Is there a geometrically intuitive explanation of the equivalence of the numbers of irreducible representations and symmetry classes? In group theory, as used in chemistry, we’re familiar with the idea that the number of irreducible representations of a symmetry group equals the number of classes of symmetry operations, e.g. three in the case of the $C_\mathrm{3v}$ group describing the molecular structure of the ammonia molecule.
There are proofs of this fact, ranging from the properties of the Great  Orthogonality Theorem and discussions of the dimensionality of the vector space of characters e.g. as given in R. Mirman’s Group Theory: An Intuitive Approach, through to more mathematical proofs relying on the application of Brauer's permutation lemma to Galois automorphism on conjugacy classes, as shown in
https://groupprops.subwiki.org/wiki/Number_of_irreducible_representations_equals_number_of_conjugacy_classes.
But given the physical nature of molecular structure and the geometric properties of the molecular orbitals corresponding to the symmetry group’s irreps, is there no geometrically intuitive explanation of exactly why the number of irreps and classes are the same?

SUMMARY
The geometrically intuitive justification for why the number of irreducible representations and classes are the same is because:

The irreducible representations of a molecule’s symmetry group correspond to a resolution of the molecule’s spatial structure (visualised even as a simple ball and stick model) into a variety of quite different, component geometrical objects – the electronic molecular-orbital components, each obeying the whole set of the symmetry group’s operations in a distinct (orthogonal) manner
As vectors in the Hilbert space of the molecule’s electronic structure, the molecular orbitals occupy (and are symmetry-bound by) the invariant subspaces of the Hilbert space that are reached by the operations of the corresponding classes of the symmetry operations of the group on any of the vectors.
Hence the number of irreducible representations of the molecule equals the number of symmetry classes of its point group.

NB A detailed reply to @Porphyrin's comment under the question, which is too long to insert as a comment in reply, is given at the end of this answer.
DETAIL
For those who, like me, think in more of a visual/intuitive than abstract algebraic manner:
In chemistry, we can easily visualise the groups of the physical symmetries of molecules as combinations of symmetry operations transforming an image of the molecule (for atoms, it is simply the group SO(3) of rotations in R3).
The irreducible representations of the group illustrate the transformation properties of the component molecular or atomic orbitals under the operations of the molecular group, and can also be visualised as corresponding directly to both geometrical objects in physical space (i.e. the spatial distributions that are modelled as the MOs or AOs themselves) and, a bit more schematically, as vectors in invariant subspaces of a Hilbert space of functions representing AOs and MOs.
I think of this in terms of the visualisable spatial structure of the molecule overall (even as a ball and stick model for the purposes of understanding its symmetries) being analysed by group theory (or, if you prefer, resolved into) a set of different component geometric/physical structures – the molecular or atomic orbitals (all having the correct symmetry properties, regardless of the approximation method used in their computation).
[Please note that I’m choosing my words very carefully as there’s much confusion in discussions of representation theory due to frequent abuses of language, especially by mathematicians and by others who are so familiar with the material that they cease to be aware of how easily students can be confused by such abuses.]
So the transformation properties of the component MOs under the overall molecule’s group operations are encoded in the irreducible representations of the group and, for most purposes, conveniently summarised in the associated character tables, listing the traces of matrix representations of the group operations on suitable basis functions).
Character tables can be considered as a shorthand version of the irreps since they contain most of the information of use to chemists.
Indeed, one can say that “the molecular orbitals are the irreducible representations” of the group, in much the same way that Wigner’s Classification leads to the rather more mysterious statement that “fundamental particles are the irreducible representations of the Poincare’ Group”; in that case, it’s the symmetries of transformations in spacetime, under the Lorentz Group plus translations, yielding the Poincare’ group).
Just as we can visualise the orbital geometries described by the irreducible representations, likewise the classes of a molecule’s symmetry operations are also easily visualised.
Therefore the answer to the original question comes by considering the other way of imagining the MOs, as vectors in invariant subspaces of a Hilbert space of functions.
The characters of the irreps will be the same for each symmetry operation of a given class; so for $C_\mathrm{3v}$, for example, all three mirror planes σ will have characters of 1 for the A1 rep, -1 for the A2 rep and 0 for the 2-dimensional E rep.

This says to me that the molecular-orbital wavefunctions (i.e. irreducible representations) each span a subspace of the Hilbert space of molecular orbitals defined or bounded by the classes. No symmetry operation will take an MO with characters = 1 for the mirror planes (i.e. those MOs unaffected by reflections in any of the three mirror planes) into any subspace in which reflections invert the sign of the MO function.
There are three such classes for $C_\mathrm{3v}$, so the MOs appear to be limited to, or bounded by, three subspaces of the Hilbert space.
I conclude therefore that the geometrically intuitive justification for why the number of irreps and classes are the same is that:

the irreps correspond to a resolution of the molecular structure into a variety of molecular orbital components, each obeying the whole set of the symmetry group’s operations in a distinct way
the same molecular orbitals are distributed across as many invariant subspaces of the Hilbert spaces as there are classes of the group.

APPENDED REPLY to @Porphyrin’s comment:
i)  I’m aware of group theory’s abstract nature, for example permutation groups; but even they can have a physical realisation. When group theory is applied to material systems, there should be some sort of physical interpretation.
ii) Some people are content to work with a purely abstract sense of groups, while others find it helpful to have a more intuitive, often geometric sense. You might be familiar with the famous mathematician Jacques Hadamard’s On the Psychology of Invention in the Mathematical Field in which his survey of leading mathematicians and theoretical physicists’ in the mid C20th found a propensity to think in visual imagery.
iii)    It’s reasonable to claim that mathematical models often usefully represent physical realities. Recall the title/quote from the 1960 lecture by one of the supreme masters of symmetry in physics, Eugene Wigner: The Unreasonable Effectiveness of Mathematics in the Natural Sciences


Am I understanding buffering capacity against strong acid/base correctly? I just want to make sure I'm understanding buffering capacity correctly.

I've been working on this problem:

  Suppose you have an acetic acid buffer, $\mathrm{p}K_\mathrm{a} = 4.74$, at the following $\mathrm{pH}$:

  1) $\mathrm{pH} = 4.00$

  2) $\mathrm{pH} = 4.35$

  3) $\mathrm{pH} = 4.70$

  4) $\mathrm{pH} = 5.00$

  5) $\mathrm{pH} = 5.40$

  6) $\mathrm{pH} = 5.60$

  Which buffer solution will have the highest buffering capacity against $\ce{HCl}$? Against $\ce{NaOH}$ ? Which is the best optimal buffer?

From what I understand, optimal buffers are where $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, since that is when the ratio between conjugate base and acid is equal to $1$. This means that there is just enough acid/base relative to each other to minimize $\mathrm{pH}$ changes whether a strong acid or base is added.

As for buffering against $\ce{HCl}$, we would want our $\mathrm{pH}$ to be as high as possible, right?

We want as much conjugate base as possible in our buffer to react with the $\ce{HCl}$, while minimizing the changes to the acid, which means we want $\frac{[\ce{A-}]}{[\ce{HA}]}$ to be as high as possible. Maximizing $\frac{[\ce{A-}]}{[\ce{HA}]}$ means $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow \infty$, which means $\mathrm{pH} \rightarrow 14$, since $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$.

The explanation for $\ce{NaOH}$ would be the exact opposite. You want as much acid compared to conjugate base as possible so that the acid reacts with $\ce{NaOH}$ while minimizing changes to the conjugate base, which means $\frac{[\ce{A-}]}{[\ce{HA}]} \rightarrow 0$, $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow -\infty$, and $\mathrm{pH} \rightarrow 0$.

Is my understanding correct? The thing that's tripping me up is that I was told the exact amount of conjugate base and acid is more important than the ratio (e.g., $0.5 / 0.9$ is better than $0.05 / 0.09$), but since there's no mention of exact concentrations in the problem, I'm guessing we just assume that we have sufficient amounts of $\ce{A-}$ and $\ce{HA}$.

Is this generalization correct: if my buffer $\mathrm{pH} <<< \mathrm{p}K_\mathrm{a}$, then does that mean it is optimal against strong bases (and vice versa)?

Thank you for your help.

Further to the answer by Poutnik the buffer capacity $\beta$ of a weak acid - conjugate base buffer is defined as the number of moles of strong acid or base $C_B$ needed to change the $p$H by $\pm1$ unit, where

$$\displaystyle \beta=\frac{d[B]}{d\text{pH}}$$

and the equilibrium concentration base present is  $\displaystyle \mathrm{[B]}=\frac{k_w}{\mathrm{[H^+]}} - \mathrm{[H^+]}+\frac{C_BK_a}{\mathrm{[H^+]}+K_a} $.

(See answer How to set up equation for buffer reaction?)  where $K_w$ is the water ionization equilibrium constant $K_w =\mathrm{ [H^+][OH^-]} = 10^{-14}$, $K_a$ is the acid dissociation constant, and $C_B$ the total concentration of buffer.

After performing the calculation $\displaystyle \beta=2.303\left[\frac{K_w}{[\text{H}^+]}+[\text{H}^+]+\frac{C_BK_a[\text{H}^+]}{ ([\text{H}^+]+K_a)^2 }  \right]$

This can be analysed to find its maximum, but can be simplified first by ignoring the first two terms, because $K_w$ is tiny as is $[H^+]$ compared to the last term close to $K_a$. Plotting this last term with $C_B=0.02$ produces the following curves

where the max buffering is seen to be very close to the $pK_A$. You can see that buffering will be ok only when the $p$H changes by $\approx \pm 1$ and when the $pK_A$ is in the range from about $pK_A $ from $\approx 4 \to 9$. The red curve is the last term in the equation, the blue curve the full equation.

To answer your first question the $p$H should be close to the $pK_A$ and as the total concentration of base ($C_B$) increases the maximum just increases but does not shift.

To find the maximum differentiating the last term only with $[H^+]\equiv x$ produces, after a little algebra, $\displaystyle \frac{d\beta}{dx}=C_BK_a\frac{K_a-x}{(K_a+x)^3}=0$ where the maximum is found when $[H^+]=K_a$


Mechanism of retro aldol condensation of hydroxycyclopropylketone The reactants were treated with $\ce{NaOH} $. How the product here obtained. Can you show me the reaction mechanism for this. I found this is somewhat  Retro Aldol Condensation. But how?

You deprotonate the alcohol, and then flip bonds to create the enolate. There are two enolates possible, so swapping bonds gives you the terminal one, which can attack the aldehyde to form the 5-ring. Elimination of water gives you the product.


Exothermic dissolution When I was learning about salts that show exothermic dissolution, I noticed that the same salts used in Daniel Cell( due to similarity of size of anion and cation) were undergoing exothermic dissolution. Is there any relation between the similarity in size of anion/cation and the type of dissolution it undergoes?

There is no easy general pattern.

The dissolution can be formally divided into 2 steps:

breaking the ion lattice into free ions in vacuum
hydration these "naked" ions by water.

Both lattice forming and hydration enthalpies are negative and their absolute value increases with increasing ion charge and decreasing ion radius.

The relation of their values and the overall enthalpy sign  depends on the particular element chemistry, that modifies hydration and lattice enthalpy values.

There are some trends, but hard to generalize.

You can compare sodium and potassium salts of big anions with delocalized charge, like $\ce{NO3-, ClO3-, ClO4-}$. Dissolution of respective potassium salts is significantly more endothermic, compared to sodium ones. Sodium ion is smaller and is more intensely hydrated with lower enthalpy of hydration.

This difference is bigger that difference of their lattice enthalpies. It is probably due anion size,  packing of bigger potassium cation makes less difference than for chlorides.


How many significant figures? In Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.. Chemistry: The Central Science (Page 35). Pearson Education. Kindle Edition, an assignment is:

  Make the following conversions: $\pu{2500^\circ F}$ to $\pu{K}$

The answer given is $\pu{1644 K}$ but I say it's $\pu{1600 K}$. What is the correct answer?

It is ambiguous. 2500 can either be 2.5E3 or 2.500E3. The number 2506 would clearly be 2.506E3. Changes of last two digits being zeros randomly is 1%. If this is a home question or on a test then work the answer to 4 figures state that the problems seems ambiguous and that your assumption is that there are only two significant figures in the given temperature and round your answer to 1.6E3.

I'll point out that your answer has the same ambiguity. Is 1600 really 1.6E3 or 1.600E3?


How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene? I get everything up until the very end. We need to identify the bond enthalpies first:
\begin{align}
\ce{C=C} &= 612\\
\ce{O=O} &= 248\\
\text{SUM of these}&= 860\\
\end{align}

\begin{align}
\ce{C-C} &= 348\\
2\cdot(\ce{C-O}) &= 720\\
\text{SUM of these} &= 1068\\
\end{align}
We know that enthalpy change is = products - reactants, which is $1068 - 860 = 208$, but the problem comes here: I know a minus sign is supposed to go there, but
how do we know that? Exothermic reaction? Is it because of the oxygen, suggesting it's a combustion reaction or not?

When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants.  In other words "reactants - products".  This is because all bond enthalpies are positive, since energy must be put into the bond to break it.  Bond formation releases energy; the minus sign compensates for this.  In short, bonds of reactant molecule must break, and new are formed to form product molecules.  So, your answer should be negative.


What is the difference between chiral molecule and chiral carbon? I read in my book, that substituted allenes do not have chiral carbons, but actually the molecule is chiral, so they show optical isomerism. But what is the difference between them( chiral carbon and chiral molecule)?

In addition to the excellent Mithoron link, heed Lord Kelvin.


What kind of carbon ions exist? I think there are $\ce{C^{4-}}$ and $\ce{C^{2-}}$ existing. How can we find the charge of ions in the elements? Should we look at the group or valence?

Many different carbon ions exist in the gas phase.

In Electronic spectra of linear carbon anions, spectra of $\ce{C_2^-}$, $\ce{C_3^-}$... $\ce{C_10^-}$ are discussed.

Dianions such as $\ce{C_7^{2-}}$ have also been observed.

In the solid phase, aluminum carbide and berylium carbide have more than 50% ionic character according to the Handbook of Refractory carbides and nitrides, and correspond to $\ce{C^{4-}}$.  Calcium carbide, which could be considered a salt of acetylene, corresponds to  $\ce{C2^{2-}}$  rather than $\ce{C^{2-}}$


"Chemicals" released in headphone with burned cable My friend works at a headphone company at the support service and today she came across someone who claimed that he threw his broken headphone away because he had heard dangerous chemicals could be released when the cable burned through (because they would be exposed to the outside world possibly) which was according to him affirmed at a local waste disposal service. As I am unable to find any confirmation for this claim online, I decided to go ahead and post the question here. I was a bit unsure whether to post on chemistry.SE, biology.SE or engineering.SE, but went ahead with this one in the end, hope it's not off topic, but if it is then my apologies.

Burning or Incineration of Plastic cause carcinogenic chemical products if without suitable catalysts... To pyrolyze plastics such as polyethylene(PE) and polypropylene(PP) of olefin series, the catalysts with high cracking property are
needed.

Dioxin- carcinogenic and antibiotic effect to the environment.
Flyash- may produce respiratory illnesses.
Sulfur dioxide
Nitrogen oxide- green plastics.
Toluene- aromatic compound that is related in the burning of cigarettes.
Carbon monoxide- main product of anerobic/ lack of ventilation combustion.
carbonaceous char, oils and gases.
paraffins, olefins, naphthenes and aromatics

Plastic and Diesel
http://www.chemtec.org/proddetail.php?pr...
http://www.p-fuel.com/index_files/faq.ht...

Thermofuel
http://www.cynarplc.com/thermo-fuel.asp

Mael is right that a  headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.so even i agree it can not be dangerous for human health to a great extent.


Amalgamation involving an alloy Just curious, would mercury form an amalgam with gold that is already part of an alloy, i.e a gold platinum alloy?

Yes, this is a traditional method of gold purification from platinum group metals.


What is a word for "atom or molecule"? What is a word for "atom or molecule"?  As in: "The entry of an atom or molecule across a cell membrane into a cell is dependent on its size and solubility."  This keeps coming up, and it's really irritating (on par with the lack of a good word for "he or she" in clinical writing talking about patients).  English majors come up with frustrating responses like "elemental particle".  I just need a regular word for normal scientific communication (intro sections of papers, science education docs, etc.).

The classic catch-all term is molecular entity, often abbreviated to just entity. There's a Wikipedia page for it, which references the IUPAC Gold Book entry. Quoting from the latter:

  Any constitutionally or isotopically distinct atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity.

  Molecular entity is used in this Compendium as a general term for singular entities, irrespective of their nature, while chemical species stands for sets or ensembles of molecular entities. Note that the name of a compound may refer to the respective molecular entity or to the chemical species, e.g., methane, may mean a single molecule of $\ce{CH4}$ (molecular entity) or a molar amount, specified or not (chemical species), participating in a reaction. The degree of precision
  necessary to describe a molecular entity depends on the context. For example, 'hydrogen molecule' is an adequate definition of a certain molecular entity for some purposes, whereas for others it is necessary to distinguish the electronic state and/or vibrational state and/or nuclear spin, etc. of the hydrogen molecule.

  Source: PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1142.

As you can see, the definition is exceptionally general. It also specifically includes atoms. Usage of the term is widespread, even if it seems a bit exotic when first encountered.

Another option which can work is chemical species (often abbreviated to just species), which is even mentioned in the definition above. Again, there is a Wikipedia page. IUPAC actually has two definitions for chemical species, but the relevant definition is the following:

  An ensemble of chemically identical molecular entities that can explore the same set of molecular energy levels on the time scale of the experiment. The term is applied equally to a set of chemically identical atomic or molecular structural units in a solid array.

  For example, two conformational isomers may be interconverted sufficiently slowly to be detectable by separate NMR spectra and hence to be considered to be separate chemical species on a time scale governed by the radiofrequency of the spectrometer
  used. On the other hand, in a slow chemical reaction the same mixture of conformers may behave as a single chemical species, i.e. there is virtually complete equilibrium population of the total set of molecular energy levels belonging to the two conformers. Except where the context requires otherwise, the term is taken to refer to a set of molecular entities containing isotopes in their natural abundance. The wording of the definition given in the first paragraph is intended to embrace both cases such as graphite, sodium chloride or a surface oxide, where the basic structural units may not be capable of isolated existence, as well as those cases where they are. In common chemical usage generic and specific chemical names (such as radical or hydroxide ion) or chemical formulae refer either to a chemical species or to a molecular entity.

  See also: chemical species (of an element)

  Sources:
  PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1096.
  PAC, 1996, 68, 2193. (Basic terminology of stereochemistry (IUPAC Recommendations 1996)) on page 2202.

In this case, there is a notion of an ensemble (a collection/multiple) of whatever is being referred to. This is why "chemical specimen" (singular) is never used.

In short, a (chemical) species is composed of a collection of (molecular) entities. Both are umbrella terms covering virtually everything, and therefore can be used in almost all contexts. Perhaps in your case, (molecular) entity is the preferred term, as it emphasizes the differences in identity of what interacts with the cells.


Which acids are created by the combination of a gas and water? Which acids are created by the combination of a gas and water?  I know that carbonic acid is created by combining carbon dioxide and water, but are there others?  Is there a specific name for this kind of acid?

I don't think there is a specific name for these acids. But I can think of at least 4 more. The gases are $HCl,\ SO_2,\ SO_3,$ and $NO_2$.


Which base is stronger? Ammonia of phosphite? $\ce{NH4+}$: $K_\mathrm{a} = 5.6 \times 10^{-10}$;
$\ce{HPO3^2-}$: $K_\mathrm{a} = 4.5 \times 10^{-13}$

My understanding is that $K_\mathrm{b}(\ce{A-}) = K_\mathrm{w} / K_\mathrm{a}(\ce{HA})$. Thus:

$\ce{NH3}$: $K_\mathrm{b} = 1.8 \times 10^{-5}$; $\ce{PO3^2-}$: $K_\mathrm{b} = 2.2 \times 10^{-2}$

Since the $K_\mathrm{b}$ for phosphate is greater than the $K_\mathrm{b}$ for ammonia, phosphite is a stronger base. However, my professor's solution lists ammonia as the stronger base. Is his solution incorrect?

Your professor could be wrong, or rather could mean it in some subtly different context that would change everything. The way you put it here, you are right. Normally the constants would tell you the truth. I see no reason to believe this case is an exception.

Besides constants, there is actually a simple way of comparing two acids: let them fight each other to the death. Let us consider $\ce{NH_4^+}$ vs. $\ce{PO_4^{3-}}$ (I know $\ce{PO_4^{3-}}$ is not what you asked; this is just an illustration). Who would end up having that proton? It turns out that ammonium phosphate is unstable, which implies that the proton would go from $\ce{NH_4^+}$ to $\ce{PO_4^{3-}}$, meaning that $\ce{NH_4^+}$ is stronger acid than $\ce{HPO_4^{2-}}$ (in accordance with what the constants would predict), and hence ammonia is a weaker base than phosphate. Now, that was phosphate, the salt of phosphoric acid, while you ask about phosphite, the salt of phosphorous acid, which by all accounts should be weaker. Hence its anion must be an even stronger base.

Simple as it may seem, there is one more thing to mention. Phosphorous acid is actually diprotic, that is, capable of ionizing only two of its hydrogens. The third hydrogen is connected directly to the phosphorus atom. Organic esters $\ce{(RO)_3P}$ are known, but that's another story. The very existence of ion $\ce{PO}_3^{3-}$ seems somewhat dubious. By any means, even if you manage to get one in water solution, it will be instantly hydrolized. This puts it in the ranks of bases too strong to exist in water, which is of course way stronger than our lowly ammonia.


Decimal places vs significant figures I was given an online assignment problem today. The answer was to be given to 2 decimal places. The answer I gave was $$P_2=0.06$$

This to me seems like 2 decimal places. However the answer was incorrect. I then wrote $$P_2=0.064$$

It was correct. I know I'm being knit-picky but I genuinely just want to know your thoughts on this.

Would you call this 3 decimal places and 2 significant figures, or 2 decimal places and 2 significant figures. In the context of the assignment it doesn't matter a whole lot but I still think he should have written 2 significant figures.

What's your 2 pence on the matter?

My verdict on $0.064$ is three decimal places, two significant figures. The underlying philosophy is: how many figures would you give if you converted this to "scientific" notation - which would be $6.4 \cdot 10^{-2}$. Since the results must be independent of representation and nobody would argue about the number of significant figures or decimal places here, this is the way to go for me.

I received some chemistry education both in the US and Germany. Unfortunately, my experience is that in Germany these two concepts are often confused, even on university level and by lecturers. All high school level books on Chemistry I looked at in the US contained a section on this and were clear and adamant.


Lemon juice and secret messages The experiment is described in the following link: http://www.sciencekids.co.nz/experiments/invisibleink.html

I carried out a small research but I was unable to find what exactly happens. In other words which particular acids, present in lemon juice, are oxidized (e.g. citric acid, malic acid etc.)? And what causes the brownish colour?

Any kind of help would be appreciated.

Lemon juice contains sugars (carbohydrates) and acid (lemon acid). Sugars turn into carbon+water when heated in presence of acid. This is a standard thin layer chromatography analysis for sugars (spray plate with acid and heat up - black spots are sugars). The reaction is

C12H22O11 (sucrose) + H$^+$ (catalytic) -> 12C (charcoal) + 11 H2O (steam).

Paper also has polysugars (cellulose), but cellulose is more robust. Eventually even paper will turn black by the same reaction.

You can use sugar + vinegar as the ink. Result will be the same. It proves that decomposition of sugars in presence of acid is the reaction responsible for revealing this ink.


Radius with Bohr model $$r_n = \left(\frac{h^2}{4\pi^2me^2}\right)\times\frac{n^2}{Z} $$

Would I be able to calculate the radius of sodium ion ($\ce{Na+}$) with the help of the above Bohr atomic model formula?

The hydrogen atom wavefunctions can be useful for multielectron atoms as a means of looking up their size (by means of a parameterization). The hydrogen atom wavefunctions form one possible basis for the definition of effective nuclear charges, see for instance here, where  Hartree–Fock orbitals are used to compute effective charges:

$$ Z_{eff} = \frac{<r>_H}{<r>_Z}$$

Here $<r>_H$ and $<r>_Z$ are the mean hydrogenic and Hartree-Fock radii (for nuclear charge Z).

The mean hydrogenic radius for ground state hydrogen is related to the Bohr radius $a_0$ as
 $$<r>_H = \frac{3}{2}\frac{a_o}{Z}$$

Expressions for $<r>_H$ are available for other orbitals, allowing $<r>_Z$ to be computed from the tabulated values of $ Z_{eff}$ as

$$ <r>_Z = \frac{<r>_H}{Z_{eff}}$$

So using the hydrogen wavefunctions in this way is a little more complicated than using the equation for the Bohr model, but just the s-orbitals can already be useful to give you an idea about the size of atoms.


Which one really determines how people feel. Heat or temperature? We use water to cool down some machines because water has high heat capacity, which means it takes more energy to increase 1 degree. Heat transfer occurs only when temperature is different. We cool down the machine because we don't want it to be too hot, but what does the hot mean? The temp or the heat? When we touch water and another liquid at the same temperature but contain different heat. We feel the heat or the temp?

Regarding the last part "Do we feel heat or the temp?" - this is more physiological then chemical, but I want to share this anyway.

We feel heat flux, i.e. how fast heat is removed or supplied to our skin. Since heat flux can be both advection in the surrounding medium (e.g. air) or conduction in the skin it depends on the situation which is dominant.

In general you can write this as
$$ Q=k A \Delta T$$
where $Q$ is the heat flux, $k$ is some transfer coefficient, $A$ is the exchange area and $\Delta T$ is the temperature difference. The value of $k$ depends on the mechanism of transfer e.g. for advection it scales with $\rho C_p v$ i.e. it depends on the heat capacity and flow velocity of the surrounding medium.

The consequence of feeling heat flux is that we can only feel temperature differences not absolute temperature and that advection gives rise to a phenomenon called 'wind chill' that causes us to feel colder when there is more wind, because the higher wind velocity increases the heat transfer coefficient.


Why is the solubility of sodium acetate trihydrate so much lower than anhydrous sodium acetate? According to Wikipedia:

Solubility of anhydrous sodium acetate in water at 0 $^{\circ}$C is 119 g/100 mL (AKA 1.495 mol/100mL)

Solubility of trihydrate sodium acetate in water at 0 $^{\circ}$C is 36.2 g/100 mL.

This makes no sense to me because it should take MORE of the trihydrate sodium acetate (about 1.495 * 136.08 g) to dissolve the same amount of sodium acetate.

So this begs two questions:

1.) Is my understanding of hydrates incorrect? Are their bonds somehow harder to solvate than the anhydrous version? I think this can't be correct because the melting point of sodium acetate trihydrate is 58 $^{\circ}$C compared to the anhydrous melting point of 324 $^{\circ}$C.

2.) Is my understanding of solubility incorrect? I thought solubility is a measure of the quantity of solute that could be dissolved in a solvent. In other words, it seems that MORE sodium cations and acetate anions can be dissolved in 100 mL of water if they came from an anhydrous crystal rather than a trihydrate crystal...
How can that be true?

The values of solubility for these salts are $100\%$ correct. Your understanding of solubility is also correct. What you didn't know (probably) is that solutions can exist in a non-equilibrium state. In case of anhydrous sodium acetate, it dissolves in water forming so-called supersaturated solution, which is metastable. Eventually, crystals of sodium acetate trihydrate will precipitate out of the solution, forming a system which is in the thermodynamic equilibrium.


Simple method to lengthen alkane? I'm working on a science project which involves the production of methane. I would like to be able to use the methane as fuel in (e.g.) cars (using octane) or gas tanks (using butane and propane) etc. I have found that there are a variety of coupling reactions out there, but a lot of them seem to involve 'aryl Grignard reagents' which are not relevant to me.

So, I'm looking for a relatively simple method to lengthen an alkane. Preferably the reaction should occur below 100 degrees Celsius.

The procedure to convert methane to longer hydrocarbons at low temperature is one of the so-called Holy Grails in catalysis (http://pubs.acs.org/doi/abs/10.1021/ar00051a001). The person who is able to invent a process wherein this is possible (at an efficient rate) would be a very good candidate for a Nobel Prize.

Why is it then so difficult to convert methane to longer hydrocarbons? Methane is a highly stable molecule. In order to form a longer hydrocarbon, methane first has to be broken down to smaller CH$_{x}$ fragments. The first dehydrogenation step of methane (to CH$_{3}$) has a very high activation energy and even using a good catalyst, this is not feasible at an appreciable rate under 100 degrees celcius.

Industrially, this process is done by first creating synthesis gas (CO + H$_{2}$) from methane by partial oxidation using pure oxygen (which is cryogenically separated from air). Thereafter, the synthesis gas is compressed to roughly 20 atm of pressure and converted to very long hydrocarbons (waxes) in the Fischer-Tropsch process. In turn, these waxes are cracked to form diesel fuels and lubricants.

I am afraid that there is no easy way of converting methane to longer hydrocarbons. Though please keep trying because basically the whole world would benefit if you succeed! :-)


Hyperconjugation in cresol In cresol both $+I$ effect as well as hyperconjugation work together to determine its acidity. When the methyl group is present in the ortho- and para-position both of them can act upon phenol.

Hyperconjugation as a rule dominates which increases the electron density of the aryl. This makes it less acidic because the conjugate base is now less stable. But when its meta-cresol, hyperconjugation no longer works.

Why is it so? It happens even though an empty $\mathrm{p}$-orbital is present near the $\alpha$-hydrogen. What else does it need?

Above are the resonance forms of m-cresol after deprotonation. As you can see, the negative charge resonates around the aryl ring, but never resides on the carbon directly bonded to the methyl group at the meta position.

Thus, the hyperconjugative effect from the meta methyl group does not affect the acidity of cresol in this isomer. Notice how the destabilising hyperconjugation effect would destabilise the resonating negative charge, if the methyl group was at the ortho- or para- position.


The need for excess acid in dehydration of alcohols Why is there a need to use excess acid in the dehydration of alcohols? My instructor said that excess acid is necessary to minimize the ether formation via the competing SN2 mechanism, but I'm not sure how this works.

Alcohols generally are very poor acids and have pka's in the range of 16-17.

Hence, if the medium is only slightly acidic, only some alcohols will protonate, after which the unprotonated oxygen (of ROH) will attack the generated carbocation, following which the H+ ion will leave making H+ act as a catalyst, resulting in the formation of ether.

But, if the solution is highly acidic, we would be generating the carbocation in full and there would be no nucleophile left to attack the carbocation.

We need to maintain the optimum pH as substitution reactions are generally favoured over elimination reactions especially in the case of primary and secondary alcohols.


Scary jelly forming on zinc anodes I'm continuing with my electrolysis experiments. Over a long time, I seek a perfect electrodes. It seems, that when you pick the right anode and cathode, it might be a lot cheaper than universal super-electrode.

The cathodes are usually OK, since (+) ions do not hurt metals. However, $\ce{O-}$ and $\ce{Cl-}$ are both very agresive and will, sooner or later, destroy any anode I found so far.

My latest experiment was Zinc. Zinc is often used as non-corrosive layer added on iron via electro-plating. So I used it as an anode in $\ce{NaCL}$ aqueous solution. The anode I used was a piece of iron plated with zinc layer.

This scary jelly is forming on it. I also do not see any bubbles at all coming from the electrode - which mean that both chloride and oxygen ions are reacting.

The anode is consumed during the process - but what is produced? The jelly does not burn. When heated intensively, it bubbles and smells like chlorine. Does it decompose?

The orange colour is caused by the iron centre of the anode being oxidized. I tried other pieces of zinc I could find - they all produced this jelly.

Keep in mind, that solution does contain following additional compounds formed from salt and water:

$\ce{NaOH}$
$\ce{HCl}$
$\ce{Cl2}$

Judging from the color, this clearly looks like more or less hydrated $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$, and it could also contain dissolved $\ce{FeCl3}$ formed by the reaction of iron with $\ce{Cl2}$ evolved at the anode. The fact that it smells like chlorine when you heat the gelatinous precipitate would support the latter assertion, but the smell could also arise from residual $\ce{Cl2}$ which was still dissolved in the water of the jelly. Heating $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$ intensively above 200 °C will give you $\ce{Fe2O3}$ by thermal decomposition (dehydration).

\begin{aligned}
\ce{2Fe(OH)3 ->& Fe2O3 + 3H2O} \\
\ce{2FeO(OH) ->& Fe2O3 + H2O} \\
\end{aligned}

During electrolysis, the zinc layer on the electrode is degraded by pitting corrosion of $\ce{Cl-}$ and oxidation in the presence of oxygen. When the coating is completely removed by the corrosion in some places, the exposed iron reacts with oxygen and chlorine to the products mentioned above. On anodes made of pure zinc, $\ce{Zn(OH)2}$ will likely be the primary product, which will form a transparent to white, jelly-like precipitate. Zinc hydroxide formed by oxidative degradation of the $\ce{Zn}$ layer would also be a constituent of the precipitate on a zinc-coated iron anode.


Where exactly do the orbital angular momentum get their numbers e.g. +1/2 I was thinking about orbital angular momentum today and  wondering how exactly do the numbers come about that quantize them. I was doing a little research on how they come about and I  came across this image many times.
How exactly does a chemist/physicist measure this to get the +1/2,-1/2 number. Are there other values that are possible?

$$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ramuno}{\mathrm{i}}
\newcommand{\ket}[1]{\,|{#1}\rangle}
$$

Classical Mechanics

In classical mechanics, the angular momentum, $\vec{L}$, of a particle with respect to a chosen origin is given by the vector product of its position vector relative to the origin, $\vec{r}$, and its linear momentum, $\vec{p}$,
\begin{equation*}
    \vec{L} := \vec{r} \times \vec{p} \, .
\end{equation*}
By expanding $\vec{r}$ and $\vec{p}$ over a standard basis $\{ \vec{i}, \vec{j}, \vec{k} \}$ of a Cartesian coordinate system as $\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$ and $\vec{p} = p_x \vec{i} + p_y \vec{j} + p_z \vec{k}$ and then writing the cross product as the formal determinant [1] and expanding it one could obtain
\begin{equation}
    \vec{r} \times \vec{p}
    =
    \begin{vmatrix}
     \vec{i} & \vec{j} & \vec{k} \\
           x &       y &       z \\
         p_x &     p_y &     p_z \\
    \end{vmatrix}
    =
    (y p_z - z p_y) \vec{i} - (x p_z - z p_x) \vec{j} + (x p_y - y p_x) \vec{k} \, ,
\end{equation}
which implies that components $(L_x, L_y, L_z)$ of the angular momentum in terms of which it can be expressed over a standard basis as $\vec{L} = L_x \vec{i} + L_y \vec{j} + L_z \vec{k}$ are given as follows
\begin{equation}
 L_x = (y p_z - z p_y) \, ,
 \quad
 L_y = (z p_x - x p_z) \, ,
 \quad
 L_z = (x p_y - y p_x) \, .
\end{equation}
The magnitude, $L$, of the angular momentum is given as usual as the square root of the scalar product of the vector with itself, $L := \sqrt{\vec{L}\vec{L}}$, or, in terms of components,
\begin{equation*}
    L := \sqrt{L_x^2 + L_y^2 + L_z^2} \, .
\end{equation*}

Quantum Mechanics

Operators

The quantum mechanical analogue of classical angular momentum is known as the orbital angular momentum and the operators for the components of orbital angular momentum can be obtained by simply replacing the components of position and momentum in classical mechanical expressions by the corresponding operators
\begin{equation}
    \linop{L}_{x} = (\linop{y} \linop{p}_z - \linop{z} \linop{p}_y) \, ,
    \quad
    \linop{L}_y = (\linop{z} \linop{p}_x - \linop{x} \linop{p}_z) \, ,
    \quad
    \linop{L}_{z} = (\linop{x} \linop{p}_y - \linop{y} \linop{p}_x) \, .
\end{equation}
and the square of the magnitude of the angular momentum operator can be expressed in terms of these operators as follows
\begin{equation}
    \linop{L}^2 = \linop{L}_{x}^2 + \linop{L}_y^2 + \linop{L}_{z}^2 \, .
\end{equation}
Then the following commutation relations of the angular momentum operators can be easily established:
\begin{equation}
    [ \linop{L}_{x}, \linop{L}_{y} ] = \ramuno \hbar \linop{L}_{z} \, ,
    \quad
    [ \linop{L}_{y}, \linop{L}_{z} ] = \ramuno \hbar \linop{L}_{x} \, ,
    \quad
    [ \linop{L}_{z}, \linop{L}_{x} ] = \ramuno \hbar \linop{L}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{L}^2, \linop{L}_{x}] = [\linop{L}^2, \linop{L}_y] = [\linop{L}^2, \linop{L}_{z}] = 0 \, .
\end{equation}

Interestingly, in quantum mechanics there exist a different kind of angular momentum, the intrinsic angular momentum, or spin, which does not have a counterpart in classical mechanics.
While the intrinsic angular momentum components operators does have a different form from that of orbital angular momentum, the commutation relations between them and the square of the intrinsic angular momentum magnitude are exactly the same.

More interestingly, the angular momentum can be treated in such a way that the treatment relies entirely on the commutation relations just introduced, and as such, valid for any kind of angular momentum.
Thus, for generality we can use the letter $J$ for any kind of angular momentum reserving the letter $L$ specifically for orbital one.
\begin{equation}
    [ \linop{J}_{x}, \linop{J}_{y} ] = \ramuno \hbar \linop{J}_{z} \, ,
    \quad
    [ \linop{J}_{y}, \linop{J}_{z} ] = \ramuno \hbar \linop{J}_{x} \, ,
    \quad
    [ \linop{J}_{z}, \linop{J}_{x} ] = \ramuno \hbar \linop{J}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{J}^2, \linop{J}_{x}] = [\linop{J}^2, \linop{J}_y] = [\linop{J}^2, \linop{J}_{z}] = 0 \, .
\end{equation}

Eigenvectors and eigenvalues

According to the commutation relations above, $\linop{J}^2$ commutes with all three components $\linop{J}_{x}, \linop{J}_y, \linop{J}_{z}$ while the components do not commute with each other.
Thus, $\linop{J}^2$ and one (and only one) of the components operators, conventionally chosen to be $\linop{J}_{z}$, admit the set of simultaneous eigenvectors, and using the so-called ladder operators technique it can be shown that,
\begin{align}
    \linop{J}^2 \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
    \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
where $\ket{j(j+1),m}$ is a simultaneous eigenvector of $\linop{J}^2$ and $\linop{J}_{z}$, $j$ could take non-negative integer and half-integer values,
\begin{equation*}
    j = 0, 1/2, 1, 3/2, \cdots \, ,
\end{equation*}
and $m$ ranges from $-j$ to $j$ in integer steps
\begin{equation}
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
\end{equation}

Then, as I already said in my comment with respect to half-integer values of $j$, for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible, but where cyclic boundary conditions are not relevant, as for the intrinsic angular momentum known as spin, the half-integer values might be appropriate.

Epilogue

I'm still not sure what this question is about, but if it specifically about the spin of an electron, i.e. "How do we now that its component in any direction is $\pm 1/2$ in the units of $\hbar$?", then we know it both from the experiment (the Stern–Gerlach one) and the theory (the quantum field theory). And yes, for electron only these values are possible, but for other particles there might be different possibilities. Thus, for instance, a photon have component of the spin in any direction  $\pm 1$ in the units of $\hbar$.

[1] Here, "formal" means that this notation has the form of a determinant, but does not strictly adhere to the definition; it is a mnemonic used to remember the expansion of the cross product.


What is the difference in dehydration of cis- and trans-1,2-cyclohexanedicarboxylic acid in anhydride formation? Is there some required stereochemistry for anhydride formation? Do both of them form the same product, or does one not even undergo dehydration?

I looked up cyclohexane-1,2-dicarboxylic anhydride's structure and found two images:

Does that mean both of them are formed?

The mechanism of formation of anhydride from a dicarboxylic acid is as follows: [1]

So as you can see, it involves a nucleophilic attack at the $\ce{C=O}$ of one of the carboxylic acids by the other one, following which internal rearrangement occurs and a water molecule is kicked out.
Stereo chemically, we are interested in finding out in which configuration can the lone pairs of oxygen from one of the $\ce{-COOH}$ attack the $\ce{π^* C=O}$ type antibonding orbital of the other $\ce{-COOH}$ group.
To achieve this, the nucleophile( oxygen in this case) has to align properly with the $\ce{π^*}$ orbital at the Bürgi–Dunitz angle. This configuration is necessary whenever a nucleophile has to successfully add to a carbonyl group. The interacting diagram will look like this for formaldehyde: [2]

(Green ball= nucleophile| Blue and Red balls= carbonyl)
We can also roughly estimate that the situation will be similar for the $\ce{C=O}$ part of the carboxylic acid group.
To see which of the two reactants in question will form the required anhydride, one can simply look at the 3-D conformers of both in the appropriate orientation, and compare to see which of the cases resembles the standard Burgi-Dunitz interaction more closely as illustrated above
Here is the cis configuration: [3]

And this is the trans configuration: [4]

1= oxygen| 2= carbonyl of acid group
As you can see, the oxygen seems much more aligned with the carbonyl in the trans configuration as compared to the cis. So, 2 might form the anhydride more easily than 1 (that is, with a lower activation energy barrier due to reduced torsional strain)
But again, it's not a strictly inaccesible barrier, as a few bond rotations in the cis configuration can also lead to a somewhat favorable configuration for the reaction. So it may ultimately lead to two different kinds of anydrides by each of the isomers like you said, 2 can lead to formation of 4, while 1 can lead to formation of 3


Difference between non-thermal and thermal ionized gases Why or when does non-thermal plasma become thermal? How do we know before hand if the ion temperature will differ from the electron temperature, on what parameters does it depend - where is the quantitative gap between the two options?

(Has it to do with the heat release in ion-recombination reactions?)

A thermal plasma has its electrons and ions at the same temperature.  A raw plasma has hot electrons and cold ions from the vast difference in masses when microwave excited.  One way to thermalize is to run the plasma through a Helmholtz pair of coils with a strong uniform magnetic field parallel to plasma flow.  The low mass electrons tightly spiral around the parallel lines of force (that, of course, are not there), colliding with the nuclei to thermal equilibrium.

(That can sound a lot better than it works.)


What is the correct expression of pressure-volume work? My book defines work in pressure-volume work as:
$$\mathrm{d}w = - P_\mathrm{ext} \mathrm{d}V$$
However before doing so it mentioned the piston to be massless.
In physics, work is defined as
$$\mathrm{d}w = P_\mathrm{in} \mathrm{d}V$$

What is the correct definition of pressure-volume work in chemistry?
Does the definition of work, $$-P_\mathrm{ext} dV$$ change if the piston was not massless in chemistry?

P.S. My question was mainly pointed towards the mathematical expression of work in Chemistry,that is, 'What is the most general work expression in Chemistry in relation with the First Law'.
Some of the users have been suggesting me to check out answers on sign convention of work. I do appreciate it but my question is different and is perhaps best interpreted in @Chet Miller answer.

At the inside piston face, the pressure exerted by the gas on the piston is equal to the pressure exerted by the piston on the gas. This just follows from Newton's 3rd law of motion.  But, in an irreversible process, the pressure exerted by the gas on the piston face cannot be obtained from the ideal gas equation (or other real-gas equation of state).  This is because the equation of state applies only at thermodynamic equilibrium.  Otherwise, the gas pressure on the piston involves viscous stresses which depend not only on the volume but also on the rate of change of gas volume.  For a reversible process, the gas pressure throughout the gas, including at the inside piston face can be determined using the equation of state, since a reversible process consists of a continuous sequence of thermodynamic equilibrium states.
In using these work relationships, it is important to specify what you are calling the system. If the system includes only the gas, and the piston is massless and frictionless, the pressure on the outside piston face is equal to the pressure on the inside piston face (and the gas pressure).  If the piston is included as part of the system, and the piston is not massless or frictionless, then the external pressure is the pressure on the outside piston face, but this is not equal to the gas pressure on the inside piston face (or, equivalently, the piston pressure on the gas).


Is a single carbon atom stable? It is well known that single atom of oxygen is not stable, and it forms $\ce{O2}$ molecule. But elements like carbon form a network of repeated bonds. As answered in another question, last atoms in such network are unstable and need to create some bonds.

Is it possible to have a (free-standing) single atom of carbon? Stable in the form of a stable powder like graphite or diamond.

If not, what is the minimum size of a carbon particle (diamond or graphite network)?

Further Clarification: I meant normal chemical stability to have a powder material at room temperature. I do not mean highly reactive carbon, quickly turning into a compound.

Is it possible to have a (free-standing) single atom of carbon?

It's definitely possible, but such an atom would be highly unstable and likely would have a short lifetime. An example of this happening is if you blasted some carbon material with a laser. Here's a really cool paper where researchers blasted carbon nanotubes and graphene with lasers and observed single-atom chains (not an individual atom in this case, but the same principle applies) of carbon exhibiting some very interesting electronic properties:

http://www.pa.msu.edu/cmp/csc/eprint/DT086.pdf

  If not, what is the minimum size of a carbon particle (diamond or
  graphite network)?

Buckyballs (60 carbon atoms) come to mind.

Tangentially related....

A few years ago, a team of Australian scientists fabricated a single-atom transistor (the basic building block of integrated circuits). The atom was Phosphorus. Granted, the atom was not free-standing, but still very impressive. Such a device is extremely difficult to fabricate, and requires cryogenic temperatures, so it's doubtful that this would ever become a viable technology.

http://www.nature.com/nnano/journal/v7/n4/full/nnano.2012.21.html


Is this enough information to solve this problem? Calculate the volume of a $\pu{0.100M}$ solution of $\ce{HNO2}$ and the volume of a $\pu{0.200M}$ solution of $\ce{NaNO2}$, that must be mixed to prepare $\pu{300ml}$ of a solution with a $\mathrm{pH=3.35}$.

Since this is a titration problem shouldn't I be given one of the volumes ?

You need to use the Henderson-Hasselbach equation:

$$\pu{pH = pK_a + \log\frac{[Base]}{[Acid]}}$$

The $\pu{pK_a}$ of $\ce{HNO2}= 3.15$

Therefore,

$$10^{\pu{pH-pK_a}} = \pu{\frac{[Base]}{[Acid]}}$$
$$\pu{10^{3.35-3.15} = \frac{[Base]}{[Acid]}}$$
$$\pu{1.58 = \frac{[Base]}{[Acid]}}$$

Implies that the numerator must be $1.58$ times that of the denominator.

The base is $\ce{NaNO2}$; the acid is $\ce{HNO2}$. The base is two times as concentrated as the acid.

This means, for every $\pu{mL}$ of acid you use, you need $\pu{1.58/ 2 = 0.79 mL}$ of base.

If,

$$V_T = V_A + V_B$$

And,

$$V_B = 0.79V_A$$

Then,

$$\pu{300 mL} = V_A + 0.79V_A  \Rightarrow V_A = \pu{167.60 mL}$$


Is the VSEPR theory correct in determining the bond angle of sulfur dioxide? Recently when we were doing the VSEPR Theory in class and discussing the bond angles in various compounds. That's when I thought of sulfur dioxide whose structure can be like this:

According to the Internet, the bond angle between the two oxygen atoms = 120°. But mathematics shows a little off. I was taught that the VSEPR Theory states that

  Lone Pair-Lone Pair Repulsion > Bond Pair-LonePair Repulsion > Bond Pair-Bond Pair Repulsion

According to this, the angle between the two oxygens should be less than 120º and I can prove it.

Imagine that instead of the lone pair there was a double bond of oxygen with sulfur. This would imply that the bond angle between the oxygens would be exactly 120° on the same plane because each one of them would repel each other with the same repulsion force. But now instead of a bond pair, we have a lone pair which means (according to the VSEPR Theory) that it would repel the two oxygens much more than the imaginary bond pair was repelling. Therefore the angle between the lone pair and the bond pairs with oxygen would be more than 120°. This would result in a smaller than 120º angle between the two oxygen atoms. Then how does the net say that the bond angle is close to 120°? If what the web says is right, it implies that the repulsion of a lone pair and 2 bonds is exactly equal to the repulsion between 2 double bonds. Is it correct?

First off, your structure violates the octet rule on sulphur and is thus not a good representation of the molecule. A better one is this one:

Figure 1: electronic structure of $\ce{SO2}$.

All methods of approximation of bond angles you have learnt or will learn about are no more than educated guesses. VSEPR gives very good approximations for e.g. $\ce{NH3}$ and $\ce{H2O}$ but fails horribly at $\ce{PH3}$ and $\ce{H2S}$. For the latter two, an estimation based on the most likely orbitals used for bonds (all p) does a much better job. To accurately determine a bond angle, one needs to perform a structure elucidation experiment such as X-ray diffraction.

This has indeed been performed with $\ce{SO2}$ (likely multiple times using different methods; I am too lazy to check the references) and the results give a bond angle $\angle(\ce{O-S-O})\approx119^\circ$. This is indeed very close to $120^\circ$ so you can say that the VSEPR method gives a good approximation of the $\ce{SO2}$ bond angle.

Figure 2: Structural data of $\ce{SO2}$. Taken from Wikimedia, where a full list of authors is available.


The solubility of different compounds relative to each other in water Which of the following has the least solubility?

  $\ce{NH4HCO3, NaCl, NH4Cl, NaHCO3}$

The answer is given as $\ce{NaHCO3}$, but I don’t understand why.

Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium.

Here's what I was told:

Solubility decreases with increase in polarisibility
It increases with increase in covalent nature.
It increases with increase in hydration enthalpy and lattice enthalpy.

Which leads me to two other questions:

How do these factors really effect solubility, especially, covalent nature?
I guess polarisibility can be linked with hydration or solvolysis, so I understand it.
How can I find the order of solubility of different compounds relative to each other in water?

First of all, let's temporarily make a distinction between ionic and covalent compounds. Generally speaking, covalent compounds are more soluble in nonpolar solvents while ionic compounds are more soluble in polar solvents. Of course ionic and covalent bonding exist on a spectrum and some compounds you would expect to be ionic are indeed more covalent in nature. This concept is often visualized using the Van-Arkel Ketelaar triangle of bonding

As described by the Fajans' Rules, highly polarizing cations will distort polarizable anions, pulling their electrons into the inter-atomic space to the extent that they essentially form ionic compounds. This is one of the reasons solubility decreases with increased polarity. Solubility does not increase with increased covalent nature. In water, it would decrease. The best example of this is beryllium. Almost all $\ce{Be^{2+}}$ compounds actually exhibit covalent chemistry, as we can see with  beryllium peroxide in the Van-Arkel Ketelaar triangle above. This is because beryllium has extremely high charge density.

In order to properly understand solubility trends of ionic compounds, we should examine the enthalpy of dissolution using a Born-Haber cycle.

The first step is vaporization of the crystalline lattice, which has a large positive enthalpy. This is affected by a few factors, but it basically represents the stability of the crystal. Generally, dipositive anions/cations will have much greater lattice energy because of stronger electrostatic interactions, and lattice match should also be considered. A significant mismatch in size will lead to lower lattice energy. In the Born-Haber cycle, the ion-ion interactions are broken when the crystal is vaporized, then ion-dipole interactions are formed when water molecules surround the gaseous ions, so the other  factor to consider in the Born-Haber cycle of dissolution is hydration enthalpy, which is negative. Generally, this is increased with increasing charge density. Sometimes net enthalpy is positive, and in these cases, entropy may also come into play. For the most part, you can think of ionic solubility as a balance between lattice energy and hydration enthalpy.

  Solubility increases with decreased lattice enthalpy increased hydration enthalpy

and you're right that

  Solubility decreases with increased polarizability

Sodium's ionic radius is 227 pm, while ammonium's is 175. Often ammonium is a good substitute for sodium, as a large, low charge density cation (for stabilizing highly polarizable anions like bicarbonate). In fact, ammonium bicarbonate is one of the only soluble non-group 1 bicarbonates. Still, ammonium is still a little smaller than sodium. As a solid, ammonium will stabilize chloride better, but bicarbonate worse.

Since bicarbonate is much much more polarizable than chlorine, the two bicarbonate salts will probably be less soluble than the two chlorides. Ammonium will stabilize chloride better by size match, so ammonium chloride will be a little less soluble than sodium chloride. Due to it's large size and low CD, though, sodium will stabilize bicarbonate better, so sodium bicarbonate will be a little less soluble than ammonium carbonate.

The way I see it, the order should be:

  $\ce{NaCl}$ > $\ce{NH4Cl}$ > $\ce{NH4HCO3}$ > $\ce{NaHCO3}$ (decreasing solubility)

Of course solubility is a pretty complex concept and these are only generalizations. To get a better handle on it, I would try looking at the enthalpy and entropy cycles of different types of salts in each of the main groups. Trends exist, but the way they work is a little different depending on the group. I think, for example, that lattice match is mostly important in group 1 salts because of their lattice types. Anion charge becomes an important factor in group 2. Trends can be established but a lot of the time they are only consistent within a given group.


Why does the Lewis definition of acids specify a PAIR of electrons and not a single electron? The Lewis definition of an acid is: "a compound or ionic species which can accept an electron pair from a donor compound."

Why does it specify a pair of electrons and not just a single electron?

I initially thought maybe covalent bonds require a pair of electrons but it says that is not the case here : https://en.wikipedia.org/wiki/Covalent_bond#One-_and_three-electron_bonds

__

For reference:

Question from textbook :
Which of the following species can act as a Lewis acid?

$\ce{NH3}$
$\ce{F-}$
$\ce{H2 O}$
$\ce{NH4+}$
$\ce{BF3}$   (answer)

I thought the answer was $\ce{NH4+}$  as it could accept an electron, but the Lewis definition specifies accepting a pair of electrons.

Thanks.

It partially has something to do with the history of acid-base definitions. We started with the Arrhenius definition which was based on the generation of $\ce{H+}$ or $\ce{OH-}$ in aqueous solution, and then to the Bronsted-Lowry definition which moved towards acceptance or donation of $\ce{H+}$. Finally came the Lewis Definition which removed $\ce{H+}$ and left us only with electron pairs, to be consistent with the past two definitions. We couldn't just say electrons because that would have included a large class of reactions called redox reactions which are clearly not the same as acid-base.

Now, with respect to your answer to the multiple-choice question:

  I thought the answer was $\ce{NH4+}$ as it could accept an electron, but the Lewis definition specifies accepting a pair of electrons.

There are two points that I would like to draw your attention to:

You're right in that $\ce{NH4+}$ is an acid (more specifically we call it a conjugate acid), but by the two older definitions.
However, $\ce{NH4+}$ cannot accept even one electron in this case. If you draw the Lewis structure, you'll find that although there is a net positive charge, all hydrogens have 2 electrons and the central Nitrogen has eight, fulfilling the octet rule.


Reversibility of the binding of carbon monoxide to haemoglobin I recently learnt the basic mechanism of haemoglobin: oxygen forms coordinate bonds with the $\ce{Fe^2+}$ ion in haem groups, and this reaction is reversible to allow oxygen to be released. I also learnt that carbon monoxide does the same thing as oxygen, however, this reaction is said to be irreversible due to larger bond strengths etc. meaning that it can't be released.

What isn't clear to me is what happens to carbon monoxide if it does bind to a haem group. If the reaction is irreversible, then how is carbon monoxide removed from the body if you were to breathe it in?

The reaction is not irreversible, but hemoglobin adheres to $\ce{CO}$ about 200 times more tightly than it does to $\ce{O2}$. Thus, part of the treatment for $\ce{CO}$ poisoning is to expose the patient to oxygen, which will slowly displace $\ce{CO}$ from carboxyhemoglobin.

For more on the chemistry involved, see this qestion.


How come uranium's relative atomic mass is 238.03 when it only contains isotopes with a mass number of 238 or less? I have always been led to understand that the mass of an element on the periodic table is the weighted average atomic mass over all naturally occurring isotopes. This seems to make sense with all the elements I have looked at except for uranium. Wikipedia's featured article on the element assures me that uranium's relative atomic mass is 238.03. The commonest isotope is U-238, mixed with 0.72% of U-235 and 0.005% of U-234, plus traces of a couple of isotopes with smaller mass numbers.

How then can uranium's relative atomic mass exceed 238? Either I have completely misunderstood something, or there is something else at work - relativistic maybe?

Approximately 99.3% of uranium on Earth is the $\mathrm{^{238}U}$ isotope, and this specific isotope has an atomic mass of $\mathrm{238.05\ u}$, where $\mathrm{u}$ is the atomic mass unit, equivalent to 1/12 the mass of a $\mathrm{^{12}C}$ atom. Including the other isotopes to obtain the average atomic mass drags the value down a little, but it still ends up being slightly larger than $\mathrm{238\ u}$.

It is interesting to point out why the atomic mass is slightly different from the mass number. The masses of a free proton or neutron are both slightly larger than $\mathrm{1\ u}$ ($\mathrm{1.0073\ u}$ and $\mathrm{1.0087\ u}$, respectively). Some of this mass is converted to energy and lost when neutrons and protons (collectively called nucleons) bind to form nuclei, which means bound nuclei will always be lighter than the sum of the masses of their component nucleons. Different nuclei release different amounts of energy per nucleon depending on how many protons and neutrons are inside the nucleus, and on their relative proportions. Very interestingly, the curve for nuclear binding energy per nucleon versus mass number starts off at zero for $\mathrm{^{1}H}$ (there can be no attractive nuclear forces between nucleons if there's only one nucleon!), then rises until it shows a maximum around nickel (specifically for the isotope $\mathrm{^{62}Ni}$) and from there on slowly decreases.

Since the atomic mass unit is standardized with respect to the $\mathrm{^{12}C}$ isotope of carbon, then in approximate terms, the nuclei with a lower binding energy per nucleon than $\mathrm{^{12}C}$ will have less of the initial free proton/neutron mass lost to energy (and thus an isotope will have an atomic mass slightly higher than its mass number), while for nuclei with a higher binding energy per nucleon than $\mathrm{^{12}C}$, more mass will be lost to energy and the resulting isotope will have an atomic mass slightly lower than its mass number.

Taking a horizontal line through the $\mathrm{^{12}C}$ isotope in this graph and looking at its intersections with the nuclear binding energy curve will show the regions where this flip happens; isotopes for atoms below carbon will tend to have higher atomic mass than mass number, isotopes for atoms between carbon and thorium tend to have a lower atomic mass than mass number, and then isotopes for atoms above thorium (such as $\mathrm{^{238}U}$) will again tend to have higher atomic masses than mass numbers. This is only an approximate analysis, as protons and neutrons don't have the same mass, so isotopes very close to the transition region don't behave perfectly smoothly.

Edit: The flipping effect I mention can be seen very nicely in the list of isotopes shown in this site. Look at the (atomic) masses for the isotopes, and see how the fractional part of the mass starts out as circa $.01$ (above the isotope's mass number), then decreases to $.001$, reaches exactly zero (by definition) for $\mathrm{^{12}C}$, then soon after flips under to $.999$ and keeps going lower all the way to $.90$ around zirconium and tin isotopes before slowly rising up to $.98$ for bismuth $\mathrm{^{209}Bi}$ and then flipping back over to $.04$ for thorium and $.05$ for uranium.

Now here's a question: I said the maximum of nuclear binding energy per nucleon happened at $\mathrm{^{62}Ni}$, so why does the fractional part of the masses in the table keep decreasing all the way to tin $\mathrm{^{116}Sn}$? It's because these masses are not normalized with respect to how many protons and neutrons there are in the nucleus. If you were to take all the isotopic masses and divide them by their respective mass numbers, you would see the minimum around $\mathrm{^{62}Ni}$, as expected (actually, the minimum ratio of isotope mass to mass number happens at $\mathrm{^{56}Fe}$ because it has a slightly higher proportion of protons to neutrons. As I said, the small difference between proton and neutron masses creates slight irregularities which I did not take into account for simplicity).


How does a mostly non-polar molecule prevent a polar head from being pulled off by other polar molecules? Linear-shaped molecules of a solute with a large non-polar section and a small polar head are not soluble in a polar solvent. Why is this? Why don't the attractive forces between a molecule of polar solvent and the small polar head of the solute molecule cause the polar head of the solute molecule to dissociate?

The typical explanation I find on google is that the non-polar section "outweighs" or "is greater than" the polar head, but I'm a little confused as to what that means.

During the disolution of a compound C  in a solvent S, intermolecular forces between molecules of C and between molecules of S are broken (which requires energy), and intermolecular forces between each molecule of C and several molecules of S around it are formed (which releases energy). As such, dissolution of C into S is only favorable if the released energy is greater than the required energy.

Let's consider octanol $\ce{CH3(CH2)7OH}$ in water. We need to break the hydrogen bonds between octanol molecules and the London dispersion forces in them to separate the molecules, and we need to break the hydrogen bond network in water to accommodate each octanol molecule. Then, we surround each octanol molecule by water molecules. Only the polar head ($\ce{OH}$) will be able to form new hydrogen bonds (and account for the broken ones), but the interaction between water and the hydrophobic tail is merely permanent dipole-induced dipole, which are weak compared to all the hydrogen bonds we had to break between the water molecules to be able to accomodate the large hydrophobic portion, so we're losing energy rather than gaining it. Thus, this will be poorly soluble in water.


Is it possible to convert methane and carbon dioxide to acetic acid? Please forgive a bumbling amateur.  I have been looking up bond strengths and it seems to me that it should be possible to combine methane with carbon dioxide to make acetic acid:

$$\ce{CH4 + CO2 ->[\color{red}{?}] CH3COOH}$$

One has to break a C=O bond (745 kJ/mol) and a C–H bond (413 kJ/mol), but one gets back a C–O bond (358 kJ/mol), an O–H bond (467 kJ/mol) and a C–C bond (347 kJ/mol) so one comes out ahead.  I know that bond energies vary slightly but this doesn't sound too far out of whack.  Wouldn't it be rather nice if one could catalyse this reaction and soak up lots of these readily available gases? Could we induce some bacterium to do it for us?

Already methanotrophic bacteria can oxidize methane to formate and the Wood-Ljungdahl pathway can take this formate and a carbon dioxide molecule and coenzyme A and produce acetyl coenzyme A so the catalysts are already there.


When an aqueous extract of a substance is said to be administered at a certain quantity, is the quoted mass analogous to one as original substance? I'm trying to read studies about the possible toxicity of the leaves of some plant often said to possess positive medicinal properties.
In the studies, performed on rats, it is usually said that an aqueous and/or ethanol extract of the plant was administered at $x \ \pu{mg/kg}$ of mice mass, and from that data possible lethal or detrimental dose is extrapolated.
What I don't understand and is not explained in the texts (therefore I assume there is standard practice I'm not aware of) is how concentrated is $\pu{1 g}$ of the aqueous extract compared to $\pu{1 g}$ of the original fresh leaves. For example, one study simply states, referring to the administered aqueous extract: "The $\mathrm{LD_{50}}$ was estimated to be $\pu{1585 mg/kg}$." Should one infer that the amount of chemical substances administered in this scenario is equal to that found in $\pu{1,585 mg}$ of fresh leaves? Or is it possibly less/more concentrated?

You probably get the $\mathrm{LD_{50}} = \pu{1585 mg/kg}$ value by reading the abstract of the paper. That's why you got confused by that value. However, it is not $\pu{1585 mg}$ of fresh leaves. It is actually $\pu{1585 mg}$ of dry leave extract. I mean dry residue obtained by removal of water from an aqueous extract of dried leaves. The authors have made a stock solution of that dry residue in  $\pu{200 mg/mL}$ concentration by dissolving in enough distilled water ($\mathrm{pH} = 6.7$). Thus, $\pu{7.9 mL}$ of that solution is lethal to a rodent weighing $\pu{1.00 kg}$. On the other hand, since they have used rodents averaging $\pu{100 g}$ of body weight, each of them need only $\pu{0.79 mL}$ of that solution to be lethal.
Hope this explain what you are looking for.

Reference:
Olufunsho Awodele, Ibrahim Adekunle Oreagba, Saidi Odoma, Jaime A. Teixeira da Silva, Vincent Oluseye Osunkalu, "Toxicological evaluation of the aqueous leaf extract of Moringa oleifera Lam.(Moringaceae)," Journal of Ethnopharmacology 2012, 139(2), 330–336 (DOI: https://doi.org/10.1016/j.jep.2011.10.008).


Can an organic molecule have a triplet ground state? Can the ground state of an organic molecule be a triplet. This would imply something like a "HOMO" formed by 2 degenerate levels and less than 4 electrons to fill them.
If yes, what are the conditions?

As Martin has mentioned, carbenes are a good starting point if you are looking for organic compounds with a triplet ground state.

In these carbenes, the HOMO is not twofold degenerate, as your question suggests. Instead, there are two singly occupied orbitals with different energy. The energy gap, though, is typically small, and since there is an energy cost associated with pairing electrons in the same orbital, the electrons prefer to stay apart.

$\ce{CH2}$ itself has a triplet ground state, but is very difficult to observe; most methods of generating carbenes actually produce singlet carbenes, and $\ce{CH2}$ is so reactive that singlet $\ce{CH2}$ reacts with other stuff before it decays. Not all that interesting, eh.

There are some examples of persistent triplet carbenes, though.1 This one, reported by Hirai and Tomioka in 1999,2 has a half-life of $\pu{940 s}$ in degassed benzene at $\pu{20 ^\circ C}$:

In general, simple alkyl and aryl carbenes adopt a triplet ground state. These particular aryl groups on the carbene provide a large degree of kinetic stabilisation, which prevents simple dimerisation of the carbene (to form an alkene); the para alkyl groups also hinder reaction at that position. A further computational study of this molecule has been carried out by the Schaefer group, which supports the analysis above.3

References

Hirai, K.; Itoh, T.; Tomioka, H. Persistent Triplet Carbenes. Chem. Rev. 2009, 109 (8), 3275–3332. DOI: 10.1021/cr800518t.
Hirai, K.; Tomioka, H. A Triplet Carbene That Can Almost Be Bottled. J. Am. Chem. Soc. 1999, 121 (43), 10213–10214. DOI: 10.1021/ja991387c.
Woodcock, H. L.; Moran, D.; Schleyer, P. v. R.; Schaefer, H. F., III. The Almost Bottleable Triplet Carbene:  2,6-dibromo-4-tert-butyl-2′,6′-bis(trifluoromethyl)-4′-isopropyldiphenylcarbene. J. Am. Chem. Soc. 2001, 123 (18), 4331–4335. DOI: 10.1021/ja003552q.


Why are acetylene and air commonly used for AAS? I see that this fuel/oxidant combination is most commonly used, but it's never stated why.

Because it gives a very high temperature: 2200-2400 $^\circ$C.
This is due to its high heat of combustion, around 1300 kJ/ mol.

If still higher temperatures are required, as for Al, Si, Ti, etc., a combination of acetylene and $\ce{N2O}$ is used, which reaches 2600-2800 $^\circ$C.


According to MO theory, do 1s orbitals in dioxygen interfere? According to MO theory, are the 1s orbitals in $\ce{O2}$ too radially contracted to interfere with each other and thus cannot form bonding and antibonding orbitals, or do they still do that?

It is always possible to define a bonding and antibonding orbital regardless of how small the overlap is. Because we solve the homonuclear diatomics in the symmetry of the molecule ($D_\mathrm{\infty h}$), the MOs we choose must respect that symmetry. It wouldn't matter if I put the oxygen atoms 10 light years apart, if I am treating them as part of the same quantum system, there are bonding and antibonding molecular orbitals. (Of course, at large distances bonding and antibonding orbital pairs becomes very near degenerate.)

That said, because oxygen has a doubly occupied 1s orbital, the corresponding sigma 1s bonding and antibonding orbitals will both be filled. Within MO theory we are always free to 'rotate' any two occupied MOs and the observable properties will all be the same.
So, there's no difference between saying that each oxygen has a non-bonding 1s orbital, and saying that there is a filled bonding and antibonding orbital. When both orbitals are filled you can't even ask the question of what the electrons are 'doing' down there, because the concept of a distinct orbital has no meaning.


What is the difference between configuration and conformation in stereochemistry? I understand conformation to be any spacial arrangement of atoms that can be achieved through rotation of a single bond. I understand configuration to be the disposition of atoms/groups in space. However, would a conformational isomer also be a configurational isomer since the groups are arranged differently in space?

I recommend you this webpage: http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html.
Configuration and conformation are two concepts are often used interchangeably, but represent very different things. The conformation regards the relative spatial orientation of a portion of a molecule relative to another. Thus, it is an aspect that is not directly related to the covalent bonds that are established within the molecule, but with their possible rotation.

When we talk about rotation around covalent bonds, we are only referring to the single bonds, as they are the only ones that can suffer rotation. Basically this concept is easily understood if we think that the bounds work as an axis...

It should be noted that when we speak of different conformations, it does not necessarily involve all the covalent bonds of a molecule, it can account only for one or few of them.
 Taken together, it is possible to convert one conformation to another without cleaving or forming chemical bonds, simply by rotating some simple covalent bonds. Thermal energy at room temperature is sufficient to rotate some simple covalent bonds.

The configuration is a concept that is related to the order by which different substituents linked to the same central atom establish covalent bonds. That means, in this case it is clearly an aspect that is a direct consequence of the covalent backbone of molecules.

To change the configuration, you must always cleave and form new covalent bonds...This costs a lot of energy!!

In conclusion, the concept conformation encompasses portions of a molecule which are not directly linked to the same atom and do not involve the covalent backbone of the molecules, while the configuration comprehends parts of the molecule which are bound to the same atom, which means that there is a direct involvement of the covalent bounds of the molecule.


Can two electron occupy the same spatial spot in a statistical way? My friend claimed that if we have a two-electron atom in the ground state, and somehow we get the "real" electronic wavefunction $\psi(\mathbf{x_1},\mathbf{x_2})$ of the system, the statistical probability of finding two electrons (spin-up and spin-down) at the same spot $\mathbf{x_0}$ is nonzero, and the repulsion potential between them looks like a cusp condition when an electron is near the nucleus.

I think if two electrons occupy the same place, then the repulsion must be infinitely large, which is not possible. So I believe the probability should vanish here. However, he argues that in the dynamic case, two electrons can never be in the same place, but statistically, the probability obtained by squaring the wavefunction is not zero.

I'm pretty confused now. Who is correct?

Update:
The link here seems to contradict the answers below. I do not quite understand the statement "while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one". Can someone give a hint?

Update 2:
Some helpful references mentioned by DavePhD:

Fitzpatrick's Quantum Chemistry "On the other hand, in the spin-singlet state, there is an enhanced probability of finding the two electrons at the same point in space (because of the final term in the previous expression). In other words, the two electrons are attracted to one"

Advances in Quantum Chemistry volume 1, page 121: for the beryllium triplet P state two opposite spin electrons being at the same point is the most probable configuration

A statement in Atkins Physical Chemistry "The other combination [opposite spins] does not vanish when the two electrons are at the same point in space".

Electrons do not occupy places. Other than that, you are mostly right. Indeed, $\psi(r_1,r_2)$ must vanish at $r_1=r_2$ because of the spin-statistics theorem, if not for other reason. Upd. It was brought to my attention that this reasoning is not quite right; see the comment below. Still, I would expect at least a downward-pointing cusp, much like the upward-pointing cusp found at the nucleus, only the other way around.

This, however, is not easy to visualize. We're not used to imagining multidimensional surfaces. And if we try to focus on one electron at a time, then we're back with the one-electron approximation, and the accurate multiparticle $\psi$ is lost to us.

BTW, the argument of infinite repulsion does not work. After all, $\psi$ doesn't vanish near the nucleus; wouldn't that imply infinite attraction? No, it would not. You integrate over the entire space and come up with some figure for energy, which is surely finite.


Electronegativity of heavier elements of Group 15 While reading about p-block I got to know that in Group 15 elements

  electronegativity value decrease down the group but amongst the heavier elements difference is not that much pronounced.

I understood that electronegativity decrease down the group as the atomic size increase but I don't understand why in heavier elements electronegativity difference is not significant.
What is the reason for this?

electronegativity value decrease down the group but amongst the heavier elements difference is not that much pronounced.

The reason for this is that in heavier elements much more filling of inner d and f orbitals starts. For example as we move down from Sb to Bi, just 6 (3+3) more electrons fill p sub-shell, and 2 electrons enter to s sub-shell. Whereas on the other hand, 24 (10+14) electrons make their way to d and f sub-shell. This results in much poor shielding of electrons in outermost shell and effective nuclear charge thus remains approximately same in heavier elements. So, they have nearly same value of electronegativity on electronegativity scales.


What does HClO2 aqueous solution decompose to in an open environment? I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substances, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.

  $$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$

Then there is

  $$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$

(Source: Inorganic Chemistry by Garg and Singh [1], Google book preview)

The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.

Can the balances in this case lead to $\ce{HClO4}?$

On the other hand, $\ce{HCl}$ recombines to make $\ce{Cl2}$:

  $$
\begin{align}
\ce{HClO2 + 3 HCl &→ 2 Cl2 + 2H2O}\\
\ce{HOCl + HCl &→ Cl2 + H2O}
\end{align}
$$

So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal or cause harms to skin?

From a consumer's point of view, $\ce{HClO2}$ seems to be a complete disinfectants for all of water, surface (by producing $\ce{HOCl}$), and air (by producing $\ce{ClO2}$). However, the issue left for concerns is the byproduct acids.

Updates: I found a list of patents about producing $\ce{HClO2}$ solutions to be used as disinfectants, which are likely the one I am talking about in this question. The patents claimed that the disinfectants are safe to human, can be used to clean food-processing utilities, and even as food additives. The exact reasons for the claims are not clear to me.

https://patents.google.com/patent/US9516878B2/en
https://patents.google.com/patent/AU2013205834B2/en
https://patents.google.com/patent/EP2999490A2/en
https://patents.google.com/patent/US20160106106A1/en
https://patents.google.com/patent/US20160113282A1/en

References

Garg, R.; Singh, R. Inorganic Chemistry; McGraw-Hill Education: New Delhi, 2015. ISBN 978-1-259-06285-8.

Per Wikipedia on Chlorous acid to quote:

  The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):

  $\ce{2 HClO2 → HClO + HClO3}$

Also, here are some interesting comments on the decomposition of chlorous acid in "Kinetics and Mechanism of the Decomposition of Chlorous Acid" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:

  Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]

  Several groups of investigators[5-7] have found
  that in the absence of chloride ion the stoichiometry of the
  decomposition of chlorous acid is given by reaction A:

  $\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)

  The stoichiometry of the decomposition of chlorous acid in the
  presence of chloride ion is given by reaction B:

  $\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$  (B)

Also, to quote:

  Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C

  $\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$  (C)

  also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."

So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.

Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.

[EDIT] To answer a comment question, as to whether is ClO2 safe, here is  a statement from the CDC, Public Health Statement for Chlorine Dioxide and Chlorite, to quote:

  If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.

Interestingly, ClO2, which is a stable free radical, has found commercial application as an odor removal agent, likely due to its interaction with organics to create volatile organic chlorides (I would suspect that the presence of light would be catalytic). These VOCs are much more of a long-term health danger (as in carcinogenic), in my opinion.


Help understanding this passage on optical activity Other than mirror images, compounds having opposite configurations do not necessarily cause opposite directions of optical rotation any more than do compounds of the same configuration necessarily cause the same direction of optical rotation.

I know what optical rotation is and I know what "molecular configuration" means but I still can't understand what I just wrote above.

I believe what is causing you confusion is the difference between optical activity and absolute configuration.

A molecule with a single stereocenter such as limonene and carvone have two stereoisomers, one rotates polarized light to the left (-) and the other to the right (+). Additionally, each one has its own absolute configuration in their chiral center, which can be R or S. The thing is, R and S configurations are not attached or are, anyhow, synonyms with (+) and (-) configurations. Carvone and limonene are good examples:

In carvone, the S-enantiomer rotates polarized light to the left and so it is named S-(+)-carvone. The S-enantiomer in limonene rotates polarized light to the right and so it is named S-(-)-limonene. The two nomenclature systems are different since they specify distinct properties.

I hope this answers your question. If not, I would be glad to help in anyway I can.


Why isn't standard Gibbs free energy always zero? I read in my textbook that one of the standard conditions we should use when calculating the standard Gibbs free energy is having a concentration of 1 M of all the reagents and the products in the reaction. But if you use the formula $G = -RT\ln K_\mathrm e$ and you use 1 M to calculate it, you would always get zero.

So that means 1 M would be the initial concentration of reagents and products before reaching equilibrium?

$\Delta G = \Delta G^\circ - RT \ln(K)$

It isn't directly $G$, but formula for change in free energy. Also, in standard conditions $\Delta G = \Delta G^\circ$, so it isn't necessarily 0.

Relevant reading on a website


if I have concentrated acetic acid (99%) what are the out put gases during electrolysis? If I use concentrated acetic acid (glacial acetic acid($\ce{CH3COOH}$)) as an electrolyte. What ions are produced at the anode and cathode?
Can I get ethanol with it ($\ce{C2H6O}$)? If not, what would be the products?

Upon electrolysis, organic acids will lose the carboxyl group and dimerize.
This is known as the Kolbe electrolysis:
https://en.wikipedia.org/wiki/Kolbe_electrolysis
In the case of electrolysis of acetic acid, ethane and $\ce{CO2}$ will be formed.


Why does carbon not form 3 pi bonds? Why doesn't carbon form  3 $\pi$ bonds? If s-s overlap is possible between 2 carbon atoms,then it will be possible for 3 p orbitals to overlap laterally .

Okay so say you have two carbon atoms and you want to make a sigma bond with their s orbitals. They won't want to since their orbitals are filled but VSEPR theory tells us that carbon can rearrange and hybridize its orbitals to allow such bonds to happen. Normally you'd have your 3 pi orbitals along the 3 spatial axis, but the s orbital, holding 2 electrons, can hybridize with the p orbital that has no electrons to get two sp orbitals that have one electron each. This sp orbital will be along an axis we'll call z. This leaves us with two p orbitals orthogonal to our sp orbitals which may overlap laterally to give two pi bonds.

If we have two p orbitals overlapping to make two pi bonds, the third p orbital leaves us a problem. If we don't hybridize the orbitals, the third p orbital is empty and cannot make a bond. Even if you had a suitable lewis acid to try to give it some electrons, the bond would't be a pi bond since the orbital overlap would be along the inter nuclear axis. If we do hybridized then we end up with the situation described earlier.

Perhaps there could be a strange configuration where the inter nuclear axis is equiangular from the three base axis and the empty p orbital is filled by some electron donor and the bonding p orbitals overlap at right angles. I wouldn't really call it a pi bond at that point and anything in that position would likely collapse into a more stable state.

Basically the geometry of the p orbitals means that we can really only have two of them overlapping at any one time in the basic sense of a pi bond. This is assuming we are not counting the cases where a single p orbital participates in bonding with more than one atom such as in graphite, though I wouldn't really call that three distinct bonds.


Calculation for buffer of pH 5 in sodium fluoride solution From a journal:

  $\pu{500 mL}$ of $\pu{50 ppm}$ fluoride at $\mathrm{pH}~5$ was prepared by mixing $\pu{0.0552 g}$ of sodium fluoride with $\pu{4.1057 g}$ of sodium acetate and $\pu{1.65 mL}$ of acetic acid.

May I know how to get the solution of calculation for amounts of sodium acetate and acetic acid which are $\pu{4.1057 g}$ and $\pu{1.65 mL}$, respectively?

For sodium fluorid calculation is

$$\frac{\pu{50 mg}}{\pu{1 L}} \times \frac{\pu{1 g}}{\pu{1000 mg}} \times \frac{\pu{41.98 g mol-1}}{\pu{18.988 g mol-1}} \times \pu{0.5 L} = \pu{0.0552 g}$$

Acetic acid is a liquid. It has a $\mathrm{p}K_\mathrm{a} = 4.75$. A mixture of acetic acid and sodium acetate form a buffer solution which means that the solution resists $\mathrm{pH}$ changes. If the acetic acid and sodium acetate are present in equal molar quantities, then the $\mathrm{pH}$ of the buffer solution will be $4.76$. Since $\mathrm{pH}$ of buffer was $5$, there are more moles of acetate ($\ce{OAc^-}$) than acetic acid ($\ce{HOAc}$).

$$\mathrm{p}K_\mathrm{a} = 4.76$$
$$K_\mathrm{a} = 1.74 \times 10^{-5} = \frac{\ce{[H^+][OAc^-]}}{\ce{[HOAc]}}$$
$$\mathrm{pH} = - \log{\ce{[H^+]}}$$


How to remove formaldehyde (smell) from clothes/jeans? Bought several jeans, then to discover they have a really nauseating smell, that is not removed by washing. Reckon it's formaldehyde, from what I can read. Anyway, assuming you have jeans with formaldehyde in the fabric, what may effectively remove/neutralize this?

I ordered $\pu{10 kg}$ sodium bicarbonate for a soak, or I plan to heat them for a long time.  But advice from chemists is perhaps more productive. So, any ideas?

Formaldehyde is indeed used for disinfection, but usually there is a subsequent procedure involving formaldehyde neutralization with $10\%$ to $25\%$ aqueous ammonia solution introduced (as a spray) at elevated temperatures. Ammonia reacts with formaldehyde producing water-soluble hexamine which, even though also has unpleasant smell of rotten fish, is then easily washed out from the fabric:

$$\ce{6 CH2O + 4 NH3 -> C6H12N4 + 6 H2O}$$

I'd suggest to wash the clothes at higher temperature ($\pu{40 .. 60 ^\circ C}$) as you would normally do, but adding some ammonia; afterwards, soak the clothes in cold water for a while (solubility of hexamine increases in cold water), wring out and air dry.

In some countries there are also ozone-chambers (basically a sealed wardrobe with an $\ce{O3}$ generator inside, offered by dry cleaning companies) where organic pollutants are oxidized; formaldehyde is no exception:

$$\ce{CH2O + 2 O3 -> CO2 + H2O + 2 O2}$$

Both vinegar and soda can also be used, though I'd think their effectiveness is going to be lower.


Make Acetic Acid from wood I would like to know if it would be possible to make Acetic Acid from Wood. I know that you can charcoal wood to get the gases, cool it, and get the liquid. This would yield a bunch of chemicals in a mixture. I also know that you can steam wood to get wood vinegar. Would it be possible to extract Acetic Acid from either the Pyrolysis Gas or Wood Vinegar? Or would I need a different method of extraction or just not be possible at all?

Acetic acid is present in 'wood vinegar' that you get by thermal destruction with a lack of oxygen.
Steam can be used as a source of heat, but to be of any help in your desired production it would have to be overheated or under very high pressure, about 100 atmospheres. It is still useful as a way to increase the yield a bit.
Likely the best way is to heat the wood to 300*C without oxygen, collect and cool the gaseous products to condense them. You will get wood vinegar.
https://en.wikipedia.org/wiki/Pyroligneous_acid
After this you can separate the products using distillation.
https://en.wikipedia.org/wiki/Distillation
About 70 % of wood mass will be converted during the thermal destruction into wood vinegar.
About 15 % of wood mass will become non-water wood vinegar.
About 5 % of wood mass will become acetic acid.
Small yield, toxic waste, energy intensity of the process, are reasons why this process is not used anymore.