IUPAC numbering of carbon atoms in a chain A chemistry handout for nomenclature of org. compounds I've been reading states a particular rule for every functional group viz.

If there is a choice in numbering not previously covered, the parent chain is numbered to give the substituents the lowest number at the first point of difference.

Can anyone explain what it means?

Consider a n-hexane compound

$\ce{CH3-CH2-CH2-CHCl-CH2-CH3}$

The Cl is different than the H of n-hexane.

Going left to right the Cl is on the 4th carbon. Going right to left it is on the 3rd carbon. So you should use the right to left numbering in this case.


Why is water considered ionic in fuel cells but otherwise covalent? If water, $\ce{H2O}$ is a covalent compound formed by sharing of electrons, why is it said (in case of fuel cells) that formation of water from hydrogen and oxygen is a redox reaction (transfer of electrons, not sharing). This would make it an ionic compound.

Any insights?
Note that, the concept of water being formed as a result of redox reaction is highlighted in fuel cells (which is the basic principle of fuel cells - generating electricity by exploiting this transfer of electrons).

I know how fuel cells work, and it is indeed a redox reaction; but my doubt is - can water be formed by both covalent/ ionic means?  (In textbooks, or in exams when asked to identify bonding present in water); when both covalent and ionic type can be achieved in its formation? Is both "covalent" and "ionic" correct while describing the formation of $\ce{H2O}$ molecule?

Water is a covalent compound. But as we know no compound is 100% ionic or 100% covalent. Since the electronegative difference (END) between hydrogen and oxygen is very high it is a strongly polarised covalent compound therefore the electron cloud is shifted towards the oxygen atom and therefore water has a very high % ionic character.


Why is the square of the wavefunction a probability density? Can anyone explain it in simple words? I tried to read about the Born rule on Wikipedia but it seems it is just describing the rule. Is there any proof that is suitable for person with no solid math background?

First, I would not call the Born rule a law as it is done in the Wikipedia article linked by OP in the question. It is, of course, the question of the definitions, but I would call the Born rule rather a postulate. A law in physics is a principle (usually universal) deduced from observations, while a postulate is an assumption, a statement that we can not prove or derive but which is assumed to be right and tested to be consistent with experimental observations.

The Born rule is a postulate, thus, there is no any proof of it; it is assumed to be right. The only thing we can do is to justify a postulate: we can make experimentally verifiable predictions on the basis of it and test them. And it happens so that so far the experimental outcomes are always in full agreement with the predictions done using the Born rule. Thus, we know it provides a correct description of physical systems.


Order of priority for including functional groups in the parent chain compared to order of priority for numbering the parent chain The following is about the IUPAC naming system.

So as far as I understand the rules for numbering a chain are as follows:
Compare the locants of functional group 1 in both cases if the are lower in one of them it wins. Compare the locants of functional group 2 in both cases if the are lower in one of them it wins. And so on for all functional groups in order of priority. And then in the end we compare the locants for all substituents in alphabetical order.

But the rules for choosing a parent chain are the same, except we firstly want to maximize its length and after that we want to maximize the number of functional group 1s, functional group 2s and so on. And after that we want to minimize the locants for functional group 1, functional group 2 and so on until any of them is better. I am not sure what we do after that though.

My main 2 questions are the following:

Is the ordering of the functional groups the same for all 3 steps (maximizing the number of functional groups in the parent chain, picking a parent chain with minimized locants of the the functional groups, and then the same but for directing it)?

Where can I find the ordering of all functional groups or at least the following functional groups: alkyl (are alkylidenes in the same category as alkyls?), halogen, OH, double bond, triple bond, and  NO2?

And I know that sometimes two or more of these are grouped together in a set (for example double and triple bonds) so an ordered list of these is not enough.

Can you please provide something like (Set of double and triple bonds)>(double bonds)>alkyls but for all of the listed? If not where can I find such a thing?

Please tell me if any of that is wrong.

There are indeed several analogous criteria for the numbering of locants and for the choice of a principal chain; however, there are also significant differences.
The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

You can find the corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) below. However, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.
Also note that there is no order of priority for substituents cited as prefixes (e.g. ‘bromo’, ‘chloro’, ‘ethyl’, ‘methyl’, or ‘nitro’). All such prefixes are considered together as a set. However, you need the seniority order of compound classes in order to choose the senior class that is expressed as a suffix.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)


Higher energy π HOMO Which is higher in energy? Cyclohexanone enol or cyclohex-2-en-1-ol?

Okay from what I understand a higher pi HOMO is synonymous with more nucleophilic. So I would expect cyclohex-2-en-1-ol to have the higher pi HOMO because in the case of cyclohexanol enol there would be inductive withdrawal by the oxygen connected to one of the sp2 carbons involved in the pi bond.

Or would it be the other way around? Resonance donation by oxygen into the pi bond of the enol makes it more nucleophilic? How can we tell how the HOMO/LUMO energies split?

from what I understand a higher pi HOMO is synonymous with more
  nucleophilic.

Yes, that's generally true.  If you have a substituent on a double bond that increases the electron density in the double bond, then the double bond (usually one carbon more than the other) becomes more nucleophilic and the energy of the HOMO and LUMO are raised.  Conversely, electron withdrawing groups attached to a double bond withdraw electron density, making the double bond less nucleophilic and lowering the energies of the HOMO and LUMO.

Comparing cyclohexanone enol (cyclohex-1-en-1-ol) to cyclohex-2-en-1-ol, you can draw a resonance structure for the former that increases electron density in the double bond (especially the $\beta$ carbon).  This electron donating resonance effect overwhelms the electron withdrawing inductive effect of the hydroxyl group.  Therefore, the double bond in the cyclohex-1-en-1-ol will be more nucleophilic and its HOMO and LUMO will be higher in energy than those for cyclohex-2-en-1-ol.


What exactly is an orbital? What exactly is an orbital? Atomic or molecular.

Is it the function that describes the behaviour of the electron?

Is it the Schroedinger's equation solution, e.g., for Hydrogen atom?

Is it the probability of finding any electron of an atom in any specific region around the atom's nucleus?

Maybe all these describe the same thing. What is it?

Terminological questions in chemistry usually are better answered using IUPAC recommended definitions and the IUPAC Gold Book defines orbital (atomic or molecular) as

  Wavefunction depending explicitly on the spatial coordinates of only
  one electron.

But to be honest, I don't like this particular definition. The problem is that a one-electron wave function is, strictly speaking, a function of not only the spatial coordinates of an electron ($\vec{r}$) but also the spin one ($m_s$). In quantum chemistry we distinguish the so-called spin orbitals, which are one-electron wave functions, from the so-called spatial orbitals, which can be though of wave functions that describes a hypothetical spin-less electron.

The connection between spin and spatial orbitals is not so trivial. In the most general case a spin orbital $\psi$ can be represented as follows,
$$
  \psi(\vec{x}) = \phi_{\alpha}(\vec{r}) \alpha(m_s) + \phi_{\beta}(\vec{r}) \beta(m_s) \, ,
$$
where $\vec{x} = \{ \vec{r}, m_s \}$ is the joint spin-spatial coordinate of an electron, and $\phi_{\alpha}$ and $\phi_{\beta}$ are two different spatial orbitals. What we have here is essentially the linear combination of two possible spin states of an electron ("spin-up" and "spin-down") each with its own spatial probability.

And since OP mentioned in comments the fully relativistic Dirac's treatment of a single electron, it is a good idea to note an alternative way of expressing the idea of a spin orbital: in a form of the  so-called two-component wave function, or Pauli spinor, that represents the state of a single electron in a non-relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})
    \end{pmatrix} \, .
$$
This is to be contrasted with a four-component Dirac spinor that describes a state of an individual electron in a relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})   \\
        \chi_{\alpha}(\vec{r})  \\
        \chi_{\beta}(\vec{r})
    \end{pmatrix} \, ,
$$
in which the last two components describe a "spin-up" and a "spin-down" states of the associated positron.

Now, back to chemistry, we note that in general chemistry we always think of each and every electron as being in a pure spin sate: either "spin-up" or "spin-down", rather than in a superposition of these two. Besides, we assume that spin orbitals come in pairs: for each and every spatial orbital $\phi(\vec{r})$ we have two spin orbitals given as follows,
$$
  \psi(\vec{x}) = \phi(\vec{r}) \alpha(m_s), \quad \psi(\vec{x}) = \phi(\vec{r}) \beta(m_s) \, .
$$


Molpro wavefunction symmetry specification I'm beginning to use Molpro software for some computations. But I got completely confused by its description of specification of wavefunction symmetry.
It tells us, that the symmetry of wavefunction is described like
$$\text{wf},\text{nelec},\text{irrep},\text{spin},$$
where $\text{nelec}$ is the total number of electrons in the molecule, $\text{irrep}$ is the number of the irreducible representation (list of numbers in Molpro docs) and $\text{spin}$ is the total spin number.
I know, that water belongs to the $C_{2v}$ point group, and I found its character  table:

I know, that 1/-1 values correspond with variance/invariance of orbitals, i.e. $A_1$ representation corresponds to the fact, that $\ce{s}$ and $\ce{p_z}$ orbitals are
not inverted by any of the four symmetry operations.
I also know, that the electron configuration of oxygen is $\ce{1s^2 2s^2 2p^4}$ and both hydrogen atoms have $\ce{1s^1}$.
But I'm not sure how am I supposed to choose the irreducible representation of the whole molecule.
In this answer the similar problem is described and $\ce{H2O}$ molecular orbital diagram is given:

The answer tells us, that molecular irreducible representation can be obtained like this:
$$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2 = (A_1)^2 \otimes (B_2)^2 \otimes(A_1)^2 \otimes (B_1)^2 = A_1,$$
which is in accordance with the first example in the Molpro docs, where $\ce{H2O}$ wavefunction is specified like WF,10,1;.

Questions

So, is this the correct way to find the irreducible representation of the whole molecule? And if it is, could you, please, explain, how to get the molecular orbital representation?
How can I obtain the total spin number?

Yes, this is the correct way to find the irreducible representation of the molecule. To obtain the energy ordering of the molecular orbitals (which determines the order in which they are filled for the ground state molecule), you need to run a molecular orbital theory computation (e.g. Hartree-Fock or density functional theory).

The spin number in Molpro is not particularly intuitive. For a molecular term symbol such as $^1B_2$, spin information is conveyed as $2S+1$, where $S$ is the total spin quantum number. In the specific case of a $^1B_2$ state, the molecule has a total spin quantum number of $0$. However, Molpro users must specify spin as $2S$.

As noted in the comments, however, specification of the irreducible representation or spin is generally unnecessary for Molpro computations. For Z-matrix inputs, Molpro determines the symmetry automatically, and in most cases this will be the symmetry the user desires. Molpro also automatically sets the spin if the user doesn't specify it. By default, Molpro uses $S=0$ for molecules with an even number of electrons and $S=1$ for those with an odd number of electrons. Since a large number of molecules that are kinetically stable under typical terrestrial conditions have an even number of electrons, the default spin should work well for many users. Molpro users who work on excited electronic states or molecules with unpaired electrons are much more likely to need to specify the spin.


Does sour milk have something else in it besides lactic acid that gives it such a horrific taste and odor? Even slightly sour milk has a horrible, nauseating taste, while plain yogurt tastes either great, or at least nowhere near as bad. A dozen sources (all of them) say that the sour odor and taste of sour milk and also of yogurt is caused by the presence of lactic acid. Does sour milk have something else in it besides lactic acid that gives it the horrific taste and odor that yogurt (which has plenty of lactic acid it it) almost never has. Or is there more than one kind of lactic acid?
Is the taste of sour milk due to lactic acid and one or more amino acids?
https://foodsafety.foodscience.cornell.edu/sites/foodsafety.foodscience.cornell.edu/files/shared/documents/CU-DFScience-Notes-Milk-Flavor-Defects-Sensory-Eval-04-10.pdf has this to say:
"CHARACTERISTICS OF MILK OFF-FLAVORS...
Typical Milk:
No criticism. Very little distinct odor, pleasantly sweet and clean with no aftertaste...
Acid:
Basic taste sensation. Sour, tart, may cause tingling sensation on tongue. “Cultured milk” or “sour” odor may be present. Cause: Growth  of  lactic  acid  producing  organisms  such  as Lactococcus lactis,  due  to  poor refrigeration, especially when temperatures exceed 70°F (21°C)."
Like everyone, the Cornell author explains the sour taste and odor of sour milk by referencing only lactic acid.
https://www.nationalgeographic.com/culture/article/when-good-milk-goes-bad is no different when talking about slightly sour milk, as opposed to the much later (very) spoiled moldy milk, which involves yeasts and molds and ...danger.
https://www.reddit.com/r/chemistry/comments/j15maj/what_does_lactic_acid_smell_like/ says pure lactic acid does not smell bad. Aha! On the other hand, it does say whether or not it smells like sour milk. What a pity.
To those that nitpick that it is subjective to say that sour milk tastes awful, I say this: 1. Taste and odor is subjective, obviously. 2. I Googled 'Sour milk tastes okay' and there was not one relevant hit so people are not in any way divided on this. 3. This is an SE question, not an academic paper.

Probably. Lactic acid bacteria produce other volatile compounds such as acetaldehyde and diacetyl that could be perceived as off-flavors. Both where fermentation is the objective and in the case of spoilage, the VOC profile can vary with the identity of the responsible bacteria but also of the milk and its treatment.
For instance, in Ref. 1 they discuss the effect of different lactic acid bacteria:

During fermentation, S. thermophilus and L. delbrueckii ssp. bulgaricus generate lactic acid and a variety of volatile organic aroma compounds. These compounds impart particular flavors to fermented milk (Beshkova et al., 1998). The overall flavor of any fermented milk product is formed by a large number of these volatile compounds. Among these compounds, carbonyl compounds and organic acids such as acetaldehyde, diacetyl, and acetic acid play a crucial role and can be used to evaluate the flavor quality of dairy products (Zha et al., 2015). Ott et al. (1997) found that acetaldehyde provided the typical flavor of dairy products such as yogurt and buttermilk. Erkus et al. (2013) also reported that diacetyl contributed the characteristic sour buttery flavor to some fermented dairy products.

The offending chemical could also be vinegar (acetic acid). Again from Ref. 1:

Acetic acid has been found previously as a volatile in yogurt by Panagiotidis et al. (2001). High levels of acetic acid contributes most to the tart flavor of yogurt and is particularly associated with a “vinegary, pungent, acidic” taste (Panagiotidis et al., 2001).

If you search for "chemical profile of spoiled milk gc" you'll find a wealth of references discussing the effect of different milk (due to species, handling or forage), milk treatment, and spoilage bacteria on the VOC profile.
In addition, what makes the milk taste bad may depend on the bacteria and on you. That makes narrowing the answer down to a particular compound impossible.
References

T. Dan, D. Wang, R.L. Jin, H.P. Zhang, T.T. Zhou, T.S. Sun, Characterization of volatile compounds in fermented milk using solid-phase microextraction methods coupled with gas chromatography-mass spectrometry, Journal of Dairy Science, Volume 100, Issue 4, 2017, Pages 2488-2500, ISSN 0022-0302, https://doi.org/10.3168/jds.2016-11528.


Questions on concentration Currently I am attempting two questions on concentration but I am not sure if I am correct. Can someone help me with these questions? These are the questions:

  1) Calculate the volume of pure water added to dilute $\pu{0.2 M}$ $\ce{HCl}$ solution in $\pu{250 cm3}$ to $\pu{0.12 M}$ $\ce{HCl}$ solution.

$\pu{0.2 M}$ solution contains $\pu{0.05 mol}$. $\pu{0.12 M}$ solution contains $\pu{0.03 mol}$. The difference between the two solutions is $\pu{0.02 mol}$ (Which is how much water was added), so

$$V(\ce{H2O}) = 0.02 \times (2 \times 1 + 1 \times 16) = \pu{0.36 L}$$

  2) Calculate the mass of $\ce{Na}$ that will form $\pu{0.15 M}$ solution of $\ce{NaOH}$ by reacting with $\pu{400 cm3}$ of water.

I know that this method is totally wrong but I'll show you:

$$\pu{0.15 M} = \frac{n}{0.4} \implies n = 0.06$$
$$m(\ce{Na}) = 0.06 \times 23 = \pu{1.38 g}$$

where $n$ just refers to the concentration ($C = n/V$).

Edit:

Thanks to @George Tian I have made another attempt for the second question.

I have calculated the number of moles of NaOH for a 0.15M solution:

c = n/v
n = 0.4 x 0.15
  = 0.06

And I calculated how much Na is needed for this reaction:

The balanced equation is: 2 Na + 2 H20 -> 2 NaOH + H2
So, 0.06 mol of Na is needed (Due to the molar ratio)

Then I converted moles to grams

Mass = 0.06 x 23
     = 1.38 g

To clarify, I've reworded your questions. Please comment if I changed the meaning:

Calculate the volume of pure water that needs to be added to dilute a  0.2M HCl solution with a volume of 250cm3 to a 0.12M HCl solution

The first step would be to work out how many moles of HCl is in the original solution. As you have stated

c = n / v

c = 0.2M

v = 250cm3 or 0.250L.

Therefore n = 0.05m

Then work out how much water is needed to make a 0.12M solution.

c = 0.12M

n = 0.05m

Therefore v = 0.42L

To work out how much water needs to be added, subtract the original 0.25L from 0.42L, which equals 0.17L

Calculate the mass of Na needed to react with 400cm3 of water to form a 0.15M NaOH solution.

As you haven't progressed much on this question, I'll give a few hints.

Write a balanced equation

Work out how many moles of NaOH is needed to make 0.15M solution

Work out how many moles of Na is needed to form that many moles of NaOH

Convert from moles to mass.

Hope this helps.


How do we know the number of atoms in an inorganic compound? How do we know the number of atoms in an inorganic compound? If there's a question like 'Write the chemical formula for sulfur dioxide', the only thing I managed to do is answer '$\ce{SO}$'. I don't know how to find out the number of sulfur atoms or oxygen atoms.

For all compounds, the systematic name of that compound should provide information on the composition and structure. Your case is an example of the nomenclature of binary covalent compounds. Consider the following example:

  dinitrogen tetroxide

The first word in this name tells us which element is first in the formula and how many of them there are. In this example the "di-" prefix tells us we have two of something.

$$\mathrm{dinitrogen = }\ \ce{N2}$$

The second word in the name tells us which element is second in the formula and how many of them there are. In this example the "tetr-" prefix tells us that there are four of something. By convention, we usually use the "ionic" form of the second element in the name even if it is a covalent compound, so "oxide" means "oxygen".

$$\mathrm{tetroxide=}\ \ce{O4}$$

Therefore, the full formula for this compound is $\ce{N2O4}$.

The prefixes are as follows (up to 10, with larger numbers being rare). Note that the trailing "o" or "a" of a prefix is dropped with oxygen to increase ease of pronunciation.

$$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$

  What about names like iodine monochloride? How many iodine atoms are there in the formula?

The "mono-" prefix is often left out if there is no compound out there that have more than one of those atoms in it. Thus, iodine monochloride is $\ce{ICl}$. Why not iodine chloride? There is another compound out there "iodine trichloride," which is $\ce{ICl3}$.


Why is the ground state electronic configuration of Group 10 (Ni group) elements so anomalous? While studying about d block elements I came across this table showing outer shell electronic configuration of group 10 elements

$\ce{Ni}$ $ $ $ $ $ $ $ $ $ $  $3d^8$ $4s^2$
$\ce{Pd}$ $ $ $ $ $ $ $ $  $4d^{10}$ $5s^0$
$\ce{Pt}$ $ $ $ $ $ $ $ $ $ $ $5d^9$ $6s^1$

(Source Concise Inorganic Chemistry, JD Lee, adapted by S. Guha Pg 563)
Now I cannot understand the reason behind this. If we consider that $\ce{Pd}$ has changed its config. to attain stability then what's wrong with $\ce{Ni}$ and $\ce{Pt}$? And should the outer shell electronic configuration of the ground state not be same in elements down a group?
Edit
To those who have marked my question as a duplicate of the one mentioned, I humbly request to explain me which answer to the above mentioned question (of which mine is marked as a duplicate of) tells about the anomalous behavior of group 10 elements and why $\ce{Zn}$ assumes $d^{10}$ config and none of the others in the group do the same and again $\ce{Pt}$ assumes a $5d^9 6s^1$ config, while $\ce{Ni}$ does not. Also please do point out  where can I find an answer to why in this particular group there is no similarity in ground state electronic configuration b/w elements of the group. Thank You.

I think the relevant comparison should be with the earlier transition metal groups where $s^2$ configurations are more predominant.

In multielectron atoms, electrons interact in such a way that for a given shell quantum number $n$, the orbitals with higher total angular momentum $l$ are raised, thus for example $3d$ ($n=3, l=2$) becomes higher than $3p$ ($n=3, l=1$).  This angular momentum effect is most important when the orbitals are well shielded with a low effective nuclear charge, where the electron-electron interactions don't get swamped by electron-nucleus interactions.

In earlier transition groups the valence shells have this low effective nuclear charge and the angular momentum effect is so strong by comparison that the $d$ valence subshell is raised above the $s$ subshell even with one higher $n$ quantum number.  So, for instance, in the fourth period we see $3d$ orbitals filled only after $4s$ in neutral atoms.  Not so much in the ions, however, where ionization leaves the remaining electrons with more effective nuclear charge, hence we often see $d$ instead if $s$ subshell electrons remaining in the early transition ions.

But by the time we get to the later transition metals the effective nuclear charge has increased even in the neutral atoms, the lower-$n$ $d$ subshell drops relative to the higher-$n$ $s$ subshell, and we begin to see a greater preference for the $d$ subshell being filled first.  By the time we get to Group 12, this transition is complete and only the $s$ electrons remain valence electrons except, possibly, in extreme circumstances.


Ammonium considered as metal Why ammonium ion has very similar properties to the heavier alkali metals and is often considered a close relative?

Ammonium is expected to behave as a metal ($\ce{NH4+}$ ions in a sea of electrons) at very high pressures. Can anyone explain?

Who needs high pressure?  Ammonium can be amalgamated into mercury like an alkali metal.  As with metals generally, there are not really any ammonium molecules in this phase; rather, the ammonium cations are intermingled with mercury cations and delocalized valence electrons.

Nor is all that high a pressure needed for unamalgamated ammonium, relative to metallizing other materials.  Bernal and Massey (1954) reported that a pressure of 250 kbar would be sufficient to metallize an ammonia-hydrogen mixture, which may be compared with the rocky material deep in Earth's mantle remaining nonmetallic even at megabar pressures.

Ammonium is easy to metallize because it has, indeed, an alkali-metal electronic structure.  The tetrahedra structure common to ammonium and methane, $\ce{CH4}$, is most stable with eight valence electrons -- there are four bonding orbitals, corresponding to the four bonds in methane, and four antibonding orbitals.  In methane the neutral molecule matches this optimal electron count, but in ammonium the neutral molecule would have to include a ninth valence electron in its antibonding orbitals, like an alkali metal atom having one to many electrons for a stable noble gas configuration. Either this extra electron breaks down the proposed ammonium molecule (which is what we ordinarily see) or, under moderately high pressure or in a metallically bonded matrix like the amalgam, it goes into the metallic bonding.


Why +M effect of -NH2 is more than that of -NHR and -NR2? Why does $\ce{-NH_2}$ shows more +M effect that $\ce{-NHR}$ and $\ce{-NR_2}$? My first conclusion was totally opposite of the reality. I thought, as the alkyl groups are attached to the Nitrogen the lone lone pair on nitrogen should be easily available. But it doesn't seem to be happening. What is the actual cause?

It is because the bulky alkyl groups provide repulsion to the lone pair such that their orientations change. Due to the change in orientation the %s character changes and the lone pair becomes less available. This is not the case with hydrogen.


Element with No Proton? On seeing this graph closely I noticed the there a element with no proton. But with no proton there will be no electron too and don't think that an lone neutron would form an element. So is that given a half-life on neutron OR there exist a element (isotope) with no electron?

A free neutron is not considered to be an element and there is no element that does not have at least one proton.

Free neutrons could be included on the graph as they are unstable and decay with a half life of around 10 minutes ($611.0\pm1.0~\mathrm{s}$) although most versions I have seen do not usually include them as they aren't really nuclei which is what the graph is about.


How is mean square displacement defined in Brownian motion? I am trying to calculate mean square displacement (MSD) of water in a simulation to calculate its diffusion coefficient. However I am little confused on its exact form. Can anyone please explain it in little detail?
My confusion is: from wikipedia first paragraph on MSD page

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

My understanding of it was that:

$$MSD =(x(0)-x_\text{0})^2/1 + \{(x(0)-x_\text{0})^2 + (x(1)-x_\text{0})^2\}/2 + \{(x(0)-x_\text{0})^2+(x(1)-x_\text{0})^2+(x(2)-x_\text{0})^2\}/3$$
... and so on

That is that each term in MSD is cumulative sum of all previous terms devided by time. However most of online programms that i saw (for example matlab programme at the last page of this link)
tries to calculate MSD as  just simple square of distance between initial position and position at time T. Can some one please explain if it is a cumulative sum of previous displacements or not.

The mean squared distance is a measure of deviation: how far away two things are.

Let's say we want to find the MSD of a particle from it's origin. What we do is find the square distance at specific time intervals, say every second. We then make a table of data:

$$\begin{array}{c|c|}
 \text{time} & \text{square distance} \\ \hline
 1 & 3.6 \\ \hline
  2 & 3.8 \\ \hline
 3 & 3.4 \\ \hline
 4 & 3.2 \\ \hline
\end{array}$$

The formula for the average is adding everything up and dividing by the number of things, so in this example:

$$MSD = \frac{3.6+3.8+3.4+3.2}{4}$$

What I've done above is stated more generally, and much more elegantly, as the formula you stated,

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

Your cumulative sum formula doesn't make intuitive sense, it instead finds the sum of the mean square distance at every time increment (there is no reason why you'd want to do that)


XSteam IF97 tables (MS Excel version) lack of inputs I am using XSteam IF97 to evaluate steam power plant cycles.
However, I miss some important input options such as pressure as a function of entropy and temperature ("p_st()") - when I assume isobaric heat addition (neglecting pressure any pressure losses) while reheating the steam. I end up looking at i-s diagram which is highly inefficient while optimizing cycle for different inputs.
Is there any way around or any more sophisticated software (in the best scenario, available for MS Excel). Or how do you work with steam power plant cycles?
(I intend to avoid any commercial software)

If you have an initial estimate for the pressure, $p_0$, then you can quickly solve iteritively for the pressure that matches the desired entropy $s=s^*$ using modified Newton's method:  $$p_{n+1}=p_n-\frac{s(p_n,T)-s*}{\left(\frac{\partial s}{\partial p}\right)_T}$$The partial derivative of s with respect to p can be obtained by finite difference at the initial guess for p.


Raman Spectroscopy: How to calculate their constants? $^1{\ce{H2}}$ is interrogated with a laser of wavelength lambda = 532 nm and vibrational Stokes transitions are observed at S(0) = 14302.206 cm$^{-1}$, Q(0) = 14638.442 and O(2) = 14992.823 cm$^{-1}$.

Estimate ωe (cm$^{-1}$)
Determine B1 (cm$^{-1}$)

I know the band origin = Q(0). I also know from rovibrational spectra that

$ ω_0 = ω_e -2ω_eX_e$.

Could someone please help me answer this question and explain?

If there are any additional resources I could read on these topics that would be great too.

In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies.

You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels are given by $E_n= \omega_e(n+1/2) - x_e\omega_e(n+1/2)^2$ (in wavenumber units) with quantum number $ n = 0, 1, ...~$, and $\omega_e$ the frequency and $x_e$ is the dimensionless anharmonicity.

In addition to the vibrational energy there is rotational energy whose levels are $E_J =B_eJ(J+1)$ for rotational constant $B_e$ (in wavenumbers) and quantum number $J=0, 1, 2,...$ and ignoring centrifugal distortion. You will also have to use different $B_e$ values for the $n=0$ and $n=1$ vibrational levels since the bond length is slightly different between (anharmonic) vibrational levels.

You know that the rotational-vibrational levels appear in groups  O, P, Q, R, S in increasing frequency and that Raman has O, Q & S bands only. In the Q band $\Delta J = 0$ between vibrational levels n to $n+1$ and is $-2$ and $+2$ for O  & S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $.

If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.

There is a figure showing energy levels at
 https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy

The Q band transition ($\Delta n =0, \Delta J =0$) in your notation is $\omega_0 = \omega_e - 2x_e \omega _e $ and the other branches calculated similarly with $\Delta J = \pm 2$ as appropriate.


How can a dilute acid dissociate more than a concentrated one? (Ostwald dilution law) According to Ostwald's dilution law, the degree of dissociation of an electrolyte is inversely related to the  concentration of it.

$$ \alpha = \sqrt{\frac{K_a}{C}} $$

This means that a dilute acid can dissociate more than a concentrated one. How is it possible? Please explain the physical meaning of this.

A dilute acid has more solvent molecules to interact with than a concentrated acid, and thereby has more frequent collisions with the solvent. You can think of the dilute acid as "drowning" in the solvent.

A concentrated acid, by contrast, interacts more with other acid molecules, and therefore less with the solvent which dissolves it.

A concentrated acid will typically have more total dissolved molecules than a diluted acid simply because there are more of them, but it will have less as a fraction of total number of acid molecules; i.e. a weaker affinity for dissociation.

In short, $\alpha$ describes affinity for dissociation, not total dissociation. "Better", not "more".

Note that a higher $K_a$ corresponds to a stronger acid, so it makes sense that stronger acids will have a higher affinity for dissociation.


Decomposition of total Interaction energy with SAPT or EDA-NOCV I want to perform energy decomposition analysis for some Hydrogen bonded complexes. I came to know that ORCA and Psi4 allows EDA-NOCV an SAPT(0,2,3) respectively. But I do not know which one would be better, and how to perform them.

SAPT(O) is an amazing tool for partitioning the interaction energies, it yields meaningful values and data. ETS-NOCV is also very good for partitioning the interaction energy, but it does it in a much different way, it yields values that correspond more to the orbital interactions, and more so the sigma and pi orbital mixing (although orca will not partition more than just the overall orbital mixing). ETS-NOCV is really good for giving qualitative orbital pictures for electron donation and back-donation. For hydrogen bonding, i'd suggest SAPT


Recommendations for textbooks with correct and up-to-date information I recently began studying chemistry, only to find out that, apparently, concepts such as “octet expansion” and Lewis structures are outdated and incorrect. And I don’t mean they’re approximations of reality, as is the case for numerous theories taught in physics, but, apparently, actually incorrect and deprecated concepts. However, commonly promoted textbooks still teach these concepts as though they are true, with no reference to them being outdated or incorrect. I find this both disappointing and disturbing, since I want to learn concepts that are actually correct. And when I purchase (expensive, recently published) textbooks, I expect to be taught correct and up-to-date concepts — especially since the people who wrote these textbooks are in an academic position to know better.

At the moment, my goal is to study both physical and organic chemistry. I was wondering if people could please recommend good textbooks in these areas that do not teach incorrect, outdated, and deprecated concepts.

I disagree with the idea to forget about Lewis formulae because of their age. If you browse across the table of contents of scientific journals, however, you see Lewis formualae are widely used.  From the first December issue of Organic Letters, for example is taken the following screen photo:

If Lewis formulae were so off and wrong, chemists would use other forms to describe molecular structures in their publications.  As so often, however, it is a question about what you want to describe, and to know what limitations of the tools at disposition inheritated.
By convention (and training), for example, chemists do not read the rings in formula about nitronaphthalene (on the left) as an alternation of isolated single- and double bonds, but conceptionalize this as to depict aromatic systems.  Counting electrons, for example, this is easier this way as with the Thiele ring which accentuates the delocalization of the $\pi$-electrons.*)

(source)

So it is better to recognize where Lewis formulae may be misleading, with some instances listed e.g., here, and some you will encounter in study, e.g. diborane

(source)

It equally is helpful to know that dashes in formulae, depending on convention, may represent different aspects:

(source)

*) In 1899, Thiele's perspective was to emphasize that benzene's molecular structure is between the two mesomeric structures expressed as Kekulé-structure and thus used the (initially dashed) cricle in the structural formula.  Chemists' perspective of delocalization of $\pi$-electrons today obviously is influenced by quantum physics.


How do you determine a chemical formula given the molecule's crystal structure? How do you determine the chemical formula of a compound given its crystal structure?

Knowledge of the crystal structure implies that you are also able to describe the unit cell of the compound. To determine its formula, count all sorts of atoms inside the unit cell, also considering fractional atoms which belong to more than one unit cell. For example, an atom at the corner of the unit cell counts as $\frac{1}{8}$ atom because it is shared between 8 adjacent unit cells. Likewise, an atom at the edge of the unit cell counts as $\frac{1}{4}$ as it belongs to 4 unit cells etc. When you obtain fractional values for the atom numbers, multiply the values of all atoms with a certain number so that all values are integers.

Let us take sodium chloride ($\ce{NaCl}$) as a simple example. Its unit cell, which is shown below, contains $\ce{Na+}$ (violet) and $\ce{Cl-}$ ions (green).

Counting the number of chloride ions in the unit cell gives

$$12 \cdot \frac{1}{4}+1=4 \ \ce{Cl-}$$

because there are 12 $\ce{Cl-}$ on the edges and 1 $\ce{Cl-}$ in the center of the unit cell. For the sodium ions, which are situated at the corners and on the faces (count as $\frac{1}{2}$ because they are shared between 2 cells) of the unit cell, we obtain

$$8 \cdot \frac{1}{8}+6 \cdot \frac{1}{2}=4 \ \ce{Na+}$$

The composition of the unit cell is therefore $\ce{Na4Cl4}$. Since the number of formula units per unit cell is 4, the formula of the compound is $\ce{NaCl}$.

For compounds in which the building blocks of the crystal structure are molecules, the procedure is essentially the same.


Is a Methoxy group meta directing in this problem? The following question was given in chemistry today , August 2020 edition , page 47 , problem 9. The answer to this problem is (a).

My Attempt

In the third step , the carbon electrophile can attack at ortho position to carbon substituent to give a 6 membered ring.
But , the O-Methyl group is ortho para directing. The resonance structures are as follows.

The given answer is , a , is possible only if we consider O-Methyl group on Benzene as meta directing group.
My Question:
Is methoxy group on ANISOLE Meta directing ?

Now that the general consensus of the Commenters is that meta cyclization is viable and that the answer is (d) and not (a), the OP and other readers should be concerned as to how an (E)-double bond can cyclize in a Friedel-Crafts (F-C) reaction. It doesn't! Prior isomerization to the (Z)-configuration is required.
(E)-Acyl chloride 1 must rotate about the red β-γ single bond and $\ce{AlCl3}$ must complex with the carbonyl oxygen forming species 2. [One may also invoke $\ce{AlCl3}$ complexation with the chlorine and rationalize the subsequent steps via an acylium ion and conjugated ketenes. The principle is the same.] Loss of a proton at the γ-position of conformation 2 provides the β,γ-(Z)-double bond of 3. Now deconjugated acyl chloride 4 is ready for F-C cyclization to intermediate 7 and onto 7-methoxynaphthalen-1-ol (9), aka (d). Alternatively, rotation about the red bond in 3 leads to conformation 5, which, upon γ-protonation leads to α,β-(Z)-conjugated acyl chloride 6. Subsequent F-C cyclization via intermediate 8 affords 9. Incidently, base is not required to produce 9 from 7 or 8. F-C conditions are sufficient.


How would I write an input Lammps command for continuous change in temperature? I'm going to study the thermal expansion of graphite. For this goal, I should change the temperature continuously. But I do not know how to code it by Lammps?

(Disclosure: I'm not a LAMMPS user, but I've used other MD packages.)

You'll probably want to use one of the thermostat settings such as the Nose-Hoover type described here in the LAMMPS documentation, but there's a more general HowTo guide on temperature control and calculation in the docs as well that looks pretty useful.

I would strongly recommend reading the sections of the manual concerning any given thermostat type you're going to use fully before running serious calculations with it, though, since there are some things to look out for, such as the tendency for setting the thermostat interval too low to cause large oscillations in effective temperature.

The Nose-Hoover type (and some of the others, but I only took a quick look) also lets you step up or down the temperature with a given interval. Depending on exactly what type of thermal expansion you're looking at, I'd just like to note: you really need to set the stepping interval appropriately to let the structure equilibrate at each temperature you want to look at, otherwise your results won't be meaningful. (Or, at least, won't have the meaning that you might want.) Relatedly, I'd also suggest that if you weren't planning to already, you run each simulation both ways: one stepping the temperature up, and one bringing the temperature down. If the behaviour matches at the temperature steps in each, you can be relatively confident that you've given the structure enough time at each temperature.


Which atom bears the negative charge in cyanide? I am confused by the MO diagram In heteronuclear species, whichever atomic orbital a molecular orbital is closer to in energy contributes more to that molecular orbital, right? In other words, if the orbital is occupied, more electron density is around the atom whose atomic orbital is closer in energy to the molecular orbital. This feature of MO diagrams can predict dipole behavior and which atom bears charge in ions. If this is the case, and we know that carbon bears the negative charge in cyanide (and hence is a carbon nucleophile), why does the MO diagram for cyanide appear to have the HOMO (the "attacking" orbital) concentrated on the nitrogen rather than the carbon atom? Based soley on the diagram and not the accompanying image of the HOMO, I thought, since the HOMO is closer in energy to the nitrogen 2pz atomic orbital rather than the carbon 2pz, the HOMO should have more contribution from nitrogen and the LUMO should have more contribution from the carbon.

As commenters have noted, charge isn't the best choice of a word here, but you're right that the HOMO is more centered on the C, and that's really your question.

The answer is a bit complicated. It relates to those dashed lines on the MO diagram. If we naively construct an MO diagram of cyanide ion, we have two favorable sigma interactions. First, the $2s$ orbitals can interact positively, which I'll call $\sigma_s$. That would be the lowest energy MO on our diagram. Second, we can have an end-to-end interaction of $p$ orbitals, which I'll call $\sigma_p$. Since that $\sigma$ interaction is stronger than a side-by-side $\pi$ interaction, we would expect the $\sigma_p$ MO to be lower in energy than the $\pi$ bonding MOs.

If we look at your MO diagram, though, we see that $\sigma_p$ is actually higher than the $\pi$ MOs, not lower as we predicted. It's the HOMO in the diagram. The reason it is higher than we naively expected is because there is mixing between it and the $\sigma_s$ (and also between their anti-bonding counterparts, but ignore that for now). These two MOs can interact positively or negatively. In the positive interaction, we get the (normalized) sum of the two MOs, which is an $sp$ orbital on C with its large lobe toward N and an $sp$ orbital on N with its large lobe towards C (see the lowest MO in the drawings below). Both large lobes have the same phase, so this is a strong bonding interaction that is the lowest MO on your diagram. Both of the starting bonding MOs are biased in favor of contribution from N, since both the $s$ and $p$ AOs of N are lower in energy than their counterparts on C. Thus, this sum orbital has a stronger N contribution as well.

In the negative interaction, we subtract $\sigma_s$ from $\sigma_p$ and again get $sp$-type orbitals on both atoms, but now the large lobes are facing away from each other. That makes the bonding interaction much weaker, so much so that it moves higher in energy than the $\pi$ bonding orbitals. Furthermore, subtraction takes away more N-character than it does C, and you end up with a greater contribution to the new orbital from C than N. As you get more comfortable with MOs, you can verify this mathematically. Another way to think about it is that the final MO is higher than the either the average of a $Cs$ and a $Cp$ orbital or the average of an $Ns$ and an $Np$ orbital, but it's closer to the C $sp$ average (which is higher than the N $sp$ average), so it has more C character by the same logic you used to predict that it would have more N character (which it would if it were just made up of $p$ orbitals).

Below is a drawing of the five occupied orbitals where you can see that the $\sigma$ ones have become $sp$-like and that the carbon contribution is greater in the HOMO. C is on the left and N on the right.


How do you calculate a half cell potential Copper(II) and Copper Sulfate? I have to complete an experiment for school where I have copper and zinc electrodes placed into an electrolyte of copper sulfate and I have measured the potential across it at varying temperatures. For example at $\pu{60°C}$ ($\pu{333K}$) I measured the potential to be $\pu{0.880 volts}$ and at $\pu{75°C}$ ($\pu{348K}$) I measured the potential to be $\pu{0.780 volts}$.

Is there a way I can prove this result? I tried using the Nernst Equation, but my reaction quotient is equal to 1, which disproves my finding. I found out about activity coefficients, but I do not know how to apply it or the Nernst Equation to half cell potentials.

Any advice or help is greatly appreciated!

Here is a figure of the famed Daniell cell:

Note the spelling of the name: it was invented in 1836 by John Frederic Daniell. At $\pu{25 ^\circ C}$, and with $n = \pu{2 equivalents/mole}$, as correct here, $(\ln 10)\frac{RT}{nF} = \pu{0.02958 V} = \pu{29.58 mV}$.


Separating lanthanides using ion-exchange method Following statements are on Separation of rare earth metals (lanthanides) via in-exchange method

  The tripositive ions are strongly sorbed by cation exchange resin: $\ce{La^{3+}}$, the largest is most strongly held; $\ce{Lu^{3+}}$, the smallest is least strongly held. Elution of $\ce{HCl}$, therefore removes $\ce{Lu^{3+}}$ first and $\ce{La^{3+}}$ last.

Generally, the smallest cation forms the strongest bond, so shouldn't $\ce{Lu^{3+}}$ be most strongly held? The most common cation exchangers, such as $\ce{RSO3^-Na+}$, where $\ce{R}$ is a cross-linked matrix are hard bases, so shouldn't according to HSAB principle also, $\ce{Lu^{3+}}$ be most strongly held?

Follow-on excerpt,

  The separation is however not good, particularly between the heavier lanthanides, but is markedly improved by the addition of complexing agents such as $\ce{EDTA}$, or ammonium citrate/citric acid buffer to the eluting solution. The citrate ion complexes most strongly with the smaller elements, i.e. with the heavier elements.

Now, the two compounds (cation-exchanger and complex) helds strongly  different elements, one smaller in size and other larger in size. How will presence of two enhance separating property? Also here ammonium phosphate is mono-, di- or tribasic?

Following graphs are also given

Please explain the meaning and significance of the graphs.

You must be reading this from really old ion exchange references. It is always good to mention the resources and references for other readers.

Ion-exchange retention can be very roughly approximated with Coulomb's law, which says that the force of interaction between two "ions" is directly proportional. The force of interaction is inversely proportional to the square of the hydrated radius of the ions. Here the word interaction means attraction or repulsion.

So you have to look up the hydrated radii of the lanthanides and try to see a trend on a simple sulfonated cation exchanger.

You second point is a misconception. The "trick" to add complexing agents works when you wish to alter the sign of the species you wish to separate. For example Cu(II) has a +2 charge, if I complex it with EDTA, the [Cu(EDTA)] has a charge of -2 Think about the Coulomb's law now. Can we use a cation exchanger now? This time we must try an anion exchanger.

One can still use a cation exchangers, in that case the complexed metallic anions will be repelled out and those ions which did not complex well will stay on the resin.


Using quantiative analysis to find empirical and molecular formula A compound of Ca, C, N and S was subjected to quantitative analysis and formula mass determination, and the following data were recorded.
A $\pu{0.250 g}$ sample was mixed with $\ce{Na2CO3}$ to convert all of the Ca to $\pu{0.160 g}$ of $\ce{CaCO3}$.
A $\pu{0.115 g}$ of the compound was carried through a series of reactions until all of its S was changed to $\pu{0.344 g}$ of $\ce{BaSO4}$.
A $\pu{0.712 g}$ sample was processed to liberate all of its N as $\ce{NH3}$ and $\pu{0.155 g}$ of $\ce{NH3}$ was obtained.
The formula mass was found to be $\pu{156 g mol-1}$. Determine the empirical formula of this compound.

My issue here is that we don't really know what the starting compound is and particularly how much of each Ca, C, N, and S is there. Without knowing that, how can you balance your reaction equation and find the number of moles needed to get the empirical formula?

The problem you gave doesn't quite work out, so I searched it up; did you perhaps mean $0.115 \space \text{g}$ of the sample were used in finding the sulfur content (you said $0.15 \space \text{g}$ instead).

If you did mean $0.115 \space \text{g}$, then we can figure out that the molecular formula of the unknown compound is $\text{CaC}_2\text{N}_2\text{S}_2$. Let me explain:

First, let's try figuring out how many moles of $\text{Ca}$ there are for each mole of the unknown compound, using the information that $0.250\space \text{g}$ of the compound was used to form $0.16 \space \text{g}$ of CaCO3. Converting everything to moles, this means we used $(0.250\space \text{g})\space/\space (156 \space \text{g/mol})=0.00160 \space \text{mol}$ of the unknown compound to get $(0.16\space \text{g})\space/\space (100.0869 \space \text{g/mol})=0.00160\space \text{mol CaCO}_3$, or $0.00160\space \text{mol Ca}$. This means that one mole of the unknown compound will contain $(0.00160\space \text{mol Ca})\space/\space (0.00160 \space \text{mol Unknown Compound}) \space\times\space (1 \space\text{mol Unknown Compound})=1\space\text{mol Ca}$

We can do this for each of the other procedures to get that for each mole of the unknown substance, we have $1\space\text{mol Ca}$, $2\space\text{mol S}$, and $2\space\text{mol N}$. Now, we can just subtract the masses of $\text{Ca}$, $\text{S}$, and $\text{N}$ from the molar mass of the unknown compound to get the mass of $\text{C}$ in the compound. For each mole of the unknown compound, we get that we have $\text{156 g}-\text{40.08 g}-2(\text{14.01 g})-2(\text{32.06 g})=23.8 \space\text{g}$ of carbon. This is just $(23.8\space \text{g})\space/\space (12.01 \space \text{g/mol})=1.98\space \text{mol} \approx 2 \space \text{mol}$ carbon per mole of unknown compound.
So, since we have $1\space\text{mol Ca}$, $2\space\text{mol S}$,  $2\space\text{mol N}$, and $2\space\text{mol C}$ for each mole of the unknown compound, the unknown compound must be $\boxed{\text{CaC}_2\text{N}_2\text{S}_2}$.


Liquid that looks like smoke when bottled? I am making a gift that will consist of a liquid in a sealed bottle. I want the liquid to have a magical appearance. Something like the appearance of smoke - varying opacity from almost clear to almost opaque but never completely mixing.
I am looking for a liquid or combination of liquids with the following properties:

Looks like smoke
Will not become homogeneously opaque over a relatively long period of time (months)
Non-toxic and non-reactive in a household setting (in case it is spilled accidentally)
Relatively
easily found / made assuming access to decently stocked pharmacies,
grocery and hardware stores
Can be coloured would be a bonus

So you want to make something that looks like smoke. What makes smoke look the way it does, anyway?

Smoke is composed of extremely fine particles, often below $\mathrm{1\ µm = 0.001\ mm}$ in diameter (check here for comparisons with other particulate materials). These particles are large enough to scatter light, which allows them to be seen, but small enough to be extremely light and stay aloft with even very weak currents in air or liquids.

One interesting and very beautiful consequence of these combined properties is that they can clearly reveal fluid flows, which is what gives smoke its wispy, delicate and complex appearance. That's what fluid flows are like all the time, we just almost never have a direct view of it.

So fundamentally your request is quite simple; all you need is a common, non-toxic source of very fine particulate. Fortunately there are good ways to achieve this in practice.

One of the most popular seems to be using fine mica dust. These non-toxic silicate minerals can be ground to form tiny, almost perfectly smooth crystalline sheets. When the particles align properly, light is reflected into a viewer's eyes giving a shiny, glittery appearance. If the mica dust is suspending in a liquid, then the many millions of particles twinkle momentarily as they tumble chaotically under the action of currents in the fluid.

Liquid mixtures containing similar materials are called rheoscopic fluids. There are many many sources online for ready-made fluids, or you can make your own from mica powder or other fine solids. Suspensions in water eventually settle, but last for a long time without degradation, needing only a shake to awaken. With a myriad of water-soluble dyes at your disposal, you can adjust the fluid to the desired colour.

Though rheoscopic fluids based on water and mica are non-toxic and quite safe, that doesn't mean they pose no hazard whatsoever. Very fine particulate can be an irritant if inhaled or exposed to eyes and mucous membranes (especially a problem if handing the dry powder). Just take a bit of care and you can have fun too! Try putting the fluid on a warm surface and see the convection currents!


What is the pressure point where deviations from ideal gas law occur? I read a lot on deviations from ideal gas law due to low temperature and high pressure and I understand it, but I am curious to what point (the value) does a pressure is considered HIGH pressure?

It depends on the gas. If the molecules have strong interactions with each other you can see measurable deviations at pressures around 5 atm. But if you have a "more ideal" gas you may have to pressurize it to much higher pressures. There is no set value.


Stability of carbocation intermediates: secondary alkane versus primary allyl? Say I have the following carbocations:

The first is secondary; the second is resonance stabilized, but primary.  Which is more stable and why?

I've been looking around in my textbooks and haven't been able to find a comparison.  This site compares alkane and alkene intermediate stability, but only with the same degrees of substitution.

There are two competing effects at work here; hyperconjugation, and resonance involving the double bond.

According to this website, a secondary alkyl cation and a primary allyl cation have roughly the same stability.

  Carbocation Stability: $\ce{CH3+ < CH3CH2+ < (CH3)2CH+ ≈ CH2=CH-CH2+ < C6H5CH2+ ≈ (CH3)3C+}$

This is not really something you can predict from qualitative considerations - it is difficult to assess the strength of resonance versus hyperconjugation - but it requires experimental evidence, or detailed computer calculations to give a definitive answer.

In addition, the effect of adding more hyperconjugated groups is not linear. The difference in stability between tertiary and secondary alkyl cations is bigger than that between primary and secondary cations - tertiary alkyl cations are actually pretty stable.

Lastly, (thanks to @permeakra for pointing this out) the stability of carbocations in influenced by the environment which they are in. In particular, different solvents affect the stability of different cations to a varying degree and so the above order of stability should be taken as a rough guideline, not a hard and fast rule.


Is there a replacement for glacial acetic acid in determining peroxide value I am not allowed to use glacial acetic acid at my school, which all procedures online for finding the peroxide value of oils use through an iodometric titration. Is there a replacement?

Your school is not allowing use of glacial acetic acid, but you insist of using glacial acetic acid in determining peroxide value, you may use a solvent system, which consists of glacial acetic acid. For example, you can use 60% ($v/v$) glacial acetic acid in chloroform. I give that example because it is commercially available online at Reagents. If you are determining peroxide value of fats, this is an ideal solvent to dissolve fat and resultant solution is soluble in water.

I'm suggesting this because there is a method recommended by International Fragrance Association (IFRA) in Belgium. Breifly, one of the IFRA analytical methods for determination of the peroxide value is described as follows:

  Transfer approx. $\pu{3.0 g}$ of the sample, accurately weighed, into a 250-mL Erlenmeyer flask with glass stopper. Add $\pu{50 mL}$ of the appropriate solvent mixture (glacial acetic acid : chloroform 3:2; glacial acetic acid : cyclohexane 3:2; glacial acetic acid : isooctane 3:2) and saturated potassium iodide solution, $\pu{1 mL}$, freshly prepared and allow to react for $\pu{60 ± 1 sec}$ and shaking thoroughly during this period. Then add water, $\pu{100 mL}$ and shake. Titrate with $\pu{0.01 mol/L}$ sodium thiosulfate solution, or $\pu{0.1 mol/L}$ sodium thiosulfate solution at higher usages, using $\pu{1 mL}$ starch solution or $\pu{0.1 g}$ of Thyodene indicator. The indicator should be added towards the end of the titration but while the pale straw color is still present. During titration shake until the blue color disappears. Carry out a blank titration under the same conditions. No more than $\pu{0.5 mL}$ of $\pu{0.01 mol/L}$ sodium thiosulfate solution should be consumed for this purpose.

Note: Use $\pu{0.1 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\gt \pu{20 mmol/L}$. Use $\pu{0.01 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\lt \pu{20 mmol/L}$.

Thyodene indicator is also commercially available in Fisher.


Why is the effect of ring activators more dominant than ring deactivators? In 3-nitroaniline, if it were subjected to an electrophilic substitution reaction, the electrophile will most likely attack at (b) or (d) as mentioned in my textbook.

To me, it seems that the only role of the nitro group is that the electrophile will not attack at (a) due to steric hindrance. So my question is, why is the -M effect of the nitro group ($\ce{-NO2}$) discarded and only the +M effect of $\ce{-NH2}$ plays a role? Nitro group is meta directing, so why wouldn't the product form at (c) as well? It is as if the nitro group doesn't even exist. In better words, why is the effect of the ring activator, $\ce{-NH2}$, more dominant than the effect of the ring deactivator, $\ce{-NO2}$?

$\ce{-NH2}$ activates the ring due to its +M effect, increasing electron density in the ring at (b) and (d). On the other hand, $\ce{-NO2^{-}}$ causes de-activation because of its -M and -I effect and the electron density would be higher at (c) relatively.

Now since in activated rings, the rate of electrophilic aromatic substitution is higher than those in deactivated rings, the effect of $\ce{-NH2}$ as ortho-para director is seen more during product formation and not of $\ce{NO2^{-}}$ majorly. The electrophile will attack at (c), but since its rate will be quite slow, it would not form the major product.


Temperature dependence of reaction rates Is it correct to say that "Increasing the temperature will always increase the rate of any chemical reaction"?

Many articles on the internet and textbooks say yes but what about biochemical reactions that take place in our body?

To my knowledge  at higher temperatures some enzymes (which are catalysts in chemical reactions in the body) become less effective which in turn slows down the reaction rate.

So is it correct to say "any chemical reaction" in the statement or do I have to think of the temperature dependence of the biochemical reactions without the enzymes?

Well mostly it does but there can be exceptions.

Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$

It has a negative temperature dependence. The rate decrease in increasing the temperature.

There is a simple reason to why this happens.
We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence , that affects the overall reaction.

The Mechanism

$\ce{NO + O2 <=> ONOO}$ Fast but reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

If Step 1 is exothermic, as a bond is being formed between the reactants.
This has a greater impact on the overall reaction and by le Chatlier's principle the equilibrium shift backward . We get a negative temperature dependence!


What is the relationship between enthalpy of atomisation and enthalpy of sublimation? The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

We know that we have to first break the metallic lattice to turn the metal into gas and hence there should be some sort of relation between the two.
My confusion is that when we split the lattice, aren't we technically subliming the substance as well?

tl; dr: In the case of a metal, sublimation and atomization can be regarded as synonymous when atomization refers to the process of converting the solid into a dilute atomic gas. In particular, the end points (states) of the process should be identical.

There are usually conventions on what you call "enthalpy of X". However the official source of chemical nomenclature - IUPAC - has not bothered to post a definition for enthalpy of atomization (or atomization enthalpy) in its Gold Book. There are entries for atomize:

  To subdivide a liquid into very small particles; methods include: impact with a jet of gas, use of a spinning disk generator, vibrating orifice generator, etc.

and for atomization (in analytical flame spectroscopy):

  The conversion of volatilized analyte into free atoms

These definitions do not refer to the same process, which leads to some ambiguity. Are you speaking of a transition from a condensed into the gas phase that preserves molecular structure, or of a chemical process that converts a sample in some phase into an atomic gas?

The wikipedia meanwhile has an entry for enthalpy of atomization which provides the following definition:

  The enthalpy of atomization (also atomisation in British English) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)

There is an entry in the IUPAC gold book for bond energy (theoretical):

  The energy required to break a given type of bond between atoms in certain valence
  states. An averaged bond energy is commonly derived by dissecting the heat of atomization of a molecule into contributions of individual bonds. For molecules with localized bonds, the heats of atomization (formation) are usually well approximated by the sum of pertinent averaged bond energies.

This reference to "heat of atomization" (which is not otherwise defined in the Gold Book) is consistent with that in the wikipedia.

If you apply this last definition then yes, the sublimation enthalpy and the atomization enthalpy of a solid metal should be identical, since both refer to the process:

$$\ce{M(s)->M(g)}$$

(assuming identical P,T end points). Note that sublimation enthalpy means standard sublimation enthalpy when referring to the process at standard pressure (1 bar) and a fixed temperature. Similarly for atomization enthalpy.

As regards the question:

  The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

Different uses of the word "atomization" render the meaning of "enthalpy of atomization" ambiguous. This is why I open my answer with an exposition on the definition of "atomization". Your problem refers to an ideal solution, but I assume here atomization refers to sublimation since reference to hydration would not seem to make sense otherwise. The question asks to break down the reaction into steps including atomization (sublimation), ionization and solvation. Ignoring the ionization step, a larger atomization enthalpy and smaller hydration enthalpy is consistent with the metal being more stable than the solubilized ion, ie the reduced form is favored (relative to standard hydrogen electrode).

A libre text and the accompanying diagram explain some of this


Number of collisions of particles in a box (or a sphere) Scenario

Imagine you randomly spread in a cubic space of volume $m^3$ (or in a sphere of volume $V_s$, as you prefer), $n_A$ particles $A$ and $n_B$ particles $B$. Particles $A$ are spheres of volume $V_A$ and are not moving. Particles $B$ have no volume (they are points) and are moving linearly (reflecting at the boundaries of the space (cube or sphere)) at a constant speed $s$.

We can assume that when a particle $B$ meet a particle $A$ (collision), we count a collision but the particle $B$ just go through the particle $A$. Whatever is the model that is easier to calculate.

Question

What function describes the probability that there is $x$ collisions in a time interval $\Delta t$. Or what is the expected the number of collisions?

If you want a number of collisions in a given time $\Delta t$, then simply from a units perspective, the other answer doesn't make sense, as it produces a number with dimensions of length. You want something that is dimensionless.

Often, physical chemistry textbooks will provide a derivation for collisions per unit volume per unit time. The proofs are a little bit involved and I'd rather refer you to a textbook instead of typing it all out. However, the bottom line is that the collision frequency is given by

$$Z_{AB} = \pi(r_A + r_B)^2\left[\langle v_A \rangle^2 + \langle v_B \rangle^2 \right]^{1/2} \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

$Z_{AB}$ is the number of collisions between A and B per unit time per unit volume
$r_A$ and $r_B$ are the radii of A and B
$\langle v_A \rangle^2$ and $\langle v_B \rangle^2$ are the mean square velocities of A and B
$N_A$ and $N_B$ are the number of particles of A and B
$V$ is the total volume

For a derivation of this I suggest reading Levine's Physical Chemistry, 6th ed, section 14.7.

For your specific case you've assumed that $r_B = 0$, $v_A = 0$, $v_B = s$. Therefore this simplifies to

$$Z_{AB} = \pi r_A^2 \cdot s \cdot \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

Obviously, the radius of A, $r_A$ is related to its volume $V_A$ by $V_A = 4\pi r_A^3 /3$. To obtain the number of collisions in a given time you just need to multiply by the total volume and total time.

$$f =  \pi r_A^2 \cdot s\Delta t \cdot \left(\frac{N_A N_B}{V}\right)$$

As you can see, the form of this equation is not all too different from the one in the other answer. The key difference is that the distance travelled $s\Delta t$ must be multiplied by a cross-sectional collision area, equal to $\pi r_A^2$, and not by a volume. The textbooks will explain this well, I will try to elaborate on this later if I have time.


Reason behind Opaque translucent transparent I would like to know the reason that some materials allow light to pass (transparent), blocks it a little (translucent), or completely reflects or absorbs it (opaque.)
I'm don't have a PhD, so answer this in a way that is easy to understand.

Opacity or transparence is a matter of light reflection.

The way light is reflected or absorbed depends on severals parameters, first the composition of your material. If it contains molecules or crystals absorbing the visible light, you'll see it black, or colored with the color corresponding to the wavelenght the less absorbed.

There is also crystallographic considerations to take care of. If your material structure is well ordered, dimension of your latice will be set so to reflect the light, then you'll see it transluscent or opaque. For something without ordered structure, you'll probably see it transparent if the visible light isn't absorbed. We'll speak about "amorphous" material then like glass for instance.

edit : if you feel easy with those concepts, I can go further in the explainations, but since you asked for a really vulgarized answer I haven't developed more.


Numbering cholesterol structure What could be the reason for clockwise numbering in c ring of cholesterol?

When two German chemists H.O. Wieland and A.O.R. Windaus were awarded Nobel prize in two consecutive years 1927 and 1928 for their research of bile acids and sterols, despite the enormous effort of structure determination, their alleged cholesterol structure was not yet correct at a time, but already had similar numbering:[1,2]

It's difficult to answer why this one exactly was chosen. But see, here you have counter-clockwise numbering in first three rings, however clockwise in the last one, and with methyl and ethyl intermission. I don't think that it's an important factor anyway. In fact, fused ring systems nomenclature usually numbers the ring system peripheral atoms continuously (through the peripheral carbons, if including the fusion carbons), and in the clockwise sense, when the skeleton is in preferred orientation. However, for natural compounds, especially terpenes, there is large variety of different parent structures with various numberings.

So, when the correct structure was determined in 30’s (I think I was R. Robinson; and later confirmed by X-ray analysis in mid 40’s), it adopted similar numbering scheme (changes were made in methyls / side chain numbering), and the first systematic nomenclature of steroids[3] (pre-IUPAC though) from early 50’s also uses it, without explanation, rather referring to tradition:

  STEROID NOMENCLATURE.

  The following proposed rules for steroid nomenclature are based on agreements reached at a Conference held at the CIBA Foundation, London, on May 30th to June 1st, 1950. It is hoped to submit them for approval to the International Union of Pure and Applied Chemistry.

  NUMBERING.

  Rule 1.—Steroids shall be numbered as in formula (I). If one or more of the carbon atoms shown in (I) is not present, the numbering of the remainder shall remain undisturbed.

  Remarks. The numbering in (I) is that generally accepted in steroid chemistry, although it does not conform to that given in the “Ring Index” for cyclopenta[a]phenanthrene (No. 2561). It has been the general, but not invariable, practice to assign the number 18 to the carbon atom of the methyl group attached to C(13), as in (I), because this methyl group is present in all natural steroids whereas the methyl group (involving C(19)) attached to C(10) is absent in members of the œstrogen series.

  (…)

…and it's used still (that's for the ring numbering. There are some changes – C-28 and C-29, not present in cholesterol, now have numbers 241 and 242).

If chemists were to decide for steroids numbering revision, an interesting option would be to choose numbering based on the biosynthesis facts. The numbering would be changed only in ring A (starting with today’s C-4, clockwise), and would be entirely continuous:

(or opposite direction)

… Based on squalene (farnesyl, or farnesene units joined tail to tail depicted in different colors, their isoprene units bold) cyclization (with a couple of methyl shifts, and also hydride shifts (not depicted)) after squalene epoxidation (bottom leftmost double bond) to lanosterol (30 carbons), which then losts three methyls via oxidation and decarboxylation, and double bond is shifted. (The correct topology of of squalene cyclization was first correctly proposed by R.B. Woodward, if I'm not mistaken, as an alternative to R. Robinson proposal[4]).

References:

Wieland Nobel lecture, PDF
Windaus Nobel lecture, PDF
Steroid nomenclature. Chem. Ind. June 23, 1951, SN 1-11.
Woodward, R. B.; Bloch, K. THE CYCLIZATION OF SQUALENE IN CHOLESTEROL SYNTHESIS. J. Am. Chem. Soc. 1953, 75 (8), 2023–2024.


Why is cyclopropane-1,1-diol stable The reason given me in my book is that it's stable due to steric relief.
The answer given on quora said "there is a lot of angle strain."

What exactly is steric relief ?

Can anyone help me elaborate how these points are contributing to the answer.Also what is the minimum angle to provide strain so that geminal diols are stable

The bonds in cyclopropane 1,1-diol are less strained than the cyclopropanone.

The preferred/lowest energy for tetrahedral bond angles is 109.5 degrees so the closer they can get to that the lower the energy. In cyclopropanes the angles are 60.
The lowest energy angle for the R2C=O system (SP2) is 120 degrees, in cyclopropanone it is 60. This is a very strained and unstable system, anything that reduces that strain is a positive change energywise.

This problem is discussed here see problem 2


Is there way to detect surface area of nanoparticle by electrochemical technique? I want to know other than SEM is there any other technique or instrument to detect the surface area of particle esp eletrochemical way.

Dynamic light scattering is commonly used to measure nanoparticle size: http://en.wikipedia.org/wiki/Dynamic_light_scattering

Whether you can measure this sort of thing electrochemically depends a lot on the type of nanoparticle. Conductive nanoparticles, you might be able to estimate the surface area by measuring adsorptive behaviours of various things, like CO for Pt and Pd: http://www.sciencedirect.com/science/article/pii/S1388248113000933 Non-conductive particles might be measured by how they impede diffusion to a surface. Electrochemical techniques could suffer from problems related to the shapes of the nanoparticles, as well.

The most straightforward ways of examining the surface features of nanoparticles would be electron microscopy or AFM.

Without further details of your application, it's difficult to recommend a specific technique. If the nanoparticles are being used as an electrode, the apparent electrochemical surface area may be all you care about, even though this is not necessarily the same as the geometric surface area.


What's the difference between HOMO-LUMO and valence-conduction bands? From Wikipedia:

  Roughly, the HOMO level is to organic semiconductors what the valence
  band maximum is to inorganic semiconductors and quantum dots. The same
  analogy exists between the LUMO level and the conduction band minimum.

Why such distinction even exists? If it's the same physical concept, why use different names?

Because convention. Yes, that’s not the answer you wished for but it’s the way it is.

Coming from an organic perspective, one would think of concepts such as linear combination of atomic orbitals or others to arrive at molecular orbitals for entire molecules. Since organic chemistry always deals with rather small molecules and since these have discreet energy levels with discreet differences, it makes sense to speak of a single HOMO, i.e. one single highest occupied orbital (and likewise for a LUMO). The larger a molecule gets, the more orbitals it has and the less the difference between HOMO and LUMO might turn out to be, but from an organic perspective it is still a molecule and must begin and end somewhere.

(Note that this applies to a lesser extent to polymers and carbon nanotubes etc. This is where the shady grey area in which unicorns live begins.)

From an inorganic perspective, if you are dealing with solid materials, they will usually be crystalline in some way; hence they have infinite extents by definition. It no longer makes sense to think of discreet orbitals but rather one realises that certain levels contain a large number of orbitals — hence bands.

If organic molecules end up being sufficiently large, there is no longer a relevant physical difference between HOMO/LUMO designations or bands. But because nobody likes throwing their traditions and customs overboard, the two different names exist.


What can create black stains on metal after putting HCl on a tap full of limestale? After putting HCl on a tap part that was really attacked by limescale, lots of gas began to form and limescale progressively disappeared, as expected.

However, the reaction left behind black soot-like deposits, on the cup holding the reaction but also on the sink (when washing away), that don't go away easily.

So, what could be this black deposit made of?

Also, the chromium-plated parts of tap began turning a bit red after exposure to HCl and the non chromium-plated parts turned black and slowly began to be covered with some kind of white powder (see the picture below).
What is the material of the tap that would explain this behavior?

The red is copper; the knob is likely brass , possibly bronze depending on age. Your knob is not zinc or aluminum or the HCl would have done more damage. Difficult to determine what is black and what is shadow. It is less likely to cause damage in the future if you use vinegar and steel wool to clean-off carbonates.


The reason behind the steep rise in pH in the acid base titration curve Most books refer to a steep rise in pH when a titration reaches the equivalence point. However, I do not understand why … I mean I am adding the same drops of acid to the alkali but just as I near the correct volume (i.e. the volume required to neutralize the alkali), the pH just suddenly increases quickly.

I've decided to tackle this question in a somewhat different manner. Instead of giving the chemical intuition behind it, I wanted to check for myself if the mathematics actually work out. As far as I understand, this isn't done often, so that's why I wanted to try it, even though it may not make the clearest answer. It turns out to be a bit complicated, and I haven't done much math in a while, so I'm kinda rusty. Hopefully, everything is correct. I would love to have someone check my results.

My approach here is to explicitly find the equation of a general titration curve and figure out from that why the pH varies quickly near the equivalence point. For simplicity, I shall consider the titration to be between a monoprotic acid and base. Explicitly, we have the following equilibria in solution

$$\ce{HA <=> H^+ + A^-} \ \ \ → \ \ \ K_\text{a} = \ce{\frac{[H^+][A^-]}{[HA]}}$$
$$\ce{BOH <=> B^+ + OH^-} \ \ \ → \ \ \ K_\text{b} = \ce{\frac{[OH^-][B^+]}{[BOH]}}$$
$$\ce{H2O <=> H^+ + OH^-} \ \ \ → \ \ \ K_\text{w} = \ce{[H^+][OH^-]}$$

Let us imagine adding two solutions, one of the acid $\ce{HA}$ with volume $V_\text{A}$ and concentration $C_\text{A}$, and another of the base $\ce{BOH}$ with volume $V_\text{B}$ and concentration $C_\text{B}$. Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A^-}$) is simply $n_\text{A} = C_\text{A} V_\text{A}$, while the number of moles of species containing $\ce{B}$ ($\ce{BOH}$ or $\ce{B^+}$) is $n_\text{B} = C_\text{B} V_\text{B}$. Notice that at the equivalence point, $n_\text{A} = n_\text{B}$ and therefore $C_\text{A} V_\text{A} = C_\text{B} V_\text{B}$; this will be important later. We will assume that volumes are additive (total volume $V_\text{T} = V_\text{A} + V_\text{B}$), which is close to true for relatively dilute solutions.

In search of an equation

To solve the problem of finding the final equilibrium after adding the solutions, we write out the charge balance and matter balance equations:

Charge balance: $\ce{[H^+] + [B^+] = [A^-] + [OH^-]}$

Matter balance for $\ce{A}$: $\displaystyle \ce{[HA] + [A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}$

Matter balance for $\ce{B}$: $\displaystyle \ce{[BOH] + [B^+]} = \frac{C_\text{B} V_\text{B}} {V_\text{A} + V_\text{B}}$

A titration curve is given by the pH on the $y$-axis and the volume of added acid/base on the $x$-axis. So what we need is to find an equation where the only variables are $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$. By manipulating the dissociation constant equations and the mass balance equations, we can find the following:

$$\ce{[HA]} = \frac{\ce{[H^+][A^-]}}{K_\text{a}}$$ $$\ce{[BOH]} = \frac{\ce{[B^+]}K_\text{w}}{K_\text{b}\ce{[H^+]}}$$ $$\ce{[A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{a}}{K_\text{a} + \ce{[H^+]}}\right)$$ $$\ce{[B^+]} = \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{b}\ce{[H^+]}}{K_\text{b}\ce{[H^+]} + K_\text{w}}\right)$$

Replacing those identities in the charge balance equation, after a decent bit of algebra, yields:

$$\ce{[H^+]^4} + \left(K_\text{a} + \frac{K_\text{w}}{K_\text{b}} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^3} + \left(\frac{K_\text{a}}{K_\text{b}}K_\text{w} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} K_\text{a} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}K_\text{a} - K_\text{w}\right) \ce{[H^+]^2} - \left(K_\text{a} K_\text{w} + \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\frac{K_\text{a}}{K_\text{b}} K_\text{w} + \frac{K^2_\text{w}}{K_\text{b}}\right) \ce{[H^+]} - \frac{K_\text{a}}{K_\text{b}} K^2_\text{w} = 0$$

Now, this equation sure looks intimidating, but it is very interesting. For one, this single equation will exactly solve any equilibrium problem involving the mixture of any monoprotic acid and any monoprotic base, in any concentration (as long as they're not much higher than about $1~\mathrm{\small M}$) and any volume. Though it doesn't seem to be possible to separate the variables $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$, the graph of this equation represents any titration curve (as long as it obeys the previous considerations). Though in its full form it is quite daunting, we can obtain some simpler versions. For example, consider that the mixture is of a weak acid and a strong base. This means that $K_\text{b} \gg 1$, and so every term containing $K_\text{b}$ in the denominator is approximately zero and gets cancelled out. The equation then becomes:

Weak acid and strong base:

$$\ce{[H^+]^3} + \left(K_\text{a} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\ce{A} + V_\ce{B}} K_\ce{a} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}K_\ce{a} - K_\ce{w}\right) \ce{[H^+]} - K_\ce{a} K_\ce{w} = 0$$

For a strong acid and weak base ($K_\text{a} \gg 1$), you can divide both sides of the equation by $K_\text{a}$, and now all terms with $K_\text{a}$ in the denominator get cancelled out, leaving:

Strong acid and weak base:

$$\ce{[H^+]^3} + \left(\frac{K_\ce{w}}{K_\ce{b}}+\frac{C_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}\right) \ce{[H^+]^2} - \left(K_\ce{w} + \frac{C_\text{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}} \frac{K_\ce{w}}{K_\ce{b}}\right) \ce{[H^+]} - \frac{K^2_\ce{w}}{K_\ce{b}} = 0$$

The simplest case happens when adding a strong acid to a strong base ($K_\ce{a} \gg 1$ and $K_\ce{b} \gg 1$), in which case all terms containing either in the denominator get cancelled out. The result is simply:

Strong acid and strong base:

$$\ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]} - K_\ce{w} = 0$$

It would be enlightening to draw some example graphs for each equation, but Wolfram Alpha only seems to be able to handle the last one, as the others require more than the standard computation time to display. Still, considering the titration of $1~\text{L}$ of a $1~\ce{\small M}$ solution of a strong acid with a $1~\ce{\small M}$ solution of a strong base, you get this graph. The $x$-axis is the volume of base added, in litres, while the $y$-axis is the pH. Notice that the graph is exactly as what you'll find in a textbook!

Now what?

With the equations figured out, let's study how they work. We want to know why the pH changes quickly near the equivalence point, so a good idea is to analyze the derivative of the equation and figure out where they have a very positive or very negative value, indicating a region where $\ce{[H^+]}$ changes quickly with a slight addition of an acid/base.

Suppose we want to study the titration of an acid with a base. What we need then is the derivative $\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$. We will obtain this by implicit differentiation of both sides of the equations by $\displaystyle \frac{\ce{d}}{\ce{d}V_\ce{B}}$. Starting with the easiest case, the mixture of a strong acid and strong base, we obtain:

$$\frac{\ce{d[H^+]}}{\ce{d} V_\ce{B}}=  \frac{K_\ce{w} - C_\ce{B} \ce{[H^+] - [H^+]^2}}{2(V_\ce{A} + V_\ce{B}\left) \ce{[H^+]} + (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}\right)}$$

Once again a complicated looking fraction, but with very interesting properties. The numerator is not too important, it's the denominator where the magic happens. Notice that we have a sum of two terms ($2(V_\ce{A} + V_\ce{B})\ce{[H^+]}$ and $(C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A})$). The lower this sum is, the higher $\displaystyle \frac{\mathrm{d}\ce{[H^+]}}{\mathrm{d} V_\ce{B}}$ is and the quicker the pH will change with a small addition of the base. Notice also that, if the solutions aren't very dilute, then the second term quickly dominates the denominator because while adding base, the value of $[H^+]$ will become quite small compared to $C_\ce{A}$ and $C_\ce{B}$. Now we have a very interesting situation; a fraction where the major component of the denominator has a subtraction. Here's an example of how this sort of function behaves. When the subtraction ends up giving a result close to zero, the function explodes. This means that the speed at which $\ce{[H^+]}$ changes becomes very sensitive to small variations of $V_\ce{B}$ near the critical region. And where does this critical region happen? Well, close to the region where $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$ is zero. If you remember the start of the answer, this is the equivalence point!. So there, this proves mathematically that the speed at which the pH changes is maximum at the equivalence point.

This was only the simplest case though. Let's try something a little harder. Taking the titration equation for a weak acid with strong base, and implicitly differentiating both sides by $\displaystyle \frac{\ce{d}}{\ce{d} V_\ce{B}}$ again, we get the significantly more fearsome:

$$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}} = \frac{ -\frac{V_\ce{A}}{(V_\ce{A} + V_\ce{B})^2} \ce{[H^+]} (C_\ce{B}\ce{[H^+]} - C_\ce{B} K_\ce{a} + C_\ce{A} K_\ce{a})}{3\ce{[H^+]^2 + 2[H^+]}\left(K_\ce{a} + \frac{C_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}}\right) + \frac{K_\ce{a}}{V_\ce{A} + V_\ce{B}} (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}) -K_\ce{w}}$$

Once again, the term that dominates the behaviour of the complicated denominator is the part containing $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$, and once again the derivative explodes at the equivalence point.


Why do Gas Laws only use Kelvin Temperature? I didn't see this in any other question, although it seems to be a fundamental question of the gas laws. I understand that Charles' Law practically has to use Kelvin temperature, because you would end up with a negative number in some cases, but can someone explain clearly what Charles did with the three gases and why the volume can only be proportional to the Kelvin temp and not the Celsius temp? thanks.

First you must know how was Charles' discovered his law?
See on wikipedia Jacques Charles

Here is a quote from Wikipedia

  Around 1787 Charles did an experiment where he filled 5 sites to the same volume with different adds. He then raised the temperature of the sites to 80 °C and noticed that they all increased in volume by the same amount. This experiment was referenced by Gay-Lussac in 1802 when he published a paper on the precise relationship between the volume and temperature of a gas. Charles' law states that under constant pressure, an ideal gas' volume is proportional to its absolute temperature. The volume of a gas at constant pressure increases linearly with the absolute temperature of the gas. The formula he created was $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

As all statements of experimental laws are completely based on the results of experiment performed and cannot be altered. Charles'law is also an experimental law, and it was discovered from the experiment performed by Jacques Charles and he found that the Volume of a ideal (close to Ideal) gas is directly proportional to absolute temperature.

You are saying that

  Why volume can't be proportional to temperature in Celcius?

But how it can be if it is not the result of the experiment.

Suppose a cricketer just scored six and you are asking why can't we write his score as four instead of six. In the same way Charles discovered that Volume is directly proportional to absolute temperature and you are saying why it is not temperature in Celcius.

If you want to know why volume can only be proportional to absolute temperature - experiment is the proof.


Naming system -"ylene" There are various compounds which have common names that end in "ylene", like acetylene and ethylene. However, how come acetylene is an alkyne whereas ethylene is an alkene. Both of these have the same "ylene" ending though. What is the cause for this discrepancy?

These are common names, and predate our knowledge about bonds. My guess is that the "ylene" comes from reduction -- reducing ethyl alcohol gives ethylene, reducing acetic acid  gives acetylene.


Why only structures corresponding to valleys on PES are taken into account? I have had to ask this question again because it was marked as a duplicate but it wasn't. Here it goes..
Suppose we want to estimate $G_f^\circ$ of a molecule. If it has rotational conformers, right way to calculate $G$ should include a Boltzmann average of them.
We will do it as:
$$\large \langle G_f^\circ \rangle = \frac{\sum_{i}G_f(i)e^{-G_f(i)/k_BT}}{\sum_{i}e^{^{-G_f(i)/k_BT}}}$$
where the sum goes over the conformers.
Question:
Why $\langle G_f^\circ \rangle$ is estimated as the sum of $G_f(i)$ of only most stable conformers? (corresponding to minima on PES)
Example
Suppose the following rotational conformers:

Consider the $y-axis$ to be gibbs free energy, we would estimate $\langle G_f^\circ \rangle$ as the sum of structures corresponding to minima (ie 60º, 180º), but there are non-minimum structures around 180º with less energy than the minima in 60º. Why aren't them considered?
EDIT
Someone could argue that a conformer in 150º won't exist because there will be a driving force to 180º, but if we take the relative energies $\frac{G_f(150º)}{G_f(60º)}$ we get that 150º must have almost the same population as 60º. There might be some criteria I am missing or maybe a conceptual error.
Any edit will be welcome.

The vast majority of thermodynamics we consider is at equilibrium, we is the issue here.

At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated.  This this because they have now force acting on them to drive them to another configuration.

An important relationship to remember with potential energy surfaces is:
$$
F = -\frac{dE}{dx}
$$
Force is the negative derivative of potential energy with respect to some coordinate.
This means that the driving force is given by the slope of the PES and is pushing down the PES.
Then a molecule with a $\phi = 150^\circ$ experiences a rotational force pushing it towards the minimum at $\phi = 180^\circ$.

Answer:

The result is that any molecule not at a minimum will be quickly driven towards the nearest minimum, so only the minima will be occupied and need to be summed over.

Side Note - Maxima

Maxima (the peaks) also experience zero force.  However this is only true at the exact peak itself.  Any point infinitesimally far away from the maxima experiences a force pushing it down the hill into a minima.

Thermal fluctuations and collisions with other molecules will continually slightly nudge the molecules around on the potential energy surface.  At maxima this will push any molecules there off the peaks and down into minima.  In minima, the force will bring the molecules back into the minima.