text:
Reaction of NaOH with methanol and ethanol?
Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):
$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$
$$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$
Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?

They react, but not "in the same way". The reaction with sodium is a redox reaction (note the changing oxidation state of hydrogen from +1 to 0), whereas that with NaOH is merely acid-base. As for the reaction with NaOH, it will certainly proceed, but in general will probably not go to completion (see also vapid's comment on the answer).

@orthocresol If an electrophile (like CS2) is added to  the solution of Methanol in NaOH, do sodium methoxide will attack to CS2 to form sodium salt of xanthate?

The proposed reaction:
$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$
According to wikipedia:

The solid hydrolyzes in water to give methanol and sodium hydroxide

The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:

In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.

This answer would benefit from quoting $\mathrm{p}K_\mathrm{a}$ values and how they influence the equilibrium.

Please do not encourage OP's abuse of notation which suggests a covalent bond between O and Na, it should be a superscript minus to represent the negative charge in the O atom.

The reaction of methanol and ethanol are based on their $\mathrm{p}K_\mathrm{a}$ values in the reaction.
Fist consider the reaction of methanol with water along with their $\mathrm{p}K_\mathrm{a}$ values.
$$\underset{\mathrm{p}K_\mathrm{a}=15.5}{\ce{CH_3OH}} + \ce{NaOH}\xrightarrow{\enspace}\ce{CH_3O^{-}Na^{+}}+\underset{\mathrm{p}K_\mathrm{a}=15.74}{\ce{H_2O}}$$
Now you can see that $\mathrm{p}K_\mathrm{a}$ value of water is a bit higher than that of methanol so the water will react with $\ce{NaOH}$ (keep in mind that the reactions are in equilibrium, and for methanol it will be shifted forwards).
Now considering the case for ethanol:
$$\underset{\mathrm{p}K_\mathrm{a}=16}{\ce{C_2H_5OH}} + \ce{NaOH}\xrightarrow{\enspace}\ce{C_2H_5O^{-}Na^{+}} + \underset{\mathrm{p}K_\mathrm{a}=15.74}{\ce{H_2O}}$$
Here it is clear from the $\mathrm{p}K_\mathrm{a}$ values that water is more acidic then ethanol and hence ethanol does not react to form water. It is not just the case with ethanol, rather all alkyl alcohols $(\ce{RCH_2OH})$ except for methanol does not reat with $\ce{NaOH}$.
If you come across a question asking that which of the given compounds dissolves in $\ce{NaOH}$, and if methanol $(\ce{CH_3OH})$ is one of the options then yes it will be your answer.

A lengthy and detailed account of the pKa of water recommends pKa = 14, and declares 15.7 to be erroneously derived. https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water%3F  The answer is currently correct in words, but the correct numbers will make it dramatically clear.

'Than' not 'Then'


text:
Energy and time considerations in chemical reactions
I'm working on hydrothermal decomposition of organic matter at high temperature and pressure (400 bar). The reactions are too complex and numerous to model individually, but proceed in a similar fashion at different temperatures.  If the reaction proceeds to the same point in 1 minute at 300degC and 1 day at 100degC, is there any way to use this information to estimate the amount of 'energy' required to do the work?  I know this is a strange question, but I want to see if the same decomposition is achieved using electrochemical energy in an ion trap. I'm not sure how I'm going to be able to compare the systems in terms of energy applied. I am quite sure that the hydrothermal decomposition depends on the force of the bonds being broken, as does the electrochemical decomposition...
Am I going down the wrong rabbit hole to calculate the specific energy of the water at this temperature?  How would time come into the equation then?

Does the Arrhenius equation not give you what you need?

Ordinarily it would - I could find an activation energy for the reaction that way... But this is a complex mixture of thousands of reactants and products, making it impossible to find something like a first-order decay constant. What I can track is the time difference to achieve the same amount of overall decay at several temperatures.

Using the Arrhenius equation will give you an effective or average activation energy that you can then use to compare with other methods.

source: Stack Exchange

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sort: votes


text:
Will O or S accept an electron more easily?
O has more electronegativity than S but S has more Electron Gain enthalpy than O. How do we decide, which one pulls the electron more easily? In other words, Electron gain enthalpy dominates or electronegativity ?

The two quantities are completely different. The electron gain enthalpy is "the energy released when one mole of electron are added to gaseous atoms of an element". The electronegativity is the tendency of an atom to attract a bonding pair of electrons.

@user34388 Exactly! But which one of them decides the fate of the pulled electron?

To me it is unclear what you are asking. What do you mean with pulling and the fate of the electron?

That should depend on the situation. For example, effects such as inductive effects in a molecule should be based more on EN than EA. You should be using EA if you're drawing a Born-Haber diagram.

@aventurin what i mean is, which element will pull the electron more easily? * this is a book question *

@AaryanDewan Then your book is unclear.

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text:
Can glutamic acid and arginine form H-bond at physiological pH?
I was wondering if, say, glutamic acid and arginine can form H-bonds at physiological pH?

According to the figure arginine has a $\ce{NH3+}$ group and glutamic acid a $\ce{COO-}$ group at physiological pH.
I know hydrogen bonding is possible amongst $\ce{NH2}$, $\ce{HF}$ and $\ce{OH}$. But $\ce{-NH2}$ is in the protonated form and the $\ce{-COOH}$ in the deprotonated form.
What will happen:

Will $\ce{-NH3+}$ donate its $\ce{H+}$ to $\ce{-COO^-}$ to form $\ce{-NH2}$ and $\ce{-COOH}$ and then form hydrogen bonds?
Will they attract each other because of the opposite charge, with resulting H-bonding
No H-bonding will occur.

(maybe the first and second option are combined, I didn't really know how to formulate it)
UPDATE
How is the H-bond possible in the following figure (I would think this is just an electrostatic interaction, this figure suggest it would a combination of both):

source of picture: Wikipedia

See this: The distributing basics of the moderate structure found in geometry, charge distribution, and ability to form multiple H-bonds make arginine ideal for binding negatively charged groups. For this reason, arginine prefers to be on the outside of the proteins, where it can interact with the polar environment. So, it can form H-bond or, preferably, dipole-dipole interaction.

As none of the chemists have picked up on this one, I will venture an answer.
First, I am answering on the assumption that this question relates to proteins as it was originally posted on Biology SE and you have previously asked chemical questions about proteins.
Second, I would have thought that you would be aware that weak interactions in proteins are usually categorized in three different types. The section on non-covalent bonds in Berg et al. states:

The three fundamental noncovalent bonds are electrostatic interactions, hydrogen bonds, and van der Waals interactions. They differ in geometry, strength, and specificity. Furthermore, these bonds are greatly affected in different ways by the presence of water.

So I find it strange that you wish to make an electrostatic interaction (aka an ionic bond) involving integral positive and negative charges into a hydrogen bond which can be regarded as a sort of poor man’s ionic bond as it involves only partial positive and negative charges. To quote Berg et al. again:

The relatively electronegative atom to which the hydrogen atom is covalently bonded pulls electron density away from the hydrogen atom so that it develops a partial positive charge (δ+). Thus, it can interact with an atom having a partial negative charge (δ-) through an electrostatic interaction.

The glutamic acid and arginine side-chains will be ionized at physiological pH because of the values of their ionization constants — the position of the equilibrium between the ionized and unionized forms. One extra hydrogen ion in the sea of hydrogen ions in the aqueous environment is not going to affect this one iota.
You should also be aware that the aqueous environment (in this case the surface of a globular soluble protein) is most unfavourable to the formation of hydrogen bonds between residues because of the competition of polar water molecules. To quote once more:

A hydrogen atom of water can replace the amide hydrogen atom as a hydrogen-bond donor, whereas the oxygen atom of water can replace the carbonyl oxygen atom as a hydrogen-bond acceptor. Hence, a strong hydrogen bond between a CO group and an NH group forms only if water is excluded.

You should read the whole of this section (which also applies to ionic bonds) to appreciate how these bonds are most important in the hydrophobic interior of proteins.
In summary, the answer to your question is “No hydrogen bonding will occur — if anything there will be ionic bonding, but in an aqueous environment this will be disrupted.”
Addendum
Well things are never that simple, and @Jan has put me in my place. His answer doesn’t provide examples, so here is one:

[Hydrogen bond between Glu472 and Arg423. The shape of the ATP-binding pocket is supported by a hydrogen bond between Glu472 and Arg423 that brings the stretch of amino acids containing residues Phe475 and Glu482 (not shown) close to other residues involved in ATP binding, such as Glu446 and Lys501 — Lánskỳ et al. Biochemistry 2004, 43, 8303-8311 (2004)]
What can be seen is a network of interactions. It should be noted that there are no measurements of bond strength to support the postulate which is based on mutational evidence.

You have described hydrogen bond using an "electrostatic" dipole-dipole interaction model. However, I would like to add that it has some features of "covalent bonding"-- it is directional and strong, produces interatomic distances shorter than the sum of the van der Waals radii, involves a limited no. of interaction partners, etc.

Thankyou @David, I'm a little bit confused because one of my books (fundamentals of general ,organic and biological Chemistry; McMurry) says this: "A peptide rich in Asp and Lys is more soluble, because it's SIDE CHAINS are more polar and can form hydrogen bonds with water?? (Sorry for the late reply)

The book says that those amino acids: Arg, Asp, Asn, Glu, Gln, His, Lys, Ser, Thr, Tyr can form hydrogen bonds using there side chains @David

@RickBeeloo — Indeed. Draw out the structures and you will see that in addition to the changed oxygen there is also a =O in the acidic residues. That is what can form a hydrogen bond. I leave you to work it out for the basics. I'm on holiday in Italy with only a phone for Internet connection. Ciao!

Yes, glutamate and arginine can form a hydrogen bond at physiological pH. Remember of course, that as soon as you enter aquaeous media, everything will be saturated by hydrogen bonds from the surrounding water molecules. But inside a protein an arginine—glutamate hydrogen bond is perfectly viable. Arginine will donate the hydrogen bond using an $\ce{N-H}$ bond and glutamate will receive towards a carboxylate.
Note, by the way, that your picture gives a wrong structure of protonated arginine. The correct protonation of the guanidine group is given in the image below (taken from Wikipedia).

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text:
If fluorine has a lower electron affinity than chlorine, why does it have a higher ionization energy?
I have read that fluorine has a lower electron affinity than chlorine despite its lower atomic radius because its electron cloud is extremely dense. If this is the case, shouldn't the ionization energy of fluorine be lower than that of chlorine, since fluorine's dense electron cloud should eagerly repel its electrons?
Even more confusing is that electron affinity can also be defined as the ionization energy of the singly charged anion form of an atom, i.e. the ionization energy of $\ce{Cl-}$ is equal in magnitude to the electron affinity of $\ce{Cl}$. Therefore, it seems that the trends for ionization energy and electron affinity should always match, which is not the case with fluorine and chlorine.

The first ionisation energy is the energy required to remove the most loosely held electron from...

The second part is a duplicate of this question.

Notice that ionisation energy and electron affinity are two only very weakly connected physical constants.

To quote chemguide:

The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.

That means, the ionization energy of fluorine is the energy of the following reaction:
$$\ce{F -> F+ + e-}$$
The ionization energy of chlorine is the energy of the following reaction:
$$\ce{Cl -> Cl+ + e-}$$
Since the most loosely held electron in fluorine is closer to the nucleus than chlorine, it takes more energy to remove the electron from fluorine than in chlorine.

The ionization energy of chloride ion ($\ce{Cl-}$) is the energy of the following reaction:
$$\ce{Cl- -> Cl + e-}$$
The electron affinity of $\ce{Cl}$ is the energy of the following reaction:
$$\ce{Cl + e- -> Cl-}$$
As a result, their energies are equal in magnitude but opposite in sign.

Why does chlorine have a higher electron affinity than fluorine?
To quote this answer:

Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

As can be seen in the second photo, this effect seems to be true for Period 2 elements.

(source: lardbucket.org)

(source: lardbucket.org)

In that case, why is it that Cl has a lower ionization energy than F, but Cl- has a higher ionization energy than F-? (This is what I was trying to get at in the 2nd part of my question.)

@jt2000 That is answered here.

This is half an answer at best, as it doesn’t compare fluorine’s and chlorine’s electron affinity.

@Jan See the comment just above you.

@user34388 An answer should be self-containing and not rely on comments or other answers to be understood. Please edit in the relevant parts. (I’m very inclined to upvote this answer but it’s just not complete enough for me yet.)

@Jan I didn't want to include it because that part would be a duplicate. I included it notwithstanding.

@jt2000 The link I provided explains why Cl has a higher electron affinity than F, which is the same as why Cl- has a higher ionization energy than F-.

@user34388 The link you provided doesn't explain why Cl has a lower ionization energy than F, it only explains why Cl- has a higher ionization energy than F.- Also, I do understand why Cl has a higher electron affinity than F, but I'm wondering why the logic used to explain this (small atomic size) does not apply to why the ionization energy of Cl is less than that of F. It seems to me that because F has a lower electron affinity because of its small atomic size, it should also attract its own electrons less, resulting in a lower ionization energy than Cl as opposed to higher.

@jt2000 As Jan said, "Notice that ionisation energy and electron affinity are two only very weakly connected physical constants. "

@user34388 Is there a particular reason why this is true, considering that they seem closely related in physical terms, their only difference being that one relates to the singly charged anion while the other relates to the uncharged form?


text:
How do you determine the pH of a solution at different volumes?
For a tube with 1 mL of 0.75 % (w/v) casein in 0.10 M sodium acetate, with 0.05 mL acetic acid, and 6.95 mL H2O, how do you determine the pH?
My original thought was to use the Henderson-Hasselbalch equation with the concentration of base being 1 and the concentration of acid being 0.05, but this does not feel right.
I think my fundamental understanding of the direct correlation between concentration and volume is not correct.

Is the concentration of the original 0.05 ml acetic acid given in your exercise?

Sorry, the concentration of acetic acid is also 0.10M

With regard to your understanding of the correlation between concentration and volume, you should know that concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.
When you dilute a sample, you change the volume $V$, but you do not change the amount of substance $n$ in your sample. Therefore, if you know the initial concentration $c_1$ in the initial volume $V_1$
$$c_1=\frac n{V_1}$$
you can calculate the new concentration $c_2$ after dilution to the new volume $V_2$:
$$c_2=\frac n{V_2}$$
You can rearrange both equations to calculate the amount of substance $n$:
$$\begin{align}
c_1\cdot V_1&=n\\[6pt]
c_2\cdot V_2&=n
\end{align}$$
Since the amount of substance $n$ is constant, you get another useful equation:
$$c_1\cdot V_1=c_2\cdot V_2$$

In your exercise, after addition of all solutions, the new volume may be estimated as
$$V_2\approx1\ \mathrm{ml}+0.05\ \mathrm{ml}+6.95\ \mathrm{ml}=8\ \mathrm{ml}$$
The initial concentration of the base (sodium acetate) is $c_{\text{base},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{base},1}=1\ \mathrm{ml}$.
Therefore, the new concentration of the base is
$$\begin{align}
c_{\text{base},1}\cdot V_{\text{base},1}&=c_{\text{base},2}\cdot V_2\\[6pt]
c_{\text{base},2}&=\frac{c_{\text{base},1}\cdot V_{\text{base},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times1\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$
The initial concentration of the acid (acetic acid) is $c_{\text{acid},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{acid},1}=0.05\ \mathrm{ml}$.
Therefore, the new concentration of the acid is
$$\begin{align}
c_{\text{acid},1}\cdot V_{\text{acid},1}&=c_{\text{acid},2}\cdot V_2\\[6pt]
c_{\text{acid},2}&=\frac{c_{\text{acid},1}\cdot V_{\text{acid},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times0.05\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.000625\ \mathrm{mol\ l^{-1}}
\end{align}$$
Finally, you can use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of the acetate buffer from the new concentrations $c_{\text{acid},2}$ and $c_{\text{base},2}$.

Thank you very much for this, this helps a lot with my understanding of the actual change that is taking place concentration wise. My only question is that performing this calculation with these new concentrations, or with the old concentrations result in the same pH yield. Am I to understand that this is because the ratio of base to acid is still the same? Does this mean that the addition of water has no effect on pH at least according to the equation used?

@Sushifish Yes, if the assumed volumes for the acid and the base are identical, this volume cancels out in the Henderson–Hasselbalch equation.


text:
Letter codes in molecular term symbols
I'm trying to find an explanation for the letter codes (X, A, B, C, etc) when you get a term symbol like this
$$X\,{}^3{\Sigma}^-_g$$
Can anyone point me to some literature that explains these letter codes?

https://en.wikipedia.org/wiki/Molecular_term_symbol#Alternative_empirical_notation In this case it is referring to the ground state (I guess probably $\ce{O2}$).

have a look also in the tables at the end of any of the Herzberg books 'Molecular Spectra & Molecular Structure'

Quoting from Hollas, Modern Spectroscopy, 4th ed. (p 236):

There is a convention, which is commonly but not always used, for labelling electronic states. The ground state is labelled $X$ and higher states of the same multiplicity are labelled $A, B, C\ldots$ in order of increasing energy. States with multiplicity different from that of the ground state are labelled $a, b, c \ldots$ in order of increasing energy.

As such, for the oxygen molecule, where the lowest three states are ${}^3{\Sigma}_g^-$, ${}^1{\Delta}$ and ${}^1{\Sigma}^+_g$ (in increasing order of energy), the triplet ground state would be labelled $X$ and the next two states $a$ and $b$ respectively.
A word of caution: it seems that this convention is not always obeyed. On the next page, Hollas writes (for $\ce{I2}$, where the ground state is a singlet)

The transition $B\,{}^3{\Sigma}_{0_u^+} - X\,{}^1{\Sigma}_g^+$ involves the $\Omega = 0$ component of the $B$ state [...] The labelling of the $B\,{}^3{\Sigma}_{0_u^+}$ state follows general usage rather than convention, which would label it $b\,{}^3{\Sigma}_{0_u^+}$.

Wikipedia also adds, although without a citation:

In polyatomic molecules (but not in diatomics) it is customary to add a tilde (e.g. $\tilde{X}$, $\tilde{a}$) to these empirical labels to prevent possible confusion with symmetry labels based on group representations.

I don't think you explained what ${}^3{\Sigma}^{-}_g$ means.

That explanation is here.

The $\ce{^3\Sigma ^-g}$ is a term symbol for the not-fully filled orbitals of a homo-nuclear diatomic molecule and comes from the $\ce{D{\infty h}}$ point group. The 3 indicates a triplet spin state, $\ce{\Sigma}$ zero orbital angular momentum, g (gerade) means orbitals have a centre of inversion, otherwise u (ungerade). The superscript - indicates -1 to reflection in plane containing nuclei, i.e. no plane of reflection. A superscript + would indicate a +1 reflection, i.e. has plane. The $\ce{O2}$ ground state is $\ce{^3\Sigma ^-_g}$ .

Thanks for mentioning your source. I totally missed the footnote on that page. It's been a while since I last had to use a lot of quantum mechanics thanks for the quick refresher.

The question didn't ask for an explanation of the term symbol anyway, but I appreciate the links/explanations in the comments. :) Just want to point out that the superscript minus indicates that the wavefunction is antisymmetric with respect to reflection in a plane containing the nuclei, not that such a plane does not exist.

@orthocresol;  antisymmetric, yes quite correct; blame my casual language.


text:
How to find maximum and minimum oxidation number of an element
I've researched around and some sources claim, that in order to find the minimum and maximum oxidation number you do this:
Maximum: the group the element
Minimum: the group of the element - 8
However I cannot get this to work for Fe? I'd appreciate a clarification. :)
It seems to work for Sulphur and what I've read about it's oxidations.
Maximum: 6
Minimum: 6 - 8 = -2
Don't metals always have a positive oxidation number?
Does this mean for Fe (iron) it's 0 to +3?
(I know this is a duplicate, sorry, however the others weren't answered and asked years ago).

That method is merely an approximation. See this wiki page for an empirical observed list of states. That method works especially poorly for transition metals, since they are so complicated.

Observed oxidation numbers for iron: $-4, -2, -1, +1, +2, +3, +4, +5, +6$.

Yes, I stumbled across with wiki page myself. However, I want to be able to explain and calculate this myself. Thanks, though :)

If you want to theoretically calculate the aforementioned list of oxidation numbers for iron, I suggest you to give up.

Oh haha, not that easy is it? Suppose I have to rely on observations made by chemists then? :) Not a bad way to go. Thanks.

Technically every oxidation state can be reached, by creating an environment that favours that state. You can see how $\ce{Na-}$ is generated: http://i.stack.imgur.com/Ced0K.gif

related http://chemistry.stackexchange.com/questions/34106/why-do-atoms-form-either-a-positive-or-a-negative-charge-but-not-both

@user34388 What do you mean by ‘every oxidation state’? I don’t think, hydrogen(II) has been or will ever be reached. At the same time, I think hydrogen($\mathrm{-III}$) is also not feasible, yet numerous elements are known to display both those oxidation states.

@Jan I'm quite sure that 20 years ago people thought it's impossible to generate $\ce{Na-}$.

But you have a point on hydrogen(II). It is theoretically impossible since hydrogen only has 1 electron.

Hydrogen can't be +2 since there is only one proton.

@user34388 My main point is that most (known) oxidation states are within the bounds of a period, i.e. I’m not aware about any oxidation states that remove core electrons or add electrons to the $n+1$ shell.

The rule you quoted is generally true for main group elements — but only if you count the groups in the older main-group/transition-metals formality. The newer IUPAC terminology counts the s, p and d element groups one after the other, so that sulphur would be in group 16. For that nomenclature, you need to subtract 10 from the group number for the maximum oxidation state.
Example: Phosphorus (group 15 by current terminology; group V by older terminology). Maximum oxidation state: $\mathrm{+V}$. Minimum oxidation state $\mathrm{-III}$.[1]
This does have a general quantum chemical explanation. These oxidation states add up to eight, which is exactly the number of electrons that typically make up the outer (valence) shell — where chemistry happens.
If we start at phosphorus($\mathrm{V}$) and want to remove another electron, we would have to remove this electron from the core orbitals. These are already pretty stable (i.e. low in energy) in the ground state, but for every electron removed they get stablised further. Thus, they end up in a very stable state and removing gets all the much harder.
Likewise, every electron added to a neutral compound will destabilise all orbitals of that atom. If three are already added to phosphous, giving phosphorus($\mathrm{-III}$), the next electron would need to be added to a distant orbital, which again is not something easy. This is why there is typically a range of eight for the chemically accessable oxidation states of main group metals.
However, I am not willing to bet anything that the s-elements of periods 4 and higher (potassium and below and calcium and below) stick to that rule; for the reason see and interpret below. So far, only $\mathrm{+I}$ and $\mathrm{-I}$ are known for alkaline metals (group 1) and only $\mathrm{+II}$ and $\mathrm{+I}$ for alkaline earth metals are known (save the elemental oxidation state $\pm 0$).

Transition metals are a lot harder. In theory, you could assume a range of either ten (d-electrons only), twelve (d and s-electrons) or eighteen (d, s and p-electrons). In fact, experimentally a range of ten has been established for chromium, manganese and iron while a range of twelve has been established for osmium and iridium. (Source: Wikipedia)
So far, the highest oxidation state has been found for iridium ($\mathrm{+IX}$). Platinum($\mathrm{X}$) has been predicted. (Source: also Wikipedia) Note that these states are well within the ‘s and d’ idea I alluded to earlier. To the best of my knowledge, there are no pieces of evidence pointing towards the ‘s, p and d’ idea.
There are a large number of unknown oxidation states for transition metals, e.g. while chromium($\mathrm{-IV}$) and ($\mathrm{-II}$) are known, chromium($\mathrm{-III}$) is not.
The notion that metals could only have positive oxidation states is incorrect. As an example, $\ce{[Fe(CO)4]^2-}$ with an iron oxidation state of $\mathrm{-II}$ is known.
All of this complicates the analysis strongly. Until much more research has been performed, you should probably not attempt to predict maximum and minimum oxidation states of these elements.

Thanks for this long, elaborate explanation! :) Just a question however, wouldn't the oxidation number of Fe in Fe(CO)4-2 be positive?

@javanewbie CO is a neutral ligand and isn't considered "non-innocent" AFAIK, so it doesn't affect ox. state. It does withdraw electrons so real charge on Fe nowhere near -2, and assigning ox. states to similar complexes is problematic.


text:
Calculating the effects in a fractionating column
Consider a fractionating column used to separate a mixture of acetone (A) and toluene (B) with a liquid molar composition of acetone called x. The feed is F, the distillate is D, the waste is W and the molar vaporation enthalpy is λ. How can one get an approximate difference between the effects in the boilar and condenser using only the given values for F, λ, x(F), x(D) and x(W)


text:
Why should all the acid be used up when making zinc sulphate crystals?
So to make zinc sulphate crystals you add granulated zinc to heated sulphuric acid. However, you have to make sure all the acid is used up and you know so when the zinc stops reacting and producing bubbles "hydrogen". You then filter out the solution and set it on a water bath to crystallise it. The question is : if all the acid is not used up in the reaction, what will happen when I start to evaporate the solution of zinc sulphate?

I don't think you have really explained the experiment correctly. I'd guess that you're using a diluted solution of sulfuric acid.
$\ce{Zn + H2SO4 ->[H2O] ZnSO4 + H2 ^}$
It would be difficult to have exactly the right amounts of zinc and sulfuric acid, so one or the other will be in excess.

If zinc is in excess the solution can be filter to easily remove the
excess Zn.
If Sulfuric acid is in excess then when all the water is gone you get zinc
sulfate which is wet with sulfuric acid. The sulfuric acid won't evaporate
on a water bath like water.


text:
What mass of water is present as a liquid when equilibrium is established?

$1.00~\mathrm{g}$ of water is introduced into a $5.00~\mathrm{L}$ flask at $50~^\circ\mathrm{C}$. What mass of water is present as a liquid when equilibrium is established? Also you know $92.5~\mathrm{mmHg}$ (vapor pressure).

$0.083~\mathrm{g}$
$0.41~\mathrm{g}$
$0.59~\mathrm{g}$
$0.91~\mathrm{g}$

I need thorough explanation for each step because I'm new to this. This question is in the gases section of the book, so I think I need to apply the ideal gas law, but I don't know.
With the help from the comment I came up with the following solution:
$$
\begin{align}
\frac{92.5}{760}~\mathrm{atm} \cdot 5.00~\mathrm{L} &= n \cdot 0.08206 \cdot 323~\mathrm{K}\\
n &= 0.023~\mathrm{mol}
\end{align}
$$
Since $1~\mathrm{mol}$ of $\ce{H2O}$ is $18~\mathrm{g}$, the answer is $0.414~\mathrm{g}$, or (2) $0.41~\mathrm{g}$.

The given temperature of $T=50\ \mathrm{^\circ C}$ and the given pressure of $p=92.5\ \mathrm{mmHg}$ approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of $T=50.000\ \mathrm{^\circ C}$ is $p=92.647\ \mathrm{mmHg}$, which could be rounded to $p=92.6\ \mathrm{mmHg}$ and not $p=92.5\ \mathrm{mmHg}$; in the opposite direction, however, the corresponding saturation temperature for a given pressure of $p=92.500\ \mathrm{mmHg}$ is $T=49.968\ \mathrm{^\circ C}$, which could be rounded to $T=50\ \mathrm{^\circ C}$.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of $T=50\ \mathrm{^\circ C}$).
Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions.
The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour.
The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about $\rho=988\ \mathrm{kg\ m^{-3}}$, the given mass of $m=1.00\ \mathrm g$ corresponds to a maximum volume of $V_\text{liquid}=1.01\ \mathrm{ml}=0.00101\ \mathrm l$, assuming that no water has evaporated. Considering that the total volume is given as $V=5.00\ \mathrm l$ (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected.
Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred.
$$1\ \mathrm{mmHg}\approx 133.3224\ \mathrm{Pa}$$
You might want to convert the given values to coherent SI units, or find a value for the molar gas constant $R$ that is expressed in a suitable unit.
The following data are given:

Temperature $T=50\ \mathrm{^\circ C}=323.15\ \mathrm K$
Pressure $p=92.5\ \mathrm{mmHg}=12332.322\ \mathrm{Pa}$
Volume $V=5.00\ \mathrm l=0.00500\ \mathrm{m^3}$

The value of the molar gas constant is $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$.[source]
The amount of vapour $n$ in the container may be estimated using the ideal gas law
$$\begin{align}
p\cdot V&=n\cdot R\cdot T\\[6pt]
n&=\frac{p\cdot V}{R\cdot T}\\[6pt]
&=\frac{12332.322\ \mathrm{Pa}\times0.00500\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times323.15\ \mathrm K}\\[6pt]
&=0.022949674\ \mathrm{mol}
\end{align}$$
(Note that this intermediate result is given with an excessive number of digits. It is deliberately not rounded in order to avoid round-off errors in subsequent calculations.)
Since the molar mass of water is $M=18.01528\ \mathrm{g\ mol^{-1}}$, the corresponding mass of vapour is
$$\begin{align}
m&=M\cdot n\\[6pt]
&=18.01528\ \mathrm{g\ mol^{-1}}\times0.022949674\ \mathrm{mol}\\[6pt]
&=0.413444803\ \mathrm g\\[6pt]
&\approx0.41\ \mathrm g
\end{align}$$
(Note that this final result is rounded to two significant digits. The number of significant digits is estimated in view of the fact that the mass of water $m=1.00\ \mathrm g$, the volume $V=5.00\ \mathrm l$, and the pressure $p=92.5\ \mathrm{mmHg}$ are given with three significant digits, but the temperature $T=50\ \mathrm{^\circ C}$ is only given with two significant digits.)
This result is the mass of vapour $m_\text{vapour}$ in the container, whereas the question is about the mass of liquid water $m_\text{liquid}$, which can be calculated from the given total mass of water $m_\text{total}=1.00\ \mathrm{g}$.
$$\begin{align}
m_\text{total}&=m_\text{vapour}+m_\text{liquid}\\[6pt]
m_\text{liquid}&=m_\text{total}-m_\text{vapour}\\[6pt]
&=1.00\ \mathrm g-0.41\ \mathrm g\\[6pt]
&=0.59\ \mathrm g
\end{align}$$
Therefore, the correct answer is option 3) $0.59\ \mathrm g$.

By way of comparison, precise engineering calculations for water and steam usually do not rely on the ideal gas law but use so-called steam tables. We may use such steam tables to check the accuracy of our estimate that has been based on the ideal gas law.
For example, for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium, we find the following values in REFPROP – NIST Standard Reference Database 23, Version 9.0:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988.00\ \mathrm{kg\ m^{-3}}=988.00\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.083147\ \mathrm{kg\ m^{-3}}=0.083147\ \mathrm{g\ l^{-1}}$

Furthermore, the database includes various other thermodynamic and transport properties, which are not required for the following calculations.
If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium can be obtained using the input “water boiling at 50 °C”:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988\ \mathrm{kg\ m^{-3}}=988\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.08315\ \mathrm{kg\ m^{-3}}=0.08315\ \mathrm{g\ l^{-1}}$

Note: The indicated values for STP are not in accordance with IUPAC recommendations.
We know that the mass balance in the container is
$$m_\text{total}=m_\text{vapour}+m_\text{liquid}\tag{1}$$
where $m_\text{total}=1.0000\ \mathrm g$ is the total mass of water (liquid and vapour) in the container (in this example, the precision of this value is arbitrarily increased to five significant digits).
Since density is defined as
$$\rho=\frac mV$$
the corresponding volume balance can also be expressed in terms of mass:
$$\begin{align}
V_\text{total}&=V_\text{vapour}+V_\text{liquid}\\[6pt]
&=\frac{m_\text{vapour}}{\rho_\text{vapour}}+\frac{m_\text{liquid}}{\rho_\text{liquid}}\tag{2}
\end{align}$$
where $V_\text{total}=5.0000\ \mathrm l$ is the total volume of the container (in this example, the precision of this value is arbitrarily increased to five significant digits).
Solving the system of the equations $\text(1)$ and $\text(2)$ yields the solutions
$$\begin{align}
m_\text{vapour}&=\frac{\rho_\text{vapour}\cdot\left(V_\text{total}\cdot\rho_\text{liquid}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=\frac{0.083147\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times988.00\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.41569\ \mathrm g
\end{align}$$
and
$$\begin{align}
m_\text{liquid}&=-\frac{\rho_\text{liquid}\cdot\left(V_\text{total}\cdot\rho_\text{vapour}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=-\frac{988.00\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times0.083147\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.58431\ \mathrm g
\end{align}$$
These results confirm that the above-mentioned estimates $m_\text{vapour}\approx0.41\ \mathrm g$ and $m_\text{liquid}\approx0.59\ \mathrm g$, which have been obtained using the ideal gas law and neglecting the volume of the liquid water, are reasonable and sufficient in order to answer the question.


text:
The mass of a carbon-12 atom is 12 amu. What percent of the mass is located in the nucleus?
The mass of a carbon-12 atom is 12 amu. What percent of the mass is located in the nucleus? and Why is the value so close to 100%?

I can only give a simple answer, so if others have better ones, great, but here's my shot:
An electron's mass is a lot less than a proton/neutron.
"When talking about relative masses, we use the term atomic mass unit (amu). Using this unit, a proton has a mass of 1.007 amu, a neutron a mass of 1.008 amu, and an electron a mass of 5.45 X 10-4 amu."
Basically, the proton and neutron is close to 1, and the electron is close to 0.
Carbon-12 means it's got a total of 12 protons and neutrons, and since carbon has 6 protons, that makes 6 neutrons. Since it's a normally charged carbon atom too, that makes 6 electrons.
Can you do the calculations yourself with that? Because now you know the masses.
Also this value should be really close to 100% because of how electrons are so much lighter than protons and neutrons; protons and neutrons are located in the nucleus and the rest of the atom is mostly empty space + some electrons, so that's why the bulk of the mass is there.


text:
Help regarding stereochemically inactive s orbital
Author JD Lee says in his book, Inorganic Chemistry

The dipole moment of $\ce{XeF6}$ is approximately = $0$ because the lone pair is present in the stereochemically inactive s orbital.

What is this stereochemically inactive s orbital? What does the statement mean?
PS: Please don't provide links to any web pages as I have gone crazy by the definitions they provide and also please try to explain it in an easy language and not some very technical chemistry as I am not an expert.

Well, it means it doesn't affect stereochemistry of compound; it's shape is almost unaffected by it's presence.

http://chemistry.stackexchange.com/questions/50187/what-is-hybridisation-of-xef6-in-solid-state http://chemistry.stackexchange.com/questions/10151/what-is-the-problem-in-predicting-the-structure-of-xef%e2%82%86-with-vsepr

S-orbital is spherical in shape , so any particular direction is not favoured, hence stereochemically inactive.

Why are we closing the old question as a duplicate of the new question? (@Mithoron)

@Jan Ortho seems keen on making new better ones...

An s-orbital is basically a pretty perfect sphere around the nucleus.
A sphere transforms upon itself no matter which element of symmetry you use. Thus, it cannot be the basis of any kind of chirality or other asymmetry.
If therefore the s-orbital is the only one that carries free electrons able to generate a dipole moment, none will be formed because it would be spherical.
(I know this answer bases quite a bit on circular — or should I say: spherical? — reasoning, but I wouldn’t know how else to write it …)


text:
Naming substituted branched alkanes with different projections

Hi all, I would like some help with naming these two compounds.
For compound 1, I know it is a 2-methyl-pentane, but I am unsure how to name the 3,4 $\ce{OCH3}$ compounds. I know that in the final name, it would be 2-methyl-3,4-di[$\ce{OCH3}$]-pentane (or vice verse depending on alphabetical order.).
For compound 2, I am absolutely lost and would appreciate any help. At first I though it would be a bicyclic cycloalkane, but I steered away from that idea because of the three bridgeheads. Is it possible to have three bridgeheads? I just learned this content yesterday in class and have scoured through my textbook but have seen no mention of a compound like this one. If you could at least give my a general idea of how the 3D structure looks, that may help me. Thank you!

The O-CH3 group is methoxy. So I think the name of the first compound would be 2,3-dimethoxy-4-methyl-pentane.

The circle in the second molecule is looking down a carbon-carbon bond. The solid "Y" are the three atoms/groups bonded to the front carbon, the "broken Y" is for the three atoms/groups on the back carbon. Can you name it now? (This drawing method is used to show stereostructure.)

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LATEX syntax. If you receive useful answers, consider accepting one.

You are right about the first one, it's called 2,3-dimethoxy-4-methylpentane. The $\ce{-OCH3}$ group is called methoxy. The second compound is a
newman projection
. It is basically the view that you will see when you are looking down the $\ce{C-C}$ bond axis. The second one should be called 2,3-difluoro-2,3-dimethylbutane.
Here's what it looks like (http://www.chemspider.com/Chemical-Structure.57499056.html?rid=29f2de9b-b762-46d4-a2ca-7452b8a5d6b6). In the newman projection you are looking down the the C2-C3 axis, and the arms show the substituents bonded to them.

The name of first compound is 2,3-dimethoxy-4-methylpentane and the name of second one is Newman projection of 2,3-difluoro-2,3-dimethylbutane


text:
Qualitative tests for sodium peroxoborate
Recently in a chemistry lab I preformed two qualitative tests on sodium peroxoborate.
The procedure for the qualitative tests involved taking a small amount(about 0.2g) dry dry $\ce{Na2[B2(O2)2(OH)4].6H2O}$ and dissolving it in water making up the solution to $10\ \mathrm{mL}$. I then divided it into to small beakers.

To one beaker I added 5 drops of 1M sulfuric acid and then add 0.02M $\ce{KMnO4}$ solution dropwise.
To the other beaker I added 5 drops of 1M sulfuric acid and then add 10% of potassium iodide solution dropwise.

What are the the balanced equations for these reactions?

It's possible that they are oxidation reactions but I wouldn't know the actual equation.

Related: Qualitative permanganate test on sodium peroxoborate


text:
How hot is elephant toothpaste?
So, I remember in high school chemistry when my teacher combined hydrogen peroxide with potassium iodide and it exploded into a yellowish foam. He said it was an exothermic reaction so it's really hot. I was just wondering exactly how hot is it? Is there a way to calculate the heat from the amount of chemicals used?

You could calculate a theoretical temperature for an adiabatic process, i.e. for a system in which there are no heat losses. However, this temperature is not achieved in a real experiment.

The enthalpy change of the reaction can be calculated using the enthalpies of formation of the compounds in the reaction.  To find the temperature change of the reaction, divide the quantity of heat released by the heat capacity of the solution in which the reaction occurred.
This won't give the exact answer.  Some of the heat will be absorbed by the surroundings including the flask.  The reaction produces gas and some of the energy released by the reaction is used up by the work needed for the gas to be released into the atmosphere.
Rough estimate:
Let's say the teacher used a 500 ml bottle of peroxide.  The solution contains about 0.5 mol of peroxide.  The enthalpy of formation of peroxide is about −190 kJ/mol.  The enthalpy of formation of water is about −240 kJ/mol and that of oxygen is 0.  So, 0.5 mol of reaction gives about 25 kJ of energy released by the reaction.  We'll assume all of it went into the 500 ml of water.  The 500 ml of water has a heat capacity of around 2100 J/K.  So, the solution gets about 12 K hotter.

This calculation seems to assume 3% hydrogen peroxide. A good elephant-toothpaste demo will use 30%, which of course produces ten times the energy. This variant of the demo can launch foam vigorously from a cylindrical container, expanding as it rises, and steaming -- I haven't measured the temperature, but I'm quite sure it's hot enough to injure, and may be near or at the boiling point.


text:
Sketch the Resonance structure
What are the resonance structure of thioformate ion and  selenocyanat ion?
Give the structure of each resonance?
With formal charge?

As can be seen from the periodic table below, oxygen, sulphur, and selenium belong to the same group, and they are all non-metals, so one should expect their chemical properties to be similar.

Having established their chemical similarity, one can transform the two ions in question into more familiar forms.
In thioformate anion ($\ce{HCSO-}$), we are not too familiar with sulphur $\ce{S}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes formate ion ($\ce{HCOO-}$), whose resonance structure is well-known:

Just transform one oxygen back to sulphur and we are done:

In selenocyanate anion ($\ce{[SeCN]-}$), we are not too familiar with selenium $\ce{Se}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes cyanate ion ($\ce{[OCN]-}$), whose resonance structure is well-known:

Just transform the oxygen back to selenium and we are done:

Copyright issues:

The first image is copyrighted by Michael Dayah.
The second image is from Wikipedia.
The third image is a derivative work of the second image.
The fourth image is taken by HGTCChem.
The fifth image is a derivative work of the fourth image.

Thioformate ion is stabilized by resonance but I am not sure about selenocyanat ion.

I think the selenocyanate ion is right.

It is just like cyanate: http://www.cureffi.org/media/2015/01/cyanate-vs-fulminate.png


text:
How to calculate the mass of water to add to copper(II) sulfate to form a 4% solution?

How many grams of water are required to make a $4.00\%$ solution with $15.0~\mathrm{g}$ of copper(II) sulfate pentahydrate?

I calculated the percent composition of water in a molecule of $\ce{CuSO4.5H2O}$ and it came out to be $36\%$. From that I calculated that there was a total of $5.4~\mathrm{g}$ of water in the sample. I'm guessing that we have to calculate how much water is required in order to make it a $4.00\%$ solution, but I don't understand why it wants it in grams. When
$$\text{Percent by volume} = \frac{\text{grams of solute}}{\text{mL of water}}?$$

For a solute a "4.00% solution" is definitely wt %. For mixing two liquids (e.g. ethanol and water) then it could either be by volume, or by weight. // Be very careful when throwing around  1 ml = 1 gram. For all practical purposes that is ok for pure water, but it doesn't work for solutions. The 4% solution of copper sulfate will be a bit more dense than 1.00 gram per ml.

A $4~\%$ solution is one that contains $4~\mathrm{g}$ of the solute in $96~\mathrm{g}$ of solvent to give $100~\mathrm{g}$ of solution. Just to get that out of the way.
To calculate the mass of the solute ($m(\ce{CuSO4})$ — not to be confused with $m(\ce{CuSO4 . 5 H2O})$) we can use at least two ways which are mathematically equivalent:

Calculate the amount of $\ce{CuSO4 . 5 H2O}$ — this is equivalent to the amount of $\ce{CuSO4}$. Then calculate the latter’s mass.
$$M(\ce{CuSO4 . 5 H2O}) = 249.72~\mathrm{g\ mol^{-1}}\\[1.2em]
n(\ce{CuSO4 . 5 H2O}) = \frac{m}{M} = \frac{15.0~\mathrm{g}}{249.72~\mathrm{g\ mol^{-1}}} = 6.01 \times 10^{-2}~\mathrm{mol}\\[1.5em]
M(\ce{CuSO4}) = 159.62~\mathrm{g\ mol^{-1}}\\[1.2em]
m(\ce{CuSO4}) = n \times M = 6.01 \times 10^{-2}~\mathrm{mol} \times 159.62~\mathrm{g\ mol^{-1}} = 9.59~\mathrm{g}$$

Use the molar weights of copper(II) sulphate and water to determine the mass percentage of copper(II) sulphate in the salt.
$$\frac{M(\ce{CuSO4})}{M (\ce{CuSO4 . 5 H2O})} = \frac{159.62~\mathrm{g\ mol^{-1}}}{249.72~\mathrm{g\ mol^{-1}}} = 63.9~\%\\[1.5em]
\Longrightarrow m(\ce{CuSO4}) = 15.0~\mathrm{g} \times 63.9~\% = 9.59~\mathrm{g}$$

In any case, we will arrive at a mass of the solute of $9.59~\mathrm{g}$. This is equivalent to the $4~\%$ mentioned at the very top. Therefore:
$$m_\mathrm{tot} = \frac{m_\mathrm{solute}}{0.04} = 240~\mathrm{g}$$
$$m_\mathrm{solvent} = 0.96 \times 240~\mathrm{g} = 230~\mathrm{g}$$
You noted that you already have $5.4~\mathrm{g}$ of water in the copper(II) sulphate hydrate present. Thus, you need to add the difference:
$$m_\mathrm{water} = 230~\mathrm{g} - 5.4~\mathrm{g} = 225~\mathrm{g}$$


text:
What is the difference between exponential factor and orientation factor in case of Arrhenius equation?

As per the book (Nivaldo J. Tro)
The Frequency Factor: The number of approaches to the activation barrier per unit time.
The Exponential factor: Number between 0 and 1 that represents the fraction of molecules that have enough energy to make it over the activation barrier on a given approach. The exponential factor is the fraction of approaches that are actually successful and result in the product.

Collision Frequency: It is the number of collisions that occurs per unit time.
Orientation Factor: Usually between 0 and 1, which represents the fraction of collisions with an orientation that allows the reaction to occur
I am unable to distinguish the between the exponential factor and orientation factor. Aren't they saying the same thing? How do they differ from each other?
Another doubt is what does the number of approaches mean? i.e. number of approaches taken by a single reactant per sec, or all reactant per sec.

The orientation factor is really an empirical add-on fix as way of trying to make the Arrhenius equation seem more reasonable. Most collisions between molecules do not lead to reaction, which is why the activation energy is there, but even when a molecule has obtained enough energy to react ( by random collisions in solution or vapour)  we may know from its structure that it can only react from one side or at one end, then an orientation factor is added.

The two are not quite saying the same thing—the orientation factor is a parameter that holds information about the likelihood the two molecules required for reaction are arranged in a way where they will able to react at all, whereas the exponential factor encodes, as you describe, the percentage of particles that have enough energy to actually engage in the reaction.
Perhaps an analogy will help here: Let's assume that we are playing a game of darts, but our game of darts will just be "who can hit the most bullseyes?" The process of hitting a bullseye will represent a successful reaction occurring.
If we stand, say, ten feet away from the dart board, then there will be some minimum energy that one has to throw the darts at in order for it to even be possible to hit the board. If I've broken my dart-throwing arm, it is very possible that I never throw a dart with enough energy to reach the board, and all my darts scatter on the ground. I need to increase the velocity of my throw in order to reach an energy where hitting the dartboard is even possible. This represent our exponential factor where the percentage of darts I throw that can even reach the dartboard are counted. Similarly, two particles need to have adequate energy for the reaction process to occur whatsoever. The only way I could get a bullseye without having enough energy would be for me to stand much closer to the dartboard, but that breaks the rules of the game! If you are curious why this parameter is exponential with temperature is related to a property in statistical mechanics where the distribution of particle kinetic energies is related to the exponential of each energy-pair's ratio.
Back to darts. Just because some of my darts reach the dartboard, this doesn't mean that they are all bullseyes. Only a fraction of the darts I throw that reach the dartboard will hit the bullseye—the arrangement of the dart relative to the dartboard it is colliding with determines whether it is a bullseye or a miss. The fraction of bullseyes I hit out of all the darts that reach the dartboard is represented by our orientation factor. Some reactions don't require super specific arrangements, and this would be like having a very large bullseye; others need very specific arrangements, and those would be like having a very tiny bullseye.
The collision frequency is, unsurprisingly, just how fast we throw the darts. The orientation factor and exponential factor basically act as fractions that determine what percent of collisions are successful. If we throw 20 darts per second, but only 16 make it to the board and only 25% of those are bullseyes, then our bullseye rate is $k = (20$ darts s$^{-1})(\frac{16}{20}$ dartboard hits$)(\frac{1}{4}$ bullseyes per dartboard hit$) = 4$ bullseyes per second. Similarly, if we have 20 collisions per second, but only 16 have sufficient energy to react, and only 25% of those with sufficient energy are oriented correctly, our reaction rate is $k = (20$ collisions s$^{-1})(\frac{16}{20}$ energetic successes$)(\frac{1}{4}$ correct orientations per energetic success$) = 4$ reactions per second.


text:
Deriving the unimolecular rate constant using the RRK model
Was reviewing some of my physical chemistry lectures and attempted a a derivation of the unimolecular gas phase reaction using a calculation based on the RRK model (Rice-Ramsperger-Kassel), and ran into some trouble half way into the derivation.
My Derivation
I have a unimolecular gas phase reaction that takes place as follows:
$$\ce{A}+ \ce{A} \rightleftharpoons^{k_a} \ce{A^*} + \ce{A}$$
$$ \ce{A^*}\rightarrow^{k_b} \ce{P}$$
Where, $\ce{A}, \ce{A^*}, \ce{P}$ represent the reactant, energetically excited reactant and product respectively.
Now, the way I understand it is that the RRK model is premised on the fact that the even though a molecule may have enough energy to react, said energy is distributed over all available modes of motion, and it has to be funnelled into a particular mode, (say particular vibrational mode to cause bond cleavage) for the reaction to proceed.
So I suppose that my reactant system has $s$ identical harmonic oscillators, and as an approximation all of them have the same frequency $\nu$. These vibrational modes are excited by $E = n h\nu$. The number of ways of distributing is $$N = \frac{(n+s-1)!}{n!(s-1)!}$$
The rationale behind this is that I have $n$ quanta that need to be distributed over $s$ oscillators. I think of these as $s$ containers, created by using $s-1$ partitions. The numerator is the total no. of arrangements of $(n+s-1)!$ and since the $n!$ and $(s-1)!$ arrangements of the objects are indistinguishable we divide by the terms in the denominator.
This is the reasoning that was presented to us in our course, and that I have attempted to reproduce
Now, let's single out a bond that will break if it is given energy (at least) $E^*= n^*h \nu$
So, after singling out this critical bond, we have $n-n^*$ quanta left, and $s-1$ oscillators (i.e $s-2$ partitions, if I am to follow the rationale described previously).
However, this is where I believe I run into a problem, and my derivation disagrees with what was presented in class, namely
$$N^* = \frac{(n-n^*+s-1)!}{(n-n^*)!(s-1)!}$$
My question is that shouldn't there be an $(s-2)$  term instead of an $(s-1)$ in the expression for $N^*$. It's been a few months, and I don't remember the reasoning presented in class, and since we have summer break right now, I can't exactly ask my professor. Am I missing something?
Also, I find this derivation a little bit convoluted, if someone can provide a alternative, or suggest a modification.
Anyway, using the expression for $N^*$ given above, I do eventually end up with the correct result.
The probability of of energy concentrating into one oscillator as described above is $$P = \frac{N^*}{N} = \frac{n! (n-n^*+s-1)!}{(n-n^*)!(n+s-1)!}$$, and assuming $ (s-1) \ll n-n^*$,
$$ P \approxeq \left (\frac{n-n^*}{n} \right )^{s-1} = \left (1 -\frac{E^*}{E} \right )^{s-1} $$
and thus, the Kassel form of the unimolecular rate constant is
$$k_b (E) = \left (1 -\frac{E^*}{E} \right )^{s-1}  k_b
\qquad \text{for} \quad  E\geq E^*$$
UPDATE I may/may not have answered my own question. I spent some time reading about combinatorics this evening and came up with this  little stars and bars exercise to work out my problem:
Say we have 20 quanta ($n$) available, and we have 6 ($s$) oscillators.
Using stars to represent the quanta, and 5 bars ($s-1$) to represent the partitions, I can draw something like this:
$\text{**|***|***|*****|*****|**}$
The no. of ways of arranging this collection is $N$ as described above.
Now, I assume that the disassociation requires at least 6 ($n^*$) quanta:
$\text{******} \quad  \text{|*|***|**|****|****}$
The left most collection is the critical oscillator (say we use exactly 6 quanta), and we hold that fixed. What is left on the right is $n-n^* = 20-6 = 14$ quanta and the 5 bars (still, $s-1$ partitions), and the total no. of arrangements is $N^*$ above.
Similarly, now if we have focus 7 quanta in our critical oscillator (we still need a min. of 6,so we have one extra), I can draw something like:
$\text{******} \quad  \text{*|*|***|**|****|***}$
Again, the leftmost collection is the critical oscillator. And our combinatorial problem is still of arranging $n-n^* = 14$ quanta and $s-1 = 5$ partitions.
I feel this works; could someone please tell me otherwise if this not the case.

I think that the equation of $P = N/N*$ lacks a ^(-1) exponent in the RHS.

@user1420303 I think you might be right, silly typo I'll fix it

Yes, that always will work. Although this does not answer your question, as you stated "after singling out this critical bond, we have $n−n^∗$ quanta left, and $s−1$ oscillators".

Your equations for N and $N^*$ are correct.
The first N is the total number of ways to distribute n quanta among s oscillators in energy range E to $E+dE$.
To calculate $N^*$ we need the total number of ways to reach a distribution that leads to reaction with $n^*$ number of quanta in the critical oscillator. This is given by the  total number of arrangements (quanta into oscillators) having energy 0 to E, which equals  $$N^*=(k+s)!/(k!s!)$$  with k quanta and s oscillators.
To obtain this last equation ( $N^*$) we sum $N=...$ equations i.e. $$N^* =\Sigma _{m=0}^k \frac{(m+s-1)!}{m!(s-1)!}$$  and is most easily seen by direct counting, say, with say $n=4$ quanta and $s=3$ oscillators. You can confirm the equation for N in the same way.
For a maximum of $k=n-n^*$ quanta in the $s-1$ remaining oscillators your $N^*$ is produced.

Hi! Thanks for your answer. I am sorry but I don't follow completely,  I wish to know why the expression for $N^$ is what it is instead of $N^ = \frac{(n-n^+s-2)!}{(n-n^)!(s-2)!}$ (which is what I thought initially). Any insight would be appreciated. Am I missing something? Should I pick up a book on combinatorics instead? haha

Hi, I've added more explanation, hopefully this will clarify things for you. The important point is that $n^*$ is the total number of arrangements from energy 0.

Indeed, I think things are a lot clearer now. I edited and updated my question a little while ago, with a direct counting exercise and the result make sense to me now. Perhaps you can take some time corroborate that?

Direct counting works just like putting any number of balls in to bags, say 4 balls into 3 bags in all possible arrangements. This gives you N. Add up all those similar arrangements from n equal to zero to 4 to get $N^$. If you try with large numbers it becomes impossible. Alternatively sum  algebraically to confirm $N^$

I find some flaws in the reasoning.
As a secondary problem, braking a bond give rise to two fragments, each of which must be treated with their correct number of normal vibrational modes.
Imagine one molecule with 11 atoms and $3N-6=27$ vibrational modes. After a bond break you can have for example: a chemical specie with 6 and other with 5 atoms, with $3*6-6=12$ and $3*5-6=9$ vibrational modes. So neither $s-1=27-1=26$ or $s-2$ make sense.
It is more important conceptually that as you are interested in the rate constant, you must concentrate the analysis in the "present" time. It is not a comparison of actual and future states. Molecules that already reacted can not be used in this context to deduce how the energy distribution in the molecules that did not react yet influence them in relation of velocity.
That is, it is still used $s-1$ because you are trying to get

The probability of of energy concentrating into one oscillator as described above

for reactive molecules.


text:
One-step oxidation of hex-1-ene to hex-2-en-1-ol

Convert the following and show a Mechanism

It has to be just in a single step and on the arrow head you can't use multiple reagents altogether. Can anyone do it? I am unable to do it in a single step.

This is my attempt ..! But it is multistep, one step is required ..!

This looks like a honework question. Please show us your attempt towards answering the question.

Singlet oxygen does the trick, if I am not wrong. As far as I can tell, I believe that with typical A-level reagents this conversion is impossible in one step.

What is A level reagent ?

Reagents taught in the A level syllabus.

Singlet oxygen conversion: http://www.chem.wisc.edu/areas/reich/chem547/2-redox%7B26%7D.htm Articles with this specific reaction: 1 (tBuOOH, some weird catalyst), 2 (O2, Au nanoparticles), 3 (O2, rose bengal, PPh3)

Is my multistep conversion correct?

You're supposed to turn an alkene into an an allylic alcohol in one step. The "classical" reagent for this conversion is selenium dioxide, $\ce{SeO2}$. Have a look at some older references, such as J. Org. Chem., 1979, 44, 4683-4689.
DOI

As kindly pointed out by orthocresol, this elegant method will unfortunately furnish the the wrong allylic alcohol. Whether in situ isomerization is possible (allylic hydroxylation in acidic medium) is feasible can't be decided until I have another coffee ;)
(Please note that a procedure using singlet oxygen will need two steps: the Schenck ene reaction furnishes a hydroperoxide which you would have to reduce.)

I was under the impression that SeO2 oxidation didn't proceed with transposition of the double bond

@orthocresol You are right and I definitely need another coffee this mormning ;-)

There is a paper which used $\ce{SeO2 + Ac2O}$ ("und Verseifen des erhaltenen Acetats") but it's old and Reaxys won't let me access it. Guillemonat, Annales de Chimie (Cachan, France), 1939, vol. <11> 11, p. 143, 188, 191, 193, 194, 196;
Comptes Rendus Hebdomadaires des Seances de l'Academie des Sciences, 1938, vol. 206, p. 1126

https://youtu.be/m51VmzjUGRA found this one . Many things about SeO2 though it is in Hindi


text:
Difference between electropositivity and electronegativity
Sodium is more electropositive than hydrogen although they are in the same group.
Why is the electronegative value of hydrogen greater than sodium (metal)?

Background:
Electropositive  implies having tendency to lose electrons to form positive ions. Electronegative implies having tendency to gain electrons to form negative ions. Electronegativity is a concept used to describe the tendency of an atom to attract electrons in a covalent bond.

Why is the electronegative value of hydrogen greater than that of sodium?

This is because sodium after losing an electron gets its octet configuration, but a hydrogen atom after losing an electron will not get the configuration of helium but it can get helium configuration if it accepts an extra electron or forms a covalent bond.
So this implies that sodium is more electropositive that hydrogen, even though they belong to same group.
Sometimes "more electronegative" is used to describe higher electronegativity and "more electropositive" is used to describe lower electronegativity. In this sense, elements are less electronegative (or more electropositive) as you go down any group in the periodic table. However, the difference in electronegativity between hydrogen and sodium is exceptionally large, and correlates with the very different chemistry these two elements show.

doesn't electronegativity mean the capacity of an atom to attract a shared pair of electrons towards itself?

@AmritanshSinghal yes

Because hydrogen is a nonmetal while sodium is a metal and tends to lose electrons because of metallic bonds present. That's why the electropositivity value of sodium is greater than hydrogen.

agree with your

The elements that can easily lose electrons to form positive ions are called electropositive elements, for example: metals. Electropositivity is the measure of the ability of elements (mainly metals) to donate electrons to form positive ions.
The elements that can easily accept electrons to form negative ions are called electronegative elements, for example: non-metals. Electronegativity is the measure of the ability of elements (mainly non-metals) to attract electrons towards itself.
Further you can sat that Electronegetivity is the property of an atom with in molecule and Electropositivity is the property of an individual atom.
Now coming to your question as hydrogen can get a stable configuration(of helium) by accepting an electron so it will be called an electronegetive element while sodium can get stable configuration(of neon) by losing an electron so it will be called an Electropositive element.

An electropositive element is an element which has a propensity to lose an electron. (e.g. metals in rancidity)
An electronegative element is an element which has a propensity to gain an electron. (e.g. non-metal in oxidation)

The caps is not necessary. You can bold with asterisks *.  i.e.: **[bolded text]**

Throughout the periodic table, there are elements within a group that have different electronegativities.
For example, Nitrogen(atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configuration of these two elements. Which of these will be more electronegativity? Why?
You can apply similar logic to your question concerning the electronegativity of hydrogen compared to that of sodium.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Can an SN1 reaction occur in compounds with the leaving group attached to a primary carbon, which can later rearrange?
Can SN1 reactions occur in compounds where the carbocation, even though is is initially $1^\circ$, can rearrange to form a more stable carbocation?
Examples cases:

Under typical lab conditions a primary carbocation is not likely to form unless there is stabilization by an adjacent double bond or benzene ring, or unless there is a neighboring group that can provide neighboring group stabilization - but then we really don't have a simple primary carbocation in such cases.  BTW, in your first example, a H shift would produce an antiaromatic (4n electrons) cyclopentadienyl cation, so this won't occur.

Well, a benzyl cation is technically a primary one, so …

Fairly sure neopentyl iodide reacts with aqueous Ag+ to give the rearranged product t-amyl alcohol

@ron Do you mind posting an answer? Thanks!

Your thinking is correct that they can rearrange and then form stable carbocation (though your first example is wrong as it would become anti-aromatic after rearrangement) ,
but the answer to your question is NO , they would be substituted by SN2 mechanism being
1 degree . The reason behind this is that the compound will not try to form a carbocation first where it is not that stable , you can think of it in this way that at the initial stage the compound doesn't know that it can rearrange later to form stable carbocation, this is something that we can see but not the compound and thus it will go by normal SN2 mechanism.


text:
White granular substance from restaurant with interesting properties - what is it?
It isn't salt or sugar.
Was in a shaker at a restaurant - caused burning/bleeding sensation after consumption. Made up of small white granules that seem moist - all sticking together. The restaurant did not know what it was.
Tested properties:
Dissolves in water and vinegar.
When dissolved in water there was little effect. In vinegar, it produced substantial heat.
When mixed with lemon juice, the powder turned orange, heated up enough to crack the plastic container it was in, and gave off smoke.
No other information - based on these observations can any suggestions be made as to what it is?

You kept some of the unknwon as evidence, didn't you?  And a witness, too? Because if you get harmed, document it, and seek both medical as well legal counsel.
Turning to the chemistry: Likely, while dissolving X in water, you felt some heat, too?  Was the colorless solution obtained, when diluted further with additional water and then applied on skin a bit "soap like", felt slippery?
Vinegar is acidic, because of the acetic acid present.  Hence, mixing / dissolving the unknown proceeds with evolution of heat is indicative for the unknown acting as base that is neutralized. Such reactions generate heat. Similar token with lemon juice, equally reacting acidic, because of the presence of citric acid (even a more acidic than acetic acid): heat evolution possibly because of the neutralization of X (as a base) with it.
An additional indication that X is something basic were the action on diluted neutral juice of red cabbage.  If basic, the addition of X would yield a blue/greenish stain, in contrast to acids turning it red.
Speculation: it could be remaining NaOH.

Because sodium hydroxide actually (NaOH) is a base, that dissolves with ease in water, it reacts with the emanation of heat while dissolved in acids.  (Even if I am surprised the lemon juice experiment heated up enough to shatter the container.)  Even while preparing a dilute solution of NaOH in water, say 1 M as commonly used in the chem lab, the heat generated is sensible across the glass wall of the container used.
Because sodium hydroxide actually is used as a cleaning agent, to remove oily residues, to remove fat and grease (in other instances, to prepare lye soap) from pans (provided they are not made of aluminum).
Because NaOH is hygroscopic, in other words, the little white grains of it quickly look wet and stick to each other when exposed to humid air.
Because in pure form, sodium hydroxide is corrosive; and aqueous solutions of it are hazardous, too (pointing to your bleeding sensation).

Thankfully, such accidents in restaurants occure not often; yet there are examples ref1, ref2.  Bars ref3 use caustic soda to rinse their pipes of beer & other beverages, too, connecting the counter with the stock, often out of sight, in the basement.

Your answer makes sense, but how do you explain the orange colour when the powder was mixed with lemon juice?

(+1) My first thought was NaOH too. The only slight doubt I have is that the dissolution of NaOH is noticeably exothermic, as anybody who has tried to prepare a stock solution can verify. :) Of course, if you just dissolve a little bit of solid in a lot of water, you wouldn't be able to tell.

Is there anything about NaOH (or perhaps inside this particular mixture) that, when mixed with lemon juice, could lead to an orange colour?

@getafix and Elkal Anwa The stated observation of orange coloration is indeed a bit puzzling. Aqueous solutions of NaOH + (acetic or citric acid alone) should remain colorless.  A lemon is more complex, a mixture.  Currently, none of the said reagents are with me to probe it.

Wikipedia doesn't state the color for monosodium citrate or disodium citrate, and Google image search is too tolerant to be helpful as well.

@JanDvorak both the Dutch as well as the German versions of wikipedia actually hold color photos of crystals of sodium citrate.  Expectedly, they are colorless.

@Buttonwood: I happened to have some lemon juice and NaOH lying around, so I tested it, and there is indeed a noticeable color change from yellow to orange. Adding clear vinegar reverses it, so it would seem that something in the juice is reversibly changing color with pH.


text:
Structure of Borax
The structure of Borax has really baffled me.
Its formula is $\ce{Na2B4O7.10H2O}$ and it is formulated as $\ce{Na2[B4O5(OH)4].8H2O}$.
2 structures of Borax given on the internet are

Can anybody explain that why the formulas mentioned on different websites are different.Something is strange about this compound because even Wikipedia does not give the structure.
It only says

there are two four-coordinate boron atoms (two $\ce{BO4}$ tetrahedra) and two three-coordinate boron atoms (two $\ce{BO3}$ triangles).

Greenwood and Earnshaw's Chemistry of the Elements, 2nd ed. (p 206) writes that the structures of the borates are rather complicated:

It is clear [...] that, without structural data, the stoichiometry of these borate minerals gives little indication of their constitution.

As such, for example, $\ce{KB5O8.4H2O}$ is more properly formulated as $\ce{K[B5O6(OH)4].2H2O}$. You've also noticed that borax, $\ce{Na2B4O7.10H2O}$ is more properly formulated as $\ce{Na2[B4O5(OH)4].8H2O}$. This means that the structure of borax does not simply consist of $\ce{Na+}$ ions and $\ce{B4O7^2-}$ anions with waters of crystallisation.
Therefore I would guess that both structures are, more likely than not, incomplete.

A further illustration is afforded by borax which is normally formulated $\ce{Na2B4O7.10H2O}$, but which contains tetranuclear units $\ce{B4O5(OH)4]^2-}$ formed by fusing 2 $\ce{B3O3}$ rings which each contain 2 $\ce{BO4}$ (shared) and 1 $\ce{BO3}$ unit; borax should therefore be written as $\ce{Na2[B4O5(OH)4] .8H2O}$.

(grey - boron; white - oxygen; black - hydrogen).

SO, is the structure you mentioned is the true anion part of Borax?Also,is the structure verified by means like spectroscopy etc?Thanks for your answer.

Yes, I assume this is the true anion. Don't know, a bit too lazy to look up.

Just wondering, if the true anion is known, then why do most of the first few structures in google image search depict the incorrect ones?And why wikipedia avoids to provide its version?(sorry if the question appears silly)

I don't know why that would be the case.

@AmritanshSinghal  This structure was verified by X-ray structure analysis, I have a paper link in my A: What is the IUPAC name of borax?

In  this structure can we say 2 B are in same plane ?


text:
Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones

In the reactions above, taken from Tetrahedron Lett. 1982, 23 (23), 2355–2358, the choice of protecting group on the α-hydroxyl group has a large influence on the diastereoselectivity of reduction.
Why does the benzyloxymethyl protecting group lead to the formation of the anti diol (consistent wiht the Cram chelate model), whereas the tert-butyldiphenylsilyl protecting group gives the syn diol (consistent with the Felkin–Anh model)?

In order for the Cram chelate product to predominate after the addition of hydride donor to a chiral carbonyl compound, which contains a heteroatom in the alpha position, this heteroatom and part of the reagent must be able to form a five-membered ring chelate. If this is not possible, one observes Felkin-Anh proudct.
In the mentioned reaction, when protecting group is not bulky it form five-membered ring chelate easily with Li+ after addition of Hydride to carbonyl resulted in cram chelate product as major product. However, when bulky protecting group is present it doesn't five-membered ring, consequently resulted in Felkin-Anh product as major product.

I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too.
Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest group prefers a conformation where it is perpendicular to the carbonyl moiety. This gives two relatively competitive lowest energy conformations, with the medium and smallest groups differing in their proximity to the carbonyl oxygen.
For the TBDPS-protected substrate shown above, the OTBDPS group would be considered the largest group. The medium and small groups are methyl and hydrogen respectively. This placement is also consistent with a modification to the Felkin–Anh model, which places electronegative substituents (such as –NR2 or –OR) perpendicular to the carbonyl (an explanation is given at the bottom of the answer). In any case, the two conformations of interest are the two shown below:

Moreover, the nucleophile doesn't prefer to attack the carbonyl carbon at a 90° angle; rather it prefers something closer to the tetrahedral angle (Bürgi–Dunitz trajectory). Naturally, the nucleophile prefers attack from the opposite side to the large group. This leaves two possibilities: for the nucleophile to attack over the small group, or for it to attack nearest to the medium group. The former is preferred, and the major diastereomer will be the syn diol:

On the other hand, Lewis basic groups such as $\ce{-OCH2OBn}$ are also capable of forming a chelate with the metal together with the carbonyl oxygen. This feature forces the $\ce{OR}$ group to be held parallel to the carbonyl. Nucleophilic attack still occurs along the Bürgi–Dunitz trajectory, and the nucleophile attacks opposite the larger of the two remaining groups. This will invert the stereochemical outcome as compared to the Felkin–Anh model previously described, and hence leads to the formation of the anti diol.

Now, to answer your question of why does the choice of protecting group matter. The previous discussion illustrates that the reason we get the Cram Chelate product is the chelation of the metal by the $\ce{OR}$ oxygen and the carbonyl. This chelation can take place easily if the $\ce{R}$ group is small, but is greatly hindered with a bulky alkyl/silyl substituent, in which case we get the product determined by the polar Felkin–Anh relationship.
The bulk of the TBDPS group does indeed prevent chelation, and you get the observed stereochemistry in accord with the Felkin–Anh rule. However, chelation is possible when BOM is the protecting group (Jan already offered an excellent explanation), and you end up with the Cram chelate product.

When there is an electronegative atom or group α to the carbonyl, there is a slight change to the Felkin–Anh model which may be applied. This ‘slight change’ is that this electronegative group becomes the equivalent of the ‘large’ group in the Felkin–Anh model.
The reason for this that the energy of the C–X σ* antibonding orbital is rather low, and so when the C–X bond is perpendicular to the carbonyl group, this can overlap with the π* MO of the carbonyl group to make a new, lower energy LUMO.
In "human" speak, this means that this conformational arrangement is more
reactive than any other. With this change in mind, the Felkin–Anh type addition now proceeds as before. However, the stereochemical outcome will generally be different from the original Felkin–Anh model.
For more information, see: How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?

The $\ce{TBDPS}$ group (tert-butyldiphenylsilyl) is a very bulky protecting group. A silicon atom is bonded to the oxygen, which already acts as a larger carbon atom. That is then substituted by three alkyl or aryl groups, one of which is the very bulky tert-butyl group. There is a lot of congestion going on next to that oxygen atom, so its ability to coordinate to Lewis acids (including via hydrogen bonds) is severely limited. Therefore, it is mainly the polar Felkin–Anh relationship that dictates the stereochemistry of the attack in the case of a $\ce{TBDPS}$ group.
On the contrary, the benzyloxymethyl (BOM) group $\ce{CH2OBn}$ used in the other example has a primary carbon attached directly to the protected oxygen. Not only is carbon a lot smaller than silicon, but it also features two hydrogen substituents ($\ce{CH2}$) and only one substituent group – an oxygen which itself only bears one other group. Therefore, the BOM-protected oxygen can still coordinate to Lewis acids easily, such as the lithium cation or aluminium as present in the example. Thus, the Cram chelate model is feasible in this case and correctly predicts the observed stereochemistry.

This is quite often the case that some protecting groups allow for chelation while others don’t. Pretty much all the silicon-based protecting groups (those that include a $\ce{O-Si}$ bond) will inhibit chelation due to their size. Typically anything that starts off as $\ce{CH2}$ will allow chelation. A non-exhaustive list of common groups of this type would be:

MOM
SEM
Bn
PMB


text:
Why is there no standard temperature?
The standard state for a gas is defined at 1 bar of ideal gas.  Why isn't there a corresponding standard state temperature?

Note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$.

The short answer
The standard pressure is not an experimental condition, while the temperature is.
The long answer
The main reason the standard state is defined is because it leads to this very useful equation
$$K_p = e^{-\Delta G^\circ/RT}$$
Say you have this reaction: $A \rightleftharpoons B + C$ One way to use this equation is to compute the free energy of 1 mol of $A$, $B$, and $C$ at 1 bar using equations derived for an ideal gas, compute $\Delta G^\circ = G^\circ (B) + G^\circ (C) - G^\circ (A)$, and use that value to predict $K_p$.
If the gasses behave like ideal gasses "in real life" then the measured  $K_p$ will match the $K_p$ computed from $\Delta G^\circ$.  You can do the measurement at any pressure you want, not just at 1 bar.* The standard state refers to the pressure you use when computing $\Delta G^\circ$.  The only thing it has to do with the experimental measurement is that it defines the units you should use for your partial pressures when computing $K_p$
$\Delta G^\circ$ does also depend on temperature, but the temperature you chose should be the same as the experimental conditions.  So the temperature is not part of the standard state definition.
But what about "Standard temperature and pressure (STP)?"
Standard temperature and pressure (STP) refers to STP conditions under which $K_p$ is measured, not the pressure used to compute $\Delta G^\circ$. I know, they couldn't have made it more confusing if they tried when they named these things.
*Of course if you do the measurements at very high pressures or low temperatures, then the assumption that the gasses behave ideally will be less valid and the measured $K_p$ will differ more from the $K_p$ computed from Equation 1.  However, that is a separate issue unrelated to the standard state because the $K_p$ in Equation 1 refers to the $K_p$ you would measure if the gasses behaved ideally at the pressure and temperature used in the experiment.

if anyone has a different take on this I would very much like to hear it

See also Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions and Standard quantities in chemical thermodynamics. Fugacities, activities and equilibrium constants for pure and mixed phases (IUPAC Recommendations 1994)

The standard in STP refers to a legal norm so that sets of data can be compared.  So, both T and P have nothing to do with the thermodynamic standards that you describe and discuss (and most probably understand a lot better than I).

The various standard conditions for temperature and pressure (STP etc.) are explained by this xkcd.

@JanJensen If we had select a different standard pressure e.g. $10$ bar would the equilibrium constant change?

@adosar yes, if the number of particles change on going from reactant to product

@JanJensen What determines which standard state we must use? For example if we had a reaction in a open vessel with pressure $10$ bar in order to calculate the equilibrium constant $K$ should we use the standard state with $1$ bar or the one with $10$ bar? What dictates which standard state should be used?

The standard state has nothing to do with experimental conditions. Sometimes I explain it in terms of units: Let's say one paper report $K_p$ for a reaction as 0.1 bar and another paper reports it as 100 millibar. Clearly using "0.1" and "100" will give different $\Delta G^\circ$ values. So we have all agreed to convert to bar units before computing  $\Delta G^\circ$ (i.e. the standard state is 1 bar).


text:
Is mg/m^3 same as ppmw (CO2)?
I'm trying to convert $\ce{CO2}$ ppm (volume) in the air into ppm (weight) in the water.
I found this formula

$\mathrm{mg \ m^{-3}}$ = (ppmv)(MW) / [($0.08205$) (¦$\mathrm{K}$)]

where,
ppmv = air pollutant concentration, in parts per million by volume
$\mathrm{mg \ m^{-3}}$ = milligrams of pollutant per cubic meter of air
$\mathrm{K}$ = atmospheric temperature in degrees Kelvin = $273.15$ + $^\circ C$
$0.08205$ = universal gas law constant in $\mathrm{L\ atm \ mol^{-1} K^{-1}}$
MW = molecular weight of the air pollutant (dimensionless)
$\mathrm{atm}$ = absolute atmosperic pressure in atmospheres
$\mathrm{g \ mol}$ = gram mole
I've read the $\mathrm{mg \ m^{-3}}$ is a subunit of ppmw, so is it correct to say $\mathrm{mg \ m^{-3}}$ is ppmw?

No. ppmw means parts per million by weight (mass). This is not the same as mg/m3, which is milligrams (mass) per cubic meter (volume).
For example, in a gas of very low density (e.g. $\ce{CO2}$ at 10-6 Pa), 1 mg/m3 could exceed the weight of $\ce{CO2}$, i.e. would be greater than 900,000 ppmw.
One measure has to do with mass in a given volume, the other measure is mass in a given mass of another substance.

How can something be more than a million parts per million?

oops... thanks for pointing that out... the answer has been corrected.


text:
Can salt and water be obtained from a reaction between a gas and an acid?
Is it possible to get salt and water (neutralization reaction) by reacting a gas and an acid?
I don't know if it's possible or not. I'm unable to find any examples in this regard.
Would anyone be able to provide an example in support of the question?

Well, ammonia is a gas.

I know that ammonia(g) and HCl(aq) produce NH4Cl (s) but I cannot find such a  reaction that  produce salt and water.

So basically you are looking for a hydroxide that is gaseous...

Well this will work .Is there any?

You can vaporize sodium hydroxide into a gas.

English is odd: "a salt" includes $\ce{CaF2}$, $\ce{NH4Cl}$ [as @ortocresol states] and $\ce{NaCl}$ as well as many other compounds. Just "salt", though, implies "table salt", $\ce{NaCl}$. So which do you mean?
Any gas that can react with an acid to produce table salt, $\ce{NaCl}$ must contain sodium, then. $\ce{NaOH}$ has a boiling point of 1,661 K, so bubbling $\ce{NaOH}$ vapor into hydrochloric acid would work. For that matter, Na boils at 1,156 K, so would be easier to do, though it also produces $\ce{H2}$ (as well as water) as a byproduct.

Fine remark brother.Well, the word salt has nothing to do with table salt; it may be inorganic or organic .In chemistry lab we utter the word salt several times for different purposes.Then your remark  construe that every time we ask for table salt!! I think  it is unimportant to explain what it is;context clearly interprets it.

Still  if you have any confusion go onto the link please "https://en.wikipedia.org/wiki/Salt_(chemistry)".

You can use a salt of $\ce{H3O^+}$ as the acid. Sulfuric acid forms such a solid salt slightly below room temperature, and even when melted at room temperature solutions having the composition corresponding to $\ce{H3O^+HSO4^-}$ retain a strong ionic character (see here). Then ammonia may be passed over this hydronium salt to give ammonium sulfate or bisulfate plus water.

If you want to get salt, then you should look for a volatile base.
$\ce{NH3}$ does our job here.
$$\ce{2NH3 + H2SO4 -> (NH4)2SO4}$$
However, I am not sure if water could be formed. If there exists something like $\ce{NH4OH}$ vapours, then sure, water will be formed.

Yeah I also think in this way but I cannot find such reaction to produce salt and water


text:
Hard and Soft Acid and Bases in terms of HOMO and LUMO
It has been given in a book and on Wikipedia that:

hard bases have highest-occupied molecular orbitals (HOMO) of low energy, and hard acids have lowest-unoccupied molecular orbitals (LUMO) of high energy.[7][8]
soft bases have HOMO of higher energy than hard bases, and soft acids have LUMO of lower energy than hard acids. (However the soft-base HOMO energies are still lower than the soft-acid LUMO energies.)[7][8]

Let us consider the energy difference between HOMO and LUMO is given by $\Delta E$. Thus,
$$\Delta E (S-S)<\Delta E(H-S) < \Delta E (H-H)$$
Hence the hard-hard interaction should have been weaker than hard-soft, which is not the case. Please explain.
P.S.: I went through the reference (Meissler and Tarr) and this question but could not find a satisfying explanation.

The idea is that hard-hard interactions also have a significant element of electrostatic attraction, which isn't accounted for by MO theory.

Can you please tell any formula that determines which path for interaction is the most feasible (say in terms of free energy)