathrm{W})-k_\mathrm{a} k_\mathrm{W} = 0$$

Results from this equation will be the reference pH values for the following comparison of pH approximations.

To calculate the pH for any monoprotic(?) base, simply replace [$\ce{H+}$] with [$\ce{OH-}$], $k_\mathrm{a}$ with $k_b$ and of course [$\ce{HA}$] with [$\ce{B}$].

Approximation 1 and 2

For very strong acids the approximation is that they are completely dissociated, which makes the resulting equation very easy:
\begin{align}
x &= [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg [\ce{HA}]_0
\end{align}
When am I able to use this approximation with an error that is smaller than $\Delta \mathrm{pH} \leq 0.1$? It's the blue area in the picture below ($\mathrm{pK_c} = -\lg [\ce{HA}]_0)$.

The simple rule is: $6.7 \gtrsim \mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5$.

Including the hydroxide ions from the autoprotolysis of water makes it a little bit more complicated and yields a quadratic equation in x:
\begin{align}
x &= [\ce{OH-}] + [\ce{HA}]_0 = \frac{k_\mathrm{W}}{x} + [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg \left(0.5 [\ce{HA}]_0 + \sqrt{0.25 [\ce{HA}]_0^2 + k_\mathrm{W}} \right)
\end{align}

This formula holds within: $(\mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5) \vee (\mathrm{pK_c} \gtrsim 7.4)$.

It's not surprising, that the range of usefulness increases over the whole are of diluted concentrations. And that the high pKa values are also covers is only due to the fact that the pH is 7 everywhere upon concentrations that are below $10^{-7}\ldots10^{-8}$.

Approximations 3 & 4

The following approximations are used for weak acids. It's the second easiest to remember, so it's used quite often. The assumption is, that $x \gg k_\mathrm{a}$.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a}} = 0.5 (\mathrm{pK_c} + \mathrm{pk_a})
\end{align}

It's application range is: $(\mathrm{pK_c} < 13.7 - \mathrm{pk_a}) \wedge (\mathrm{pK_c} < \mathrm{pk_a} - 0.6)$.

It is obvious, that the approximation is useful in a small area where the acid is not too diluted and it's pKa value is not too low. Seems to fit well for weak acids.

Again, what happens if $k_\mathrm{W}$ gets involved involved?

\begin{align}
x &= \frac{k_\mathrm{W}}{x} + \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a} + k_\mathrm{W}}
\end{align}

Again, the range expands over very weak acids and strong diluted solutions as expected: $(\mathrm{pK_c} < \mathrm{pk_a} - 0.6) \vee (\mathrm{pK_c} > 14.2 - \mathrm{pk_a})$.

Approximation 5

The last approximation is for strong acids that are not diluted. The amount of protons in solution is equivalent to the amount of dissociated acid but this time with taking $k_\mathrm{a}$ into account. As only the influence of the autoprotolysis of water is neglected, it is "the most accurate" approximation for high to medium concentrations. This can easily be seen as it includes the scope of both approximations, 1 and 2. It is again a quadratic equation in x.

\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}\\
\mathrm{pH} &= -\lg(-0.5 k_\mathrm{a} + \sqrt{0.25 k_\mathrm{a}^2 + [\ce{HA}]_0 k_\mathrm{a}})
\end{align}

It can be used if: $(\mathrm{pK_c} < 6.7) \wedge (\mathrm{pK_c} < 13.7 - \mathrm{pk_a})$.

Now there are all those approximations, with all their respective acceptable ranges but it's quite hard to compare five diagrams with another. I made two further pictures that show 1) which approximation is the best per (pkc,pks)-pair and 2) which approximation is the easiest to remember per (pkc,pks)-pair.

1) The best -- which function yields the lowest error in comparison with the exact equation?

2) The easiest -- which function yields the lowest error in comparison with the exact equation and if there is more than function with $\Delta \mathrm{pH} < 0.1$, which one is the easiest to remember?

So ... the pka ranges kind of resemble when to use which function but in the end it's a little bit more complicated. If $k_\mathrm{W}$ is neglected, the low-error-area (LEA) is quite small, only approx. 5 is able to yield acceptable values for a big range of concentrations and pka values. Approximations 1 and 3 should be used with care, as their LEAs are quite fancy. For approximations 1 and 3 it's often better to switch to their low-concentration counterparts 2 and 4 to be on the safe side.


Why does a silanized glass surface become hydrophobic? What is the physical reason that a glass surface becomes hydrophobic by silanization? (e.g. is it due to increase in surface roughness due to the long chains which is created on the glass surface? or ...)

Water has many, rather strong intramolecular bonds called hydrogen bonds, which influences many of its properties, for example being fluid even at high temperatures.

Wheter some thing is hydrophilic/hydrophobic is influenced mostly by the availability of places where hydrogen bonds can happen, these are either
1. bonds between hydrogen and an element with high electronegativity (e.g. O; N)
2. an orbital already filled with 2 electrons.

Because water molecules are also polar, charged or polar particles are also more hydrophilic. (e.g. salts)
Hydrophobic substances simply lack those features.

A good example illuminating this is that even small alkanes are quite insoluble in water whereas starch, a large molecule with many hydroxy groups is still soluble.

Of course water "tries" to minimize  its  surface energy for example by forming spherical (least surface are) drops. If a surface has many possibilities for  bonds on the other hand, water wets it.

A silanized surface has many long alkane chains, and therefore few possibilities to form bond, ergo it is quite hydrophobic. (In contrary to the normal glass, which  has many atoms with (formal) charge.)


Why can two carbon atoms not form more than triple bond with each other? Why two carbon atoms don't form more than triple bond with each other? Please tell me simply as I am a student of 10th class.

Unfortunately, an answer to this requires a quick dip into orbital theory. You may have heard of s, p and d orbitals; carbon as a main group element of group 14 has the electronic configuration of $[\ce{He}]\,\mathrm{2s^2\,2p^2}$, meaning two of its electrons are in an s orbital and two in p orbitals in its atomic ground state. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. (This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.)

The bonding (or antibonding) orbitals that are generated by this process are labelled σ or π orbitals. Note that these are the Greek letters corresponding to s and p: a σ orbital is basically an s orbital streched in one direction and a π orbital is a p orbital streched in the same manner.

Due to a number of constraints that you will learn about soon enough™, a carbon–carbon triple bond $\ce{C#C}$ will consist of one σ bond and two π bonds as shown in the image below.

Figure 1: $\ce{H-C#C-H}$ showing the σ and π orbitals required to make the respective bonds. Image taken from jahschem.wikispaces.org.

See how two of the three possible p orbitals on each carbon (figure 1’s left half) overlap to form two π bonds; the third possible p orbital takes part in both the σ bond to the other carbon and that to the hydrogen — it has been linear combined with the s orbital to form an sp hybrid orbital outside the scope of this answer.

The main point of interest here is that it is, for reasons of symmetry, not possible for more than two p orbitals of a single atom to form π bonds to a single other atom at any one time. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds.

The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding. Once you have a better understanding of orbitals, it may be worthwhile to look up quadruple bonds of transition metals such as in $\ce{[(H2O)Cr({\unicode{x3bc}-}OAc)4Cr(H2O)]}$ or dirhenium octachloride. The latter is particularly interesting because the fourth — δ, in analogy to d — bond can be cleaved by irradiation causing a structural change that clearly shows there must have been a quadruple bond previously. To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though.

And finally, $\ce{C2}$ does exist; however, explaining it requires orbital theories on a much higher level than in the scope of this answer.


Effect of inductive effect on boiling point,melting point and dipole movement In my book the following was written:-

  Inductive effect is a permanent effect operating in the ground state of the organic molecules and hence is responsible for higher melting point, boiling point and dipole movement of polar compounds.

How is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds?

Why is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds only?

Not quite sure about melting and boiling points but things like polarity ,dipole moment and inductive effect are kind of interconnected.Inductive effect is a consequence of electronegativity differences.More inductive effect leads to more dipole moment but that also depends from one case to another.For example each of the $\ce{C-Cl}$ bonds in $\ce{CCl4}$ is highly polar but net dipole moment is 0 due to the fact that Dipole moments are added through vector addition and $\ce{CCl4}$ is tetrahedral.


Can glass be deionized? Is there a chemical treatment that could remove sodium and calcium ions from the surface of soda-lime glass to turn it into quartz glass, increasing the hardness?

You can indeed "harden" glass by exchanging ions on the surface, but not because it turns into quartz glass.

Quartz glass is mostly so robust because it has an extremely small thermal expansion coefficient, and is therefore mostly free of internal stress due to uneven cooling. Ordinary glass already gets a lot more durable when you temper it, thereby relieving those internal stresses.

Generally, glass breaks under a very small extensional strain. It has a huge $E$ modulus and can take a lot of stress, but when you bend it, so the extensional stress and deformation is largest at the surface, where also you always have small scratches etc.), it shatters.

The trick is to put the glass surface under a large compressive stress. That way it can take a lot more pull before the critical strain (deformation) is reached and breakage occurs.

There are basically three ways to do that: Firstly, you can coat the hot glass with another glass that has a lower thermal expansion coefficient. That way the surface gets compressed during cooling. Tricky, and expensive.

Then, the most common way, is to rapidly cool the still nearly molten surface by e.g. blowing cold air on it. The outer shell is already cold, the inner part takes longer to cool and still shrinks, thereby compressing the surface.

The last, chemical trick (making supremely solid glassware) is to exchange sodium ions in the surface layer (100nm or so i think) by potassium ions. This was done in the eighties in the GDR for household glassware, under the brand name "SUPERFEST", by soaking the glasses in molten potassium nitrate (at 400 °C) for twenty minutes or so. The ions are larger, but the silica network lattice does not yield below at least 500°C, so the surface gets compressed by the exchange.

You know you had a glass that was treated in one of these ways when it does not break into large shards, but basically crumbles into small pieces or even dust. The classic example for the latter is Prince Rupert's drop, of the former any car windshield. (It seems the Na/K exchange does not (necessarily) do that, see your mobile phones "Gorilla glas".)


Should carbon tetrachloride really be considered an organic compound? Wikipedia says:

  Carbon tetrachloride […]  is an organic compound with the chemical formula  $\ce{CCl4}$.

From an organic point of view, carbon tetrachloride can be called the tetrachloro derivative of methane and is described in many organic textbook along with $\ce{CH3Cl}$, $\ce{CH2Cl2}$ and $\ce{CHCl3}$.

From an inorganic point of view, carbon tetrachloride can be considered chloride of a group 14 element (along with $\ce{SiCl4}$, $\ce{GeCl4}$).

Shouldn’t carbon tetrachloride be considered an inorganic compound? It neither contains any $\ce{C-H}$ bond nor is it found in any living organism.

Tetrachloromethane is its organic name. But it is generally not called by this name. Carbon tetrachloride is its better name and inorganic compounds are named like that.

Simple carbon compounds such as the tetrachloride blur the boundary between organic and inorganic. You should know that presence in natural systems is not an indicator of a compound being organic; many organic compounds are completely artificial, though they may be derived from natural feedstocks.

There is no strict criteria for organicity (is there a better word?). Consider perfluorinated carboxylic acids, they have no C-H bonds but are considered organic.


Is there a database which give reaction constant? For instance, I have a chemical reaction like :
$$\ce{Fe^{2+} + $0.5$O2 + H^{+} ->Fe^{3+} + $0.5$ H2O}$$

And I would like to use Action-Mass law, but I need to have equilibrium constant for this reaction. And this for many other reaction.
That's why I'm wondering if there exist a nice database with this information ?

The information you are looking for is out there, but sometimes you might have to work a little for it.

The example reaction you post is a redox reaction. If we split the redox reaction into its two half-reactions, we can look up their reduction potentials (easier to find sometimes than equilibrium constants) and use the Nernst equation to calculate the equilibrium constant. I use Wikipedia for my list of standard potentials, but only because the values are identical to those in textbooks that I own.

  Oxidation half-reaction

$$\ce{Fe^{2+} -> Fe^{3+} + e-} \ \ \ E^\circ_{ox}= -0.77 \ \text{V}$$

  Reduction half-reaction

$$\ce{O2 + 4H+ + 4e- -> 2H2O} \ \ \ E^\circ_{red} = +1.229 \ \text{V}$$

  Full balanced redox reaction

$$\ce{4Fe^{2+} + O2 +4H+ -> 4Fe^{3+} + 2H2O}$$

  Cell potential

$$E^\circ_{cell}=E^\circ_{ox} + E^\circ_{red} = -0.77 \ \text{V} + 1.229 \ \text{V} = 0.459 \ \text{V}$$

  Nernst Equation

$$E^\circ_{cell}=\frac{RT}{zF}\ln{K_{eq}}$$

$$k_{eq}=e^{\dfrac{E^\circ_{cell}zF}{RT}}$$

In this equation, $z$ is the number of electrons transferred, $T$ is the temperature in kelvins, $R$ is the ideal gas constant, and $F$ is the Faraday constant. At 0 Celsius (273.15 K):

$$K_{eq}=e^{\dfrac{0.459 \ \text{V} \cdot 4\text{ mol}\cdot 96.49\times 10^3 \frac{\text{C}}{\text{mol}}}{ 8.314\ \frac{\text{J}}{\text{K}\cdot \text{mol}}\cdot 273.15\ \text{K} }}=7.565\times 10^{33}$$

If you can look up the enthalpy change or free energy change for the reaction (many of which are available at webbook.nist.gov/chemistry, then you can use a variant of the van 't Hoff equation

$$\Delta_r G^\circ = -RT\ln{K_{eq}}$$
$$K_{eq}=e^{\dfrac{-\Delta_R G^\circ}{RT}}$$


Predicting sign of enthalpy of allotropic transition of phosphorus How to predict the sign of enthalpy during the transition
$$\ce{P(s, white) -> P(s, red)}?$$
$\ce{P(s, red)}$ is more stable than $\ce{P(s, white)}$ due to release in angle strain.
Since during formation of red phosphorus from white phosphorus bond breaking and bond formation both takes place, how to predict sign of enthalpy?

You can use the fundamental relation between energy and enthalpy. At constant temperature and pressure:
$$\Delta H_m = \Delta U_m + P\Delta V_m = \Delta U_m + PM (\frac{1}{\rho_{red}}-\frac{1}{\rho_{white}})$$
where the subindex m labels molar properties and M is the molar mass. Taking values of the density posted in the Wikipedia indicates that red phosphorus has a significantly higher density than white (~2.25 vs 1.82 g/mL). This means the volume decreases and the surroundings do work on the system during the reaction. From this data one can approximate the value of $P\Delta V$ at standard pressure (1 bar):
$$\Delta H_m=   \Delta U_m -\pu{0.32 J/mol} $$
In addition the difference in strain would indicate that $\Delta U_m<0$. This would mean that the enthalpy change should be negative, as can be verified by looking up thermodynamic tables. The Wikipedia page describing the allotropes of P (or alternately the CRC handbook) provides a value  of $\Delta H_m=\pu{-17.6 kJ/mol}$, so the correction for the volume contraction can be seen to be negligible (and does not influence the conclusion that $\Delta H<0$). Interestingly, even though white phosphorus is less thermodynamically stable it is considered the standard state of P.


Is it possible to melt diamonds into a liquid? Is it possible to melt diamonds into a liquid? I mean if you heat diamond at open air it will start to burn around 700 degrees Celsius, reacting with oxygen to produce carbon dioxide gas. In the absence of oxygen it will transform into graphite, a more stable form of carbon, long before turning into graphite. My question is, is it possible to melt diamonds into liquid? If so, then, how? If not, then why?

Liquid carbon does indeed exist, but perhaps surprisingly, relatively little is known about it. It only exists above around $4000\ \mathrm{K}$ and $100\ \mathrm{atm}$, which are not trivial conditions to sustain and probe. There certainly are many theoretical studies into the properties of liquid carbon, though. You can find a phase diagram for carbon here, though it is likely just a rough guide. The specific bonding properties of carbon atoms in the liquid phase still seems uncertain to some extent because carbon is very versatile and can create very dissimilar structures (easily seen comparing its two most common phases, graphite and diamond). This source suggests that bonding in the liquid varies continuously between linear acetylenic chains, graphite-like and diamond-like structures, depending on the exact conditions.

Edit: Now I realize that obtaining liquid carbon and melting diamonds are two slightly different things. Liquid carbon is trivial to obtain compared to melting a diamond. As you mention, heating diamond at relatively low pressures will cause it to convert into graphite before it could melt. The only way to inhibit that would be to exploit the slight difference in density between graphite and diamond; as diamond is denser, the increased pressure will stabilize it slightly, decreasing the tendency for diamond to revert back to graphite.

You'll need a lot of pressure though! Rigorously speaking, according to the phase diagram above, to obtain liquid carbon directly from melting a diamond would require a pressure of at least $10\ \mathrm{GPa}$, which is about $10^5\ \mathrm{atm}$. To put that into perspective, it's about $1/30$ of the pressure at the centre of the Earth. Amazingly, we actually can reach such absurd pressures in the lab by compressing two diamonds against each other. However, if the conditions required conditions for melting diamond (and sustaining it in molten form) are reached, it would probably compromise the integrity of the instrument and cause catastrophic (possibly explosive) failure, so I don't think anyone will be purposefully melting diamonds any time soon! Other techniques could be used (laser inertial confinement, light-gas guns, etc), but I believe most would only transiently produce liquid diamond.


Enthalpy of formation and reaction equations I'm disagreeing with my chemistry teacher over this enthalpy of formation problem, so can someone tell me if I am right or why I am wrong?

  The enthalpy of formation of gaseous sulfur trioxide is $-396~\mathrm{kJ/mol}$. What is the enthalpy of the reaction represented by the following balanced equation?

  $$\ce {2 SO_3 (g) -> 2 S (s) + 3 O2 (g) }$$

  (A) $-396~\mathrm{kJ}$   (B) $396~\mathrm{kJ}$   (C) $792~\mathrm{kJ}$   (D) $-792~\mathrm{kJ}$   (E) $198~\mathrm{kJ}$

I got (C) by flipping the sign of the enthalpy of formation given (because the equation given proceeds in the opposite direction from the formation of sulfur trioxide) and multiplying it by two (because there are 2 moles of sulfur trioxide).
My teacher and most of my other classmates said that it was (B) instead. They said that I shouldn't multiply by 2 because the equation given is at it's lowest realistic coefficient values.
I don't understand this for a few reasons:

the enthalpy of formation is defined as the enthalpy change per mole of substance
in previous problems, we would change the chemical equation to make the product only have one mole

For example, here's one problem:

  For which of the following reactions is the heat of reaction equal to the heat of formation of the product?
  \begin{align}
\text{(A)} &&\ce{N_2 (g) + 3 H_2 (g) &-> 2 NH_3 (g)}\\
\text{(B)} &&\ce{1/2 N_2 (g) + O_2 (g) &-> NO_2 (g) }\\
\end{align}

We narrowed it already down to (A) and (B).

My teacher said the answer was (B), because while both equations had the elements in their elemental form, the first reaction produces 2 moles of product while the second produces 1 mole of product. Thus the fraction coefficients were necessary.

As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By "enthalpy of the reaction" it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, but even if it did, the most common convention for balancing equations is to use the smallest whole numbers for stoichiometric coefficients, which is what is written there and would give the answer you got. You can calculate an enthalpy change of reaction for any reaction conditions you like and the value you get will change depending on the quantities of the substances involved, but if the question is implicitly assuming moles for coefficients, you still get the same answer.

Incidentally, there's no reason you couldn't use some other definition of "standard" in which you set the coefficients such that the smallest ones are equal to 1, though I can't remember seeing anyone do it this way. IUPAC convention is that whenever a quantity referring to some standard state is used, the conditions of the standard should be specified (though chemists like to ignore IUPAC).


Why is Cr2O3 amphoteric but CrO not? I faced a question where I had to find the amphoteric oxides among some given oxides. The answer key says $\ce{Cr2O3} $ is amphoteric but $\ce{CrO}$ isn't. But why is that?

Basic oxide are those oxides that dissolve in water to form soluble hydroxide. Basic oxides are very soluble for this reason. Acidic oxides are that oxides that dissolve in water to form strong acids. Generally, covalent oxides form acidic oxides as the element to which oxygen is bonded is electronegative. As charge increases, electronegativity increase and thus oxide become acidic. This also applies for elements having multiple oxides. As oxidation no. increases, charge increases and thus oxides become acidic. This link shares a good information on this topic:

  Since the acidity of a cation rises rapidly with charge, d-block
  elements which exhibit a wide variety of oxidation numbers may have
  one or more oxides that exhibit only basic properties and one or more
  oxides that exhibit only acidic properties. The higher the oxidation
  number the more acidic the corresponding oxide. Chromium is an example
  of such an element.

  $$
\begin{array}{|c|c|c|}\hline
\text{Oxide}&\text{Oxidation number}&\text{Category}\\\hline
\ce{CrO}&\ce{Cr^2+}&\text{basic}\\\hline
\ce{Cr2O3}&\ce{Cr^3+}&\text{amphoteric}\\\hline
\ce{CrO3}&\ce{Cr^6+}&\text{acidic}\\\hline
\end{array}
$$

The amphoteric character of chromium(III) is stated in its Wikipedia article:

  Chromium(III) oxide is amphoteric. Although insoluble in water, it
  dissolves in acid to produce hydrated chromium
  ions, $\ce{[Cr(H2O)6]^3+}$ which react with base to give salts
  of $\ce{[Cr(OH)6]^{3−}}$. It dissolves in concentrated alkali to yield
  chromite ions $(\ce{[CrO2]^{-}})$.

So, the rule of thumb is as oxidation number increases, charge on metal increases, acidic character increases. Also, basic oxide is soluble in water, acidic oxide is insoluble in water. This can be used to differentiate between acidic and basic oxide.