Why does salt dissolved in water taste like salt? If it's just Na+ and Cl- ions When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$. It stays in this form until the water evaporates and then the ions go back to normal $\ce{NaCl}$. So why does water with salt in it still taste like salt? I am asking because if the molecule $\ce{NaCl}$ is broken down into $\ce{Na+}$ and $\ce{Cl-}$ then how does it have the characteristics of $\ce{NaCl}$?

When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different).

There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See here

When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$.

No, not quite. Dissolving something in water does not break the chemical bonds of the molecule*. You still have $\ce{NaCl}$ molecules floating about. That would leave you with elemental sodium and elemental chlorine, and those two fellows are a bit like drunk bikers.

Mixing sodium, chlorine and water will produce all sorts of things like sodium hypochorite ($\ce{NaClO}$, best known as bleach), sodium hydroxide ($\ce{NaOH}$ - caustic soda) and some lesser-known chlorine compounds. Many are rather dangerous, and I don't think we would have survived as a lifeform if a bit of salt on our food turned into bleach (which destroys DNA) after a Perrier chaser.

exception: some molecules will react with water rather than simply dissolve in in it. Depends on a number of factors.


How to explain the periodic trends in boiling points in groups? Observing the trend of boiling points of the compounds listed, choose the appropriate terms to fit into the blanks:
\begin{array}{lr}
\hline
\text{Compound} & \text{b.p.}/\pu{°C}\\
\hline
\ce{H2Te} & −2 \\
\ce{H2Se} & −41.5 \\
\ce{H2S} & −60.7 \\
\ce{H2O} & 100 \\
\hline
\end{array}
The ___ the compound, the stronger the ___ present between molecules. The stronger the ___, the higher the boiling point of the compound?
More polar, smaller, London dispersion forces, larger, less polar, dipole-dipole and ion-dipole forces

I noticed that the electronegativity for each compound increases from top to bottom, so does higher electronegativity implies higher boiling point?
I've also tried putting more polar the compound, the stronger the dipole-dipole and ion forces. No luck, I still got it wrong.

I think it is wrong to say there should be a trend as we study the trends of individual atoms, not molecule and compound itself. Boiling points depend on the intermolecular force which depends on the polarity of the whole molecule, which in turn depends on the constituting species, so there should not be specific trend.
It is only the matter of comparison between the species in a molecule. Moreover, there is not a single type of intermolecular forces which bonds molecules such as dispersion, metallic or hydrogen: all have different strength.


What are glues/adhesives for cardboard boxes/crates made of? Open cardboard boxes/crates, often for foodstuffs, are often glued to hold their shape. What is the glue/adhesive made of?

I ask to know if it is biodegradable, such that it is suitable for composting/gardening.

My preliminary research seems to indicate polyvinyl acetate (PVA), which seems to biodegrade fast enough for my application.

See example box (from CTHOE at Wikimedia):

Sorry if this is the wrong Exchange, it seemed the most suitable for my question.

Because the glue is not in direct contact with food you can use basically any glue. And cardboard is easy to glue so there is no limitation. Packaging should be cheap and therefore I would not expect a biodegradable glue.


Charge on Nitrogen in NCl3 What is the charge on $\ce{N}$ in $\ce{NCl3}$? I want to know if it is positive or negative. I feel it should be positive but I am not sure.

First of all charge of atom is not well defined (maybe except AiM). Formal charge of both is zero, but that does not help us to understand chemistry. Above mentioned reaction shows that nitrogen is more negative, and if you do simple QM computation both Mulliken and NBO gives ~-.44 for N and ~+.147 for Cl. If you want to have a bit feeling about it, think that way: both have similar electronegativity but Cl is bigger so it's easier to polarize.

Charge on an element is given by the oxidation number hence the total charge on the compound is equal to 0 then the charge on nitrogen + chlorine must be equal to 0.

N+3(valency of $\ce{Cl}$) =0

N+3(-1)=0

N-3=0

N=+3

Hence the charge on nitrogen in $\ce{NCl3}$ is +3


Laser cutting polyurethane I have a laser cutting machine and have always read that PVC (or types of vinyl in general) are very dangerous to cut because it releases chlorine gas. Chlorine gas = toxic (plus the $\ce{HCl}$ thing), makes perfect sense. This is well-known information, and it appears all over informational sites about laser cutting materials.

But I now have someone telling me that cutting polyurethane foam is way more dangerous since it produces $\ce{HCN}$ (hydrogen cyanide). Now this information is in a few limited places and mentioned kinda as a footnote, but it is not nearly as widespread as the Chlorine gas risk that is talked about elsewhere. (I know that the chlorine gas presence during cutting of PVC effectively makes $\ce{HCl}$ gas which clearly does damage to the laser system itself which might explain the prevalence of the information about it)

I know that polyurethane is made with isocyanates, which I'm wondering if that is leading to some people seeing that word and getting confused and assuming that there is cyanide in the foam.

What is confusing about this is that polyurethane is generally considered a very "laser friendly" material to cut (stated all over the place on laser hobbyist and professional websites under acceptable materials). Producing the same gasses that were used in gas chambers seems pretty darn dangerous and like there should be red flags up everywhere.

Can anyone here speak to the chemistry of this, firstly how much if any HCN would be produced by using a 10000nm laser (meaning intense heat based, but nonflaming cutting). And if properly ventilated with an industrial air handling system designed to move the full volume of air from the cutter outside many times per minute what are the risks involved? Does HCN bond to surfaces and stick around? Or is it fully ventilated as a gas?

Clearly, I'm not a chemistry expert, so looking for someone who knows more to help me understand what is real and what is just scary sounding jargon.

From SDS:

Special hazards arising from the substance or mixture: carbon oxides ($\ce{CO}$, $\ce{CO2}$) nitrogen oxides ($\ce{NO}$, $\ce{NO2}$ etc.) hydrocarbons, isocyanate vapours and hydrogen cyanide can be released in case of fire.
The polyurethane polymer formulation greatly changes so the fumes that can form at high temperatures (above the melting temperature). A suction and exhaust system is required. To measure the exact composition of the fumes very specific analysis are required.


Applications of magnesium sulfate monohydrate MgSO4 · H2O How is magnesium sulfate monohydrate $\ce{MgSO4 * H2O}$ used for in chemistry and how does it work as a valuable laboratory drying agent?

I could not find too much information in my book or the internet besides the basics (chemical formula, physical or chemical characteristics, etc.), but I did find what it can be used in. For example, fertilizer, feed additives for livestock, nutrient flavor enhancer, pharmaceuticals, leather and fur production, coloring fabrics, fermentation in beer, paper production, sewage treatment, cosmetics, but nothing about how or why chemistry labs would use it. What is its purpose?

Each time I tried to look it up other products that sounded similar but not the same appeared such as $\ce{MgSO4 * 7H2O}$ , manganese sulfate, manganous sulfate, etc.

Magnesium sulfate is frequently used in the laboratory, especially after aqueous work-up.
Aqueous work-up is a common technique in the lab to get rid of residual impurities after completion of a reaction.
For this the organic solution of the reaction is cooled to room temperature and mixed with water. Specific impurities will then diffuse into the aqueous phase and can be separated with a separatory funnel.
Unfortunately some residual water will stay in the organic phase and this can have negative impact for following characterization of the compound.
Therefore magnesium sulfate is added, which is able to catch the residual water in its crystal lattice. After filtration of the solid magnesium sulfate, the water is reduced to a non-significant amount and the compound ready for further investigation.


Meaning of charges in a group? In my class, my professor said that since hydrogen is in group one, it has a charge of 1+. Does this refer to its ionic charge or to its oxidation state?

To go from being confused to less confused, here are some facts.

Hydrogen has one valence electron which can participate in reactions so it is a group 1 element.
Hydrogen is not a group 1 alkali metal, nor is it a group 1 compound, so simply stating hydrogen is in group 1  is ambiguous and confusing.  It is a group 1 element.  The word element needs to be included for the statement to make any sense.
Elemental hydrogen has no charge at standard conditions because it exists as dihydrogen gas, $\ce{H2}$.  If hydrogen gas were to be ionized it could hypothetically break into two ions, one with a +1 charge and one with a -1 charge.  If no ions are involved then the concept of ionic charge is irrelevant.
Elemental hydrogen, again as dihydrogen gas, $\ce{H2}$, has a zero oxidation state.  The oxidation state refers to a hypothetical charge of an element within a compound. If no compound is involved then oxidation states are irrelevant.
Elemental hydrogen has an oxidation state of +1 in most compounds and -1 in metal hydrides because of an agreement within the International Union of Pure and Applied Chemistry (IUPAC). -- http://goldbook.iupac.org/O04365.html --  Oxidation states, as many of the concepts in basic chemistry, should be viewed as a tool to gain a rudimentary understanding of how compounds form.

The utility (usefulness) of the Periodic Table is beyond question. I've used it a couple of times during my career as a chemist - but ONLY a couple of times (in 40+ years). It is unquestionably most useful in school, for those learning the similarities (and differences) between the properties and characteristics of the elements. So, it is a "guide", not a "rule book". What you wrote isn't quite correct - possibly your professor is being sloppy - in a philosophical sense. H is in Group 1 because it usually ionizes to the cation (+1 charged ion) and not the other way around (it doesn't ionize to +1 because it is in Group 1!!). For individual atoms there is no difference between its ionic charge and oxidiation state. When atoms are combined into groups (via covalent bonding (which is by definition (at least partial) sharing of electrons) things get more ambiguous. Even IUPAC confuses the terms Oxidation State and Oxidation number. Ionic charge can be assigned to a single atom in a group, but it almost never is an integer number (some charge sharing is going on). The most straightforward was to assign charge is via crystallographic or quantum mechanical computational methods, neither are useful as you stand at the lab bench with a new compound wondering if it is charged and if so, how. OK, I've digressed a long way from your question. "Oxidation state" is generally done by assuming that all bonds are pure ionic (which is almost never true, virtually all bonds are a mix of ionic and covalent). "Formal charge" assumes just the 'opposite', that all bonds are 50:50 shared between the two atoms (which is only applicable to bonds between identical atoms each of which is in an identical environment (eg H2, N2, and the C-C bond in H3C-CH3). So, the Periodic Table is justified BECAUSE IT WORKS but now we usually explain it by explaining how atoms are built up (Aufbau Principle) and electrons populate the orbitals (maybe with a bow to Quantum Mechanics). Perhaps your professor meant: if you look at the Periodic Table, and see that H is in G1, then you should assume it will bond with an oxidation state of +1. (That is, you predict this based on the PT as a "guide"). As an example of where this works, consider HCl. In aqueous solution it exists, they tell you, as H+ and Cl-. So, the guide works. (I won't mention the fact that H+ doesn't 'exist' in solution and what we mean by H+(aq) is H(+)n(H2O) (n = 1, 2, 3...) nor that HCl in the gas phase is covalent) If you just assume H is ionized to H+, you'll get the STOICHIOMETRY right, which is an important goal for beginners. (in some compounds, H exists as the hydride H(-), this is especially true in the Sun's atmosphere - google it!) So there will be situations where the periodic table lets you down, but you gotta start somewhere. Consider H in H2O. According to your professor, it will be composed of two H+ cations and one O= anion. So, it should conduct one heck of a lot of electricity. But pure water is quite nonconductive (if it is very pure). It is correct to claim that the oxidation state of the Hs in H2O are both +1, it is very wrong to claim that they are both (really) ionized to +1. (It is also correct to say their formal charge is 0, each H has one H-O bond (2 electrons in a covalent bond) and has no other electrons. So,(remembering each covalent bond is split ½:½ for determination of formal charge), 2 * ½ = 1 it has 1 (formal) electron (-1) and its nucleus is +1 giving a formal charge of +1 -1 = 0. ) Let me try to bottom-line this. A particle has an ionic charge. That particle might be an ion or an atom or a molecule (actually if charge is non-zero we could also call it either an ion or a molecular ion). When you rub your feet on a carpet YOU pick up a charge; you've got a charge and NOT an "oxidation state". Oxidation state can be used for single atoms, (in which case its same as ionic charge, so why bother?) but is generally used for atoms in bigger assemblages → which means oxidation state is most useful in dividing up the electrons in a group particle (in a compound ion). The Periodic Table can be useful (as a guide) for both.


What reaction takes place here? Upon reading the given question, I realized that:

The first reaction is an aldol condensation reaction between the given compound and formaldehyde.

So the product after 1st step is:

Now, the second part has thrown me off. The current ketone has no $\alpha$ hydrogen left. Condensation reactions are out of the question.
Cannizzaro reaction can not take place as this is a ketone (and not an aldehyde) also the medium is now acidic.
However, I saw one possibility of formation of hemi-acetals.
(I did not think about full acetals, as such a structure was not present in the given options to this question)

Which gave me the following product:

However, the answer provided seems to be not this.
The given answer is this:

Can someone give a suggestion that where I want wrong and/or how could I reach the answer and also what was wrong in my thinking as well?

This question is taken from the JEE(A)-2016 [Paper - 2]

I think a cross Cannizzaro with the Ketone and Formaldehyde takes place. The ketone give the alcohol and Formaldehyde formiate. In a second step a full acetal will be created with formaldehyd and now the Diole.
Sorry I cannot add comments here. So use answer function.


Dilution of whisky – the molecular perspective Not a chemist. I'm trying to understand this abstract. In particular:

  We found that guaiacol is preferentially associated with ethanol, and, therefore, primarily found at the liquid-air interface in mixtures that contain up to 45 vol-% of ethanol.

So guaiacol is amphipathic which means that is a largish molecule in which one end is hydrophilic and the other hydrophobic (or lilophilic). At least that's my understanding from Wikipedia. So, how did they arrive at the "therefore primarily found at the liquid-air interface" part? At the surface the molecule attaches one end to the alcohol and one end to the air?

Sorry for any naive interpretation of chemistry on my part.

Edit:
Just run across this article discussing the matter.

What the paper says is that in that range of concentrations, that the excess of alcohol at the air-liquid interface is enough to extract some of the guiacol from the bulk solution. At higher bulk concentrations of ethanol, there's no particular benefit for the guiacol to separate to the interface (there's enough ethanol in the bulk to keep the guiacol satisfied, so to speak). I forget whether the paper determined if there was a concentration (proof) below which the guiacol did not segregate, but if there was, it would be because the slight increase in alcohol concentration at the interface wasn't attractive enough. In other words the guiacol only is preferentially concentrated at the interface when the difference between the concentrations of alcohol at the interface to the concentration in the bulk makes it worth-while for it. (Worthwhile here means the difference in energy is large enough). Here's an analogy (My family would now roll their eyes, my analogies are lousy). A little kid is deciding whether to get up from his video game and go trick-or-treating. There's no candy in his house, and maybe he'll get a half dozen or a dozen pieces if he hits 20 or 30 houses/doors. Nah, he says, not worth it. But what if he's in a rich neighborhood? If he hits 10 doors, he's sure to get at least 2 dozen, maybe 3 or 4 dozen pieces of candy. Well, if his home hasn't much candy then he'll probably go, but if his home is chock-full of candy (rich neighborhood, remember?) then he may decide it's not worth it either. The choice he makes depends not just on the difference in the absolute amount of candy he will end up with, but also depends on how much he values an extra couple of pieces. It's non-linear, or as Liberals might say "context sensitive". (added: just to be clear, going trick-or-treating is analogous to concentrating at the air-liquid interface, not going is analogous to staying in the bulk phase.)


nomenclature of citric acid I named Citric acid : "3-hydroxypentan-1,3,5-trioic acid"
but when I searched , I found that the right name is "3-carboxy-3-hydroxypentane-1,5-dioic acid"

Can anyone explain to me why my nomenclature is wrong?

The "-oic" ending means to convert the carbon into a COOH group, not that there's an additional COOH group added to that carbon.

Look at your main chain. You have five carbons, of which carbons 1 and 5 are fully oxidized, hence "pentane-1,5-dioic acid". If the "COOH" group was counted as an independent entity adding an additional carbon, then it would be a three carbon chain, to which was added two COOH groups.

That's why it can't be a 3-oic acid - that would imply that the third carbon of main chain has been oxidized to a COOH group (which is preposterous), not that there's a COOH group that's been added. Instead, the COOH group (carboxy) has been appended to the three position, and the molecule is named accordingly.

2-hydroxypropane-1,2,3-tricarboxylic acid is the correct name. For tricarboxylic acids we are finding longest carbon chain with excluding carboxyl groups. Then we are naming it with all fuctional groups but without carboxyl groups. At the end position carboxyl groups and say tricarboxylic acids.


According to the Molecular Orbital Theory, what makes a compound aromatic? For example, benzene has 6 $\pi$ electrons. Which means the p orbitals will be occupied, and only the bonding molecular orbitals.
But 1,3-cyclopentadiene has 4 $\pi$ electrons, which means 2 of the 3 molecular bonding orbitals will be occupied. Why does this make it non-aromatic?

There's 4 criteria for aromaticity:
1. The compound must be cyclic. As in a ring structure of some kind.
2. The compound must be planar. All the atoms making up the ring must be on the same plane. The hybridization of bonds causes this which is why benzene is planar and cyclohexane is not.
3. The compound must be conjugated. Therefore it must alternate between single bonds, double bonds, lone pairs, or empty orbitals. For example, two double bonds in a row mean it is not conjugated. But a single bond connected to an atom with an empty orbital still satisfies the rules of conjugation. Another example, benzene alternates single bonds and double bonds, so it is conjugated.
4. Must satisfy Huckel's rule. So the number of pi electrons must fit in the equation 4n+2, with n being a whole number or 0. So 6=4n+2 works but 4=4n+2 would mean n=1/2 so that does not work.


Why is salt needed when using vinegar to clean pennies? Let's say you have a solution that is vinegar and it has salt (NaCl) dissolved in it. Then you place old dull pennies in it. Then you look at it 5 minutes later and the pennies are clean and looking like new again.

I understand acid can etch the copper compound but why is the salt necessary for this process and what exactly does it do?

The copper surface is oxidized, which slows down the formation of copper acetate. When you add chlorides to the solution, the chloride penetrates the oxide layer and creates a hole in that layer that allows acetate to react with the copper. To find more information on the effects of chlorides on the surface of copper look up copper corrosion and chlorides.

Copper(II) oxide is basic and reacts with acids to form a copper salt and water.

A strong acid, like HCl, reacts faster than a weak acid - like acetic acid (which is the main acid component of vinegar).

If one adds NaCl to vinegar, you get HCl:

$$\ce{CH3COOH +NaCl<=>CH3COONa +HCl}$$

Copper(II) chloride dissolves in the water.

So, the salt is added to speed up the process.


Is Bond Formation "Strictly" Exothermic? Is bond formation "strictly" exothermic? The IUPAC definition of exothermic doesn't make any reference to bond formation. However, I have seen the aforementioned statement before - that bond formation is "strictly" exothermic.

I suspect that the answer is "it depends" and that "it depends" at least on how accurately bonds are characterized. For example, the first reaction below is endothermic, while the second one is exothermic:

$\ce{2NO -> ONNO}$.

$\ce{2NO_2 -> O_2NNO_2}$.

Lewis structure analysis for the first reaction at best misleads because one might suspect the nitric oxide molecule as having a bond order of 2.

However, the lone electron is actually delocalized. So the actual $\ce{N-O}$ bond order should be 2.5.

In the $\ce{ONNO}$ molecule however there is no delocalization of electrons and the bond order of the $\ce{N-O}$ bond is 2. So the $\ce{N-O}$ bond is weakened, and even the formation of the $\ce{N-N}$ bond cannot compensate for this weakening.

Also, the above example reminds me of certain unstable homonuclear species. Such as $\ce{Ne_2}$, which according to MO theory has a bond order of 0 - i.e. there is no bond. So I suppose that bond formation is not always exothermic. On the other hand, if the bond order is 0, is a bond really "formed"?

Bond formation is alway strictly exothermic in the sense of the change of enthalpy.

  exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.

A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive.

  bond-dissociation energy, $D$ The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.

Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to "Why are noble gases stable" and to answers and comments within.

Sometimes it takes energy to cause a reaction which places the products in a higher energy state than the reactants (think graphite to diamond). This is endothermic and it meets the definition of a chemical reaction since bonds are broken (sp2) and new bonds are formed (sp3). I disagree with @Martin saying that ALL bonds are STRICTLY exothermic. As @DavePhD hints, you need to consider the starting state when discussing whether the reaction is endo or exothermic; if you start with diamond and make graphite, it's exothermic, but if you start with graphite and make diamond, it's certainly endothermic.


Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable? Predict major product:

In this question first I've protonated the OH group and then water is removed to form a carbocation. Now I have to expand the ring, but I'm confused which ring to choose: 3-membered ring or 4-membered ring?

Note:  I can make an alkene (which would be the final product of the reaction) by removing an alpha hydrogen if it is known which ring to expand.

It is clear to me that no matter which ring you open the first, you'd get the same product as A. Michael Lautman gave the correct product , first opening the cyclopropyl ring. Michael's mechanism is fair, but I'm reluctant to say it is the most preferable path because it gives you bicyclobutyl $2^\circ$ carbocation as an intermediate, which needed extra 1,2-hydride shift to get stabilized as $3^\circ$ carbocation (total of 5 steps). However, if you have opened the cyclobutyl ring first, you'd get the same final product A, through a relatively low energy cyclopropylcyclopentane carbocation (total of 5 steps) even though it is also needed a 1,2-hydride shift (in relatively more stable cyclopentyl $2^\circ$ carbocation when compared to $2^\circ$ carbocation in cyclobytyl ring in Michael's mechanism) to get stabilized as $3^\circ$ carbocation. Complete mechanism is depicted below:

This mechanism is supported by following reference:

G. K. Surya Prakash, V. Prakash Reddy, G. Rasul, J. Casanova, G. A. Olah, “The Search for Persistent Cyclobutylmethyl Cations in Superacidic Media and Observation of the Cyclobutyldicyclopropylmethyl Cation,” J. Am. Chem. Soc. 1998, 120(51), 13362–13365 (https://doi.org/10.1021/ja9828962).

The cycloproyl ring may expand as it is more angle strained. So after ring expansion a comparatively more stable carbocation will be formed.


What is the smell of a glass of water in contact with fresh air? When you pour water into a clean glass in a small confined room with no or little fresh air circulation, the only smells you perceive are those of the glass and the water (or the impurities from both of them).

However, expose that very glass to a bit of fresh air, and voilà, it starts to smell different, a bit like the (wet) shells of mollusks in the sea or from a sweetwater source (it is not a "fishy" smell in the sense that is smells like fish but rather one reminding of the sea, lakes or rivers).

I found this smell with wet glass, ceramics, porcelain, but apparently not really or not at all with plastic or metal (maybe the smell of metal somehow obstructs perception, or maybe it alters or destroys the chemical compounds).

So what is this smell actually?
What chemicals are there causing this smell?

Why does it apparently need contact with fresh air (maybe oxygen radicals or something) to emerge? And it seems to be volatile so that the smell weakens with time (or maybe you just get used to it).

I hope at least some here know what I am talking about... if not, maybe try it out.

Some people are apparently more susceptible to this smell than others (a friend of mine is very susceptible to it while others have no idea what this fuss is all about).

It also works when you get the glass hot and fresh from a dishwasher, though it is recommended to try with a non-scented detergent.

And you absolutely have to try it with non-chlorinated water.

Update:
I've come across two substances that may be related to the phenomenon, Trimethylamine and Dimethyl sulfide, however, I don't know how they smell, so I can't really say if any of these are responsible for the odors, but from the Wikipedia articles, trimethylamine sounds more fishy than "mollusky" while dimethyl sulfide is described to be cabbage-like.

I do have a very similar experience! What the OP calls the mollusk smell, I call it slightly rotting "seaweedy" smell. It has been a very unpleasant experience for us in our new home.

Similar to OP's experience, this seems to be more intense with glasses and ceramics right after the newly washed dishes get exposed to air when the windows are open. But I also detect it with plastic cups as noted below.

In order to eliminate other variables such as dishwasher problems and tap water quality, I bought bottled water and disposable plastic cups. I did the following experiment.

WITH CLOSED WINDOWS

Water from bottle poured in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle poured in a cup and couple of sips taken, sitting for for 15 minutes : No smell

WITH OPEN WINDOWS

Water from bottle in a previously unused plastic cup, sitting for 15 minutes : No smell

Water from bottle in a cup and couple of sips taken, sitting for 5 minutes : Seaweedy smell

So I'm about to conclude that something in the outside air combines with the saliva (and/or chemicals/enyzmes from the detergents) to produce the seeweedy smell! The same thing happens with the newly washed dishes out of the dishwasher, and hand washed dishes after they get exposed to air with the windows open.

Yes, I know this doesn't make any sense to people that haven't experienced it before! It sounds like nonsense, but it's quite annoying every time I take sips from water cup.

I'm almost on the verge of calling up the EPA about it. There must be something in the air, but what...

The only smell you're perceiving is evaporated water molecules. Since higher temperature increases the rate of evaporation it makes sense the smell of water vapor is more prevalent when the glass is hot. Fresh air has a much larger affect on your ability to smell than it does on the rate at which water evaporates.

So really it's the fresh air that let's you smell the water vapor more easily not the water actually changing composition and smelling differently.


Why does calcium chloride cause a greater freezing point depression in water than magnesium chloride? So I saw this equation for the freezing point depression, when you add a solute: $\Delta T= K_\mathrm fmi$

Where,

$\Delta T =$ The freezing point depression
$K_\mathrm f =$ The freezing point depression constant (Cryoscopic constant)
$m =$ Molality
$i =$ van't hoff factor.

I read online that $\ce{CaCl2}$ would be better than $\ce{MgCl2}$, but I cannot understand why in the terms of this formula since $i$ for both of them is the same and $K_\mathrm f$ seems to be independent of the solute since I saw values of $K_f$ of water online. (without any solute given)

So, could someone answer why this happens? Because it seems to me that the $K_\mathrm f$ could be the only factor that would explain this. Specifically I would like to know if the charge or the electronegativity of the dissolved ions would make a difference.

As quantified in Fig. 17 of Manual of Practice for An Effective Anti-Icing Program (FHWA-RD-95-202):

By weight percent, MgCl2 depresses the freezing point of water to a greater degree than CaCl2, up to about 24%.  This is consistent with the freezing point depression formula, as Mg has less mass than Ca.

Beyond 24%, CaCl2 is superior.

This is because the eutectic point of MgCl2-water is at 21.6 percent, whereas the eutectic point of CaCl2-water is at 30%.

The freezing point depression formula is only valid at reasonably low concentrations.  It is certainly not valid beyond the eutectic point.

My most recent chemistry lab experiment was determining the effectiveness of various deicers. We tested the freezing point of $\ce{NaCl}$, $\ce{KCl}$, $\ce{CaCl2}$, and $\ce{MgCl2}$. I was told by my professor that $\ce{MgCl2}$ produces an exothermic reaction. When $\ce{MgCl2}$ is dissolved in water, the temperature actually increases. This could be why the freezing point of $\ce{MgCl2}$ isn't as low as that of $\ce{CaCl2}$.


How to cheaply extract iodine from an alcoholic solution? I have an alcoholic solution of iodine. My goal is to extract the pure iodine without using rare or expensive compounds. How could I do this?

Alcoholic solution of iodine

What firstly comes to mind is that you meant "iodine tincture".
This question must have been frequently asked, since answers are available out there with different degrees of reliability. You must first follow the safety instructions:

Do your experiment somewhere that the noxious fumes won't be able to do harm to anyone. It's best to work under a fume hood.
You're gonna be dealing with strong acids. They're very corrosive when in contact with skin, metals or anything you can imagine (except glass).
Iodine tends to sublime. If not contained wisely, you'll get irritated skin, along with damages to your eyes (See about iodine hazards here).

In short, what you should do is:

Pour the tincture in a beaker, or any container. The important thing to note is to have "control" over what's happening in there.
Add water to the solution. The volume of the water you add should preferably be equal to that of tincture.
Add $\ce{HCl}$ to the solution, $\approx \pu{5-7 ml}$.
Add bleach to the solution, preferably equal or more than that amount of $\ce{HCl}$ you added.
Wait...wait...wait. For a relative amount of time, you should leave the solution for the reactions to happen.
You've past the hard part! Congratulations! Now, filter the dilute solution with a coffee filter.

Additional tips:

Iodine crystals sublime, so you shouldn't leave them in open air for long.
You should at least be wearing appropriate eye and skin protection. If you don't want them, do not do the experiment!

Wikihow has additional tips in case of any failures.
Sources:
http://www.wikihow.com/Extract-Iodine
http://www.sci-spot.com/Chemistry/iodine.htm

Your best bet is to evaporate the alcohol from the solution to get Iodine. Iodine boils at $184.3\,^{\circ}{\rm C}$. But the vapor pressure of Iodine causes it to melt anyway. Note: Iodine does actually melt. To resolve this, take a watch glass and put it onto whatever you are using to heat the tincture. This will make the alcohol drip down the center of the watch glass as the iodine deposits/solidifies on the sides. To vent pressure, I recommend you to use a filtering flask and let some of the liquid condense in another flask. You can then re-evaporate. the liquid to get more Iodine. It should be impossible to to get 100% of the Iodine in the solution out, but is is pretty close.


The intuition behind carbon's covalent bonds As an introduction to hybridization, my textbook discusses that "The formation of covalent bonds often starts with the excitation of the atoms." Taking carbon as an example, the author says:

  Carbon  forms four covalent bonds. Yet if we consider the elctron configuration in the carbon atom, we would not predict this as it has only two singly occupied orbitals available for bonding. ($1s^2\, 2s^2\, 2p_x^1\, 2p_y^1$)

I don't understand why carbon's bonding is counterintuitive. Why don't the 4 added electrons just go into the $p_x$ and $p_y$ orbitals then into some newly formed $p_z$ orbital? Why does the existence of "only two singly occupied orbitals available for bonding" complicate things?

As an extension, why is the excitation of the s-orbital electron to $p_z$ needed?

Thank you for your answers and please try to respond keeping in mind that I am a high schooler (more specifically, part of the IB Diploma program).

The most naive answer would guess that since every orbital wants two electrons and there are two orbitals with only 1 electron in each, then 2 more electrons are needed.

However carbon typically combines the one $2s$ orbital and three $2p$ orbitals to form four equivalent $2sp^3$ orbitals each with one electron. The four hybrid orbitals have a tetrahedral shape. So after hybridization, the carbon atom wants to create four bonds to fill the four electron shell vacancies.

My answer is very similar to MaxW's answer, but maybe it can still provide a
bit of insight.

Consider methane as our current molecule:
1 carbon atom reacts with 4 hydrogen atoms to form 1 methane molecule.
Most/Many textbooks explain this by hybridization of three p orbitals
and one s orbital to form four energetically equivalent (degenerate) sp3
orbitals. This is the valence bond model.

"[Carbon] has only two singly occupied orbitals available for bonding."

The author suggests that since only two electrons are available carbon should
form two bonds. This would leave the last carbon p
orbital completely empty, though, and, tbh, this would to me be completely
counterintuitive. To say it in a sloppy way: Chemical bonds/orbitals are
all about being either completly full or completly empty (half-filled is also
possible sometimes). Leaving one of the three p orbitals completly empty is
absolutely not what any chemist would expect in such a situation. To suggest
that carbon would only form two bonds is actually the really counterintuitive
thing here (at least for me). It would even make more sense that carbon loses
its two p electrons in order to have noble gas configuration, but ok... let's
leave that aside

"Why don't the 4 added electrons just go into the px and py and some new pz?"
The thing is that every bonding orbital can contain maximum two electrons. A bond between two atoms is only established, if every atom provides one electron each. If one carbon p orbital is empty then it would not make sense to put in two electrons from hydrogen. If two hydrogen electrons are put in the same bonding orbital it only means that the two hydrogen atoms are now bonded and form an H2 molecule.

So since every electron from the 4 hydrogen atoms needs an electron from
carbon, the best way is to take the two electrons from the 2s
shell. In that way carbon can provide its two 2s electrons for two bonds with
two hydrogen atoms and the 2 p electrons can bind with another two hydrogen
atoms. This will give four bonding orbitals, which are all completely full.


Determine the structure for C6H13Cl using the following data. The H NMR for C6H13Cl states that there are 3 different types of protons:

6H, triplet
4H, quartet
3H, singlet

What will the structure look like?

Here's a useful hint:  when you have a triplet and a quartet with the area ratio given as above, it indicates the presence of ethyl groups.  Since the total area under those peaks is 10 H, there are two of these  groups equivalent to each other.  Only one isomer with the given formula does that.

The molecule you're looking at is 3-chloro-3-methylpentane.

Symmetry causes H NMR to integrate 2x for the 6H and 4H signals.


Difference between Lindlar and Rosenmund catalysts Is there any difference between the Lindlar and Rosenmund catalysts? I've checked around, and it seems the same compounds are used to make both. Is there a difference in their reactivities or are they essentially the same?

This is a rather interesting question because these names actually refer to classes of reactions (specific to certain reagents and products), and aren't constrained by specific proportions of substances or even the identity of these substances.

$\hspace{4cm}$

A Rosenmund catalyst is used to reduce acyl chlorides to their corresponding aldehydes, and is typically composed of palladium supported on $\ce{BaSO4}$, and poisoned by sodium acetate, N,N-dimethylaniline, thiourea, thiophene, dibenzothiophene, ethyldiisopropyl amine, or, most commonly,
quinoline.

A Lindlar catalyst is used to reduce alkynes to cis alkenes, and is typically composed of palladium supported on $\ce{BaCO3}$, $\ce{CaCO3}$, or carbon, and poisoned by lead acetate and quinoline.

The broad list of poisons which cause nearly identical catalytic activity of the supported palladium in both Lindlar and Rosenmund catalysts tells that the poisons are not involved in the catalytic process itself, but rather are affecting the surface of the catalyst. Interestingly, Lindlar-active palladium shows no activity for the conversion of benzoyl chloride to benzaldehyde under various conditions. This demonstrates that despite similar catalyst preparation procedures, the two reactions are catalyzed by different active sites.$^{[1]}$

$[1]$ Mcewen, A. B.; Guttieri, M. J.; Maier, W. F.; Laine, R. M.; Shvo, Y. The Journal of Organic Chemistry 1983, 48 (23), 4436–4438.

Lindlar catalyst: we see it was used in the preparation of alkenes from alkynes, where an alkyne is partially reduced with a calculated amount of dihydrogen in the presence of palladised charcoal $(\ce{Pd/C})$, and it is deactivated with poisons like sulfur compounds or quinoline give alkenes.

Alkyne+H2  Pd/C     R-H-C=C-H-R'(cis-Alkene)
                     --------->

Palladium on barium sulfate $(\ce{Pd-BaSO4})$ is sometimes called Rosenmund catalyst. This is used in Rosenmund reduction, where acyl chlorides leads to the preparation of an aldehyde.

These both catalysts have palladium in them but the compound used with it is different in both above cases. Thus we can conclude that these catalysts look similar, but they are not the same.


Oxygen or carbon with a positive charge? (comparison of canonical structures) Which of the contributing structures of the resonance is more stable?

I'm watching a video lecture by a professor of my college where he puts this question to the class. The class unanimously says B. At first I thought they were wrong, but then the professor agreed with them.

Is this true? The reason he gave was that oxygen's octet would be complete in B, hence it contributes more. But I don't understand this.

1) Isn't the octet complete in A too? It has 2 covalent bonds - 4 electrons (2 from sharing) and 2 lone pairs. Adding up to 8.
2) Oxygen being more electronegative than carbon, would pull the electrons towards itself, leading me to think that it would be more comfortable with a lone pair rather than another bond. Also, won't it prefer to lose out that '+' charge which the bonding's burdened onto it?

Am I wrong?

Another thing which made me believe that Prof. may have made a mistake while talking about the more relevant contributor, is that when he added Br$^-$ to the compound, he added it to the canonical structure A. And don't you explain reactions with the most relevant contributor, i.e. B?

Answer

The presence of an extra bond in the canonical structure B, along with the completion of the carbon valency makes it more stable than A. The reaction with Br$^-$ is best explained looking at A, though, because the incoming nucleophile ends up bound to carbon, not oxygen.
Thanks, everyone!

Which of the contributing structures of the resonance below is more stable?

Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities.

What the question is asking is "Which contributor is more important in describing the structure and behavior of the hybrid?" Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).

Structure A is more stable. Reasons are:

1) The O has a complete octet with 2 electrons per bond around it(based off of octet rule, not formal charge or oxidation state) + 2 lone pairs.

2) The O is not positive. O prefers to be negative if charged at all because of its electronegativity being so high.

3) Carbon is not so determinant on charge. It is fine being positive, negative, or neutral.

4) Each carbon that isn't positively charged has its octet from C-C-C or O-C-C + 2 H's.


Is nitrogen of aniline sp² or sp³ hybridized? I found a thread which discussed this which totally confused me. (refer to the part of the link which talks about $\ce{sp^2/sp^3}$)
The question "How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?" is helpful, but doesn't have aniline specifically. I understand that if the lone pair is in resonance, it becomes $\ce{sp^2}$.
Can someone help me out?

The nitrogen in aniline is somewhere between $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized, probably closer to the $\mathrm{sp^2}$ side.  We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$ hybridized).  However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be $\mathrm{sp^2}$ hybridized).

We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is $\mathrm{sp^2}$ hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is $\mathrm{sp^3}$ hybridized the inversion barrier is typically around $\pu{4-5 kcal/mol}$.

(pyramidal inversion diagram)

In aniline this barrier is very low, somewhere around $\pu{1-2 kcal/mol}$. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being  planar ($\mathrm{sp^2}$) than pyramidal ($\mathrm{sp^3}$).

As the non-bonding electron pair of nitrogen could partially contribute in the resonance with the other  $\pi$ electrons in the cycle. (I will  prove this point indirectly further using an experimental data), the nitrogen atom hybridization is a linear combination of sp2 and sp3.

Now if we consider the nitrogen atom in the pyridine, it  is sp2 hybridized. Because the nitrogen atom has two neighboring atoms (2 carbon atoms) and one non-bonding electron pair, i.e. three neighboring electron pairs. But, the non-bonding electron pair can not contribute in the resonance with the other four $\pi$ electrons in the cycle. In fact, it is  perpendicular to the aromatic $\pi$ system.

The pKa for the pyridine systems is 5.2, for aniline it is 4.6.  So pyridine is a stronger base.  We can link this to the availability of the lone pair in pyridine, as it is not delocalized ( perpendicular to the aromatic $\pi$  system.  While in aniline the nitrogen lone pair can be delocalised using resonance with the adjacent $\pi$ system.


Does hydrogen peroxide exhibit intramolecular hydrogen bonding interactions? We know that intermolecular hydrogen bonding is seen in hydrogen peroxide $(\ce{H2O2})$. But is intramolecular H-bonding, i.e. hydrogen bond between an oxygen atom and a hydrogen atom attached to the adjacent oxygen atom possible? How strong will it be, compared to intermolecular H-bonding and why?

Preamble:

I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives.
I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$.  In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric).  For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital.  I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.

Methods:

The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure A). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions.  The basis set is denoted henceforth as haTZ.  The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$.  The frozen-core approximation was invoked for all computations.
A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density.
This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure B).
All computations were performed with the G09 (Rev. D01) software package.

Results:

Structure A is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure B is a transition state with one imaginary mode of vibration.

The NBO analysis of structure A reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity.  The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.

The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure A.  The occupation number for this antibond, 0.00351, indicates a very small overlap.  The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$.  We note that this is merely an estimated measure of the strength of the interaction.

Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure B), the $n\rightarrow\sigma^{*}$ interaction disappears.  The intramolecular hydrogen bonding  may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure A) as the transition state, structure B, does not contain this interaction.

Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward.  Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex.  Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number.  To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond.   The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$

Conclusions:

An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds.  These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex.
The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.

(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)

ADDENDUM:

I have tried to avoid calling this intramolecular interaction a hydrogen bond.  Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$.  If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond.  Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.

Upon reflection of these issues, I conclude that there is, in fact, no intramolecular hydrogen bonding in H$_2$O$_2$.

My molecular models set has led me to believe that hydrogen peroxide can conform to form 2 bent hydrogen bonds and one bent covalent bond between the $\ce{OH}$ groups.

This means that there can be a triple bond between the 2 oxygen atoms. The triple bent bond (i.e., 3 banana bonds) of which 2 are of the hydrogen bond type and one is the Linus Pauling bent-covalent bond type, like in ethylene ($\ce{CH2CH2}$), tau bonds, banana bonds, bent bonds, and indirect angular bonds.


Pi electron stacking, how does it work? I've come across the term base-pair stacking (with reference to B-DNA) in my school text book, and I had posted a question in that regard on Bio.SE.

I've also seen a similar (albeit brief) version of my question on Chem.SE.

I looked up the term online, and after checking out a couple of links, I'm led to believe that this term stacking refers to the forces of attraction between aromatic rings on account of the de-localized pi-electron clouds on either side of their (the aromatic rings) planes.

Now I'm not sure if I skimmed over some important details when I read up on 'stacking', but I'm unable to find a simple and solid explanation as to how these interactions are brought about.

If 'stacking' is the interaction between pi-electron clouds, then shouldn't these interactions be repulsive (on account of 'like-charges repel')? But I'm told they're attractive interactions.

I thought about it for a bit, and I realized that if you consider the fact the the pi-electrons, despite being de-localized all over the plane of the aromatic ring, can only be found at one particular location at any given instant of time, resulting in a 'partial' negative charge being formed there. The other regions of the electron cloud probably acquire a 'partial' positive charge on account of the fact that the electron is not present there at that instant. It's these opposite, partial charges formed that results in the attraction. (Essentially the same idea that governs London Dispersion Forces)

But the thing thing is I'm not sure if my 'hypothesis' is correct in the present case. If this isn't what happens then I've no idea what makes 'stacking' attractive interactions.

So my question stands: How exactly does stacking give rise to attractive interactions between aromatic ring? Why don't they result in repulsive interactions?

You going along the right track. The stacking attraction occurs when two molecules with $\pi$ orbitals come face to face with one another, typically their separation is 0.34 nm. However, for the interaction to be attractive the two molecules have to be displaced slightly, this prevents direct electron repulsion between $\pi$ electrons on one molecule with those on the other. The attraction is in general between the $\pi$ electrons on one molecule and the $\sigma$ framework on the other$^1$ and vice - versa.

The model assumes that a $\pi$ orbital can be split into three charged parts, separated by a distance $\delta$. The lobes of the orbital have charge -1/2 each and at the position of the atom's nuclei is placed a charge of +1. Thus there are also three interactions $\sigma - \sigma , \pi - \sigma $ and $ \pi - \pi$ and three positions of charge, making 9 interactions in all for each atom. The interaction is taken to be electrostatic, i.e. Coulomb and each term has the form $E_{ij} = q_iq_j/r_{ij} $ for atoms i and j and is summed over all atoms.

In addition a van-der-Waals term is added to the calculation of the total interaction, this has the form $E_{vdw} = \Sigma _{ij}(A_{ij}exp(-\alpha_{ij}R_{ij})-C_{ij}/r_{ij}^r))$ where A, $\alpha$ and C have standard values$^2$. In solution the value of this energy is much reduced vs. its value in the crystal making the $\pi-\pi$ interaction the dominant one.

The charges are shown schematically below for carbon atoms. For other atoms, e.g. nitrogen the charge is given as 1.5 with 1.5 electrons associated with them. (All images are from ref 1 )

charge distribution

optimal geometry for Zn porphyrin $\pi-\pi$ interaction. The molecules are displaced (in x and y) relative to one another at a separation of 0.34 nm.

The contour plot for $\pi-\pi$ interaction and van-der-waals (vdw) energy as a function of their centre to centre offset as shown above (energies in $\pu{ kJ mol^{-1}}$). The black squares are geometries from $\pi-\pi $ stacking in crystals. The shape of the contour plot with or without the vdw interaction is the same as with just the $\pi-\pi$ interaction alone, i.e this interaction determines geometry but  the vdw generally lowers the energy and more clearly show regions of low energy as negative values.

$^1$ The model was developed by Hunter & Sanders J.Am. Chem. Soc. 1990, v112, p 5525-5534

$^2$ Van der waals parameters Caillet & Claverie, Acta Crystallogr. sect A1975, v31, p448

The stacking that occurs between the base pairs in the DNA double helix occurs between a "donor" base, having relatively higher electron density, and a "acceptor" base, having a relatively lower electron density. Such donor-acceptor pairs can be formed, for example between Adenine and Guanine, Adenine being the donor because of an electron donating amine group and guanine being the acceptor molecule, because of a electron withdrawing C=O group. Such interactions contribute significantly to the stability of DNA, along with the usual H bonding and London dispersive forces, as you have observed.


Why P2O5 does not show oxidising properties while N2O5 is the strongest oxidising agent among pnictogen oxides? Stability of oxides of higher oxidation state decreases with increasing atomic number
Then why P2O5 is thermally stable while N2O5 not.
Infact N2O5 is the strongest oxidising agent among pnictogen oxides because N2O3 is most stable.
Please Offer a lucid explaination to understand this...

A very easy half-answer would base this on electronegativity.

Nitrogen’s electronegativity is $\approx 3$ while phosphorus’ is $2.19$. Oxygen’s is, of course, $3.5$. So the electronegativity difference between nitrogen and oxygen is much lower, meaning that oxygen is much less able to stabilise nitrogen’s high oxidation state.

Contrarily, phosphorus’ low electronegativity means oxygen can stabilise its high oxidation state rather well and the compound is much less likely to act as an oxidating agent by itself.

For the same reason you have Na3PO4 (not NaPO3) and NaNO3 (never Na3NO4). Nitrogen doesn't have an orbital to form five bonds, while phosphorus has. This explains why PF5 exists, but NF5 doesn't. In N2O5 n=o bonds have character similar to bonds in NH4+. But in NH4+ nitrogen is surrounded by electron donating H atoms, so it is happy to donate it's lone pair of electrons. In N2O5 it is surrounded with electronwithdrawing O atoms and is not to happy to give away its lone pair. thus it is happy to transfer one oxygen atom elsewhere (in other words - to oxidize something).

tldr N is the only element in the group that doesn't have p orbital.

"NaPO3" is actually a trimer Na3P3O9


How come water would not affect the results of a titration? We were told to rinse the buret with the base so that the concentration won't be messed up, but when it comes to the flask that is going to hold the acid, there is no need to rinse, because it would not affect the result if it has water in it.

Why is that? Wouldn't the acid be diluted a little bit and that throws off the accuracy?

The volume and concentration of base equals the number of moles of base. This is use to determine the number of moles of acid in the flask. Since you are interested in the number of moles of acid in the flask, why would adding water (which doesn't have any acid) change the number of moles of acid?

If you were to calculate the concentration of acid from the titration, and you add distilled water to the erlenmeyer flask, then yes, you would affect the [H3O+(aq)] since increasing volume with same number of moles decreases the concentration of the solute (or analyte) and thus increases the pH of the solution (because pH is -log(hydronium conc.)), ultimately affecting the color of the indicator since the indicator is affected by the pH of the solution. Since you're using the indicator to determine of the titration has reached its endpoint and therefore the predicted equivalence point, if the indicator is affected by the dilution, the entire result of the titration will be afflicted with error if you were to add distilled water into the erlenmeyer.


Why transition metal chloride complex have lower reduction potential than metal aqua ions? Why transition metal chloride complex have lower reduction potential than metal aqua ions? For example:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &-> Au(s)} &\quad E^\circ &= \pu{+1.52 V}\tag{1}\\
\ce{[AuCl4]-(aq) + 3 e- &-> Au(s) + 4 Cl-} &\quad E^\circ &= \pu{+0.93 V}\tag{2}
\end{align}
$$
Is it because the chloride complex formation constant is very high (higher than aqua ions) so the reduction of $\ce{Au^3+}$ is less favored?

The  anion  $\ce{AuCl4−}$ is  the  most  known  complex  of  gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we are dealing with:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &<=> Au(s)} &\quad E^\circ_\mathrm{cathode} \tag1\\
\ce{[AuCl4]-(aq) + 3 e- &<=> Au(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} \tag2
\end{align}
$$
If we substract the equation $(2)$  from the equation $(1)$, we get:
$$\ce{Au^3+(aq) +  4 Cl- (aq) <=> [AuCl4]-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$
The equation $(3)$ is the reaction for the formation of $\ce{AuCl4−}$ with formation constant $K_f$. If you applied the Nernst equation for the equation $(3)$, then at $\pu{25 ^\circ C}$:
$$E^\circ_\mathrm{cell} = \frac{RT}{nF} \ln Q = \frac{RT}{nF} \ln K_f = \frac{0.0592}{n} \log K_f \tag4$$
This last solution of the equation $(4)$ (where $n = 3$) is because, when at equilibrium $Q = K$. If we know the $K_f$, we could have find the $E^\circ_\mathrm{cell}$, and hence $E^\circ_\mathrm{anode}$ (which is for $\ce{AuCl4−}$). Unfortunately, I couldn't find $K_f$ for $\ce{AuCl4−}$ ion formation. Thus, I'd show how big $K_f$ for $\ce{AuCl4−}$ by applying the given values of $E^\circ_\mathrm{cathode} = \pu{+1.52 V}$ and $E^\circ_\mathrm{anode} = \pu{+0.93 V}$ on the equation $(4)$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f \ \Rightarrow \log K_f = \frac{n}{0.0592} E^\circ_\mathrm{cell} = \frac{3}{0.0592} (1.52 - 0.93) = 29.9$$
$$\therefore \ K_f = 10^{29.9} = 7.91 \times 10^{29}$$
It is also in similar way you can find the $E^\circ$ for redox half-reaction of $\ce{PtCl4^2−}$ since we know its formation constant ($K_f = 1.0 \times 10^{16}$):
$$
\begin{align}
\ce{Pt^2+(aq) + 2 e- &<=> Pt(s)} &\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\
\ce{[PtCl4]^2-(aq) + 2 e- &<=> Pt(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} = ? \tag6
\end{align}
$$
If we substract the equation $(6)$  from the equation $(5)$, we get:
$$\ce{Pt^2+(aq) +  4 Cl- (aq) <=> [PtCl4]^2-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$
From equation $(4)$ where $n = 2$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f = \frac{0.0592}{2} \log( 1.0 \times 10^{16})  = 0.474$$
Since, $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.474 V}$, and $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,
$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$
$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- <=> Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$
This value is in good agreement with the experimental value, which is given as $\pu{0.755 V}$ in electrpchemical series of CRC Handbook of Chemistry and Physics.
Source of formation constant of platinum complex ion

This difference of redox potentials is a consequence of Nernst's law. The second equation is nothing else as the first one, where the concentration of the $\ce{Au^{3+}}$ in a solution of $\ce{AuCl4^{-}}$ is extremely low. This sort of comparison could be used to calculate the equilibrium constant of the equilibrium $$\ce{Au^{3+} + 4 Cl- <-> AuCl4^{-}}$$
Here, Nernst's law can be written :  $$\ce{Au -> Au^{3+} + 3 e-}$$
$$\ce{E = 0.93 V = 1.52 V + \frac{0.059 V}{3} . log [Au^3+]}$$ so that the concentration $\ce{[Au^{3+}]}$ in $\ce{AuCl4^-}$ is given by  $$\mathrm{log}\ce{[Au^{3+}] = \frac{3}{0.059 V}(0.93 V - 1.52 V) =  - 30}$$ This means that in a $1$ molar solution of $\ce{AuCl4^-}$, the concentration of the ion $\ce{Au^{3+}}$ is only $\pu{1E-30M}$. But the redox reaction happening in $\ce{AuCl4^{-}}$ is the same redox reaction that occurs between $\ce{Au}$ and $\ce{Au^{3+}}$. Just the concentration of $\ce{Au^{3+}}$ is extremely low in $\ce{AuCl4^-}$ solutions.


Diethyl Ether reaction mechanism I was trying to determine the reaction mechanism behind the production of diethyl ether. Looking at the ethyl alcohol + sulphuric acid method, I have come to the following conclusions for the reaction steps:

The ethanol nucleophilic oxygen gets protonated by the addition of sulphuric acid. This is more favourable because of the high density of electrons surrounding the oxygen. The bond broken between the carbon and oxygen is also needed to form the bond with hydrogen.

Water is then released and the ethyl alcohol conjugate base is formed (H3C-CH2+). The remaining ethyl alcohol molecules can either react with the HSO4- or the ethyl conjugate... Since HSO4- is more stable than the ethyl conjugate, ethyl alcohol reacts with the ethyl conjugate. Carbon is more electronegative than hydrogen and is an electrophile. It is attracted to the oxygen from the ethyl alcohol and replaces the H, thus forming diethylether, H+ and H2O where the sulphuric acid essentially acts as a catalyst!

Am I correct in presenting the mechanism this way? If the sulphuric acid still remains in the solution, why does it not protonate the diethyl ether? Is it because the stability of the diethyl ether conjugate base is unstable?

Thanks

Your mechanism would create a primary alkyl cation, namely the ethyl cation ($\ce{CH3CH2+}$). Primary carbocations are extremely unstable and only generated if there are stabilising effects present such as in the benzylic or allylic case — and these are then often not considered primary cations but rather allylic or benzylic ones.

Rather, the protonation of ethanol ($\ce{CH3-CH2-OH}$) creates an oxonium ion as you suggested in the first step:

$$\ce{CH3-CH2-OH + H2SO4 -> CH3-CH2-OH2+ + HSO4-}$$

The $\ce{CH2}$ carbon is then attacked from behind in a nucleophilic $\ce{S_{N}2}$ reaction liberating water in the same step. It is important to note that the attack of another ethanol molecule and the liberation of water happen synchronously; there is no first or second.

$$\ce{CH3-CH2-OH2+ + CH3-CH2-OH -> CH3-CH2-OH+-CH2-CH3 + H2O}$$

Finally, the oxygen which is now part of a protonated ether loses the proton to give the final product diethyl ether.

In the case of a nucleophilic substitution on a substrate like tert-butanol, or even better, tert-butyl bromide, one would consider the leaving group to be liberated first and a carbocation to be created; however, tertiary carbocations are orders of magnitude more stable than primary ones.

You got it all wrong. First of all the reaction written by you concisely is

$$\ce{diethyl~ ether + H2SO4 -> CH3-CH2-OH + CH3-CH2-HSO4}$$

Your first step is correct that oxygen gets protonated due to acid. Then you said that  water is then released and the ethyl alcohol conjugate base is formed $\ce{H3C-CH2+}$. This is the point where you got it wrong. The reaction goes under SN2 mechanism( I am not going into the details of it) and there is no formation of intermediates as you gave it as $\ce{H3C-CH2+}$ So no carbocation inter mediates are formed only transition stage is formed.
Your next question was for what if $\ce{H2SO4}$ is in excess.
Here it goes like this

Ethanol formed is again protonated by $\ce{H2SO4}$ by the same reason you stated.
Hence now there are $\ce{CH3CH2OH2+}$( where +ve charge is on oxygen) and $\ce{HSO4-}$ anions.
These $\ce{HSO4-}$ again attack on $\ce{CH3CH2OH2+}$ by SN2 mechanism.
Now you will get salt of the 3rd step and water as by product.


What is the meaning of the word root "aceto"? What is the meaning of the word root "aceto"?

I believed that "aceto" meant a methyl attached to a carbonyl group. For example, as in acetamide. Then, why do we name $\ce{CH3C\bond{3}N}$ as acetonitrile, even though it does not have any carbonyl group?

The root "acet-", according to wiki, is derived from latin acētum, meaning vinegar, i.e. diluted acetic acid. Consequently, direct derivatives of acetic acid, like acetamide and acetonitrile are called so. Also, acetone called so because it can be produced from acetic acid (directly or by dry distillation of calcium acetate)

What isn't obvious is why acetonitrile  is considered a derivative of acetic acid and why nitriles in general are considered derivatives of carboxilic acids. Partially it's tradition and partially it's because compounds with carbon skeleton and oxidation state of carbon atoms unchanged are often rather easy to convert into each other.

Acetamide (ethanamide) is an organic compound with the formula CH3-CO-NH2. It is the simplest amide derived from acetic acid.

Acetonitrile is the chemical compound with the formula CH
3-CN. This colorless liquid is the simplest organic nitrile

In organic chemistry, acetyl is a moiety, with chemical formula CH3-CO. It is sometimes represented by the symbol Ac.


What causes electrons to move from zinc to copper? Here is something that no video in youtube about electrochemistry can explain me about galvanic cells.

Suppose we have $\ce{Zn}$ metal immerse in $\ce{Zn^{2+}}$ in one side,  then $\ce{Cu}$ metal immerse in $\ce{Cu^{2+}}$ in the other side. Then, if we link the $\ce{Zn}$ metal to $\ce{Cu}$ metal with some conductor wire, electrons will flow from the $\ce{Zn}$ metal to the copper wire, but this will stop quickly without a salt bridge.

My question is why do the electrons move from $\ce{Zn}$ metal through the wire to the $\ce{Cu}$ metal in the first place? What causes that?

There is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.

From Wikipedia:

  Any junction of dissimilar metals will produce an electric potential related to temperature.

In this case, we don't care about the temperature effects.

When a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode.

So the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves.

I'm not sure what starts the reaction, nor why the zinc dissolves easily.

Electron move from zinc to copper as zinc is more reactive metal than copper and thus more readily loses the electron to turn into zn2+ ion . Thus electrons move from zinc to copper


Zinc screw turning black I tried one experiment. I started with ordinal yellow zinc screws.
The screws were sumberged in bath of two parts of $4.7\%$ solution of $\ce{NaClO}$ and one part $\ce{NaOH}$. The screws turned silverish in color. Then, I submerged them into vinegar. Vinegar turned them black. I am now wondering what actually happened. The bath of $\ce{NaClO}$ and $\ce{NaOH}$ is used to blacken the steel, but it doesn't work with zinc. The vinegar can be used to remove zinc.
But why it was black after that bath?
I was expecting that it would need another bath in $\ce{NaClO}$ and $\ce{NaOH}$ to become black.
Another control screw which wasn't in first bath was just stripped of zinc layer and got a steel color.

These screws are steel with thin cover of zinc with colored oxide layer, probably done by anodizing, I think. At first you dissolved zinc oxide with NaOH. This removed the colored layer and exposed zinc, making it look metallic. Then you reacted zinc with vinegar. Zinc itself is more reactive with acids. The zinc acetate that could be forming is transparent, so that doesnt explain the black color. If black color is present everywhere in solution, it could be that zinc was not pure, and as zinc was being chipped away by vinegar impurities colored the solution black. Or, if black was present only on the screw itself it could be that steel formed iron oxide on its surface. In both cases black is likely non zinc metal oxide. Most likely copper or iron.


What kind of mechanism is adopted here? The relative reactivities of acyl compounds towards nucleophilic substitution is acyl chloride> anhydride>ester>amide.
I would like to know how  this kinetic property is decided. Is it on the basis of better leaving grp or is there another explanation for it?,

It is likely to be due to the different amounts of partial positive charge which exist on the carbonyl carbon.

We can consider for example the rate of reaction of a given amine (such as aniline) with the acid chloride, acid anhydride and an ester such as the methyl ester. The partial positive charge argument would explain this reactivity series.


How energy is released when chemical bond is formed? In books it is commonly written that whenever a chemical bond is formed the energy of the molecules/atoms gets lowered and hence energy is released. This is generally explained by diagrams like this

But I'm really unable to understand this process, I accept that before bonding the potential energy of the molecules were more and after bonding the potential is less and hence this difference of energy must be released but how does this lowering of potential energy happens? Let's just take an example of electrostatic situation (and as far as I know its the electric potential energy only that get's lowered and hence some energy gets released), say I have an electron at the origin and a proton at $x=x_1$ and then somehow it is moved to $x=x_2$ like this

   We know from equation of electrostatics that potential energy of proton at $x=x_1$ is $$ U_1 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_1} \\
\\
\textrm{and at} ~x=x_2 \\
U_2 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_2}$$
Since, $x_1$ is greater than $x_2$ (from figure) therefore $U_1 \gt U_2$. So, we energy at $x_2$ is less therefore there must be some release of energy, but how? If we let the field to do the work then no energy will be released because our proton gonna get the kinetic energy, if we do the work to move it from $x_1$ to $x_2$ then according the to the laws of physics I will do the negative work and hence energy will transfer from the proton to me. So, how does energy ever get released?

How in chemistry the energy gets released by the formation of bonds? I know that we can say without intricacies that potential energy gets lowered and hence the difference of energy is the energy that we get, but how does this happens? If we have some cation and then we put some anion in the reaction mixture then, of course, cations will get attracted towards the anion and hence will gain kinetic energy and according to me that's the only possible way to lower the potential energy. Is that kinetic energy that we measure as heat and say that the reaction is exothermic?

Any help or clarification will be much appreciated.

Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))

In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy $E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at $r_e$, the release of a photon with energy $h\nu=D_0+E_\text{kin}$ will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)

This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore $\text{H}_2$ molecules are unlikely to be formed via radiative association.

Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision.

When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.

You may simply say that energy is needed to break a bond. Breaking a bond is endothermic. So the inverse is exothermic. Energy is released when a bond is formed.

Apparently you want to just discuss what is happening when an electron approaches a proton from far away (x1 in your drawing) to a shorter distance (x2). The electron is supposed to have no kinetic energy at the beginning, in x1. When arriving at x2, the difference in potential energy has been transformed into kinetic energy. When the electron arrives at the x2 position, it has a great velocity. It will not stay here. It will continue its way until it will nearly touch the proton at the abscissa x3 < x2. At this point, all its energy is potential again (no kinetic energy) : it will be repelled back to higher x values. It will go back exactly to the initial point x1, where the kinetics energy was zero, then start again the same oscillation to x3 and back, again and again. The only way of stopping this oscillation is a collision with a third body, to which the electron gives a part of its kineticenergy. This will stabilize its movement. After a couple of collisions, the electron will get to the stablest position at x = 0.5 Å, as it is in a stable H atom.


How do biochemists seperate nucleotides from DNA to add to the PCR How do chemists separate single nucleotides from DNA strands?  How do they add those nucleotides to e.g. a PCR reaction in order to replicate DNA?

I ask this because in my microbiology class, my teacher didn't know the exact answer.

Do you cleave off nucleotides by hydrolysis?  Do you somehow build nucleotides?

You just buy the nucleoside triphosphates from your favorite supplier, they're commercially available.

As far as I know the NTPs you can buy are still produced biologically and not synthesized (see this example from Sigma Aldrich, which states a microbial source). So you grow a large amount of bacteria, lyse the cells to get at the contents and then digest the DNA or RNA with an appropriate enzyme that cuts it into single nucleotides. To get the triphosphate from that you have to phosphorylate the monophosphates again with the appropriate kinase.

According to sigma aldrich the nucleotides used for PCR are now produced synthetically.

The low yield and high cost to produce them biological makes it impractical. When produced synthetically they get a high yield at low cost.

They are added to the pcr mixture in a pre prepared solution and usually transferred via micropippette or a machine.

The new gene can be built from these nucleotides.

Using gene therapy techniques the new DNA can be incorporated into a genome.

One, of multiple techniques, uses an adenovirus vector with the new DNA packaged inside to transport and  insert the new DNA to the cell in the form of a nucleic acid polymer which incorporates the genes into the chromosomes.

Experiments were done to compare the efficiency of the synthetic genes to the biological ones with the need for biological nucleotides as a control. Now they know the gene built from synthetic nucleotides shows the same expression as the biological gene of the same code synthetic nucleotides are used. However microbial produced nucleotides can still be used.

Ref:Wikipedia, Gene therapy. Sigma aldrich - Technical support ph - 1800 800 097


Why do atoms generally become smaller as one moves left to right across a period? It seems to me that the addition of electrons and protons as you move across a period would cause an atom to become larger. However, I'm told it gets smaller. Why is this?

As you move from left to right across a period, the number of protons in the nucleus increases.  The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).

As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons.  These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.

Its all due to the increase in Zeff which is equal to the net pulling force minus the shielding effect.


Calculate the wavelength of the radiation released when an electron moves from n= 5 to n=2 Calculate the wavelength of the radiation released when an electron in a hydrogen atom moves from $n = 5$ to $n = 2$.

How can I find this wavelength?

The Rydberg formula (for the hydrogen atom) is usually given as

$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$

Here, $R$ is the Rydberg constant; take care of the unit ($\mathbf{m^{-1}}$  or $\mathbf{cm^{-1}}$) here! Wikipedia quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.

$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.

So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$

and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the Balmer series, this value is indeed correct.

Start by using Rydberg formula ν~=1/λ=R(1/n2f−1/n2i) =1.10x10^7 m^-1 (1/2^2 - 1/5^2) = 2.31x10^6 this positive value means it is absorbing that energy instead of releasing it in the form of a photon.


How would I find R/S configuration in this case? I know about the rules to assign R/S configurations when the orientation of each bond to the central Carbon atom is given- the whole, give them a priority order based on their Atomic number/masses, move the group with the least priority as far as possible from the eye, and then draw an arrow joining the other groups in order of their priority. If the arrow goes clockwise, it's an R configuration, otherwise it's an S configuration. But how would I do the same when the orientation of none of the groups are given? The answer key says that compounds I and II have R orientation- but when I simply apply the steps above to the  compound I, I get S. How do I do this?
https://imgur.com/a/K57fCJL is the question

While assigning priority in I, $\ce{-CHO}$ is second priority group instead of $\ce{CH2OH}$ as in $\ce{CHO}$ there is a double bond between $\ce{C}$ and $\ce{O}$, so we can consider the duplicate form, i.e. $\ce{C}$ is bonded to two $\ce{O}$ atoms and $\ce{O}$ is bonded to two $\ce{C}$ atoms. So, the rotation is counterclockwise, but $\ce{H}$ is in front of the observer, so we consider the opposite rotation, i.e. clockwise. Thus it is R configuration.

Instead of using the rule that we have to consider the opposite rotation (while moving the fingers from the more priority group to the less priority), when the least priority is in front of the observer. You can visualize it this way, when you see the the compound I, drawn on paper, then $\ce{H}$ and $\ce{OH}$ is towards you and $\ce{CHO}$ and $\ce{CH2OH}$ is away from you and the rotation is counterclockwise. But when you see the same compound below the paper, the $\ce{H}$ and $\ce{OH}$ is away from you and $\ce{CHO}$ and $\ce{CH2OH}$ is towards you, but now the rotation is clockwise. So now we consider this rotation to say that it is in R configuration.

The figures are basically the Fischer projections of the corresponding molecules. The actual representation of Fischer projections is below.

Hence the 'pushing $\ce{H}$ as far as possible' thing is done when $\ce{H}$ is on dashed bond (into the screen). Just use the rules of Fischer projections and change the molecule to the form which you feel most comfortable to determine R/S configurations.

I think you have a problem with assigning priority order in these groups. Actually, we assign priority according to the atomic number. Let's say you have two $\ce{CH3Br}$ and $\ce{CH3Cl}$. Now look upon the first atom of both the group. If both the group have the same atomic number then move to next. If there is a double-bonded atom, like $\ce{O}$ so if it is like $\ce{=O}$, then we would treat it like $\ce{-O-O}$. And then same priority rule according to the atomic number will continue, and clockwise rotation will give R and anti-clockwise rotation will S. If you jump hydrogen while moving clockwise or anti-clockwise, then R is converted to S and S is converted to R. There are some more rules for some typical example. Since you have asked for this question so I haven't mentioned them. You can get the answer.


Which singlet dihalocarbene has the strongest carbon-halogen bond? There are three singlet carbenes - difluorocarbene, dichlorocarbene and dibromocarbene. Which has the strongest carbon halogen bond?

Obviously backbonding is a point to be considered, as carbene is an electron deficient species. As fluorine is the most easiest to overlap with, the $\ce{C-F}$ bond will be strongest.

But I am not able to get that whether it is possible for the singlet carbene to overlap or share electron density with empty d-orbitals on chlorine and bromine atoms?

And which is more probable, backbonding from halogens to carbene empty orbital, or the other way round (from carbene filled orbital to vacant d-orbitals)?

Singlet carbene is zwitteronic, meaning the $\mathrm{sp^2}$ orbital is doubly occupied resulting in a negative $\mathrm{sp^2}$ orbital and a slightly positive empty $\mathrm{p}$ orbital. Therefore it attracts electron on one orbital and repulses them around another orbital. The huge reactivity of the singlet carbene comes from this zwitterionic character.

Below is the picture electrostatic potential of on the left a singlet carbene with doubly occupied sp2 orbital and the right triplet carbene.

It can be argued in many ways how a halogen stabilizes the singlet state of carbene.

Two explanation usually used are:

Electronegative species increase the $\mathrm{s}$ character of σ orbital (Bent's Law).
Electron donation from the electron pairs into the empty carbene $\mathrm{p}$ orbital.

Though the second one is usually used, I suggest you follow the first (more physical). Adding electron negative substituents increases the s character of the doubly occupied sigma in-plane orbital of carbene stabilizing it, significantly decreasing the energy of the system. Following this concept you can see that flourine should provide the maximum stabilization and so on.

$$\ce{F} < \ce{Cl} < \ce{Br} \ldots$$


What about the aromaticity of cyclopropenylidene? Here's a compound which we have to tell whether it's aromatic or not.

Drawing pi-orbital for the 3-membered ring, I think there are only 2 pi electrons delocalised all over the ring. The orbital having non-bonding electrons on the carbon having sp2 hybridization(carbon no. 1) is not having the correct orientation with the pi-orbital of the ring so these electrons are not delocalised over the ring. 2 being a Huckel's number, I think that this compound should be aromatic.

But the answer key says that it is ANTI-AROMATIC! Where am I going wrong? Perhaps the ring is not planar?

EDIT: The carbon no. 1 doesn't have a negative charge so it is not an anion. It is neutral.

This is a problem of counting lone electron pairs as part of the delocalized pi system.  In pyrrole you count the lone pair on the nitrogen, in pyridine you don't.  What's the difference?
In pyrrole you have the lone pair left over after you have given every ring a ligand (the outer hydrogen atoms).  With the sigma bonding requirements thus satisfied, the lone pair goes into the pi system.  In pyridine the nitrogen is missing its ligand, so the lone pair must replace that in the plane of the ring and can't go into the pi system.
Which situation corresponds to your compound depends on what compound is meant, and there seems to be confusion on this point.  Cyclopropenylidene ($\ce{C3H2}$) and cyclopropenyl anion ($\ce{C3H3^-}$) differ only by a carbon-hydrogen bond, and the shorthand method commonly used to render organic structures thus fails to distinguish them.  The problem statement should be checked to ascertain which of these compounds is meant because they give different outcomes, and the answer key is right for only one of them.
If it is cyclopropenylidene
In cyclopropenylidene, the carbon with the lone pair is missing a ligand like pyridine, so the lone pair has to replace the missing ligand in the plane of the ring.  That leaves two pi electrons to be delocalized which makes the ring aromatic.  Cyclopropenylidene is not only aromatic by the electron count, it is a sufficiently preferred three-carbon structure to be seen in interstellar space.
If it is cyclopropenyl anion
Cyclopropenyl anion has all the ligands in place, so the lone pair goes into the pi system like pyrrole.  Unfortunately in this case, adding that lone pair gets you four pi electrons in the ring and that's bad.  In fact it turns out to be really had, even worse than larger (and thus more flexible) formally antiaromatic rings like cyclooctatetraene.
Cyclopropenyl anion goes through some serious contortions to get rid of this antiaromaticity.  See this question.

The answer key is right.
This is a simple compound,planar and conjugated.It has 1 lone pair (carbene) delocalised therefore contains 4 elctrons (2 pi and 2 from LP). Huckels rule for Anti Aromatic compounds say,if a compound is planar,conjugated and cyclic and has 4n pi electrons,properly delocalised then the compound is Anti-Aromatic.
For non aromatic,the compound must be localised or non conjugated.
For Aromatic, 4n+2 pi electrons with proper delocalisation.
Even if there was a cyclopropenylidene anion,it would still be an anti aromatic.
And it would be Aromatic,if it was a cation it would then follow 4n+2 pi where n =0, ie; 2pi system with proper delocalisation.
SO THE COMPOUND IS ANTI-AROMATIC.


Why does the pH of soft drinks get lower as they go flat? I am doing an investigation about the chemistry of soft drinks and as part of this I observed the effect of time on the pH of soft drinks fully expecting the pH to increase as the carbonic acid, a weak acid, decomposed and CO2 escaped. Instead, time after time, the pH decreased, not by much though in fact it hardly made a difference to the acidity of the drinks but it the pH still decreased every time.

So I did a control experiment with soda water and found that the pH did increase when the CO2 escaped the drink. I originally did it with Coke and Pepsi, which aside from carbonic acid also have phosphoric acid (Pepsi has citric acid as well) so the increase in acidity probably has to do with an interaction between the the weak carbonic acid and the strong phosphoric/citric acid. Can anyone help me to explain this?

I think you said it... phosphoric acid. Carbonic acid is "buffering" phosphoric acid... test seltzer water (composed of water and carbon dioxide with no other additives).

Another way of thinking about this is to consider carbonic acid (nominally a weak acid) to actually be a base. Looking at the pKa (the pH at which half of the acid is dissociated) is the best way to describe this.

The pKa of Carbonic acid is 6.35

The pKa Phosphoric acid is 2.16

pH is basically a measure of free (dissociated) H+ in water... so as CO2 is removed, it is not available to act like a buffer and bind with H+... so the free H+ increases.

You can just use the three reversible reactions involved in the amount of carbon dioxide dissolved in a fizzy drink, use Le Chatelier's principle and solve for the changes in pH:
\begin{align}
\ce{CO2(g) &<=> CO2 (aq)}\tag{1}\\
\ce{CO2(aq) + H2O(l) &<=> H2CO3(aq)}\tag{2}\\
\ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)}\tag{3}
\end{align}


Electrochemistry - Understanding Diffusion Potential I have a big problem trying to wrap my head around this for several hours and I cannot seem to make sense of it.
I have to link this scheme with the 3 colored potential profiles on the bottom of the picture. Considering the electrolyte AK, which splits into A- and K+, that it is 10x more concentrated in compartment I (left) and that the 3 different cases are the assumed:

1) Mobility of A- >>> than mobility of K+
2) Mobility of K+ >>> than mobility of A-
3) Both ions have an equal mobility

Now, I know that in the 1) case, anions will diffuse through the membrane and work toward balancing their concentration. However, this will generate a potential difference because K+ cannot move and compartiment I will be left with a higher potential. Now, I can imagine that if we set our reference (left electrode) at an arbitrary number, let's say 0, then we can assume compartiment II being negative potential in respect to I. This leads to the constant value of the Nernst eq. for same metal electrodes and electrolyte with 10x concentration difference of 59 mV. Since II is considered lower potential VS I in case n°1, should I assume that the potential difference at the membrane's level is -59 mV (green profile)? And another +59 mV between the electrodes because of Nernst equation, which would bring to a final measured potential of 0V. Is this a legit reasoning?

Thank you in advance.

Let's discuss the problem in a qualitative way. In such a cell, the metal will get dissolved in the right-hand side of the cell, where the concentration is low, so that $[M^{z+}]$ increases in this compartment, which becomes the anode. In the left-hand side, the ions $[M^{z+}]$ are discharged  and deposited as a metal layer on the electrode, which is the cathode.

The only problem is what is going on in the solutions. If the anion $A^-$ has a high mobility, it will flow from left to right through the membrane to compensate the chang of concentration of the metal ion. And after some time, the two solutions will have the same intermediate concentrations, and the cell will not deliver any current.

But if the ion $A^-$ has a low mobility, the cation $[M^{z+}]$ has a higher mobility and will easily cross the membrane from right to left. It will take a much longer time for the concentrations to be equalized in the two compartments. The reason is that a much smaller amount of ions $A^-$ will cross the membrane from left to right. But at the end, the two solutions will have the same concentrations.


Will strong concentrated acid destroy litmus paper and pH paper if they are dipped in acid? I am curious to know what will happen if I have litmus paper and pH paper, and I dip them into strong concentrated acid like conc. $\ce{H2SO4}$ to test its acidity or basicity. Will acid destroy papers? if YES, then how will I determine the acidity? do I need something else to carry my experiment?

Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.