Saturated vs unsaturated fats - Structure in relation to room temperature state? I'm sure most of us have heard that saturated fats are solid at room temperature, and unsaturated fats are liquid at room temperature. I'm wondering how this relates to their chemical structure -- saturated fats contain only single bonds between carbons, yet to qualify as an unsaturated fat a C=C double bond must exist.

Since a double bond is stronger than a single bond, and the length of the C=C double bond is shorter than that of the single bond, why is it that the fat containing a double bond is a liquid and saturated fats are solids at room temperature? Seems like the double bond would inhibit movement and the resulting substance would be less like olive oil and more like butter.

This has to do with the chemical structure of the fatty acids. Single bonds create a straight molecule, which can be easily packed together to form a solid, while the double bond introduces a kink in the structure.

  Fats, which are mostly from animal sources, have all single bonds between the carbons in their fatty acid tails, thus all the carbons are also bonded to the maximum number of hydrogens possible. Since the fatty acids in these triglycerides contain the maximum possible amount of hydrogens, these would be called saturated fats. The hydrocarbon chains in these fatty acids are, thus, fairly straight and can pack closely together, making these fats solid at room temperature. Oils, mostly from plant sources, have some double bonds between some of the carbons in the hydrocarbon tail, causing bends or “kinks” in the shape of the molecules. Because some of the carbons share double bonds, they’re not bonded to as many hydrogens as they could if they weren’t double bonded to each other. Therefore these oils are called unsaturated fats. Because of the kinks in the hydrocarbon tails, unsaturated fats can’t pack as closely together, making them liquid at room temperature.

The source website is no longer accessible, but biochemistry textbooks will state the same thing.


Consequences of using reagent preserved with sodium azide on gold film I am running experiments using reagents preserved with 0.1% sodium azide, which are applied to a gold film.  Unfortunately, there are very few options for using reagents without sodium azide.

I understand that sodium azide reacts with gold to form heavy metal azides, which are explosive.

I would like to know:

How long will it take for sodium azide to react with gold?
Generally, what is going to happen to the gold film?
Any other info you can think of

http://www.ehs.neu.edu/hazardous_waste/fact_sheets/sodium_azide/

"Sodium azide reacts with heavy metals (such as silver, gold, lead, copper, brass, or solder in plumbing systems); and metal salts to form an accumulation of the highly explosive compounds such as lead azide and copper azide."

Assume reaction is fast.  How many micromoles of gold/cm^2 constitute your film?  How many micromoles of azide/volume are in your application?  Will the gold be corroded through?

Is the gold there as a convenience or as a necessary surface?  A brief soak in dilute decanethiol solution plus rinse will cover the surface with a monomolecular layer of hydrocarbon, each chain being anchored by its sulfur.  The surface will be passivated and probably hydrophobic. 1-hydroxy-10-decanethiol will give a hydrophilic surface.  Is it worth the bother and expense to diddle with the gold surface?

Gold is selling for $(USD)42.15/g today.  How much gold is in your annual discard pile? When my school dumped a load of ancient undergrad combustion bombs I went through them.  They originally shipped each with a platinum crucible.  One remained.  Half a Troy ounce of platinum was a good hour's tax-free wage.


Work done in an irreversible process I was given a problem to identify whether the given statement is true or false.

And the statement was:

"Consider an ideal gas which is expanded from $(P_{1},V_{1},T_{1})$ to $(P_{2},V_{2},T_{2}).$ Then, the work done on the gas is maximum when it is compressed back irreversibly from $ (P_{2},V_{2}) $ to $ (P_{1},V_{1})  $ against a constant pressure $P_1$".

I concluded that the statement is false, as:

It is nowhere mentioned what is greater $P_1$ or $P_2$ as there is no information on how the expansion process was done.
Even if I assume that $ P_1 > P_2$, still during irreversible compression I can vary the pressure such that the pressure goes beyond $P_1$ in the middle of the process but finally returns to the required final state $ (P_{1},V_{1})  $ , which would mean that I am doing more work than what I would be doing if I keep the pressure constant.

However the answer says that the statement is true.

Where am I wrong?

It is tricky wording. Usually when you compress something you want to do so as efficiently as possible. The minimum amount of work would be required if you compressed this reversibly.

If for some reason, as in the question, you want to waste your energy and overkill the work required (i.e., do things the hard way), then compress the gas IRREVERSIBLY. that will make it take the maximum amount of energy to get the desired compression.

The engineer in my cringes at this :)


how are colloids and suspensions solutions? my definition of solution is a homogenous mixture of a solute and solvent.

and in my textbook they call colloids and suspensions solutions . how is this possible if both of them are heterogenous mixtures.

The definition of solution is a little bit cloudy. IUPAC define a solution in this way:

  A liquid or solid phase containing more than one substance, when for
  convenience one (or more) substance, which is called the solvent, is
  treated differently from the other substances, which are called
  solutes.

And indeed this is a quite general approch that takes in account that we usaually call solutions systems (such as metallic solution) that are not compose of single molecules or ions dissolved in a solvent. Besides this approch chemists generally call solution a system with the folowing charachteristics:

It is stable.
It has only one phase.
It has to be an homogeneous mixture of two or more substances.
The solute can't be distinguished from the solvent (no scattering, can't be filtered)

And so I would not define colloids and suspension solution because they have a dispersed phase and often they are not stable.


What does "Anal. Calcd." stand for written out? I'm wondering what the abbreviation "Anal. Calcd. stands for exactly. It is put in front of calculated (expected) values for elemental analyses. "Analytically calculated" makes no sense for me, but maybe it's more of a language problem for me in this particular case. According to a quick Google research, some are listing it in their lists of abbreviations as "Analytical Calculated". Using the adjective instead of an adverb doesn't make it any more correct to me.

They are two distinct terms, that just happen to frequently be found next to each other. Their definitions can be found in the Journal of Organic Chemistry Guidelines for Authors (see page 29).

Anal. - abbreviation for combustion elemental analysis.

Calcd - abbreviation for calculated.

It is not actually a single phrase, as you've suggested here. Combustion elemental analysis is what people often mean when referring to 'getting an analytical done'. Writing Anal. signifies the type of experimental data that is about to be listed, just as you might specify 1H NMR, IR or MS. A common alternative to combustion elemental analysis is high resolution MS, which is often written in experimental sections as HRMS (ESI) m/z calcd

The format for presenting elemental analysis is journal specific, and therefore varies. The format you're asking about is consistent with the ACS publication recommendations. RSC guidelines accept the the data in a slightly different format.


Mechanism for acid-catalyzed dehydration of 2,3-dimethylcyclopentanol How do you get from 2,3-dimethylcyclopentanol to 1,2-dimethylcyclopent-1-ene?

I know the -OH group leaves as water (being protonated by acid) forming a carbocation. Then the weak nucleophile (water) attacks the beta-hydrogen to form what I think is 2,3-dimethylcyclopent-1-ene. But why is it 1,2-dimethylcyclopent-1-ene?

I know Zaitsev's rule wants the more substituted alkene, but forming the 1,2-dimethylcyclopent-1-ene would leave the carbocation hanging there, wouldn't it (the carbocation would still be unsatisfied)?

Update:


Effect of nuclear spin statistics on the IR spectrum of CO2 Specifically, I am talking about $\ce{C^{16}O2}$. With a nuclear spin of zero meaning, it is a boson - it's total wavefunction must be totally symmetric. The total wavefunction is made up of an electronic, vibrational, rotational and a nuclear spin contribution. For this molecule the electronic and vibrational (in ground state) wavefunctions are symmetrical. There is no antisymmetric nuclear spin wavefunction so only the even J rotational levels exist.

Now consider the degenerate bending vibrational mode. Excitation of this mode leads to excitation into the excited vibrational state of this mode and the IR spectrum has a corresponding P, R, and crucially a Q branch (due to additional angular momentum from the degenerate bending mode).

I have been asked whether there are similar J restrictions in the excited state as there are in the ground state. I initially jumped into assessing the symmetry of the contributions to the wavefunctions and concluded that the vibrational mode is antisymmetric and thus the corresponding excited state wavefunction is antisymmetric leaving the possibility of only odd J values in the excited state. However, there is a Q branch so this must be incorrect since the Q branch indicates the presence of J=0 in the excited state.

Why is there seemingly no restriction on the excited state?

Because the degenerate bending mode has vibrational angular momentum, it is a vibrational $\Pi_\text{u}$ state and just as in electronic $\Pi$ states there exist levels with $+$ and $-$ parity. As a consequence, the degenerate bending mode has $\Pi_\text{u}^\pm$ symmetry and the vibrational wave functions can be both symmetric and antisymmetric.

See also Herzberg Vol 2 (p372 and Fig 99 on p373).


Are β-sheets formed from one or more polypeptides? I know that a β-strand forms bonds with another β-strands and so on and thus a β-sheet is formed, but are those strands from the same polypeptide? And does their distance matter? Do they have to be close or it is not necessary? Thanks in advance.

A β-sheet is formed as the result of hydrogen bonding between the β-strands in the secondary structure of a protein. The hydrogen bonding occurs between the carbonyl oxygen and the amide hydrogen of adjacent strands, and can take on either a parallel or an antiparallel arrangement. As can be seen in the diagrams, antiparallel arrangement of the polypeptides results in more stable interactions between the strands, but both arrangements are possible and can be found in nature. Regardless of the type of arrangement, β-sheets can be formed by either interactions between different polypeptides or intramolecular hydrogen bonding within a single polypeptide. In fact, intramolecular hydrogen bonding creating β-sheets is really common. One example of it occurring in β hairpin motifs.


Why would the specific heat capacity of water be used as the specific heat capacity of zinc sulfate? I could not find the specific heat capacity of zinc sulfate and saw that people use water's heat capacity instead. Why is that?

If the substance is dissolved in water, it's very typical (and more correct) to use the heat capacity of water in lieu of the pure salt, acid, base, etc.

The reason is simply that, when in solution with water, there is much more water than the solute. It's an approximation, but a decent one. This is common practice in freshman-chemistry lab courses which I've taught.

It's difficult to find heat capacities for pure salts such as zinc sulfate because the means by which we measure it is typically through a reaction, and typically a salt is dissolved in solvent (e.g. water) in order to perform a reaction. That is to say, reactions of solids are just less common than liquids, gases, and solutions.

That said, the NIST database reports a $c_{p}$ for zinc sulfate of 99.01 Jmol$^{-1}$K$^{-1}$ at 298 K.

Looking at the original paper, I found this plot of $c_p$ as a function of temperature. At room temperature, you can see it is roughly 100 Jmol$^{-1}$K$^{-1}$.

Stuff like this can be easily obtained from citrination.com search engine.


Balancing a chemical equation, but with an incorrect mole ratio I'm wondering if it's possible to balance a chemical equation with an incorrect mole ratio.  I have the unbalanced equation $\ce{AgNO3 + Cu -> Ag + Cu(NO3)2}$, I have a $2:3$ ratio of $\ce{Cu}$ and $\ce{Ag}$ that I must plug in.  Is this possible, because it doesn't look like it.

The balanced reaction is $$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$

Firstly, it's not necessary that all the reactants are depleted in the reaction.

There is the concept of a "limiting reagent" -- the reagent that gets depleted first and drives the extent of the reaction.

In this case, the limiting reagent is $\ce{AgNO3}$, and you will have half a unit of $\ce{Cu}$ left over.

In the general case, the limiting reagent need not be the one in the least quantity. For example, for $\ce{2A +B ->C +D}$ , if $\ce{A:B}$ is 3:2, then $\ce{A}$ is still the limiting reagent as it will get depleted first.


Removing the hydrazine byproduct from the mitsunobu reaction Looking to remove the hydrazine byproduct left after the mitsunobu reaction. Using DIAD and polymer bound triphenyl phosphine. The reduced diad is coeluting with my product on silica. My thoughts for getting rid of it are trying an alumina column, acidic wash (I have an N-Boc so it can't be too acidic), or switching to a different azodicarboxylate.

Acidic wash is not going to help. As you said you can try a DIAD alternative for example DCAD can give good results (sry no advertising intended here). There is also the option of taking your contaminated product to the next step(s) and hope you can get rid of the impurity there. I would take option A if I have to make a bunch of similar analogues by a Mitsunobu and B otherwise for not wasting too much time


Field Effect and it's effect on basicity on amines What is the order of basicity between $\ce{Me2NH}$, $\ce{Me3N}$, $\ce{MeNH2}$ in a protic solvent.

It has something to do with field effect which I could not understand and could not find anywhere.

Can someone please explain?

The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation.

The electronegative nitrogen is able to isolate electron density from the alkyl groups. So having more electron-releasing alkyl groups to stabilize the positive formal charge created by the abstraction of a proton from the protic solvent might seem like a good thing.

On the other hand steric hindrance due to bulky alkyl groups reduces basicity.

Also, having more alkyl groups reduces solvation due to hydrogen "bonding." Consider ammonia, $\ce{NH_3}$. This molecule can form four hydrogen bonds. On the other hand, a tertiary amine that has been protonated can form only "good" hydrogen bond; one might argue that the hydrogens on the alkyl groups can form hydrogen "bonds", but these are obviously going to be weaker "bonds" compared to the ones among ammonia molecules.


Clarifications on thermodynamic cell efficiency The thermodynamic efficiency of any cell (especially Fuel cells) is given as $$\frac{\Delta G}{\Delta H} \times 100$$

I understood this partly, that since $\Delta G$ is the useful work obtained in the ideal case, the efficiency expression must have it in the numerator. But I fail to see the intuition (or maths) behind the $\Delta H$ term in the denominator. I have two questions regarding this:

On applying this equation to ordinary (not fuel cells), for a general reaction $$\ce{A^+ {(aq)} + B {(s)} -> A {(s)} + B^+ {(aq)}}$$ the $\Delta G$ term is given as $-nF E_{\text{cell}}$ where $E_{\text{cell}}$ depends on the concentration of the two ionic species as given by the Nernst equation $\left(E=E^\circ-\frac{RT}{nF}\ln\frac{[\ce B^+]}{[\ce A^+]}\right)$. On the other hand, is the $\Delta H$ term independent of the concentration? and if not, how would it (and consequently efficiency) depend on the ionic concentrations?
For fuel cells, suppose the cell reaction is such that the $\Delta S$ is positive and since $\Delta G=\Delta H -T\Delta S$, therefore $|\Delta G|>|\Delta H|$ since the second term in the equation for $\Delta G$ becomes negative due to positive entropy change. Would this predict an efficiency greater than 100%?

${\Delta G}$ and ${\Delta H}$ are normally expressed on a per mole basis.  Your assertion that ${\Delta G}$ is related to concentration is not precise.  It is related to the ratio of concentrations, and as such the incongruity with ${\Delta H}$ is not troubling.
If you found a fuel cell with a gain in entropy, according to your equation you would have more than 100% efficiency.  I seriously doubt such a situation exists.


Relation between electron gain enthalpy and electron affinity The correct relation between electron gain enthalpy $(Δ_\mathrm{eg}H)$ and electron affinity $A_\mathrm{e}$ at any temperature '$T$' is
A) $Δ_\mathrm{eg}H = -A_\mathrm{e} - \frac{5}{2}RT$
B) $Δ_\mathrm{eg}H = \frac{A_\mathrm{e}}{RT}$
C) $Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT}$
D) $Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT} + \frac{1}{A_\mathrm{e}^2}$

I have read that electron affinity could be taken as electron gain enthalpy at absolute zero, i.e. 0 K. So, maybe we could use Kirchoff's law for temperature $T$, but why is $C_p$ taken and not $C_v$ or something else, is it from the definition?

You are correct, at absolute zero $Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e}$ for the gas-phase act of gaining an electron:

$$\ce{X(g) + e-(g) → X-(g)}$$

$$A_\mathrm{e} = E(\ce{X(g)}) - E(\ce{X-(g)})$$

The term $5/2RT$ arises from the so-called "electron convention" when electron is treated as ideal gas with corresponding heat capacity $C_p$ from Boltzmann statistics and the element of enthalpy

$$H_T(\ce{e-}) - H_0(\ce{e-}) = \frac{5}{2}RT,$$

resulting in exact relation

$$Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e} - \frac{5}{2}RT.$$


Enolate Ion formation I was wondering why when acetaldehyde is treated with $\ce{OH-}$, the $\ce{OH-}$ will attack the methyl group instead of the carbonyl carbon, forming an enolate anion. I would have though that the $\ce{OH-}$ would have attacked the carbonyl carbon due to it having a  relative positive charge.

Additionally, my professor said during lectures that form x is favored over form Y. Is this because a negative oxygen is able to delocalize the negative charge (and be protonated more easily) more efficently than a $\ce{CH2-}$ group? If so,  is the x form so dramatically favored that it would not make sense to draw any mechanisms involving the y form?

Enolate vs hydrate formation

Both the enolate and the hydrate form.

Aldehydes (and ketones) react with water in the presence of acid or base to form hydrates. However, for more aldehydes in aqueous solution (even dilute solution), the hydrate is only a minor component of the mixture (~1%).

The other 99% of the aldehyde molecules are perfectly able to form the enolate, but only a small number of them do because the difference in pKa between most alpha hydrogens on aldehydes and ketones (~21) and water (15.7).

So, in actuality most of the aldehyde molecules at any given time are neither enolates nor hydrates. Now, if a reaction should occur to consume enolate, then Le Châtelier will come out and nudge the system to make more enolate.

  "Favored" resonance contibutors

The two resonance contributors $x$ and $y$ are not separate compounds. It is not possible for one structure or another to be "favored", though we often use that language because we can relate to comparing the structures of two distinct compounds, like the cis and trans isomers of an alkene. What we really mean when we say "favored" is that the resonance contributor is more important for describing the structure and behavior of the ion than the other. The true structure of the enolate ion is some linear combination of resonance contributors $x$ and $y$ (as well as some other minor things). Thus, the true structure might be like a weighted average over the contributors with $x$ having higher weight.

For example:

$$enolate = \frac{2}{3}x + \frac{1}{3}y$$


Migration of ethyl group in pinacol rearrangement For this substrate, will the methyl or ethyl group migrate preferentially? I know that the migratory aptitude of Et is greater than that of Me, but I saw a lecture where the prof said migrating Et will break the bonds and that only shifting of Me is possible. Is it right? What will be the product formed?

The two possible compounds resulting from the pinacol rearrangement of your substrate, 3,4-dimethylhexane-3,4-diol, are as follows.

A quick Reaxys search throws up a paper in which this pinacol rearrangement was investigated.1 I quote directly:

  Four different investigators have studied the pinacol-pinacolone rearrangement of 3,4-dimethyl-3,4-hexanediol. It was finally demonstrated by Nybergh that, contrary to the conclusions reached by the three previous investigators, the rearrangement involved the preferential migration of an ethyl group rather than a methyl group.

The paper describes how the authors carried out the reaction and analysed the mixture to determine the ratio of the two products shown above. In 1956, there was much less technology than we enjoy now, so no NMR and no HPLC. It turns out that they had to hydrogenate the products to make a mixture of the corresponding alcohols and figure out the ratio by IR spectra. The result: $70\mathrm{-}73\%$ of 3-ethyl-3-methylpentan-2-ol (corresponding to the ethyl shift), depending on the form of the starting pinacol used.

Unfortunately, I can't access the (very old) journals that claim that the Me shift happens preferentially.

References

(1) Reeve, W.; Karickhoff, M. The Pinacol Rearrangement of meso- and dl-3,4-Dimethyl-3,4-hexanediol. J. Am. Chem. Soc. 1956, 78 (23), 6053–6054. DOI: 10.1021/ja01604a028.


Is ΔS of a system related to temperature and change in enthalpy? I often see the equation $\Delta S_\text{universe}$ = $\Delta S_\text{surroundings}$ + $\Delta S_\text{system}$, where $\Delta S_\text{surroundings}$ can be re-expressed in terms of the enthalpy change of the system as

$$\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$$.

This leaves me wondering why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$ (Maybe as $$\Delta S_\text{system} \overset{?}{=} \frac{- \Delta H_\text{surroundings}}{T}$$ although this would make $\Delta S_\text{universe}$ equal $0$).

Instead, $\Delta S_\text{system}$ is usually given as a standard value, seemingly independent of temperature. Is there a reason why we can only express the change in entropy of the surroundings, but not the change in entropy of the system, as related to the change in enthalpy and the temperature of a reaction?

If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system.  So the heat received by the surroundings is $-\Delta H_{system}$.  Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible."  So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$.  The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.


Seashell and grape juice My daughter is working on a 5th grade science fair experiment.  She soaked white seashells in various drinkable liquids to test the effects of those liquids on teeth (teeth are represented by seashells). We were expecting to find loss of mass and discoloration of the seashells, but we were completely surprised by the reaction observed in the grape juice seashell.  The pictures attached are not great at showing what I mean, but besides the discoloration, the seashell shimmers!  There seem to be crystals all over it. It appears like it is covered in glitter.  Has anybody heard of such chemical reaction?  I researched this, but cannot seem to find anything to explain it to my daughter.  We don't know if the juice simply ate away at the outer protective layer of the seashell to expose the sparkly inside (not likely) or are we looking at a chemical reaction (more likely).

I warn you that I've virtually no knowledge of the composition of grape juice - I didn't even remember that it contained tartaric acid, for example. My admimission of my ignorance is to guard you from letting my technical vocabulary  fool you into thinking that just because I might sound knowledgeable, that what I say is correct concerning grape juice and sea shells.

Calcium tartrate, when pure, is colorless (its crystals are "white"). When made from grape juice, I would expect that the crystals are colored. This would be due to impurities which co-precipitate & are incorporated into the crystal structure. I was able to find some pictures of crystals of calcium tartrate using Google which were colored (purplish red in one case, orange-red in another) and which might be consistent with what you saw. See also https://en.wikipedia.org/wiki/Tartrate.

On the other hand, grape juice contains other organic acids which might also form colored crystals. Two families in particular I am thinking about: tannins and anthrocyanins. Both are polyphenols, which means they have two or three -OH (hydroxyl) groups attached to one aromatic ring (the simplest example of an aromatic ring is benzene).

Sheen is generally not a property of a chemical compound's pure crystal. The prototypical example is the gemstone opal, which is composed of microscopic silica spheres packed into a regular array. The point here is that "sheen" is more likely due to microscopic particles included in the crystal rather than an inherent property of a pure crystal. There are exceptions to this, however.

The reason I mention the polyphenols is that they typically form very highly colored complexes (salts) with iron. Often they are colored brown, purple or black (and the that their crystals might have a sheen.) It appears to me that the shell has turned purple-black due to the polyphenols reacting with trace quantities of iron. A way to check this would be to take an iron nail (ungalvanized, unplated, uncoated) and let it rust in some grape juice and see what you get - if its just the typical orange-red rust, then there probably aren't many polyphenols present. But, note that in anaerobic (low oxygen,eg. in unmixed water) conditions, iron will rust to a black color, which will turn slowly red-brown when allowed to air dry. Wine is well known to contain tannins, which come from grape skins.

Anthrocyanins are known to change color with changes in acidity (pH). So, you could check to see if the crystals change color in water with added acid (vinegar, say) and base (ammonia, say). With supervision, Draino in water is also an option; it contains a very strong base, sodium hydroxide, but it definitely shouldn't be used by a child - it can cause severe burns to the skin and worse to the eyes. Most powdered detergents also are basic.

Grape juice contains large amounts of tartaric, malic, and citric acids. Neither of the latter two are likely to be the source of your crystals, imho. Two other experiments you could try are: 1. Add some rubbing alcohol (40% or even better 70%) to a bit of the juice. You might get some crystals (sediment).
2. Put a shell into some Cream of tartar in water. (The Cr. of T. will be more soluble with some vinegar or lemon juice added, making it more acidic). If the crystals you saw were indeed calcium tartrate, then this should give you the same thing. When I saw the picture of the black seashell, I immediately assumed tannins (and iron) were present, fwiw.


Are the following expressions for base dissociation constant equivalent? Given a weak base such as $\ce{C2H5NH2}$, I usually see the expression for $K_\text{b}$ as

$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{OH-}]}{[\ce{C2H5NH2}]}$.

Are the expressions

$K_\text{b} = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$,
$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$

equivalent? Do they represent the same process (if so, why do the latter two not work with calculations?) or fundamentally different ways of dissociation? Which one represents what's happening molecularly most accurately? Given the Brønsted-Lowry definition of a base as a proton acceptor I'd think that the second one is most accurate, but I rarely see it used.

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$
This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$
here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.


Does it sometimes take energy to create a chemical bond? Wikipedia says so:

  Bonds can also be broken apart. Since most bonds require energy to form, they also give off energy when they are broken. But before most bonds break, the molecule has to be heated. Then the atoms start to move, and when they move too much, the bond breaks. Molecules that require less energy to break than they give off when broken are called fuels. For example, a candle will just sit there and nothing happens. But when you use a match to light it, it will burn for a long time. The match brings the energy to break the first bonds, which release enough energy to break the bonds below them, until the candle has burned down.
  But what I have learned is bonds release energy when formed.

"Molecules that require less energy to break than they give off when broken are called fuels." is the most imprecise sentence in the paragraph, because fuels don't give off energy when their bonds are broken, but when these broken bonds form new bonds (with oxygen) that are much more stable and release more energy than it took to break the original (carbon-hydrogen) bonds.

A candle is stable in air, even tho it does not have complete stability in an oxygen-containing atmosphere. But if you put sufficient energy into a few bonds, the excited atoms will combine with oxygen, which gives off enough energy to excite a few more bonds, which etc., etc. It's this new reaction, which is overlooked in the paragraph, which provides the increased temperature during the reaction.

The energy that "bonds require to form" is an activation energy which destabilizes a molecule enough so that it can reorganize with other reagents (like O2) to form more stable products. This activation energy may be supplied initially from outside sources (the match), but in a continuing reaction, the exotherm from the reaction provides the activation energy and then some.


If an electron moves closer to the nucleus in a given value of n, does it lose energy? If you look at a radial probability chart for a given electron sublevel (excluding n=1), there is a probability of an electron in that sublevel being closer to the nucleus. The electron should lose energy if it is closer to the nucleus because it experiences a greater force of attraction to the nucleus, but how is this possible if all electrons in a given sublevel have the same energy?

You are trying to apply macroscopic understanding to the world of quantum mechanics. This thinking is doomed to fail. The energy of an electron occupying a specific orbital is that orbital’s energy. Full stop. Unless it changes its orbital it does not change energy.


Crystal structure database search for space groups I'm usually using databases like the Springer Materials site or the free Crystallography Open Database to get crystal structure data. But none of these offers the possibility to search for all entries with a defined space group. I recently prepared a new compound and solved and refined the crystal structure. I would like to compare it to compounds known by literature to give the structure a name. It's quite common to compare structures to already existing ones. But I can't find any website that would give me the possibility to look for all entries with the same space group without knowing the elements.

Does anyone know a search engine for this specific problem?

You could use COD SQL interface. For example this little Python script returns all entries in space group F-43m that contain Si:

#!/usr/bin/env python
from mysql import connector
con = connector.connect(user='cod_reader', host='sql.crystallography.net',
                        database='cod')
cursor = con.cursor()
cursor.execute("""select file, formula from data
                  where formula LIKE '% Si%' AND sg = 'F -4 3 m'""")
for n, formula in cursor:
    print n, formula

The script prints entry number and formula for each matching COD entry:

$ python cod.py
1010995 - Si C -
1011031 - Si C -
1011254 - Al13 Cl F9 H9 O29 Si5 -
9000281 - Al13 Cl F3.84 H14.16 O34.16 Si5 -
...


Intuition for why catalyst affects both forward and reverse reactions equally? I have looked for quite a long time on the site however all the answers involve mathematical explanations involving Arrhenius's equation. I would really like an intuitive reason (or one using collision theory?) as to why adding a catalyst should increase both reactions equally.

It seems that if you add a catalyst it should decrease the activation energy by a constant for both forward and reverse reactions. Thus it should decrease the forward activation energy proportionately more than the reverse reaction then the forward reaction should go forward, and shift equilibrium. But it doesn't. Why not?

I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ is a function of the ratio of the forward and reverse rate constants. For a simple reaction $\ce{A <=>[k_1][k_{-1}] B}$, the equilibrium constant $K=\frac{k_1}{k_{-1}}$.

If we assign the activation energy $E_\mathrm{A,fwd}$ to the forward reaction and $E_\mathrm{A,rev}$ to the reverse, we have

$$k_1 = Ae^{-E_\mathrm{A,fwd}/(RT)} \quad \text{and} \quad k_{-1} = Ae^{-E_\mathrm{A,rev}/(RT)},$$

so

$$K=\frac{Ae^{-E_\mathrm{A,fwd}/(RT)}}{Ae^{-E_\mathrm{A,rev}/(RT)}}.$$

Using simple rules of exponents, this reduces to

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}}.$$

Now if we subtract a quantity $E_\mathrm{cat}$ from both activation energies, the same derivation gives

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{cat} - (E_\mathrm{A,fwd} - E_\mathrm{cat})} = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}},$$

so $K$ is unchanged.

Intuitively, it may help to think about the fact that the reverse reaction is necessarily composed of exactly the same microscopic states as the forward reaction, just happening in reverse order (by the principle of microscopic reversibility). The catalyst can stabilize one of more of the states on that reaction pathway, but cannot influence what direction that stabilized state reacts in. So making a microscopic state more easily achievable in the forward direction necessarily makes it more easily achievable in the reverse direction.


Where do negative ions migrate from the salt bridge to which electrode in the voltaic cell? In the voltaic cell with $\ce{Mg/Mg^2+}$ and $\ce{Ag+/Ag}$, I am not sure whether the ions move from the salt bride to the $\ce{Mg}$ cathode or the $\ce{Mg^2+}$ cathode. I distantly recall my teacher saying the cathode (in this case) is a metal, and so it can't be $\ce{Mg^2+}$? I am confused, and please help ASAP. Final is tomorrow.

Somewhat related question:
At the cathode during electrolysis, what is its charge and what type of reaction occurs?

Is the correct answer: positive charge, and reduction?

Oxidation half reaction (happens at the anode)

$$\ce{Mg -> Mg^2+ + 2e-}$$

The electrons get removed via the wire, and the magnesium cations enter the aqueous solution, giving it a net positive charge.

Reduction half reaction (happens at the cathode)

$$\ce{Ag+ + e- -> Ag}$$

The electrons are supplied via the wire, and the silver cations leave the aqueous solution to be deposited on the cathode, giving the solution a net negative charge.

Salt bridge

The salt bridge provides anions (negative charged particles) to the solution of the oxidation half cell, and cations (positively charged particles) to the solution of the reduction half reaction.

Silly mnemonics

Oil rig: oxidation is loss, reduction is gain (of electrons)

Red cat: reduction occurs at the cathode.

An ox: oxidation occurs at the anode.

  Where do negative ions migrate from the salt bridge?

They don't migrate to any electrode (neither cathode nor anode). Instead, they migrate to the solution of the solution of the oxidation half cell.


Ethene to ethanol: electrophillic additions involving ethyl sulfates http://www.chemguide.co.uk/mechanisms/eladd/symh2so4.html

I was reading my textbook and this website. I just cannot find the mechanism for what happens when we go from CH3CH2OSO3H to ethanol by the addition of water and heat.

I understand the electrophillic addition part.

But, what happens next? Is there some kind of Sn2 reaction to make HSO4- the leaving group? Then, why does the EtOH2+ molecule give out one of its acidic protons to HSO4- and not to H2O which is a more powerful base?

I hate when my textbook leaves out the details and I obsess on them :(

Yes, I believe you are correct. Water attacks at the sulfate containing carbon by an SN2 mechanism. You are also right that sulfuric acid is a stronger acid than hydronium (or protonated ethanol), which means that in solution the bisulfate anion will predominate. Often people (including textbook authors) will draw a mechanism that goes to neutral products. Most likely water will carry out the final deprotonation (especially given that water is the solvent).

Incidentally, most likely this reaction is self catalyzed, as one sulfate protonates another, activating the sulfate for SN2 reaction by a weak nucleophile.


A thermos bottle containing milk is shaken vigorously I have a conceptual physical chemistry question here.

  A thermos bottle containing milk is shaken vigorously. Consider the milk as the system. Will the temperature rise as a result of the shaking?

My answer to this is yes, but I'm having trouble finding an equation that explains this result.

The best i have is the molecular kinetic energy formula that $$\mathrm{ke_{avg}= \frac{1}{2}mv^2=\frac{3}{2}kT}$$
So a change in kinetic energy is $$ \mathrm{\Delta ke = \frac{3}{2} k \Delta T}$$

Is this the right approach?

A conceptual approach might work better, since the shaking system is not going to convert mechanical energy into thermal energy as neatly as something like Joule's Paddle Wheel.  As the bottle is shaken, the liquid in it is swirled around into more or less randomly oriented vortices.  These cancel one another out and slow to rest via friction with their container, and the energy they contained has to go somewhere.  As you wrote, it winds up increasing the average velocity, and thereby temperature, of the particles in the milk.

This basic mechanical to thermal energy conversion is the source of the conversion factor from joules to calories.  Joules are used to measure mechanical energy and calories were used to measure thermal energy.  It was the paddle wheel experiment I linked above that first related the two to one another.

The challenge presented by trying to relate the work done on the bottle of milk to the thermal energy that winds up going into the milk is less mathematical than practical.  On any given shake we would first put energy into the bottle (move it upward, let's say) then take energy back out (slow it to a stop) before putting energy back into the bottle (moving back downward) and taking energy out again (slowing to a stop a second time.)  This repeats until we get tired of shaking, and if we measured the difference in all of the energy inputs and outputs that ought to equal the thermal energy added to the milk, but actually doing that for real would be a pain.


Why is electron gain generally exothermic? I understand that we need to supply energy to counter the nuclear attraction when we remove electrons, and that is the reason why ionization energy is endothermic. However, why does an atom release energy when we add an electron to it?
With respect to ionization energy, why is energy required to remove electrons of elements like $\ce{Be}$, $\ce{Li}$ that naturally want to lose their electrons?
My teacher told me that elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?

Energy of electron in vacuum is zero. If it can attach (however weakly), the energy is gained and process is therefore exotermic. It is not an atom which releases energy, it is the whole system.
Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined).
In some cases, adding of extra electron to neutral atom is not an exothermic process. See Periodic table of electron affinities. For example, a helium atom has filled shell with $n=1$. The next electron would be accepted to the shell $n=2$ which has much higher energy, and therefore is not stable. For the half-filled orbitals, similar reasoning applies, but spicing it with Hund's rule.


What is the product of the reaction of carbon with sulfuric acid? I don't know if $\ce{C}$ or $\ce{CuO}$ react with $\ce{H2SO4}$. Some sources say that:
 $$\ce{C + H2SO4 -> CO2 ^ + 2SO2 ^ + 2 H2O}$$
(It seems to be similar to $\ce{Cu}$ reactions with $\ce{H2SO4}$).
Is the reaction possible? If the answer is yes, in what conditions does the reaction take place?

And I was pretty sure that $\ce{CuO + H2SO4 -> CuSO4 v + H2O}$

But my chemistry book says that none of the reactions is possible.
I'm little bit confused, some help would be apreciated.

Carbon reacts with sulfuric acid to produce carbon dioxide and sulfur dioxide gas along with water. Sulfuric acid should be a concentrated, heated solution.

$$\ce{C + 2H2SO4 ->[\Delta] CO2 + 2SO2 + 2H2O}$$

More insight about this reaction can be found here.

Copper(II) oxide, a black solid, and dilute sulfuric acid react to produce copper(II) sulfate, giving a characteristic blue colour to the solution. From this solution, blue copper(II) sulfate pentahydrate crystals may be obtained if desired.

$$\ce{CuO + H2SO4⟶CuSO4↓+H2O}$$

It is a general "insoluble metal oxide is reacted with a dilute acid to form a soluble salt" type of reaction. It is a standard class experiment.

More insight about this reaction can be found here.


Why does activated carbon preferentially adsorb anions? Brands like Brita and Pur (in the U.S.) have made a name for themselves for the ability of their product (essentially a large-pore filter with activated carbon/charcoal) to extract the added chlorine from tap water.  I visualize $\ce{C}^*$ as a vast network of channels in which ions are "trapped".  I had assumed that most cations and anions were trapped effectively, but it seems to favor chlorine and iodine.

According to Wikipedia:

  Activated carbon does not bind well to certain chemicals, including alcohols, glycols, strong acids and bases, metals and most inorganics, such as lithium, sodium, iron, lead, arsenic, fluorine, and boric acid.

Why would it not bind effectively to fluorine?  Is it an issue of atomic radius?  Does $\ce{F-}$, with a radius of 0.136 nm sneak by, while $\ce{I-}$ and $\ce{Cl-}$ at 0.181 and 0.216 nm, repectively, get caught up in the matrix?  Why do the cations get passed through?

This answer applies to carbon filtration of water.

According to most of the sources I found, activated carbon binds to most substances through London dispersion forces (from the Wikipedia article). This should mean that it adsorbs larger molecules and non-polar molecules preferentially since they would have larger dispersion forces. As your quote indicates, it does not bind polar molecules like "alcohols, glycols, strong acids and bases" well. This source also lists neutral, non-polar compounds, along with organics and compounds with low water solubility as being effectively removed.

I think that you may have misread or misinterpreted the sentence about binding "chlorine and iodine" as saying that it traps chloride, iodide, and (by extension) other anions well. I can't find data to support this statement. It traps molecular iodine well, and as a large (in terms of electrons), non-polar molecule, that makes sense. But it removes molecular chlorine $\ce{Cl2}$ through a chemical reaction by reducing it to the chloride ion $\ce{Cl-}$ which is then soluble in water and flows through the filter.

I did find a couple of sources (1, 2) that said "activated charcoal" has a slight electropositive charge that helps it to attract negatively charged species, which would fit with your statement that it preferentially attracts anions. (However, they don't list what kind of negative species they're talking about and nitrate, sulfate, and fluoride ions were specifically excluded). The idea that it will remove ionic compounds seems to be contradicted by the statement in at least a couple of sources (one given here) that carbon filters don't remove minerals, salts, and dissolved inorganic compounds


Why does the Birch reaction not yield fully saturated products? Why isn't benzene reduced all the way down to cyclohexane when performing a Birch reaction? (Why is it not reduced further?)

Similarly for the reaction with sodium dissolved in liquid ammonia, why isn't an alkyne reduced fully to an alkane?

The mechanism of the Birch reaction includes single-electron transfers from the sodium dissolved in ammonia to the aromatic compound. These transferred electrons will attempt to occupy the lowest possible orbitals.

In the case of benzene, we have a delocalised $\unicode[Times]{x3C0}$-system. Delocalisation can be understood here to mean higher occupied orbitals and lower unoccupied orbitals. Therefore, the electron will find an orbital at a lower energy and it will gladly hop in to create the radical-anion intermediat.

Once this is protonated, the remaining radical system can still be seen as a delocalised pentadienyl radical. Thus, the energy level a new electron will occupy is still rather low and another anion can be created. Further protonation yields the product of the Birch reaction, a cyclohexa-1,4-diene — most importantly, a non-conjugated system. This can be regarded as a pair of isolated double bonds.

The energy of the unoccupied $\unicode[Times]{x3C0}^*$ orbitals in an alkene seems to be just too unfavourable for a free electron to occupy them. I don’t have the facilities to do calculations (and neither am I firm enough to be able to do them) so I can’t provide you with values to compare, but it has to be the case otherwise the reaction would continue. If it were not so, there would be no reason for the reaction to stop.


Photochemical reduction of benzophenone: why inverted flask? I've come across an interesting protocol for photochemical reduction of benzophenone to benzhydrol (diphenyl methanol) with sodium alcoholate in alcohol, the 2nd protocol in https://prepchem.com/synthesis-of-diphenylmethanol/. The reaction proceeds via ketyl radical initially to benzopinacol, which is immediately cleaved by a catalytic amount of sodium alcoholate to the benzhydrol and benzophenone. The latter repeats the reaction until it's used up. The alcohol acts as a reducing agent and solvent.
I'm particularly interested in the following part:

the flask is filled to the neck with more of isopropyl alcohol ... The flask is tightly stoppered and inverted upside down in a beaker.

I want to know why is the flask inverted?
My best guess is to avoid exposure to air, since that would interfere with the reduction. But then why isn't it enough to just stopper the full flask?
Might this have more to do with the quality of exposure to the light source (neck and stopper producing a shadow in the upright position)?
Also, is there some official term for this inverted flask technique?

Notes (extra info for the interested):

The ketyl radical (which is actually a radical anion) is brightly colored, which provides a convenient way to monitor the reaction. Upon exposure to air, the color disappears, presumably because the radical anion gets oxidized back to the ketone.
Any strong enough base should work for cleavage of the pinacol. In a related synthesis where benzopinacol is the desired product, a small amount of acid is added, because the glass itself is alkaline enough to cleave the product. The sodium alcoholate is convenient in that it's the conjugate base of the alcohol. It's formed by reaction of the alcohol with sodium metal.
Cleavage of the pinacol proceeds by deprotonation to the pinacolate anion. This rearranges to the ketone and alkoxide, which protonates to the alcohol, regenerating the base.

I think the reason for inverting the the round flask is to capture sunlight very efficiently. A round bottomed flask filled with a fluid is like a ball lens. Sun rays can be considered parallel and see how they focus. If light is coming from all directions, every part will be illuminated well as the Sun moves. See ball lenses here ball lenses

The clue comes from the instruction in the procedure given in your link "Then the flask is exposed to the best available sunlight. "


Symmetry representations for acetonitrile (CH3CN) Can anybody share the symmetry labels for acetonitrile ($\ce{CH3CN}$), please?
I mean to assign the symmetry labels to orbitals of acetonitrile

I am afraid I disagree with Geoff's assignment of acetonitrile to the point group $C_{\infty v}$ and "approximation" with $C_{2v}$, so I will make another attempt to answer this question. The way I have structured this answer, the resources I used, and the level of detail included follow a similar question on this site.

Molecule structure

First you need to know the geometrical structure of your compound. This can be found easily from Wikipedia. The image is reproduced here.

Assigning the point group

Then you need to find the point group of acetonitrile. This can be done by following flow charts, one of which is available here. While it may be a little difficult at the beginning, with some practise you can look at a molecule and determine its point group accurately, so it is worth working through the flow chart for a range of molecules until you are confident with this. There are other arrangements of this flow chart, but all flow charts should give you the same result, so use the one provided in your course if possible.

With the flow chart I have linked to, we go through these steps (if you have a flow chart organised differently, you will encounter the same steps, but likely in a slightly different order - check this is the case!):

molecule is not linear
molecule contains only one $C_3$ axis
molecule DOES contain a proper rotation axis which is $C_3$ (rotation over methyl group)
the highest rotation axis is $C_3$ but there are no perpendicular $C_2$ axes
the molecule does NOT contain a horizontal reflection plane ($\sigma_n$). This can sometimes be confusing -- in general, a horizontal plane is perpendicular to the highest rotation axis.
the molecules DOES contain 3 vertical reflection planes ($\sigma_v$).
therefore the point group is $C_{3v}$.

"Orbitals"

In your question, you said that you need to "to assign the symmetry labels to orbitals of acetonitrile". This is a little unclear, because you can mean that you want to assign symmetry labels to atomic orbitals of acetonitrile, or molecular orbitals of acetonitrile. I am going to assume that you need the molecular orbitals here. But how do we find the molecular orbitals?

Molecular orbitals are constructed by making combinations of atomic orbitals. Which means that, a logical next step is to consider all the valence atomic orbitals (VAOs) on all the atoms of acetonitrile.

Atomic orbitals and symmetry sets

To help us talk about VAOs, I have put numbers on all the atoms.

Carbon, atoms 1 and 2: $2s$, $2p_x$, $2p_y$, $2p_z$

Nitrogen, atom 3: $2s$, $2p_x$, $2p_y$, $2p_z$

Hydrogen, atoms 4, 5 and 6: $1s$

When we try to tackle these types of problems, it is common to construct symmetry sets of atomic orbitals that would transform in the same way under the operations of the point group of the molecule. You will also notice that you cannot use any of the operations of the group to transform certain atomic orbitals on the same atom into each other (cannot transform $2s$ to $2p$, cannot transform $2p_x$ to $2p_z$, etc), so these should be considered as separate sets. The $p_x$ and $p_y$ orbitals are mixed by the rotations and reflections of the group, so we should consider them as one symmetry set.

In acetonitrile, there is only one nitrogen, so we have to consider as separate sets:

$2s$
$2p_z$
$2p_x$, $2p_y$

Carbon 1 and carbon 2, though both carbons, cannot be transformed from one to another, and your chemical intuition should also tell you they are different carbons because they have different bonds attached to them, so they are also in separate sets of their own. The three hydrogen atoms, on the other hand, are related by rotation (a $C_{3v}$ rotation on the principal axis, so they can form one set of 3 hydrogen 1s orbitals.

Assigning symmetry labels to each set of valence atomic orbitals

The usual trick at this point is to simply read off the orbitals that have corresponding symmetry labels, e.g. the $p_x$ orbital would transform as $x$ on the character table. This definitely works for any atom that lies on the origin point of the point group, but it is not completely obvious where you would take the origin point for putting a (right handed) axis system on acetonitrile. The "middle" of this molecule would appear to be somewhere between the two carbons, and that may seem a little awkward.

Remember that in all cases, the choice of origin and axis system is actually arbitrary, however you should pick the most convenient one. Now, for acetonitrile you can pick any point on the principal rotation axis ($C_{3}$) and it would work, because that is an inherent feature of the $C_{3v}$ point group. If this is not obvious, I suggest looking at the answer I referenced previously, and also working through any other examples that you can find.

Luckily for us, $C_{3v}$ has quite a small character table (i.e. it does not have many symmetry elements), so we can work through it and check. The character table of the point group $C_{3v}$ is shown below and can also be found here. The character tables for many common point groups can be found here.

$s$ orbitals that lie on the principal axis

Now let's consider the $2s$ orbital or carbon atom 1. It should be fairly obvious that it transforms back to itself under all the operations of the point group (unchanged under identity $E$, any of the $C_3$ rotations, any of the $\sigma_v$ reflections), and so it transforms as $A_1$. It should also become very obvious that this is also true for the $2s$ orbitals of carbon atom 2 and nitrogen atom 3, even though these atoms are not symmetry-related, their atomic orbitals can have the same symmetry under the point group. This will be important when we come to construct molecular orbitals.

$p$ orbitals that lie on the principal axis

For p orbitals, we need to impose an axis system. Conventionally, the principal axis is taken as the $z$ direction. You have to follow these conventions when doing symmetry analysis.

Let's consider the $2p_z$ orbital or carbon atom 1, which lies along the principal axis. This again transforms as $A_1$.

The $2p_x$ and $2p_y$ orbitals are a little tricky. They remain identical under the identity operator $E$, giving a character of 2. For the reflection operator $\sigma_v$, one of the three $\sigma_v$ planes would keep one identical while inverting the other one (giving zero), and the other two $\sigma_v$ planes give characters of zero, so the overall character for $3 \sigma_v$ is zero. The $C_{3v}$ is the difficult one, where you have to put vectors on the axis, rotate them, work out the geometry with trigonometry, form the $2 \times 2 $ matrix and work out the trace (i.e. the character) is -1. Or you can use the fact that the character tables show (x,y) transform as the irreducible representation $E$.

Again, the same symmetry labels apply for all the atomic orbitals for all atoms that lie on the principal axis.

Set of hydrogen $1s$ orbitals that do not lie on the principal axis

For these hydrogen $1s$ orbitals we examine explicitly how they behave under all operations of the point group.

$E$: 3  (all three $1s$ remain unchanged from themselves)
$C_v$: 0 (all three change and do not remain at the same position)
$\sigma_v$: 1 (for any one operation, only 1 of the orbitals remain the same, the rest change)

This is a reducible representation. You need to reduce it to find the symmetry labels for the hydrogen $1s$ orbitals.

Symmetry labels for all valence atomic orbitals

For the whole molecule of acetonitrile, the valence atomic orbitals give the irreducible representation $4 A_1 \oplus 4 E$. From the discussion above you should be able to deduce the exact symmetry label for any atomic orbital on any atom of the molecule.

Molecular orbitals

You know that you need to generate as many molecular orbitals as atomic orbitals, and only the atomic orbitals of the same symmetry can mix. Molecular orbitals use lower case letters, so you expect two types of symmetry labels based on the irreducible representations of the VAOs. You should be able to say how many molecular orbitals of each type you expect, remembering that the irreducible representation $E$ is two-dimensional and indicates a degenerate pair of VAOs.

Symmetry analysis cannot give you the energy of these molecular orbitals, however, so you cannot know how they are ordered relative to one another.


Why the colour of KMnO4 at the end point of titration may disappear after some time? In the redox titration of $\ce{FeSO4}$ with $\ce{KMnO4}$, the colour change of the solution at the end point is colourless to light pink. Most laboratory manuals and books advice us to check whether the light pink colour stays or not for 30 seconds. If the colour stays after 30 seconds, then end point is reached. If the colour disappears before 30 seconds, then the titration is not finished, so more $\ce{KMnO4}$ should be added.

As far as I understand, at the end point, all the reducing agent is depleted. So, the colour of $\ce{KMnO4}$ stays. And, if I have interpreted my book correctly, the colour of $\ce{KMnO4}$ is to stay for more than 30 seconds so that we can say that the end point is reached.

This means, that the colour of $\ce{KMnO4}$ might disappear after 30 seconds even though the end point is reached. But why? All reducing agents are depleted. Aren't they?

Now, I have found some reactions which might possibly cause the discharge of colour after the end point:

\begin{align}
\ce{3 MnSO4 + 2 KMnO4 + 2 H2O &-> 5 MnO2 + K2SO4 + 2 H2SO4}\\
\ce{3 MnSO4 + 2 KMnO4 + 8 H2SO4 &-> 5 Mn(SO4)2 + K2SO4 + 8 H2O}\\
\ce{4 KMnO4 + 2 H2O &-> 4 MnO2 + 4 KOH + 3 O2}
\end{align}

I am not sure which of them occur in reality.
So, which of the reactions are the reasons for the 30 second rule?
And why is it 30 second anyway? Why not 40 second or 1 minute?

The most likely reason for the colour of permanganate to vanish after a few seconds even though you thought you have the end point is: you have not reached the end point.

Acid/base reactions are typically extremely fast reactions; their rate is not even limited by the speed of diffusion since protons can be shuffled along via the Grotthuss mechanism much more rapidly. Thus, you can assume each acid/base reaction to instantly reach completion.

With practically all other reactions that is not the case. They are typically slower and can be much slower. The reacting species must meet each other in solution which can take time. At the beginning of the titration, there are still many iron(II) ions floating around ready to be oxidised by any incoming permanganate. Towards the end, iron(II) is getting more scarce. So a permanganate ion may linger around much longer before it meets an iron ion. You should give it a certain time in case it does find one. The value of $\pu{30s}$ is arbitrary. Ten seconds or two minutes would probably do equally well (except that ten seconds may or may not be slightly on the short side). The book/method just needs to present any ballpark value.

So the reaction occuring that causes the discolouring of the solution is indeed exactly the reaction you are observing: the oxidation of iron(II). It is merely slowed down so it can take a few seconds to proceed to completion when close to the equivalence point.


How to attach soft tubing to glass I plan to do some amateur chemistry with some basic equipment. I've watched hours of youtube videos for how to bend and polish glass tubing and push it through rubber stoppers, and I've seen diagrams of such tubing attached to rubber or silicone tubing, but nothing showing how to do that. I've even searched through catalogs of equipment and seen lots of specialized gadgets with hose barbs and such, but nothing basic. I just want to do a simple thing: have a rubber stopper with a glass tube in it, and attach that directly to a flexible tube. Nothing fancy, just one tube connected to another. I see it in diagrams like http://www.mctcteach.org/chemistry/C2224/labman/vacfilt/filter7.htm , but it gives no explanation. What am I missing?

Heat up the soft tubing to make it easier to fit over the connection. Most labs have a heat gun that is used for this (amongst other uses). A hairdryer will serve, or put the end of the tubing in hot water. Many glass connection have what is called an olive which you have to push the tubing over while hot, when it cools it contracts giving a good seal (particularly important for the water lines of condensers which may be left overnight)


How to find the pH of acetic acid after adding sodium hydroxide? $3\cdot 10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ are added to a $1\,\mathrm{l}$ solution of $2\cdot10^{-2}~\mathrm{M}\ \ce{CH3COOH}$. Find the $\mathrm{pH}$ of the solution.

I don't understand why my solution is wrong:

I said this, $\ce{NaOH + CH3COOH -> H2O + CH3COONa}$, so I'm left with $2\cdot10^{-2}~\mathrm{mol}$ of $\ce{CH3COONa}$ and $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$.

Now, from those $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ I get the same moles of $\ce{OH-}$ (from dissociation).

Then, $\ce{CH3COONa -> Na+ + CH3COO-}$, and $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$

Setting up the equilibrium equation: $K_h = \frac {\text{Products}}{\text{Reactants}}$, I get (where $K_h= K_\mathrm{w}/K_\mathrm{a}, K_\mathrm{a} = 1.8\cdot 10^{-5}$):

$$K_h=\frac {(x)(1\cdot 10^{-2}+x)}{x-1\cdot10^{-2} }=5.55\cdot 10^{-10}$$

Which yields $[\ce{OH-}]= x= 1.11\cdot 10^{-9}$ and so $\mathrm{pH}=5.04$ which is wrong. I know the answer should be around 12 from the answer sheet.

What am I doing wrong?

If you add $3 \cdot 10^{-2}~\mathrm{mol}$ $\ce{NaOH}$ to $1~\mathrm{L}$ of $2 \cdot 10^{-2}~\mathrm{M}$ $\ce{AcOH}$ you end up with a solution that contains $2 \cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{AcO-}$ and $1\cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{OH-}$. So the pH is 12.


Does 6-methyl-2-(methylamino)pyrimidin-4-ol show aromaticity? Does this structure show aromaticity? Or does this not work with nitrogen?

Each Nitrogen has 5 valence electrons. They are using 2 of those in sigma bonds to their neighbouring atoms, 2 in an sp2 lone pair in the plane of the ring, leaving only 1 electron to contribute to the delocalised pi system.

Each carbon is also contributing 1 electron to the delocalised pi system, and has 3 involved in sigma bonds.

This means each member of your 6 membered ring is contributing 1 electron, 6 in total. This means it is aromatic, as all 3 bonding orbitals are fully occupied.


Effects of potassium hydroxide & carbonate on metals? BACKGROUND

I am in the process of restoring an old computer. The problem with the computer is that it has a NiCad battery fitted to the motherboard, which powers a real-time-clock circuit. The NiCad battery has leaked onto the motherboard. The leaked material has caused quite extensive damage, which I am cleaning and repairing.  See here and here.

I am curious about this leaked material. Here is what I (think) I understand so far:

NiCad batteries use potassium hydroxide as an electrolyte, and this
is the material which has leaked from the battery.
The potassium hydroxide forms potassium carbonate when it reacts with
the carbon dioxide in the surrounding air.

MY QUESTIONS

Some white(ish) powder has formed on the battery terminals themselves. Is this potassium carbonate?
The effect of the battery leakage has been quite widespread corrosion of metals, mainly electronic components. The corrosion has spread much further than the leakage site. How?
Why does the material attack metal? Do both materials (potassium hydroxide and potassium carbonate) attack metal, or does one of these have more of an effect on metals than the other?
Are any other chemicals likely to have formed that I have not considered? (e.g. there are green/blue deposits to be found in areas affected by the leakage)

CONJECTURE

Having pieced together some information from reading on the internet, I have formed a bit of an opinion about the process which has unfolded here. And I'd like to run it by you all to see what you think about it:

When the battery started to leak, the potassium hydroxide reacted with the air and formed potassium carbonate, as a white powder on the battery terminals.
When all of the carbon dioxide in the air was depleted (due to closed computer cover), the potassium hydroxide stopped forming potassium carbonate. And this is when it began to attack the metals.

THANK YOU IN ADVANCE!

The purpose of my questions is purely to satisfy my curiosity. If someone takes the time to respond, and share their knowledge, then I respect that and I am grateful. So I just wanted to say thank you in advance!

Your conjecture is at least partially correct, in that there was a mix of $\ce{KOH}$ and $\ce{K2CO3}$. However, there is only ~400 parts per million (ppm) $\ce{CO2}$ in air, so at first, the $\ce{KOH}$ was free to react with metal (and, possibly, the green lacquer on the PC board), until enough $\ce{CO2}$ entered the cabinet to form the carbonate. It is possible some raw $\ce{KOH}$ is still present, under a film of carbonate, so be careful. Note that there is ventilation in a PC cabinet, so $\ce{CO2}$ is never really depleted in it.

Far more damage, though, may be due to electrolytic corrosion. Wikipedia has a photo similar to yours, showing "Severe PCB corrosion from a leaky PCB mounted Ni-Cd battery". The liquid electrolyte creates bridges across the bare IC pins and across the PC board traces, and the continual current from the power supply and the not-quite-dead battery cause metal to be removed from one place and deposited in another.

After removing the battery, rinsing many times with distilled water, and tooth-brushing the board, you may find that surface traces need to be rebuilt (such as by soldering a bit of braid across a gap). This is likely a multilayer board, and if an internal trace is gone, it would be difficult to find and bridge the gap. Note that any such patches will affect timing, due to the impedance mismatch, so you may find that after all repairs, crashes occur with some frequency.

Best wishes for your restoration project. It's nice to know that at least one venerable computer is not going to landfill!


Finding direction of the reaction using Affinity of the reaction I encountered following reaction:

$$\ce{CO + 1/2 O2 -> CO2}$$

I am provided with the composition of the various gases at $\pu{1000K}$ and am asked the direction of the reaction by finding the affinity of the reaction. I know the way to find affinity is:

$$A= \mu_\ce{CO2}-\mu_\ce{CO} -\frac{1}{2}\mu_\ce{O2}$$

I know the change in chemical potential considering ideal mixing. But how do I go about finding the absolute value of chemical potential, rather than the change, because the expression requires the chemical potential itself rather than the change? I thought of writing $A$ as $-\Delta G$, but for that also I need some reference value of $G$ and from there I can integrate $\Delta H-T\Delta S$, but again I can't figure out how reference state is chosen?

Please suggest something.

Firstly, since $\overline{\mu_\alpha} = \overline{g_\alpha}$, the two approaches that you propose are basically the same: $\overline{\mu_\ce{CO2}}-\overline{\mu_\ce{CO}} -\frac{1}{2}\overline{\mu_\ce{O2}}$ is the same as the change in $G$ per mol of forward reactions.

Let's suppose that we can calculate the true, absolute value of the molar Gibbs energy at some reference physical and chemical state, labelled $^*$. The molar Gibbs energy at any arbitrary state is then

$$ \overline{g} = \overline{g}^* + \Delta \overline{g}^\text{chem} + \Delta \overline{g}^\text{phys} $$

Where the to $\Delta g$ terms account for the fact that the state of interest may have a different chemical and/or physical state than the reference state. If we substitute this back into your expression, we find

$$\begin{aligned} A &=\overline{\mu}_\ce{CO2}-\overline{\mu}_\ce{CO} -\frac{1}{2}\overline{\mu}_\ce{O2}
\\
&=\overline{g}_\ce{CO2}-\overline{g}_\ce{CO} -\frac{1}{2}\overline{g}_\ce{O2}
\\
&=(\overline{g}^*_\ce{CO2}+\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\overline{g}^*_\ce{CO}+\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\overline{g}^*_\ce{O2}+\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys})
\end{aligned}
$$

If we collect the $g^*$ components, their sum is

$$ \sum \overline{g}^* = \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^* $$

But then, by definition of the reference chemical state,

$\overline{g}^*_\ce{CO2} = \overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO2}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)
$\overline{g}^*_\ce{CO} = \overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)

Substituting this back:

$$\begin{aligned} \sum \overline{g}^* &= \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= (\overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}) - (\overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}) - \frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= 0
\end{aligned}$$

(The term cancels to zero because the reaction is balanced)

Thus $A$ has no dependence at all on the absolute value of $g$. We can cancel the sum of the $g^*$ components and write

$$ A = (\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys}) $$

In practice, we typically define state $^*$ by:

Physical State: $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$
Chemical State: pure elements in their most stable form at $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$

and then re-label/re-name the chemical component of $\overline{g}$:

$\Delta \overline{g}^\text{chem} \rightarrow \overline{g}^\circ_f$ the "Gibbs energy of formation" (change in Gibbs energy associated with forming the substance from pure elements at the reference physical state)

To actually evaluate this, you'd

Look up $\sum \overline{g}^\circ_f$ for this particular reaction (which might be labelled as $\Delta G$) in a table, if one is available. If one isn't, you could look up the individual species' $\overline{g}^\circ_f$ (if you have tables which present them), or calculate them from the species' $\overline{h}^\circ_f$ and $\overline{s}^\circ$ (again, from tables).
For each species, evaluate $\Delta \overline{g}^\text{phys}$ (the difference in $\overline{g}$ associated with changing the species' physical state from $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$ to the $T$ and $P$ of interest) using some property model (equation or table) for each substance in turn

My Notation: For some extensive quantity $X$

$x$ represents $X$ per mass
$\overline{x}$ represents $X$ per amount


Calculating Buffer Capacity I completed a titration of Ammonium Acetate buffer solution and to it I added $\pu{2M}$ Hydrochloric Acid.

I measured the initial $\ce{pH}$ of the buffer solution before any acid was added and I used methyl orange as an indicator which has a $\mathrm{p}K_\mathrm{a}$ value of 3.
My independent variable was the temperature of the buffer.

I was just wondering how I would calculate the buffer capacity. Is it $\Delta\ce{pH}$/Volume?

In addition to Martin's answer, there is at least an old recommendation
by E. B. Sandell and T. S. West in Pure Appl. Chem., 1969, 18, 427-436 (DOI), which states:

  Buffer capacity or buffer index. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or strong base required to change the pH by one unit when added to one litre of the specified buffer solution.


What constraints are imposed on a wavefunction by the symmetry of the system? As a follow-up to my answer here, I'd like to ask what exactly does it mean for a wavefunction to "respect the symmetry" of the system.

The original context is: immediately after ionisation of $\ce{CH4}$, the wavefunction of $\ce{CH4+}(T_\mathrm{d})$ cannot be adequately described by removal of an electron from an $\mathrm{sp^3}$-type hybrid orbital, as the resulting Slater determinant $|\phi_1\phi_2\phi_3\phi_4\overline{\phi}_1\overline{\phi}_2\overline{\phi}_3|$ does not reflect the tetrahedral symmetry of the system immediately after ionisation ($\phi_i$ being the four hybrid orbitals).

Intuitively, it is clear that this is related to the fact that the four C–H bonds must be equivalent. Mathematically, how is this formalised, i.e. what is the criterion that the wavefunction must obey?

As a side question, how does the criterion above apply to a boron atom, for example? Is the wavefunction $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$ acceptable on symmetry grounds? My gut feeling is that it's not (because the electron density needs to be spherically symmetric?), but I might really be overthinking this.

I use quantum chemistry about as much as a lumberjack uses math (that is, just enough not to fell a tree on one's own head), so my answer will be short and hopefully useful as a tl;dr to someone else's more comprehensive take.

In quantum chemistry, if two operators commute, their values are simultaneously observable. In other words, there is a basis of functions which are eigenfunctions to both operators at once. For an opposite example, think of coordinate and momentum; they do not commute, hence the Heisenberg uncertainty and stuff.

When our problem has some symmetry (including, but not limited to, geometrical symmetry like $T_d$), this means that symmetry operators commute with its Hamiltonian. Therefore the eigenfunctions of the latter must also be eigenfunctions of the former. In other words, the application of symmetry operator to any of our states is equivalent to a multiplication by some number. That's the constraint you were after. Since a symmetry operator of a finite point group must have finite order (by applying it a few times, you get back where you started), the said number must be a root of unity.

That's about the size of it. The numbers for different symmetry operators can be chosen in a multitude of ways, each corresponding to some representation. The character tables tell you how many functions of each representation do you have. Since the functions of different representations are orthogonal to each other and don't interact, this helps us to split a huge problem into smaller chunks.

With methane, the qualitative result could have been envisioned in a simpler way, almost without any calculations. Think of a carbon atom with four $sp^3$ orbitals. They are all equivalent and orthogonal to each other. Now bring in the four H atoms. This will produce four $\sigma$ orbitals (and $\sigma^*$ too, we just don't care about those at the moment). But wait, those $\sigma$ orbitals are no longer orthogonal, so they must interact, if only a little. And because of symmetry, each two of them overlap by exactly the same amount. So it is as simple as Hückel can be: find the eigenvalues of the matrix $$\begin{pmatrix}
 0&-1&-1&-1\\
-1& 0&-1&-1\\
-1&-1& 0&-1\\
-1&-1&-1& 0
\end{pmatrix}$$
The symmetry approach I outlined above still applies; it is just that the system became so small that one can manage without it. The thing can be solved by hand. Or it can be noted that we're dealing with a complete graph, and their spectra are well known (just remember that math people have it upside down, for they use $\mathbf1$s instead of $\mathbf{-1}$s). Either way, the eigenvalues are $\{-3,1,1,1\}$, which kinda explains the 1:3 thing.

So it goes.


Is there a treatment for distilled water to become drinkable? It is my understanding that, drinking distilled water may not be recommended. Its not recommended indeed. Or maybe its not not recommended... Maybe not not indeed........ Anyway....

Well, apparently, in some places, they actually acquire distilled water from various sources (desalinization, atmospheric moisture, etc), and sell for people to drink.

My question is: What are the methods used to treat distilled water, such that it becomes more like the "tap water" we drink? (or, perhaps, drinkable/safer in the sense of the not-recommended-links (that is, assuming such links are true ofc)).

The content of mineralic ions in tap water is usually rather small compared to the amount that you take in with your daily food. Especially Calcium content varies wildly between sources, and nobody recommends calcium supplements in cities with soft, low calcium content water.

The common DI water you have from a tap in a laboratory building or buy in the supermarket is not safe for drinking, firstly and mostly because it is not tested. But who would want to drink it anyway? It is more expensive, often has a taste that varies between lame and slighly foul, and you do usually have a regular drinking water tap to use instead.

The only way you could run into problems is when you have little food, drink a lot (hot environment), and have nothing to drink but distilled/rain water. On the other hand even well water likely has not enough sodium to balance what you loose through sweating.

To answer your question: A big problem with distilled water is its corrosiveness to common household tubing. Regular tap water coats everything in a layer of calcium carbonate, which makes even lead pipes relatively safe. In a standard copper pipe, DI water will first slowly dissolve the oxide layer, and then a lot of electrolytic corrosion will take place.

So, to use distilled water as tap water, you either have to make everything out of PE (doesn't really work if you want hot water), or add calcium carbonate. I believe this is what's done in areas where desalinated water is fed into the communal water supply. You just have to mix in some amount of local well water.


Bond length comparison between two carbon atoms Why is the bond length of double and triple bonds between two carbon atoms shorter than the single bond length between two carbon atoms?

In the case of a carbon-carbon single bond, 2 electrons are shared in the bond connecting the two carbon atoms. With a carbon-carbon double bond, 4 electrons are shared between the two carbon atoms and 6 electrons are shared in a triple bond.  Having additional electrons between the two atoms 1) improves the bonding overlap (makes the bond stronger) between the two carbon atoms and 2) better screens the two carbon nuclei form each other.

Both of these factors, better bonding overlap and better nuclear screening, will allow the two carbon atoms to approach closer together.  Consequently, the more electrons (or the more bonds) between two carbon atoms, the shorter the distance between them.


Halides of group 15 elements Why are the pentahalides of group 15 elements more covalent than the trihalides?

Fajans' Rule is  used to predict whether a chemical bond will be covalent or ionic.

Polarization will be increased by:

1) High charge and small size of the cation

2) Ionic potential

3) High charge and large size of the anion

4) The polarizability of an anion is related to the deformability of its                                            electron cloud (i.e. its "softness")

5) An incomplete valence shell electron configuration

In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.


Odd looking carbocation stability order I am studying introductory organic chemistry, suddenly this one started looking odd,

The carbocation stability for three compounds as given by my book is shown below,

Let me call those as compounds A>B>C as given in image. It is very well known that 'N' and 'O' are electron withdrawing group(shows -I effect) and they can also show +M effect( some texts call it as Resonance effect ) by delocalising their lone pair over the vacant p-orbital of the carbocation.

Now all i want to know is, How to decide whether +M effect or -I effect will operate in this case? If I consider only inductive effects, the order would be C>A>B   which is wrong in this case.

One part of my text book says that +M effect > -I effect (in more than 99% of cases). But then too the next question arises, why +M effect of 'N' is more than that of 'O'?

I got stuck in this one and i am now asking whether such questions are feasible without conducting experiments.

Explanation with the principles which are familiar to my level would be nice.

How to decide whether +M effect or -I effect will operate in this case?

The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$.

In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ density to carbocation. On the other hand, in inductive effect, induction of charge takes place, which partially transfers $\ce{e-}$ density through $\sigma$-bond. Hence, it is weaker.

  why +M effect of 'N' is more than that of 'O'?

As $\ce{N}$ has lower electronegativity than $\ce{O}$, hence it has greater $\ce{e-}$-donating tendency, therefore greater $\ce{+M-\text{effect}}$


Extraction of hydrogen and oxygen from water The apparatus that I am using is a glass container, with two electrodes,two glass cylinders, water, and a battery. In the glass container will be filled with water and two electrodes kept inside the glass cylinder and then dipped in water.

The two electrodes are then connected with the positive and negative terminals of the battery and the container is sealed.  As the current is started the water will be separated into hydrogen and oxygen. After extraction of the gas, the two gases are passed separately.

In this process, will the hydrogen gas start burning immediately as soon as produced? If yes, then what to do to make it pass without burning?  The separated gas is passed through glass pipes, will the extraction work properly?

No, it won’t burn immediately. As your electrodes are generally separated by some distance, you should be able to isolate the produced oxygen and hydrogen gas by catching the respective bubble streams under a submerged test tube or the like. Alone, neither oxygen nor hydrogen will burn, however combining the oxygen and hydrogen gas will produce an explosive mixture. Mixing the hydrogen with air will also produce an explosive mixture as the atmosphere is about 22 % oxygen.

To get the hydrogen to burn, you need to supply the necessary activation energy, which can be supplied by a spark or flame. After this, the reaction will be self-sustaining due to its exothermicity. There are in fact entirely self-contained apparatus which can be purchased from scientific teaching shops that can electrolytically split water and then combust the hydrogen and oxygen using a piezoelectric ignition system, the ensuing tiny explosion causing water to spurt out of a small hole. Wikipedia mentions that the autoignition temperature of hydrogen in air is 500 degrees Celsius.

On a small scale, this reaction is quite safe, however you should always take safety precautions such as wearing protective eyewear and having fire-suppression equipment at hand.

Here’s an illustration of a reasonable water-splitting setup. Remember that your water should contain an electrolyte such as dissolved salt. Keep in mind that if your potential difference is high enough you will start to produce small quantities of chlorine gas and aqueous sodium hydroxide, both of which are hazardous in high concentrations.

I should draw attention to the fact that the $\ce{H2}$ headspace is about twice the volume of the $\ce{O2}$ headspace. This comes down to the stoichiometry of the reaction, and also means that the $\ce{H2}$ volume will increase at twice the rate of $\ce{O2}$. The wires I’ve illustrated connect to the terminals of a battery or lab PSU.

Addendum: Here’s a nice guide on how to construct and safely operate a water-splitting/combusting apparatus using household equipment, aimed at secondary school students.


Why is Borax used as a primary standard? It's given in my book that borax is a useful primary standard for titrating against acids. I have heard that primary standard is a pure reagent and does not gain any weight by absorbing moisture. (Hence any error during weight measurement is prevented).

I got this from a website that came first on searching "primary standard"

  Primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has a high molecular weight .

But borax (represented as $\ce{Na2B4O7.10H2O}$) has water of hydration.

Yes borax is a hydrate, but the reason it is favored for titration is that it gives consistent results, which to most scientists, is far more important than accuracy.

The $\ce{ .\!10H2O}$ hydration number will remain constant and will not fluctuate to $\mathrm{11}$ or $\mathrm{9}$ within normal temperature and humidity ranges. This means that per mass the same amount of sodium ions are dispensed every time and gives the same results time after time and experiment after experiment even with different batches of powder. Also the tetraborate conjugate base is very weak, which is ideal for a titration agent as well.


Can 25mL of 10% phosphoric acid substitute for 10mL of 25% solution? Could I substitute $25~\mathrm{mL}$ of $10\%$ for $10~\mathrm{mL}$ of $25\%$ phosphoric acid solution?

This is for a measurement of free $\ce{SO2}$ in a sample of wine. The phosphoric acid is added to the wine sample and then more steps take it through a color change as it goes from acid to base. Finally a reagent is added dropwise to measure the amount of free $\ce{SO2}$ in the wine.

The full test is set out at More Wine.com as "Test for free SO2 in wine".

Not reliably. Acid strengths measured in percentages are usually given as percent by weight, and if that's the case just swapping the volume numbers around will not provide you an accurate quantity of acid.

You first need to be sure of how the percentages are expressed.  Assuming they are percents by weight, you would then need to look up the density of phosphoric acid at the two percentages and work out the calculation so that you were adding the same number of moles of $\ce{H3PO4}$.


What is the cause for Crystal Field Splitting Parameter? Taking the example of octahedral complexes, the theory says that since the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are closer to the ligands than the $d_{xy}$ ,$d_{yz}$ and $d_{xz}$ orbitals, they reach a higher energy level than the $d_{xy}$ ,$d_{yz}$ and $d_{xz}$ orbitals.(due to greater interaction with the ligands) I can't understand why the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are closer to the ligands. I agree that they are along the bonding axes, but with some rotation in space I can have other d-orbitals along the bonding axes. For example, if I rotated the octahedral arrangement of ligands about the z-axis, then the $d_{xy}$ orbital could be along the bonding axis instead of the $d_{x^2-y^2}$ orbital.So why can't other combination of d-orbitals (like the $d_{xy}$ and $d_{z^2}$ ) reach a higher energy level instead of the $d_{z^2}$ and $d_{x^2-y^2}$ during splitiing?

According to this theory ligands are considered point charges. These points acquire positions as those you've presented above. This leads to some d orbitals being closer to the ligands. That means that a higher amount of energy is needed for an eletron to be placed there, because it would be closer to the negative charged ligand.

However, which orbital is going to be more affected, is a matter of an "agreement" you may say. In order to be easier to explain situations, structures etc it is inclined to use a conformation such as the one you've shown. It would be much more difficult to discuss matters if everyone considered different naming and configurations.


What happens to iron(III) sulfate (ferric sulfate) in aqueous solution? Iron(II) sulfate (ferrous sulfate) is a salt that, in an aqueous solution, dissolves and yields $\ce{Fe^{2+}(aq)}$ and $\ce{SO4^{2-}(aq)}$. It then decomposes to ferric sulfate shortly thereafter.

But what happens to the iron(III) sulfate (ferric sulfate) in solution? Is the $\ce{Fe_2(SO_4)_3}$ compound preserved?

In aqueous solution $\ce{Fe2(SO4)3}$ dissociates into $\ce{Fe^{3+} (aq)}$ and $\ce{SO4^{2-} (aq)}$ ions. The sulphate ions will be solvated by hydrogen bonding with the water and the iron ions will form the hexaaquairon(III) complex, $\ce{[Fe(H2O)6]^{3+}}$.

As @Mithoron mentioned, the $\ce{Fe^{3+}}$ ions cannot easily be further oxidised so no additional reaction takes place.


Why is free radical chlorination exothermic whereas bromination is endothermic? In the energy profile diagrams below, it can be clearly seen that the free radical chlorination of an alkane

$$\ce{RH + Cl2 -> RCl + HCl}$$

is exothermic. However, the corresponding bromination

$$\ce{RH + Br2 -> RBr + HBr}$$

is endothermic. Why is there a difference?

First off, we can see that both reactions are exothermic If we do the thermodynamic analysis.

Given that your starting materials have the same heat of formation:

$$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$

$$\ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}}$$
$$
\begin{array}{cccccc}
\hline
\text{Product 1} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \text{Product 2} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \Sigma\ \Delta \text{H}^\circ_f\  [\pu{kj mol^{-1}}] &\text{Result}\\
\hline
\ce{HBr} & -\text{36.45} & \ce{Et-Br} & -\text{97.6} &-\text{134.05}<-\text{84} &\text{Exothermic} \\
\ce{HCl} & -\text{92.31} & \ce{Et-Cl} & -\text{137} & -\text{229.31}<-\text{84}  &\text{Exothermic}\\
\hline
\end{array}
$$

Back to your question:

  Why is there a difference?

The graphs you are looking at only consider the reaction:

$$\ce{R-H + X^. ->  R^. + HX}$$

For which we get:
$$
\begin{array}{cccccc}
\hline
\text{X}
& \Delta \text{H}^\circ_f(\ce{X^.})
& \Delta \text{H}^\circ_f(\ce{RH})
& \Delta \text{H}^\circ_f(\ce{R^.})
& \Delta \text{H}^\circ_f(\ce{HX})
& \Delta \text{H}^\circ_{rxn}
& \text{Result}\\
\hline
\ce{Br} & -\text{96.94}^{**} & -\text{84} & -\text{119}^\dagger &-\text{36.45} &\text{25.49}&\text{Endothermic} \\
\ce{Cl} & -\text{121.29}^{**} & -\text{84} & -\text{119}^\dagger & -\text{92.31} &-\text{6.02}&\text{Exothermic}\\
\hline
\end{array}
$$
*all units in $[\pu{kj mol^{-1}}]$ all values from Wikipedia unless otherwise stated
** Table of Bond Dissociation Energies
$^\dagger$ Values from NIST WebBook

Here the enthalpy of radicalization of the alkane is different for chlorine or bromine, but the halogenation in total is exothermic


How to dissolve iron powder? I know iron can react with many solutions, but when I tried to mix iron powder together with permanganate or dilute sulphuric acid, the iron powder would not react and instead, it just float on the surface of the solution. I tried to stir it for several minutes but it seems doesn't work.

It is pure iron powder, can be attracted by magnet

How can I dissolve iron powder and let it become $\ce{Fe^2+}$ or $\ce{Fe^3+}$ ? (Just dissolve it is ok)

I did some brief research and I saw that sulfuric acid can dissolve iron:

Fe(s) + H$_2$SO$_4$(aq) → Fe$^{2+}$(aq) + SO$_4$$^{2-}$(aq) + H$_2$(g)

Of course, this would leave sulfate ions and perhaps some precautions would need to be taken because of hydrogen gas production. If you absolutely cant have any sulfate ions, you could precipitate them out by adding barium or silver ions to the solution, but that would require extra work and introduce new ions to the solution. Also note, the iron ions should be aqueous in +2 state but I read that if there is oxygen in the system (which there naturally would be), some iron(II) can oxidize to iron(III).


Is acetone saturated or not? This was a question in a chemistry exam:

  How many moles of hydrogen is required to convert 1 mole of this compound into a saturated compound?
     $\ce{CH3COCH3}$ (acetone)

Some teachers said that it's saturated so the answer is zero.
Others said that one mole is required to break the bond between $\ce{C}$ and $\ce{O}$ and convert it into an alcohol.

So which answer is correct and please mention a reference if available.

Typcially, the terms saturated and unsaturated only apply to the carbon chain in itself. A saturated compound would be one whose systematic name ends in -ane while unsaturated ones would end in -ene or -yne — grossly simplifying. Going by this rather simplistically strict definition, at first sight acetone (propanone) is saturated. But.

You may have heard of the keto-enol tautomery. That is, a ketone is always in equilibrium with its corresponding enol form, for acetone that would be propen-2-ol — an unsaturated compound. It is shown in the scheme below. The first indication of the distinction being difficult.

Furthermore, acetone can be reduced by molecular hydrogen to isopropanol. That means that given the correct conditions, it can basically react much like an unsaturated compound would. It also reacts with elemental bromine similarly to how an unsaturated hydrocarbon would, although only the first mechanistic step is identical; only one carbon is brominated in the product.

There is also the measure of double bond equivalents, that is often used in the process of structure elucitaion. Taking the molecular formula of a compound (which can be deduced well from high-resolution mass spectrometry) and performing a simple calculation allows one to arrive at a number of double bond equivalents for a compound. The formula is:

$$\tag{1} \text{DBE} = \frac{2 \ce{C} - \ce{H} + \ce{N} + 1}{2}$$

Where $\ce{C}$ is the number of carbon atoms (including other tetravalent atoms such as silicon), $\ce{H}$ is the number of hydrogen atoms (including other monovalent atoms such as chlorine) and $\ce{N}$ is the number of nitrogen atoms (including other trivalent atoms such as phosphorus).

However, having 1 double bond equivalent or more does not automatically mean that a molecule is unsaturated, no matter by which definition. Take cyclohexane, $\ce{C6H12}$: It has one double bond equivalent which is caused by the fact that it is cyclic. It will not react with hydrogen or with bromine and no sane chemist would call it unsaturated. So this measure is not of much help.

Concluding, I can only say that both sides of the discussion are wrong. It is wrong that acetone will not react with hydrogen to give isopropanol. It is equally wrong that that makes acetone an unsaturated compound. The concept of unsaturation simply does not extend well to $\ce{C=O}$ double bonds.

The exam question should not be graded.


Why is the boiling point of CH3COOH higher than that of C2H5OH? Why is the boiling point of $\ce{CH3COOH}$ higher than that of $\ce{C2H5OH}$ ? Both are polar molecules held by hydrogen bond.

Yes you are correct. The molecules are polar in nature and are bound by intermolecular hydrogen bonding. But you must pay attention to the extent of polarization in both the molecules. Consider the alcohol. Here the carbon bearing the $\ce{-OH}$ group is the only polarizing group present. But for $\ce{CH3COOH}$ , the carbonyl carbon is polarized by an $\ce{-OH}$ group as well an $\ce{=O}$ group attached to it, thus increasing its effective polarization more than the alcohol. Therefore $\ce{CH3COOH}$ has greater boiling point.