Determine from pKa if acid or base I'm studying medicine, not chemistry, but I hope you can help me anyway.

I just had to solve a question where it is given that the drug Propranolol has a $\text{p}K_{\text{a}}$ of 9.5 and the un-ionized form is non-polar.

Is there any way to say if propranolol is an acid or base just from the $\text{p}K_{\text{a}}$?

Hope to hear from you, and sorry if it is a stupid question.

Based on the structure, this drug is weakly basic. It has an amine moiety, which if protonated, would likely have a $\mathrm{p}K_{\mathrm{a}}$ value of around 9.5.

We are taught (assuming the $\ce{pK}$ values are about aqueous solutions)
\begin{align}
\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} &= 14&
\text{so} &&
\mathrm{p}K_\mathrm{b} &\approx 14 - \mathrm{p}K_\mathrm{a}.
\end{align}
In this case, $\mathrm{p}K_\mathrm{b} \approx 3.5$, which is not a strong base ($\mathrm{p}K_\mathrm{b} < \approx 2-3$) but it's close. In medical literature a "drug" (i.e. a medicine) is a family of chemical compounds (or mixtures of compounds) where the exact chemical structure is often ignored. Propranolol may refer to the nonpolar compound, or one of several salts (such as Propranolol.HCl). There's no way to know without knowledge of how the drug is actually provided.

Anyone studying medicine should also study $\mathrm{p}K$ values, you may have done that in high school. As far as identifying the organic chemical groups which may ionize, the good news is that there are only a few, so you can learn them. At least, only a few at physiological pH.
Knowing that a group can ionize isn't as good as knowing it does ionize, but it's probably the best you can do from basic chemical knowledge. Experience will help some, but generally you'll have to rely on the literature since the exact electronic (and perhaps steric) effects that contribute are complex and difficult for the non-expert to analyze.


Why is the electronegativity of indium greater than that of thallium? Why is the electronegativity of indium greater than that of thallium?

One possibility which I had assumed was the poor shielding effect by the d and f orbitals which leads to an increase in the effective nuclear charge and hence the anomaly in the 13th group. But with thallium having a f orbital (which indium doesn't) which has a much poorer shielding ability than 'd', why does indium have more electronegativity?

Going down any group of the Periodic Table the general trend is that electronegativity decreases. The relative electronegativities of In and Tl are simply a reflection of this fact. That's the end of the story; you don't need to bring in anything more complicated.

You are correct that the 4f electrons are poorly shielding, and thus the 6p elements (Tl, Pb, Bi, Po, At, Rn) would likely be more electronegative than otherwise expected. But there's no reason why this should make them so electronegative so as to reverse the general trend. None of these elements are more electronegative than their 5p counterparts.

Stuff like these are one of few things in Inorganic Chemistry that have an explanation in High School(unless you are willing to touch research material). Getting to the point this is a result of poor shielding by electrons in the f orbitals. Shielding effect/capacity decreases in the order s>p>d>f.After Indium the electrons are put in s,d,f orbitals and the latter two have little effect in shielding, also considering that the size difference between the two atoms is very small the outer electrons of Thallium are attracted more strongly by the nucleus than in Indium and hence the electronegativity is higher in the former.


When does hydrogen peroxide act as a reducing and when as an oxidising agent? My chemistry books details out both, the oxidizing and reducing properties of hydrogen peroxide, $\ce{H2O2}$.

  Its reducing properties:

  It reduces $\ce{Cl}$ to $\ce{HCl}$
  It reduces $\ce{Ag2O}$ to $\ce{Ag}$
  It reduces $\ce{K3[Fe(CN)6]}$ (potassium ferricyanide) to $\ce{K4[Fe(CN)6]}$ (potassium ferrocyanide)
  Etc.

  Half equations for reduction:

  In acidic medium: $\ce{H2O2 -> 2H+ + O2 + 2e-}$
  In basic medium: $\ce{H2O2 + 2OH- -> 2H2O + O2 + 2e-}$

  Its oxidizing properties:

  It oxidizes sulphites to sulphates
  It oxidizes arsenites to arsenates
  It oxidizes nitrites to nitrates
  It oxidizes potassium ferricyanide to potassium ferrocyanide
  Etc.

  Half equations for oxidation:

  In acidic medium: $\ce{H2O2 + 2H+ + 2e- -> 2H2O}$
  In basic medium: $\ce{H2O2 + 2e- -> 2OH-}$

I was unable to find any pattern in all of the reactions I studied above (except for potassium ferrocyanide, for which the medium triggers the electron addition/removal) to recognize when hydrogen peroxide behaves as an oxidising agent and when it behaves as a reducing agent.

Let us say that $\ce{H2O2}$ reacts with some ion $\ce{Y+}$. If $+1$ is yttrium's highest oxidation state then it will have no other option but to to get reduced to one of its lower oxidation states, and hydrogen peroxide acts as reducing agent. If this is not the case it will get oxidized to some higher oxidation state, and hydrogen peroxide will act as oxidizing agent. This is because $\ce{H2O2}$ is both an oxidizing and reducing agent.

Hydrogen peroxide is a redox substance. Its behaviour as oxidising agent and reducing agent depends upon nature of substance that reacts with hydrogen peroxide. If a substance is oxidisable then H2O2 will acts as oxidising agent.


How can chlorine have an oxidation number of 3 in chlorite? I am studying redox reactions right now and I'm a bit confused about something:
$\ce{ClO2^-}$ is a compound ion and in a compound ion the sum of oxidation state of participating ions has to be equal the charge on compound since $\ce{ClO2^-}$ has a whole charge of $-1$ the oxidation state sum of $\ce{Cl}$ and $\ce{O2}$ must be $-1$, right? Oxygen has an oxidation state of $-2$ so $\ce{O2}$ will have
\begin{align}
2\cdot -2 &= -4\\
-4 + x &= -1\\
x &= -1 + 4\\
x &= 3\\
\end{align}

Chlorine has to have an oxidation state of $+3$.
But if you look at the periodic table chlorine is in 7th period which means it has 7 electrons in valence so it gains one electron while reacting.
Getting oxidized by definition means losing electrons so an oxidation state of $+3$ means chlorine is losing 3 electrons. I just can't get my head around why would chlorine do that? Or am I calculating the oxidation state wrong?

It is indeed in the +3 oxidation state. Keep in mind, though, that that does not mean chlorine lost three electrons. That polyatomic ion is not made up of smaller ions.

Chlorine can indeed have an oxidation state of +3. Infact, it has multiple oxidation state to a maximum of +7. This is because chlorine is able to engage its d-orbital by hybridization, in which an electron in a lower energy level of may be an s-orbital or any other orbital below the d-orbital, reaches an excited state where it is promoted to a higher energy level which is the d-orbital. The d-orbital has an orbital orientation of 5. I hope this helps.


Determination of percentage of Methanol in an Alcohol based hand sanitiser Note: I am looking for advice on a chemical test - not a sermon.
I expect that this is entirely on topic in this forum, downvotes and sermon comment notwithstanding.
More than about 5% of Methanol in alcohol based hand sanitisers indicates that corners have been cut during manufacture and that the product should be avoided.
I'm looking for an easy test to determine the approximate percentage of Methanol in an Ethanol / Methanol hand sanitiser mix. Many hand sanitisers may contain small percentages of Methanol (typically in the 1% - 5% range) and this is acceptable. Anything over 5% I'd like to avoid. Much over 5% (say 10% +) I'd like to be very clearly distinguished. If actual approximate percentage can be indicated, so much the better.
Ideally I'd like a test with either low cost, simple to use, readily available non-hazardous reagents (one can hope) or failing that, something that can be prepackaged for easy use by the unskilled.
In web searches I found a range of tests for Methanol but most were not very quantitative and most were not intended to work in an ethanol-methanol mixture.

Background: Added:
Some guy y'know said:

Using disinfectants outside of a heathcare environment is still nonsense in the year 2020. Using disinfectants of questionable provenience is simply hazardous. Don´t do it.

The use of hand sanitisers is well covered in recommendation documents from regulatory authorities in many administrations and by internationally relevant bodies. Sanitisers are sold under regulatory guidelines which if followed would guarantee their safety when used appropriately. In the US many out of specification brands are being sold which are indistinguishable from those which comply. Several of the largest US sellers have been selling out of spec product. There is no way to determine provenance based on labelling.
While the use of soap and water is often a good or even superior solution it is incorrect to suggest that the use of sanitisers in general applications is nonsense.
In this instance I am interested NOT in their efficacy or application but re their specific bonafides with respect to a crucial component which is known to be used in excessive and illegal amounts in some cases.
I'm an engineer. Sanitisers are sold by a wide range of manufacturers and repackagers. I read labels, note sources, note claimed contents and viscosities and residence times and more.

Russell, agree with your down voting frustration but it is a common problem of any public forum. The bad news is that "simple" distinguishing test do exist, but they generate more toxic products (including a carcinogen) in a so-called iodoform test. Good for school labs only. The problem is not in distinguishing methanol versus ethanol, the problem is the matrix (as analytical chemists call it), i.e., these alcohols are present in a complex mixture.
The only reliable test test to determine methanol in a complex mixture like a hand sanitizer is gas chromatography. You need to consult a professional certified service to ask them to analyze the sanitizer. There is no home-kit or simple test to differentiate the two in a mixture.

It is likely that without sophisticated equipment that one cannot accurately determine methanol concentration.
That being said, my speculation on a possible process, for possible use outside of the lab, could be based on this source outline of relevant chemistry, to quote:

Methanol influences the oxidizing capacity of the atmosphere through reaction with the hydroxyl radical (Jiménez et al., 2003), its main sink, leading to the formation of water vapour and either CH3O or CH2OH radicals, which both react with O2 to give HO2 and formaldehyde (H2CO) (Millet et al., 2006 ). The photo-oxidation of formaldehyde, a key intermediate in the oxidation of numerous volatile organic compounds , leads to the formation of HO2 radicals and carbon monoxide (CO). ...

As such, consider an experiment consisting of placing a test sample of the hand sanitizer in an expandable transparent plastic bag along with some B-12 mixed with 3% H2O2 (which was shortly before made alkaline by adding Na2CO3 until CO2 evolution ceased) in direct sunlight. Periodically shake.
My expectation: The B-12 is a photocatalyst that turns clear with sunlight exposure. Some hydroxyl radicals, along with CO and O2 are produced in the reaction from the breakdown of the H2O2 and importantly, the CH3OH (methanol) in direct sunlight in the presence of B-12. The alluded to HO2 radical above at pH>5 (from the added alkaline Na2CO3) is now in the form of gaseous superoxide ($\ce{.O2-}$).
Perform a parallel run employing ethanol mixed with an ethanol-based hand sanitizer. The chemistry on photolysis differs and is discussed here.
Comparatively, I would expect a faster reaction with methanol over ethanol (accelerated loss of the B-12 coloration) and more/faster gas creation. If I am wrong on the latter point, not material, as it is still likely quantitatively different.
You may be able to use the measured timing of the volume of gas along with the timing of coloration lose, to assess relative methanol presence.
Note: Unfortunately, there is more than alcohol added to hand sanitizer mix (like a moisturizer, perhaps glycerine) which may make the comparative analysis even harder.
[EDIT] Now, some may have suspected that my suggested photolysis experiment was a bit far afield from classical analytical chemistry. Wrong, this experiment is definitely more so, but even greener, likely easy to implement at home, and decidedly fun. Select two (or more) small nearly identical green plants, treat the leaves of one with a glycerine/methanol mix and with the other substitute ethanol. Place each plant in a container which has been fulfilled with a fixed amount CO2 gas, but still should have some remaining air (plants apparently need to breathe O2 at night, but during daylight, on net, produce much more oxygen from CO2) where the very top part of the plant should be exposed to air. Place the open mouth vessel with the plant in sunlight with a thin transparent plastic seal. The plant demonstrating a greater consumption of CO2 and corresponding formation of pure oxyen can be tested by placing a glowing splint into the vessel. Record the seconds and/or the splint's position in the vessel until the burning bright oxygen flame is reduced/extinguished. This may produce a relative metric to judge the relative presence of methanol given its apparent photocatalytic acceleration of a leaf's ability for oxygen formation in sunlight.
Here is a supporting Science Direct source, to quote:

The results revealed that photosynthesis was greatly improved by methanol, as indicated by higher photosynthetic rates and stomatal conductance. The enhancement effect of methanol on photosynthesis was maintained for 3–4 days. Different methanol concentration treatments also increased intercellular CO2 concentration and transpiration rates.

As the methanol treatment is effective for days, repetition of the oxygen generation/sunlight treatment, after replenishing the CO2 supply, may be viable. Also, for very low added amounts of CH3OH, conducting a four-day trial run is recommended as time, assuming a constant rate of photosynthesis, is a multiplicative factor.
Finally, as to whether this can actually work, in practice, and could quantify variations in the concentration of CH3OH, remains to be demonstrated, AND I PLAN TO TEST IT.
By the way, the inspiration of my original photolysis path was based on my acquaintance in the photolysis literature of the reputed abilities of methanol, so the latter photosynthesis reported results with CH3OH should not be too surprising to some.


Why is double bond stronger than single bond? If a pi-bond is weak due to its sideways overlapping, then how can we say that breaking a double bond is a bit difficult? Is it because of the slight increase in energy as compared to a single bond to break it?

A double bond is composed of a π-bond and a σ-bond. In order to break a bond such as this you must provide the molecule with the energy needed to break the σ-bond alone plus the one needed to break the π-bond alone. In other words, you must provide the sum of those two amounts of energy. Thus, it goes without saying that the sum is a greater number than the two energies alone.

Double bond is more stronger than single bond because,

Energy required to break double bond is 614 J while in breaking single bond is 349 J, thus the energy to break double bond is more than single bond so it is stronger than single bond.
Double bond is restricted to rotation while single bond rotate freely so, double bond is stronger.
Bond length in double bond is 1.34 angstrom while in single bond is 1.54 angstrom  where bond length is inversely proportional to bond energy so less bond length, more bond energy is required to break thus it is more stronger bond.


Where does the energy of solar radiation hitting the atmosphere go? I was reading about solar radiation, and there was a part that says that the atmosphere absorbs most of the radiation emitted by the sun. I wonder if when the atmosphere absorbs the energy, where does it go since energy doesn't disappear or that's what I've been taught in high school. I'm guessing it transforms to heat. Correct me if I'm wrong, and a explanation would be very appreciated.

It is broadly true that some solar radiation ends up heating the atmosphere. But it isn't true that the atmosphere absorbs most of the heat emitted by the sun.

Broadly there are several ways that radiation is absorbed:

Some of it (but not much of it) is absorbed directly by certain molecules in the atmosphere (carbon dioxide and water vapour are biggies), directly heating them.
Some of it is reflected by clouds and ice and other parts of the earth and reemitted to space (quite  a lot of it)
Some radiation is absorbed by plants and used to drive photosynthesis (important but not a lot of absorbed radiation)
A lot of it is absorbed by the surface of the earth (land and sea) which is the part that contributes most of the warming effect of sunlight (of course this warmer sea and earth then warms the atmosphere above it either directly be contact or by being absorbed before it is transmitted back into space, though a lot of it is emitted to space).

So, in the end, a lot of the radiation ends up as heat, but mostly via heating of earth and sea not by directly heating the atmosphere. On the other hand, the atmosphere does end up much warmer as a result. And the absorption by the atmosphere does make the difference between the earth being very cold or very habitable as a planet. The details of this are very complicated which is why people build very big computer models when they want to understand global warming which depends of the complicated details of how the atmosphere absorbs fractions of the heat.

Not sure why you think this is a chemistry problem: see https://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif, for example. Note that this energy diagram ignores both biological energy flow (e.g. photosynthesis, decay, burning) as well as geothermal energy flow (e.g. volcanoes) as well as processes which take hundreds, thousands, and millions of years (cooling of core of Earth). An isolated system is one which, by definition, energy is constant (although the form that that energy takes may and often does change). In common experience, the Law of Conservation of Energy and Mass holds. In more exotic situations we have to include strain or density as also part of the conserved quantity, but unless you're dealing with relativistic velocities, or gravities far heavier than that found on Earth, or with densities far greater than those found on and within several hundred kilometers of the surface of the Earth, you can ignore that and just remember that Mass is energy, and energy is conserved. The other exception is at cosmological scales where space is expanding but our best models suggest that each cubic meter of vacuum (of space) has a constant energy density. This means that as space expands, the total energy of our Universe is increasing. This is theoretical, and is disputed among experts in the field.


How to obtain calcium carbonate? I read calcium carbonate is the main component of seashells and other animals' shells (such as snails), and also limestone. How can I prepare pure $\ce{CaCO_3}$?

Could someone describe the process and the chemical equations involved? I saw this video and apparently that's one way. Are there others? For example, is there a way to obtain calcium carbonate from bones?

Please do tell me about the risks, but don't worry about my safety, since this is not going to be done in real life, but in a Minecraft mod instead. It is important to know the risks though, because I would like to make those risks part of the game, to make very clear for my players that, if its risky, the same stuff that happens to Steve/Alex will happen to them. (I intend to make it realistic gruesome, meaning I won't exagerate a bit, but either will I minimize the risks because it's "just a game". I'm 14, so I know better, but I know Minecraft is played for kids as young as 7, so I really don't wan't little toddlers thinking Chemistry is all fun and games).

Most of the $\ce{CaCO3}$ in commercial use today is mined not synthesized. The one exception I know a bit about is the ultra-small particle sized materials. However, few of those are pure, they have surface treatments which constitute a significant portion of their total weight. Pharmaceutical and perhaps food grades are also synthesized.

The major synthetic pathway is circular: limestone is mined, then it is heated which produces lime and carbon dioxide. The lime is then mixed with water and $\ce{CO2}$ is added (bubbled in/through) and $\ce{CaCO3}$ precipitates. It is then dried.

Purification depends on the source; how impure it is to start with. It is generally difficult to separate magnesium carbonate from calcium carbonate, and given that limestone and aragonite as well as dolomite ($\ce{CaMg(CO3)2}$) are taken out of the ground (i.e. are 'dirt' cheap) there's usually no need.

As far as the risks of making it, well you could burn yourself, adding lime too fast to water could splash and it's infamous for dissolving bodies in murder mysteries, as well as being used in real life to help in outhouses and gardens where organic material is desired to rapidly decompose. Meaning you don't want to get it in your eyes, nor do you want it left on your skin for very long. It can, theoretically, fume, while being decomposed and that wouldn't be good to breathe. And if a centrifuge or mechanical press is used to squeeze the water out of it (first step in drying), then you have the assorted mechanical dangers. Its synthesis is probably one of the safest chemical syntheses you could undertake.

$\ce{CaCO3}$ decomposes with fairly mild heating (I think around 840 °C), so any material which is predominantly that (like seashells, bones) could be used to make the lime.

Bones are mainly made up of Collagen and Calcium, in the form of Hydroxyapatite (Ca10(PO4)6(OH)2), though there's traces of other minerals, such as Mg, Bo, Si, Al, etc.

If you put bones in Hydrochloric acid (HCl), you'd get a solution predominantly composed of Calcium Chloride, via the same reaction occurring in nature (which is the way osteoclast dissolve Calcium from our bones in the context of bone remodelling).

Be careful as (1) HCl is dangerous to handle and can cause severe burns to eyes, skin, and lungs, especially if using a concentrated solutions thereof / in a closed environment, (2) The reaction will release H2 which makes up for an explosive mixture with Oxygen in air.

From that solution you could precipitate calcium carbonate by adding Na2CO3, via the following reaction:

Na2CO3(s) + CaCl2(s) + H2O(l) ⤑ CaCO3(s) + 2 NaCl(aq)

NaCl is soluble so you could just rinse it away (poor your solution & precipitate through a coffee filter for instance and rinse with water).

The result would be CaCO3, probably at min >98% purity, though you'd have traces of a few other carbonates in there.


How can I tell if a molecule is polar or not? Consider $\ce{PCl3}$, $\ce{CH2Cl2}$, $\ce{HCN}$, and $\ce{SiO2}$. Which of these molecules are polar substances?

I appear to be lacking on the concept of "polar substance". I was just studying about Lewis structures and this question popped up in the book, but I'm not quite sure how is this related.

I think that a chemical bond is polar when one atom is more electronegative than another. But my basic understanding tells me that a chemical bond is non-polar only when the two atoms are the same (like $\ce{F-F}$ or $\ce{O-O}$). But all the above options have different atoms, so the answer would be "none", which is clearly wrong because that's not one of the options.

How can I tell if a molecule is polar or not?

Every sufficiently asymmetric molecule will be polar, but some more than others

The polarity of molecules is related to the polarity of bonds within the molecule, but just having polar bonds is not enough to create a polar molecule.
 Consider, for example, CCl4 and CHCl3. Carbon tetrachloride has 4 fairly polar bonds but they form a regular tetrahedron and the polarity of the individual bonds cancel each other out to leave a molecule which has no net dipole moment and is non-polar. But the chloroform molecule (which swaps a hydrogen for a chlorine) is fairly polar (~1.1 Debye–the unit used to measure dipole moments–usually abbreviated to D) because the symmetry is lower and the three C-Cl bonds now all point away from the C-H bond and their dipoles don't cancel out.

Similarly carbon dioxide is a linear molecule so the dipoles of the (fairly polar) C=O bonds point in exactly the opposite direction and so cancel out. Water, however is bent so the dipoles of the polar O-H bonds  do not cancel out leaving a polar molecule with a significant dipole moment (~1.8 D).

Saturated hydrocarbons are usually thought to be non-polar for most meaningful purposes. But those that lack symmetry still have measurable (but tiny) dipole moments. Propane $\ce {CH3CH2CH3} $ and isobutane $\ce {CH3CH(CH3)CH3} $, for example, both have small but measurable dipole moments (0.081 D and 0.132 D) so are, strictly, polar. In practice, though, these molecules have such low polarity that the effects of their net dipole are dominated by other non-polarity related intermolecular forces and can be though of as non-polar.

But the critical message is that the most fundamental feature that determines polarity is symmetry or the lack of it.

Also note that we are talking about discrete molecules here. One of your initial examples, $\ce {SiO2}$, is a network solid not a discrete molecule and the ideas won't apply.

To find answer to the question that you encountered you can follow these steps

Step 1 : 1st you have to understand what is hybridization and how to find hybridization of molecules , also each hybridization has different structure .

Step 2 : Once you have determined the structure successfully you can use the formula for calculating dipole moment

μ = q*d  and this formula shows dipole moment for a single bond .

A more generalised form :

Here are some examples which can help :

For multiple bonds use concept of vector addition/subtraction.

Remember that polar compounds means  the compounds that have a net dipole moment , but Polar Solvent means that the compound has high dielectric constant .


Is vapour pressure of a liquid solution constant at a given temperature, no matter the size of closed container and amount of liquid taken? Does vapour pressure of a liquid solution depend on the size of closed container, amount of solution taken, given that, temperature is kept constant? Or will it change, if some extra gas is added to the container at constant temperature?

The equilibrium vapor pressure of a compound at a given temperature is an intrinsic property of the compound, meaning it does not depend on variables like volume of the container, quantity of the compound, etc.  Other examples of intrinsic properties include density and melting point. None of these intrinsic properties depends on the quantity of the material or it's physical surroundings.

YES, the vapor pressure will change if gas is added to a closed system.

Vapor Pressure is Temperature and Pressure dependent. So adding gas to a closed system increases the pressure, in turn changing vapor pressure.


Why is the H−C−H bond angle in сhloromethane larger than 109.5°? Symmetrical tetrahedral molecules (like $\ce{CH4}$) have a bond angle of $109.5$. Those with lone pairs in place of one atom (like $\ce{NH3}$) have bond angles less than $109.5$.

The standard explanation for why is that the lone pairs are "larger" than the hydrogens, creating repulsive forces that push the hydrogens down, decreasing the bond angle.

However, this explanation fails when confronted with $\ce{CH3Cl}$. According to the above logic, the $\ce{Cl}$, which is much bigger than the hydrogens, should repel the hydrogens just like a lone pair, making the $\ce{H-C-H}$ bond angle smaller than $109.5$. However, the $\ce{H-C-H}$ bond angle is actually $110.5$ (see here, I could not find a more authoritative source).

Why is this (or is my source wrong)?

Chlorine is more electronegative than carbon which is further more electronegative than hydrogen and thus chlorine pulls the bond pair towards itself, polarizing the bond. This decreases the electronic repulsion between the $\ce {C-Cl}$ bond pair and the $\ce {C-H}$ bond pair, decreasing the $\ce {H-C-Cl}$ bond angle. Consequently, the $\ce {H-C-H}$ bond angle increases to 110.5°.

Cl is more electronegative than C and H. Hence Cl will pull the electron toward itself. Again the C is also more electronegative than H. Hence C will pull the electron towards itself. Hence repulsion from the Cl side will less. The repulsion between bonds of C-H will be more and hence the angle


Do 1 mole of both dioxygen gas and its atoms occupy 22.4 litres at STP? The ideal gas law states:

  One mole of an ideal gas will occupy a volume of $\pu{22.4 liters}$ at STP.

Does this mean that both

$\pu{1 mole}$ of $\ce{O}$  would occupy $\pu{22.4 L}$ (or if this doesn't usually occur in nature, say \pu{1 mole }of $\ce{He}$ or any another monoatomic gas)
$\pu{1 mole}$ of $\ce{O2}$ would occupy $\pu{22.4 L}$

I am assuming the answer is yes based on:

reading the ideal gas law definition, it wouldn't make sense if it was true only for $\ce{He}$ but not for $\ce{O}$.
in class when we are finding the volume of e.g. $\pu{100 grams}$ of $\ce{Cl2}$, it was always $\frac{\pu{100 grams}}{(2\times\pu{35 grams} \text{[molar mass of $\ce{Cl2}$]}} \times\pu{22.4L}$  and not $\pu{1\times35 grams}$

But why? Wouldn't $\ce{O2}$ contain twice the amount of $\ce{O}$ matter and hence take up twice the volume?

Does this mean that both

1 mole of $\ce O$ would occupy $22.4~\mathrm L$ (or if this doesn't usually occur in nature, say 1 mole of $\ce{He}$ or another monoatomic gas)
1 mole of $\ce{O2}$ would occupy $22.4~\mathrm L$

Yes, it means exactly that.  And you're right, a stable gas of $\ce O$ atoms is a pretty exotic thing, so $\ce{He}$ is a much better example.

Wouldn't $\ce{O2}$ contain twice the amount of $\ce O$ matter

Yes and no. The $\ce{O2}$ gas would indeed contain twice the mass of the $\ce O$ gas, but it would contain the same number of particles ("molecular entities") as the $\ce O$ gas.

and hence take up twice the volume?

Nope.  There is a lot of empty space in gases. A lot. So much, in fact, that molecules have to get pretty darn large in the gas phase$^\dagger$ before they start to get "cramped" and "insist" on taking up more space (viz., occupying a larger volume).
To give you an idea of just how much room there is, consider what's called the van der Waals volume of $\ce{O2}$.  This is an estimate of the total volume actually occupied by one molecule of $\ce{O2}$, and has a value of about $0.053~\mathrm{nm}^3$ (that's 'cubic nanometers').  For one mole of $\ce{O2}$ molecules, that works out to just under $32~\mathrm{cm^3\over mol}$ of volume actually occupied by the oxygen molecules themselves.
However, as you note, that mole of $\ce{O2}$ gas actually expands to fill a full $22.4~\mathrm L$ of space at STP. So, that space is actually only about ${32~\mathrm{cm^3}\over 22.4~\mathrm{L}} = 0.14\%$ full of $\ce{O2}$.
So, even though one $\ce{O2}$ molecule does take up more space than one $\ce O$ atom, compared to the amount of space available, they're both really, really tiny.  And, thus, the fact that the two $\ce O$'s in $\ce{O2}$ are bound so closely to one another means that they both behave essentially identically in the pressure they exert when they fill up a container of a given size.
$^\dagger$ At ambient temperature and pressure, anyways.

Yes. Because STP is the standard temperature which is 0 degree celcius or 273 K nad standard pressure which is 760 mm Hg. So, if there's no given of volume, you can use the standard volume which is 22.4 L. No matter what kind of element is that, if we say STP, the number of volume at standard is 22.4 L


Which has more metallic character: aluminium or magnesium? Which element has more metallic character - aluminium or magnesium?

I know that metallic character decreases along a period (from left to right) and increases down a group.

Aluminium comes after Magnesium in the same period (period 3) in the periodic table. So, I thought Magnesium would have greater metallic character.

However, I read in a book that Aluminium has greater metallic character. Is this true? If this is true, why?

Two types of metallic character
In fact, there are two type of metallic character if you look at the metal from the chemical point of view or if you look at the metal from the physical point of view:

So it really depends on how do you define metallic character

Chemical metallic character
Since metallic character in chemistry is defined as:

the tendency of an atom to lose electrons.

We can understand that it is directly related to electronegativity.

chemical property that describes the tendency of an atom to attract
electrons (or electron density) towards itself

Basically, it is the opposite (that would be Electropositivity...)! An so it increases from right to left. The picture below shows the electronegativity but can be used for the purpose.

I think the reasoning that the book wants you to do is not remembering all the subshell but just the basic chemistry principles.
So from the chemical point of view $\ce{Mg}$ has a greater metallic character compared to $\ce{Al}$.
Physical metallic character
The physical metallic character is defined taking in account the physical properties of the metal. A metal, from the physical point of view, has a high electrical conductivity, high thermal conductivity, and high density.
These properties are harder to predict:

In this case, however, $\ce{Al}$ has a more metallic character compared to $\ce{Mg}$.

As metals can easily remove an electron .Aluminium is more metallic than magnesium because the last electron in Mg enters in s orbital  and in Al it enters in p orbital and s orbital is more penetrating towards nucleus so it is difficult to remove  an electron in Mg.


How can either reactants or products be "favored" at equilibrium? From Atkins' Elements of Physical Chemistry (6th ed.):

  $$\Delta G_\text{r}^\circ = -RT \ln K \tag{8}$$
  This fundamentally important equation links thermodynamic quantities – which are widely available from tables of thermodynamic data – and the composition of a system at equilibrium. Note that:

  If $\Delta G_\text{r}^\circ$ is negative, then $\ln K$ must be positive and therefore $K > 1$; products are favored at equilibrium.
  If $\Delta G_\text{r}^\circ$ is positive, then $\ln K$ must be negative and therefore $K < 1$; reactants are favored at equilibrium.

It says products/reactants will be favored at equilibrium, but I thought at equilibrium neither are favored?

At equilibrium the rates of the forward and reverse reactions are equal. This means that you can have more product than reactant, or more reactant than product. It just means that the rates of formation of each are equal and so there is no net change in the amount of each. The amount of products and reactants at equilibrium can be favored, their amounts just won't change once they reach equilibrium.
Depending on the temperature and pressure, the composition of the system at equilibrium can change.
Recall that in the reaction $$\ce{aA + bB<=>cC + dD}$$
The equilibrium constant is: $$K=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$
These concentrations refer to the concentrations of $\ce{A}$ and $\ce{B}$ at equilibrium.
Therefore in a situation where $\ce{A<=>B}$ and $\Delta G_\mathrm{r}^\circ\lt0$, then $K>1$, meaning that there is more of $\ce{B}$ than $\ce{A}$ at equilibrium (and vice versa).

If you are interested to know why $\Delta G_\mathrm{r}^\circ = -RT \ln K$ gives the standard change in Gibbs energy for a reaction, then you should note that the general form of the equation at any point in the reaction is given by
$$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + RT \ln Q$$
Where $Q=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ at any point in the reaction. At equilibrium, there is no more Gibbs free energy, as the system no longer tends to the formation of products or reactants. Therefore at this point $\Delta G_\mathrm{r} = 0$, and the original expression given in Atkins is what you get. Also we define this special case of $Q$, saying that at equlibrium $Q=K$, the equilibrium constant.

Source
You might find the above image helpful. The y-axis shows the amount of Gibbs free energy, $G$. The reaction will want to tend towards the state where it has the lowest amount of Gibbs free energy (the local minimum). The slope of the curve can be therefore seen as $\Delta G_\mathrm{r}$. Therefore the point where the system has the least Gibbs free energy will be when $\Delta G_\mathrm{r} = 0$ could occur anywhere in the extent of reaction. This shows that the equilibrium concentrations need not be equal, but shows that the reaction will not progress any further when $\Delta G_\mathrm{r} = 0$.

It's not at equillibrium; it occurs when we are going to achieve equillibrium. More the value of $K$, the reaction will get driven towards the product side and more amount of product is formed easily (as more negative value of $\Delta G$ suggests). If $K\lt1$ then $\Delta G$ becomes negative and the reaction will drive towards reactants side, opposite to what happens for $K\gt1$ case.


Solutions of Group 1 and Group 2 metals in Ammonia When we add Group-1 and Group-2 metals to liquid ammonia, they dissolve to form metal cations and solvated electrons.
$$\ce{M + NH3(liq) -> M+ + e-}$$
Now, when the G-1 solutions evaporate, we get the metal back, but in the case of G-2 solutions (except Beryllium), they give a metal complex, $\ce{[M(NH3)6]}$.

"Evaporation of the ammonia from solutions of Group 1 metals yields the metal, but with Group 2 metals evaporation of ammonia gives hexammoniates $\ce{[M(NH3)6]}$ of the metals. These slowly decompose to give amides."
Page No. 335, Chapter 11-'Group 2 - the alkaline earth elements', Concise Inorganic Chemistry

and,

"The alkaline earth metals except beryllium form similar solutions [to alkali metals] readily, but upon evaporation, a solid 'ammoniate', $\ce{[M(NH3)x]}$ is formed."
Page No. 249, Chapter 9-'Chemistry in Aqueous and Nonaqueous Solvents', Inorganic Chemistry: Principles of Structure and Reactivity

What is the reason for this contrasting behaviour, and do all Group 2 metals(except beryllium) exhibit such a property?
Source:
1. Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998
2. Huheey, James E., Keiter, Ellen A., Keiter, Richard L.; Inorganic Chemistry: Principles of Structure and Reactivity, 4th Edition; Pearson Education Inc.; 1993

As per P. Bahadur's Objective Chemistry 1st Year Program for IIT-JEE and All Other Engineering Entrance Examinations, under the chapter "s-Block Elements - The Alkaline Earth Metals", under Solubility in Liquid Ammonia, it reads:

  Like Alkali Metals, alkaline earth metals also dissolve in liquid ammonia and form coloured solutions. When such a solution is evaporated, hexa-ammoniate, $\ce{M(NH3)6}$ is formed, which slowly decomposes to amides.
  $$\ce{M(NH3)6 -> M(NH2)2 + 4NH3 +H2}$$

(I noticed that it is almost exactly the same as JD Lee's Concise Inorganic Chemistry text.)

So far, it doesn't specify whether this metal $\ce{M}$ is only Beryllium or is the representation of the other alkaline earth metals as well... Since this is the case, I can safely assume that this applies to all the alkaline earth metals, and not just beryllium.

Another point mentioned in the book by Bahadur, under the chapter "s-Block Elements - The Alkali Metals", under Complex Ion Formation is that the Group 1 metals, i.e., the alkali earth metals have a larger size, low nuclear charge and hence don't tend to form complexes too easily. On the other hand, the Alkaline Earth metals are smaller, have higher nuclear charge, and hence, have a greater tendency to accept electrons, forming complexes. In Bahadur's Book, under "s-Block Elements - The Alkaline Earth Metals", under Formation of Complexes, he writes that:

  However, $\ce{Be^{2+}}$ on account of smaller size forms many complexes such as $\ce{BeF3-}$, $\ce{BeF4^{2-}}$ and $\ce{Be(H2O)4^{2+}}$. Of the others, only $\ce{Mg}$ and $\ce{Ca}$ show much tendency to form complexes in solution and these are usually with oxygen donor ligands.

$\ce{NH3}$ is a ligand, and hence acts as a powerful complexing agent. This is why the $\ce{NH3}$ coordinates with these metals, forming hexa-ammoniates.

Source:

Bahadur, P.; Objective Chemistry 1st Year Programme for IIT-JEE & All Other Engineering Entrance Examinations; GRB Books; 2016
Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998

I can't answer your two questions. As to your first question, which if I'm understanding it, is "Why are Group I and Group II metals different?" - I don't understand it. Why are red rubber balls and blue rubber balls different? Why are giraffes and crocodiles different? We actually can get into the details of what their differences are, but that's so obvious as to suggest that that is not what you have attempted to articulate. As a broad rule, Science isn't very good at answering "Why" questions, it is much better at answering "How" questions. I suspect you know why I and II elements are different. While I am not a big fan of the Periodic Table, we can see that I elements have an s electron, and II elements have two (full sub-shell) s electrons (in their valence shells). Would you predict, knowing just that, that they'd behave the same? I sure wouldn't.
As to your second question, (which I also can't really answer), well either Huheey, et al are wrong or not. I couldn't find much on-line about it, especially for Ra and Fr. What little I did find was decades out of date and itself admitted that the chemistry wasn't (then) well understood. (Cotton & Wilkinson, Adv. Inorg. Chem. 1972(!!)) From that is claimed that A. The II ammines are only "moderately stable". B. Dilute solutions of I & II metals in NH3 are blue, more concentrated solutions (e.g. 3M) are yellow/gold. This and other evidence suggests that in more concentrated solutions metal clusters rather than lone M+ ions exist. C. M+n + ne- is in equilibrium with  M(NH2)n + (n/2)H2 and D. the M° coordination compounds are non-stoichiometric and only approach M(NH3)6.
I re-emphasize that this info is 45 years old and I know that a lot of progress has been made since then, especially in the area of M° coordination cmpds.
As an aside, I was tempted to "answer" your question by pointing out that everything has an opportunity cost. That is, the "reason" why the group I metals didn't form M° hexammines would be either because either the metal and (liquid? gaseous?) ammonia has a more negative free energy OR the amide is somehow kinetically retarded, contradicting the claimed equilibrium OR because the amide has a more negative free energy - compared to the M° complex's crystal structure. Opportunity cost. But since that is obvious (I hope!) I didn't think it would be helpful. (It's a tautology)


Determining the mechanism followed when alcohols undergo a reaction with sodium bromide and sulphuric acid $$\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}$$
Due to sulphuric acid, the $\ce{-OH}$ group will get protonated and convert into $\ce{-H2O+}$ which is a great leaving group in comparison to the hydroxyl group. Therefore I assumed it would readily form a carbocation and then $\ce{Br-}$ could attack to form the alkyl bromide.
That is, it would follow the SN1 pathway (since the reaction is happening in 2 steps, first the formation of carbocation and then the attack of $\ce{Br-}$) but I also read somewhere that we might need to have a look on the substrate, if it is primary then it would follow SN2.
I am a bit confused after so many conditions and situations being introduced so I was hoping if someone could give me a brief and crisp explanation about this, and is there any particular factor that needs to be focused on before judging this mechanism?

yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall reaction follows SN1 (as it is acid-catalyzed) hence we would compare the rate of reaction in terms of RDS (FORMATION OF CARBOCATION) and hence the more stable the carbonation is the faster the reaction proceeds.


Why do laws of thermodynamics "outweigh" any other law ever formulated? "It is the only physical theory of universal content concerning which I am convinced that within the framework of the applicability of its basic concepts, it will never be overthrown." - Albert Einstein
"Nothing in life is certain except death, taxes and the second law of thermodynamics". - Seth Lloyd
“If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation." - Arthur Eddington
"“Every question or effect has the right to exist if it does not contradict the second law of thermodynamics”. - Boris Pavlovich
“A good many times I have been present at gatherings of people who, by the standards of the traditional culture, are thought highly educated and who have with considerable gusto been expressing their incredulity at the illiteracy of scientists. Once or twice I have been provoked and have asked the company how many of them could describe the second law of thermodynamics. The response was cold: it was also negative. Yet I was asking something which is the scientific equivalent of: Have you read a work of Shakespeare's?” - C.P. Snow

I find these particularly interesting but there are many others which mark the supremacy of the laws of thermodynamics over any other law ever formulated by man.
I have just begun to learn the laws of thermodynamics and it is indeed very different and "interesting" but so was Newton's laws of motion, Faraday's laws, Heisenberg's uncertainty principle, Einstein's theory of relativity, then why are laws of thermodynamics considered so immutable and supreme to other laws. What is it that makes all these great people hold them as so impactful and perpetual and makes these laws outweigh others?

The laws never outweigh each other, nor do they have to. Newton's laws of motion, Maxwell's equations of electromagnetism, Einstein's theory of relativity are all equally fundamental, in that they've never been overthrown within their respective frameworks of applicability. Sure, these frameworks have their limits: Newton fails at near-light speeds where Einstein takes over, Maxwell and Einstein fail at quantum limit. So what? Thermodynamics has its limits as well. If you don't believe this, then good luck trying to apply it to a single molecule.

If anything, the laws of thermodynamics are less fundamental, in that they are true only statistically, that is, with less precision than the rest. With relatively little effort, you can measure the mass of a body to six-digit precision. Can you do the same to its temperature?

The scientists quoted were remarking that they believed the second law most strongly of all. The reason is that the second law corresponds to real world experience. Scientists for instance believe that perpetual motion machines don't exist and accept that as evidence the second law must be true. (If a perpetual motion machine were possible then you could use such a machine  to create "free" energy by hooking it up to an electrical generator.)

The notion is that humans can't see the lines of force in an electric or magnetic field. We can do experiments to detect relativity, but it outside our day to day experiences. So those theories are abstract in nature.


How to determine the order of nucleophilicity for given chemical species? How to determine the order of nucleophilicity for given chemical species?

Like I came across this question, to rearrange $\ce{RCOO-, OR-, OH-, H2O}$ (alkyl acetate, alkoxide, hydroxide, and water) in decreasing order of nucleophilicity and it ranked $\ce{OR-}$ first.

Shouldn't it be $\ce{OH-}$ since $\ce{OR-}$ would be more sterically crowded?

Within a group of nucleophiles with the same atom, the nucleophilicity decrease with decreasing basicity of nucleophile.

Decreasing basicity means the decreasing affinity of electron pair for a proton. The decreasing order of nucleophilicity is shown in the figure.

But the relationship between nucleophilicity and basicity can be reversed by steric effect. Less basic but steric unhindered nucleophile therefore have higher nucleophilicity than strong basic but strically hindered nucleophile.

OR¯ is NOT RO¯ !!!
Clearly, R¯ is an unknown species and nothing else can be said about it.
OR¯ would be an unusual structure, given oxygen's electronegativity compared to carbon's. In the case of RO¯, the R "group" would have to be very large and bulky to dramatically sterically affect (hide) the O¯ from other molecules. So for simple organic compounds, the effect will be quite small. As already said pKa of the conjugate acid would be an excellent guide to nucleophilicity (in the relevant solvent).


How there are 7 enolizable hydrogens in 2‐methylcyclohex‐2‐en‐1‐one? 49. How many enolizable hydrogens are there in the following compound?

(a) 2
(b) 4
(c) 5
(d) 7

I know that enolate ion is the anion formed when an α-hydrogen in the molecule of an aldehyde or ketone is removed as a hydrogen ion.
In the above question I find only two α-hydrogens which can participate (denoted with red color):

But the answer is (d) 7.
Where am I getting wrong?

Actually there are exactly 7 alpha hydrogen in the given structure.  Just look that there is also a double bond and it is also considered as functional group.  So, you should count alpha hydrogen w.r.t  double bond and they are 5 .

Finally 2 + 5 you get the required answer = 7


Breaking/reducing water adhesion with an electric field Does anyone know if the presence of an electric field can help to reduce water adhesion to a surface?

On the question relating to breaking/reducing water adhesion with an electric field, I suspect, the answer is that magnetic fields can have an impact.
Support, here is a 2008 paper published in Journal of Applied Physics, The surface tension of water under high magnetic fields, to quote in part:

The surface tension of water-air interface was measured by means of a surface-wave resonance method under magnetic field up to 10 T. It was found that the surface tension increased linearly with an increase in the square of the magnetic field.

Of note, background on the concept of surface tension, per Wikipedia, to quote:

At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion).
There are two primary mechanisms in play. One is an inward force on the surface molecules causing the liquid to contract.[1][2] Second is a tangential force parallel to the surface of the liquid.[2] The net effect is the liquid behaves as if its surface were covered with a stretched elastic membrane.

So, per the cited paper, surface tension increased linearly with the square of the magnetic field. Per Wikipedia, surface tension relates directly to cohesion, as opposed to adhesion.
As such, the question likely correct cites increased magnetic fields as reducing water's adhesion (as it apparently relates to promoting surface tension/cohesion).


Obtaining mercury from mercury(II) nitrate I was thinking about obtaining mercury from mercury(II) nitrate. Is it possible?
I've tried putting chunks of Fe into it, I've got the logic from the voltaic table (where Fe is on the left side of Hg, so I think it would get it out of the compound). Can anyone please correct me and show me the right way? I've tried searching on Google, but found nothing.

I'm just a rookie in chemistry, and was interested in it when my teacher announced about a talent show which I'll perform some cool chemistry there (I'm going to perform using fires, I've tried the colored fires and the alcohol carboy). I'm also planning to make a mirror, if anybody has some ideas please comment below.

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury".
Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though).

Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in aqueous solution unless acidified.
Mercury(II) nitrate quickly hydrolyses to poorly soluble yellow mercury(II) oxide upon dilution or addition of alkali:

$$\ce{Hg(NO3)2(s) + H2O(l) -> HgO(s) + 2 HNO3(aq)}\label{rxn:R1}\tag{R1}$$

which can be converted back to nitrate by adding excessive amount of nitric acid, which in turn won't leave any chance for iron not to be oxidized, so you end up with both metals in solution.

I didn't want to dive deep into the discussion of how mercury(II) salts hydrolyze and leave it simple, but after receiving critique from Maurice claiming $\ce{Hg(OH)NO3}$ to be the “real” product, I think I allow myself to add a paragraph or two.
Raman spectroscopy and x-ray scattering studies in late 1960 demonstrated that hydrolysis of mercury(II) salts produces series of polynuclear oxo-bridged species of types $\ce{Hg2OH(H2O)2^3+},$ $\ce{Hg3O(H2O)3^4+}$ or $\ce{Hg4O(OH)(H2O)3^5+}$ [1, 2].
Formation of mercury(II) hydroxide nitrate $\ce{Hg(OH)NO3}$ as a hydrolysis product was taught in 1940–1950s era, and it stuck in several textbooks published later probably because it's listed in all editions of Pauling's General chemistry up to 1988.
However, not only it's oversimplified (granted, reaction \eqref{rxn:R1} is also an oversimplification in a sense it's a boundary case), but is also an incorrect concept.

The most up-to-date summary of what is really happening when $\ce{Hg(NO3)2}$ is dissolved in water can be found in Mercury handbook [3, p. 115]:

  $\ce{Hg(NO3)2}$ solutions are only stable in the presence of a certain amount of nitric acid, which prevents hydrolysis.
  $\ce{Hg(NO3)2}$ quickly hydrolyzes in excess water and produces a precipitate of
  $\ce{Hg3O2(NO3)2 · H2O}$ or, when boiled in dilute solutions, forms mercury(II) oxide $(\ce{HgO}).$

Regarding the reduction of mercury(II), there are two pathways: $\ce{Hg^0}$ may be obtained from mercury(II) nitrate using a dry or a wet method.

Speaking of a dry method, the most straightforward way to obtain metallic mercury from nitrate is to heat $\ce{Hg(NO3)}$ in a distillation apparatus (b.p. of mercury is 357 °C).
Above $\pu{400 °C}$ it readily decomposes:

$$\ce{Hg(NO3)2(l) ->[\pu{400 °C}] Hg(g) + 2 NO2(g) + O2(g)}\label{rxn:R2}\tag{R2}$$

\eqref{rxn:R2} is a brutto-reaction; the nitrate first decomposes to the red mercury(II) oxide at lower temperatures (which, in turn, decomposes to the elements):

$$\ce{2 Hg(NO3)2(l) ->[\pu{350 °C}] 2 HgO(s) + 4 NO2(g) + O2(g)}\tag{R3}$$

Wet method suggest mild conditions and a reaction in solution.
For example, formic acid (also used in silver refining) being a strong reducing agent would cause a precipitate from ammoniacal mercury(II) nitrate solution.

Since nitrogen dioxide, mercury vapors as well as mercury salts and oxides are highly toxic, the reactions must be carried in a fume hood which makes it a poor fit for a talent show.
Considering the possible risks and your level or preparation (no offence), I'd strongly advise to take extreme caution doing mercury chemistry and avoid public demonstrations until you become more experienced.

Note: chemical reactions are adopted from [4, p. 310–312]

References

Cooney, R.; Hall, J. Raman Spectra of Mercury(II) Nitrate in Aqueous Solution and as the Crystalline Hydrate. Aust. J. Chem. 1969, 22 (2), 337. https://doi.org/10/b6t3h2.
Johansson, G.; Haugsten, K.; Rasmussen, S. E.; Svensson, S.; Koskikallio, J.; Kachi, S. An X-Ray Investigation of the Hydrolysis Products of Mercury(II) in Solution. Acta Chem. Scand. 1971, 25, 2787–2798. https://doi.org/10/bn5j2g. (PDF)
Kozin, L. F.; Hansen, S.; Kit, M. Mercury Handbook: Chemistry, Applications and Environmental Impact; RSC Publ: Cambridge, 2013. ISBN 978-1-84973-409-7.
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Contrary to Andselisk's opinion, mercury nitrate $\ce{Hg(NO3)2}$ is not decomposed into $\ce{HgO}$ in water. Mercury nitrate is hydrolyzed into a basic salt, whose formula is $\ce{Hg(OH)NO3}$, and the equation is:

$$\ce{Hg(NO3)2 + H2O -> HNO3 + Hg(OH)NO3}$$

To get pure mercury from mercuric nitrate, Andselisk recommends heating their substance to 400 °C. This technique will produce huge clouds of nitrogen dioxide which is brown, toxic and corrosive. An easy way of avoiding this drawback is to mix thoroughly mercury nitrate $\ce{Hg(NO3)2}$ with an equal mass of sodium carbonate $\ce{Na2CO3}$, and heat this mixture. It will produce carbon dioxide $\ce{CO2}$ instead of nitrogen dioxide $\ce{NO2}$. The equation of the reaction is:

$$\ce{2 Hg(NO3)2 + 2 Na2CO3 -> 4 NaNO3 + 2 Hg + 2 CO2 + O2}$$

It should be mentioned that, above 380 °C, $\ce{NaNO3}$ will be slowly decomposed. So it is not recommended to heat to such a high temperature. The reaction starts at 150 °C.
This operation should be done in a retort. The mercury will condense in the spout of the retort, and may be recovered. But mind! Mercury is toxic. And all operations with mercuric compounds are forbidden by the law.


How is Al2Cl6 covalent and Al2O3 ionic? I encountered a question where the type of bonding for various oxides are prompted. Two of which are $\ce{Al2Cl6}$ and $\ce{Al2O3}$.

I thought that chlorine atoms are more electronegative than oxygen atoms, so if one of them is ionic and the other is covalent, surely chlorine would be ionic, because it may be able to "tear off" the electrons from aluminium atoms.

This is evidently not the case according to the answer key of this question. I would like to know an explanation for the ionic/covalent properties of these bonds.

You are right that the difference in electronegativity is responsible for the ionic bonds in aluminum and oxygen, and normally that would be the case with chlorine too, but $\ce{Al2Cl6}$ is a special molecule.

The structure of $\ce{Al2Cl6}$ can be viewed as two aluminum atoms covalently bonded to four chlorine atoms each. Two of the chlorine atoms bridge between the two aluminum atoms. The following is a possible representation:

A special type of bonding called 3 center 4 electron bonding is present in the structure. These bonds result from a combination of a filled p orbital and two half filled p orbitals. This causes there to be a filled bonding and a filled non-bonding orbital. The bond orders between each bridging chlorine and aluminum atom are 0.5; the structure of 3 center 4 electron bonds makes it so that the bonding orbital is delocalized over both the bonds. $\ce{Al2Cl6}$ has two of these bonds and they are responsible for the covalent nature of the molecule.

Firstly let's take $\ce{AlCl3}$, since we know that aluminium has a charge of +3, provides chlorine with three electrons (1 to each chlorine). The chlorine being in the second period has a larger atomic radius (in comparison to oxygen). This even gets bigger when it forms an anion by receiving the electron from aluminium. The aluminium's cation being smaller than its original atom has a high polarizing power which attracts and distorts the electron cloud of the chloride ion (has a high polarizability) and forms a covalent bond.

So is not in the case of $\ce{Al2O3}$, as the oxide ion's atomic radius is not large enough for the aluminium cation to distort it thus it stays in ionic bonding.

In short, $\ce{AlCl3}$ does form an ionic bond at first but it's a really small transition state for like nanoseconds, so it quickly forms a covalent bond by the polarizing process.


What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8? I am trying to solve the following exercise:

  What volume of a concentrated $\ce{HCl}$ solution, which is $36.0\%$ $\ce{HCl}$ by mass
  and has a density of $1.179~\mathrm{g~mL^{-1}}$, should be used to make $5.00~\mathrm{L}$ of an $\ce{HCl}$
  solution with a $\mathrm{pH}$ of $1.8$.

I found the respective volumes that $\ce{H2O}$ and $\ce{HCl}$ contributed.
I then found the molarity of the $\ce{HCl}$ aqueous solution given.
I used $10^{-\mathrm{pH}}$ to get the amount concentration of $\ce{HCl}$ needed, I then proceeded to find the amount of substance of $\ce{HCl}$ by multiplying the concentration needed by $5.00~\mathrm{L}$. I then took the amount of substance of $\ce{HCl}$ needed and multiplied by the density given to get an answer of $2.4~\mathrm{L}$.

Am I correct?

Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $\mathrm{pH}$ of $-0.74$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.

We can solve this by making the calculations step by step. Let's start with the concentration of $\ce{H+}$, which will be the same as the concentration of $\ce{HCl}$, we will achieve:

$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$

Since we're going to prepare 5 liters, this means the amount of substance of $\ce{H+}$ should be

$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$

To transform this into amount of $\ce{HCl}$, we need the acid's molar mass ($\pu{36.46 g//mol}$) to convert from amount of substance to mass:

$$\require{cancel}%
g_{\ce{HCl}}
= \mathrm{\frac{36.46\ g}{\cancel{mol}}}
  \times 0.07924466\ \mathrm{\cancel{mol}}
= 2.88926 \ \mathrm{g}$$

Now that we have the amount of substance of $\ce{HCl}$, we calculate the volume, which is where the density comes in:

$$V_{\mathrm{HCl}}
= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}}
= 2.450602 \ \mathrm{mL}$$

Now the last step: Our stock solution isn't pure $\ce{HCl}$, it's diluted in water. So actually the volume we need is:

$$V_{\mathrm{HCl}}
= 2.450602 \times \frac{100}{36}
= \pu{6.807229 mL}
\approx \pu{6.8 mL}$$

This might look that it is not much, but remember $\ce{HCl}$ is a very strong acid. For example, to prepare $\pu{1000 mL}$ of a $\pu{0.001 M}$ solution you only need $\pu{0.086 mL}$ of concentrated $\ce{HCl}$!

You can check this result by imputing the given density, weight percentage, final volume and $\ce{H+}$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.

It's simple. We all know that density = Mass/volume.
The density of your hcl is 1.179

Mass of hcl in one molar solution which is the molar mass is 36.5

but the percentage purity of the hcl is 35% according to you. That means, 35% of that 36.5g is in 1 molar/dm3.

Which means, your mass will be 35% of 36.5
=36.5*35/100
35.5*0.35=12.775

Therefore, 1.179=12.775/v
v=12.775/1.179
v=10.84 (approx)

This is the volume of your conc hcl needed to prepare 1M in 1 dm3

But you want to prepare 5 dm3
5 dm3=10.84*5=54.20

The simple formula that can guide you is

Molar mass times percentage purity everything over density

MM x pp

D
The volume u get using the above formula will prepare 1M in 1 dm3 so, you can now multiply by the number of molar concentration u want to prepare.
Spasiba


Why is hofmann product prefered here? In the following reaction, Hofmann elimination is preferred.

This reaction takes place by E1 mechanism. I thought the Zaitsev product would be major due to 1 extra hyperconjugating structure. In the book, it is given that it is due to sterically compressed alkene, so Hoffmann product is major. I am unable to understand why the sterical hinderance would lead to the instability in alkene as the three methyl groups is one bond away from alkene.

The Hoffmann product is formed as a major product generally in four cases. They are:

When the ${\gamma}$ carbon is 4°
When the leaving group is poor leaving group
When the leaving group is bulky leaving group
When the elimination is internal elimination (${E_i}$ mechanism)

Case 1:

It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product

Case 2:

It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)

Case 3:

It happens  in  the Hoffmann's exhaustive methylation reaction where the bulky leaving  group  is a 3° amine group

Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters

Lastly I want to add that these are general results the yields may vary based on reactant.

Look at the carbon which has more α-hydrogens and thus will be stabilized more by resonance when it is attacked by the $\ce{OH-}$ ions. So the carbon atom on the right hand side that has two methyl groups has more α-hydrogens with it and is stabilized by resonance and  $\ce{OH-}$ is attacked there. Hence the major product has carbon with the double bond.


Stability of F- Cl- Br- I-? In aqueous state If we talk about hydrolysis F- will be more heavily hydrolysed so it should be more stable in aqueous state.
But I know that HI is a stronger acid than HF. So the stability of conjugate acid ie I- should be more than F-. As dissociation takes place in water.

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium
$$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$
that can be expressed also as:
$$\ce{acid1 + base2 <=> base1 + acid2}$$
The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous environment. Acids we consider as strong acids ($\ce{HCl}$, $\ce{H2SO4}$, $\ce{HNO3}$ ) are much stronger acids than $\ce{H3O+}$. Therefore for $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, the equilibrium is practicaly fully pushed toward right:
$$\ce{HX(aq) + H2O -> X-(aq) + H3O+(aq)}$$
On the other hand, $\ce{HF(aq)}$ is a weak acid, weaker than $\ce{H3O+}$, with dissociation constant $\mathrm{p}K_\ce{a}=3.17$, for which the equilibrium is rather on the left side:
$$\ce{HF(aq) + H2O <<=> F-(aq) + H3O+(aq)}$$
Highly concentrated $\ce{HF(aq)}$ is rather a strong acid:
$$\ce{2 HF(aq) + H2O <=>> HF2-(aq) + H3O+(aq)}$$
and liquid HF is very strong acid ( H0 = −11 ) :
$$\ce{3 HF <=>> HF2- + H2F+}$$

Fluorine ion and chlorine ion are very corrosive in nature, so In water they easily form the hydrohaloacid as order of acidic strength in water is  $hi>hbr>hcl>hf$ hence the stability order will be $F^{-} >Cl^{-} >Br^{-} >I^{-}$ which also shows stability of their acids.


Why do the amino groups in EDTA deprotonate first, in contrast to amino acids? We know that in amino acids, carboxylic acid groups deprotonate first than the amine groups. For example, in alanine, $\mathrm{p}K_\mathrm{a,\ce{COOH}} = 2.34$, $\mathrm{p}K_\mathrm{a,\ce{NH3+}} = 9.69$ according to this page.

But according to the structures of EDTA as searched from Google, and from this extract from p.325 of Analytical Chemistry, 8e by Christian, et al.:

  At sufficiently low pH, the nitrogens can also be protonated and this diprotonated EDTA can be considered a hexaprotic acid. However, this occurs at a very low pH and EDTA is almost never used under such conditions.

the hydrogens of these amino groups seem to be deprotonated first before those of the four carboxylic acid groups. Why is this so?

The carboxylates deprotonate first followed by the amines.  The pKa values for the carboxylates ar 0.0, 1.5, 2.0, and 2.66.  The amines have pKa values of 6.16 and 10.24.  Ref. Harris.  Quantitative Chemical Analysis.


Does isopropyl chloride undergo SN1 or SN2 with sodium acetate? What's the reaction of isopropyl chloride with sodium acetate? (The solvent is acetic acid)
Is it SN1 or SN2?

My thoughts:

It's not that difficult to approach isopropyl chloride's carbon, for there isn't much steric crowding. It's just a 2 degree alkyl halide with two methyl groups. This favours SN2.
The carbocation formed by the 2 degree alkyl halide, here in this case, is stabilised by the +I effect of two methyl groups and also six hyperconjugative structures. This should favour carbocation formation, hence SN1.
Acetic acid is a polar protic solvent, and can solvate sodium acetate.
Acetate ion isn't a great nucleophile (or is it?). I'm not sure, here. I know that acetic acid is a weak acid, so acetate ion is a strong base but I believe its nucleophilicity is suppressed by solvation.
Chlorine isn't an amazing leaving group.

So, does SN1 take place or SN2? Could someone also give some experimental data to support their reasoning?

Here's something I found, it says that both SN1 and SN2 can occur. Now which one occurs here, is probably decided by the nucleophile? I'm not sure. Anyway, thought I'd put this table up here for reference:

Source:
Organic Chemistry - Clayden - Oxford University Press - 1st edition - Page 424

Edit: Someone in the comments section said that you can't predict which mechanism is major, and have to look for experimental facts. So, for the sake of the completeness, where are the facts? Could someone please post the relevant details as an answer?

Actually in these type of reactions, where the reactant can both possibly undergo $\mathrm{S_N1} $ and $\mathrm{S_N2} $, the reactions actually proceeds through a mixture of both type of reaction mechanism. In these type of reactions, rate is related as $$\text{Rate} = k_1 [\ce{R}] + k_2 [\ce{R}][\ce{X-}]$$ So, there is also the stereochemically inverted product and also racemised product.
Fraction of inverted product is $k_2/(k_1+k_2)$ and the fraction of racemised product is $k_1/(k_1 +k_2)$. But the actual values of $k_1$ and $k_2$ is experimentally determined and probably very difficult to determine by logical approach only.

In this case the transition state of the reaction is much polar than ground state if the reaction follows the SN1 Pathway. So in presence of polar medium like acetic acid the transition state will be more stabilised than ground state and hence activation energy of the reaction will decrease. And the rate of the reaction will be faster. But if the molecule follows SN2 pathway the transition state of the reaction will be less polar than ground state. As a result in presence of polar medium like acetic acid the ground state will be more stabilised than transition state.As a result the rate of reaction  will decrease. So by analysing all aspects we can predict in acetic acid medium isopropyl chloride will wish to follow up SN1 reaction readily than SN2.


Why are hydrogenation reactions exothermic? I learned that all reactions that yield hydrogen are endothermic (such has reforming) and reactions that use up hydrogen are exothermic (FCC cracking, hydrogenation, etc.). But why is this so? I know what exothermicity/endothermicity means, but I haven't found a specific explanation for the correlation with hydrogen generation or consumption yet.

Let's take a look at an example of a hydrogenation reaction, namely the hydrogenation of ethene to give ethane

$$\ce{CH2\bond{=}CH2\ (g) + H2 (g)-> CH3\bond{-}CH3 (g)}$$

And now let's determine the enthalpy change of this reaction. By Hess's Law, the enthalpy change of the reaction is essentially the same as the enthalpy change of breaking all the bonds of the reactants and forming all the bonds in the molecules of the product. We can see that

1 $\ce{H\bond{-}H}$ bond has been broken
1 $\ce{C\bond{=}C}$ bond has been broken
1 $\ce{C\bond{-}C}$ bond has been formed
2 $\ce{C\bond{-}H}$ bonds have been formed

The other bonds remain unchanged by the end of the reaction

The following is the list of bond enthalpies of the aforementioned bonds, obtained from the Internet

$\ce{H\bond{-}H}$ bond: +432 kJ/mol
$\ce{C\bond{=}C}$ bond: +614 kJ/mol
$\ce{C\bond{-}C}$ bond: +347 kJ/mol
$\ce{C\bond{-}H}$ bond: +413 kJ/mol

Breaking bonds gives a positive enthalpy change since it is an endothermic process, and vice versa for bond formation. We can now estimate the enthalpy change of the reaction

$\Delta$$H$ = 2(-413) + (-347) + 614 + 432 = -127 kJ/mol

As indicated by the negative value of the enthalpy change, the reaction is exothermic.

In fact, the enthalpy changes of hydrogenation reactions of other alkenes are also similar (though more substituted alkenes may have a lower magnitude of heat of hydrogenation, but still the reaction is exothermic). You may notice that for other alkenes, the way you calculate the enthalpy change is the same since they involve the same bond breaking and bond forming processes aforementioned (any substituents bonded to the carbons with the $\ce{C\bond{=}C}$ bond remain unaffected throughout the reaction and will still be present in the product being bonded to the carbon).

To simply put, the energy released from bond forming is larger than the energy required (absorbed) for bond breaking, and that is why you get net energy released from the reaction.

Only one bond is broken, as in H-H bond in hydrogen gas. But two bonds are made, that are C-H bonds. One extra bond is made. As we know bond making is exothermic, so more energy due to that one bond made will be released than absorbed.


Enthalpy of a reaction changes with temperature, does this mean bond strengths change with temperature? During chemical reactions, the bonds between atoms break or form to either absorb or release energy. The result is a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy.

Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the potential energy of the reactant bonds.

But we know that enthalpy of a reaction depends on the temperature according to Kirchoff's equations. So does this mean that bond energies also change with temperature.

Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?

And since the enthalpy of a reaction depend on the constant pressure at which reaction is carried out, does this mean bond strengths also depend on pressure?

No: bond strengths as such do not change with temperature. The potential energy surface holding the atoms together, i.e. all the bonds, is constant for a given molecule.

However, all molecules have vibrational and rotational energy levels and molecules also have bulk kinetic energy due to thermal motion in a gas or liquid. As the temperature increases the bulk thermal energy increases, i.e. molecules move faster, but also more vibrational and rotational energy levels become excited, which means that the vibrational and rotational quantum numbers increase. This means that less extra energy is needed to break a bond if the temperature is already high, compared to that needed at low temperature, simply because the bond is already has a lot of internal (vibrational/rotational) energy.

Not all molecules have exactly the same energy but instead there is an exponential distribution of energies (the Boltzmann distribution) where the higher the energy is the fewer molecules are there that have this energy.

The subject of  Statistical Mechanics describes these molecular effects and also forms the foundation for Classical Thermodynamics, i.e. what we usually call 'Thermodynamics', and which was developed before atomic/molecular theory.

The connection between quantum mechanics and classical thermodynamics is not straightforward. Or perhaps I should say it isn't simple. Quantum Mechanics, as you may know, isn't an easy subject. Richard Feynman famously said:"... no one understands quantum mechanics." So, looking too hard for intuitive understanding of the way the world works isn't likely to be productive. As has been widely discussed since it was invented/discovered in 1925, you can't use your intuition to understand it, you have to use the math. Does temperature affect bond strength? Temperature is a measure of molecular motion. Will motion affect bond energy? It depends, if the two atoms involved in the bond have sufficiently different momentum, the bond will break. So as you raise temperature of a material, you get closer to the point of disintegration. How about pressure? Well, you may know that water ice has various crystal structures depending on T and P? You can't explain that if you assume there's no change in energy and entropy of the system at different P or T. But chemicals are generally assumed to be at their LOWEST energy state, and compressing them (ignoring the PV work) will result in more repulsion between electrons, right? So you'll have to overcome that repulsive energy hence compression adds to potential energy.


Acidity order of nitrophenols As in other posts I have seen and learnt that, if one is given four compounds, namely para-nitrophenol, ortho-nitrophenol, meta-nitrophenol and phenol and is told to arrange them in order of acidity, then one has to take into account the $\text{-R}$ and $\text{-I}$ effect.

Now, for the inductive effect, the greatest intensity is observed at the ortho-position, followed by meta-, and then para-.

For the resonating effect, the intensity is greatest for para and lowest for ortho, while meta has no such effect. Now keeping in mind, that $\text{-R}$ and $\text{-I}$ both tend to withdraw electrons away from the phenol group (and thus increases the acidity), and the $\text{R}$ effect has more contribution than the $\text{I}$ effect, we can thus deduce that para-nitrophenol is most acidic, followed by ortho-nitrophenol, then meta-nitrophenol and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

The facts are the proof of my theory, where the $\mathrm{p}K_\mathrm{a}$ for:
$$
\begin{array}{c|c}
\text{Compound} & \mathrm{p}K_\mathrm{a} \\
\hline
para\text{-nitrophenol} & 7.16\\
ortho\text{-nitrophenol} &7.2\\
meta\text{-nitrophenol} & 9.3\\
 \text{phenol}   & 9.95
\end{array}
$$

So far so good.

But one of my colleagues, when I asked the problem, tried to give the theory in different way. He said, in para-nitrophenol, hydrogen bonding is strongest, so it can't be most acidic, which in his view port should be least. I knew this is not the case, but I just can't explain or point out where he's been wrong. Can anyone of you make it little easy for me to understand it?

You have correctly identified that m-nitrophenol should be the weakest acid among the isomers as the resulting anion cannot be stabilised by the $\text{-M}$ effect of nitro group owing to its meta location.
The acidity order of the remaining two nitrophenols can be explained in this way:

Hydrogen bonding is possible in both of them. But note that:

$$\text{H-bonding} \propto \delta^+ \text{charge on $\ce{H}$ atom}$$
Inductive effect (here, $\text{-I}$) is distance dependent. So $\delta^+$ charge on the $\ce{H}$ of ortho-nitrophenol's -$\ce{OH}$ is going to be more than the $\delta^+$ charge the $\ce{H}$ atom of para-nitrophenol's -$\ce{OH}$.

Also,  quoting from this comment:

Usually the $\mathrm{p}K_\mathrm{a}$ of a substance is determined in dilute solution. So just due to dilution, there won't be many/any inter-molecular H bonds, but there can still be plenty of intra-molecular ones.

The answer above the comment relates the strength of these two types of hydrogen bonds to kinetic motion of molecules:

Inter-molecular interaction, such as inter-molecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.
Intra-molecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

These three factors help us conclude that:

Strength of intra-molecular hydrogen bonding in o-nitrophenol $\gt$ Strength of intermolecular hydrogen bonding in p-nitrophenol

Thus, p-nitrobenzoic acid is more acidic than the o-isomer.

...and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

This reasoning is incorrect. Phenol has no group that stabilises the negative charge on the resulting phenoxide ion through $\text{-M}$ or $\text{-I}$ effect (a deactivating group). Therefore, it's the weakest acid of all under investigation.

More the stability of phenoxide after removal of $\ce{H}$ more will be  acidity.

ortho-Nitrophenol phenoxide ion is stabilised by  -R and -I effect where both -I  and -R effect are maximum.
meta-Nitrophenol phenoxide ion is stabilised by  -I effect only.
para-Nitrophenol phenoxide ion is stabilised by  -R and -I effect where both -I  and -R   effect are minimum.

Thus according to stability of anions o-derivative will be more acidic than p-derivative which will be more acidic than m-derivative.

But due to H-bonding in ortho-nitrophenol it is reluctant to remove hydrogen due to which it's acidity decreases and order becomes:

p-derivative $\gt$ o-derivative $\gt$ m-derivative $\gt$ phenol
(acidity in decreasing order)

Source- Jagdamba Singh: A logical approach to Modern Organic Chemistry. Revised edition 2009.  Pg 151


Structure of pyrophosphorous acid What is the structure of pyrophosphorous acid?

I asked this question here because the answer which I found on web and other resources are inconsistent which each other.
For example, following data is given regarding the structure of pyrophosphorus acid in my textbook, which is not consistent with the number of oxygen atoms.

On checking here,and here there are no double bonds of phosphorus with oxygen, however here there are. And here both the structures are given on the same website.

Oxyphosphorus compounds, all of which contain phosphorus-oxygen linkages, are the most dominated subset in Phosphorus Chemistry. You may find good review of oxyphosphorus compounds in Ref.1. In particular, most of these commonly known as phosphates are described in Chapter 3 of Ref. 1 (Pages 169-305) which states that:

  Oxyphosphorus compounds may be defined as compounds which contain phosphorus-oxygen linkages. They may contain up to six oxygen atoms linked to a central phosphorus atom. Pyramidal derivatives are represented by phosphite esters (la), tetrahedral compounds by orthophosphate salts (lb) and esters (lc), trigonal bipyramidal compounds by pentaoxyphosphoranes (Id), and octahedral compounds by a few hexaoxyphosphorides (hexaphosphates) of type (le). In addition, there are the comparatively rare 2-connected angular phosphenites (If) and 3-connected planar phosphenates (lg).

The term normal phosphates is used to described the compounds, in which only $\ce{P—O}$ linkages are present (e.g., structures 1a-g). Compounds such as 1b containing discrete $\ce{PO4^{3-}}$ anions are known as orthophosphates. If the neutralizing cation is $\ce{H+}$, then it is called phosphoric acid or orthophosphoric acid.

On the other hand, if some of the oxygen linkages are replaced by other atoms or groups (e.g., $\ce{H, NR2, CR3}$, etc.), the compounds can be termed substituted phosphates. For example, if one $\ce{P—O}$ linkage is replaced by a $\ce{P-H}$ linkage, the compound series are called phosphites (2a) while when the two $\ce{P—O}$ linkages are replaced by two $\ce{P-H}$ linkages, that series is known as hypophosphites (2b).  Similar to phosphates, if the neutralizing cation is $\ce{H+}$ in 2a and 2b, then they are traditionally called phosphorus acid (Wikipedia1) or hypophosphorus acid (Wikipedia2), respectively. The IUPAC names of these acids are phosphonic acid and phosphinic acid, respectively. Meanwhile, substituted phosphates containing a single $\ce{P=O}$ linkage (e.g., 2c) are usually referred to as phosphoryl compounds.

When $\ce{PO4^{3-}}$ anions are linked together with sharing oxygen atom in common, they are called condensed phosphates (e.g., pyrophosphate, 3a). If the neutralizing cation is, again, $\ce{H+}$, then it is called pyrophosphoric acid or diphosphoric acid (IUPAC name). On the other hand, one can expect when the two anions linked together with sharing oxygen atom in common are phosphites, they would be called pyrophosphites, 4a (Wikipedia3).  However, recent research showed that there are some controversies on the structures, not only on pyrophosphites, but also on phosphite itself.

Phosphorous acid, $\ce{H3PO3}$ (systematic IUPAC name is phosphonic acid), exists in two tautomeric forms: An one with tricoordinate about phosphorous $(\ce{P(OH)3})$; and another with tetracoordinate about phosphorous $(\ce{HP(O)(OH)2})$ (See Scheme below), which is the favored tautomer between two (Ref.2).

The $\ce{P(OH)3}$ form of two tautomer is rarely isolated or directly obtained, yet its presence in the equilibrium has been repeatedly postulated from kinetics studies (Ref.3,4) and thermodynamic studies (Ref.5,6).

According to most literature, the $\ce{P(OH)3}$ form of tautomers is possible only in triesters such as 1a $(\ce{P(OR)3})$. Even in such cases, there is a strong tendency to rearrange to the tetracoordinate species, $\ce{RP(O)(OR)2}$ (Ref.5). Hence, it is safe to say that pyrophosphites has high tendency to have a structure resembling of 4a (See first scheme).

References:

D. E. C. Corbridge, In Studies in Inorganic Chemistry, Volume 20: Phosphorus – An Outline of its Chemistry, Biochemistry and Uses; Elsevier B.V.: New York, NY, 1995, Pages 1-1208 (https://www.sciencedirect.com/bookseries/studies-in-inorganic-chemistry/vol/20).
F.A. Cotton, G. Wilkinson, C.A. Murillo, M. Bochmann, Advanced Inorganic Chemistry, 6th Edn.; Wiley-Interscience: New York, NY, 1999, “Chapter 10: The Group 15 Elements: $\ce{P, As, Sb, Bi}$,” pp. 380-443.
G.A. Haight Jr., M. Rose, J. Preer, “Reactions of chromium(VI) with phosphorus(III) and phosphorus(I). I. Dihydrogen phosphite, phosphorous acid, and hypophosphorous acid,” J. Am. Chem. Soc. 1968, 90(18), 4809-4814 (DOI: 10.1021/ja01020a011).
R.O. Griffith, A. Mckeown, “Kinetics of the reaction of iodine with phosphorous acid and with phosphites,” Trans. Faraday Soc. 1940, 36, 766-779 (DOI: 10.1039/TF9403600766).
J. P. Guthrie, “Tautomerization equilibria for phosphorous acid and its ethyl esters, free energies of formation of phosphorous and phosphonic acids and their ethyl esters, and $\mathrm{p}K_a$ values for ionization of the $\ce{P—H}$ bond in phosphonic acid and phosphonic esters,” Canadian Journal of Chemistry 1979, 57(2), 236-239 (https://doi.org/10.1139/v79-039).
J. P. Guthrie, “Carbonyl Addition Reactions: Factors Affecting the Hydrate–Hemiacetal and Hemiacetal–Acetal Equilibrium Constants,” Canadian Journal of Chemistry 1975, 53(68), 898-906 (https://doi.org/10.1139/v75-125).

As you can see that the textbook has not considered the P-O-P and thus has one less oxygen. Just remember that in any oxo acid if the number of oxygen is one more than twice the number of phosphorous/sulphur (like in the case above where the number of O=(2*P or S)+1), then it has a bridge bond and if two more than twice the number of phosphorous/sulphur (O=(2*P or S)+2) then it has a peroxide bond like in peroxodisulphuric acid (H2S2O8).


Is there a difference between hydrated oxides and hydroxides? My guess is that they are different.

I think hydrated oxides would have a lattice with the cation and oxide anions as lattice points and molecules of water trapped in interstitial spaces, while hydroxides would have the cation and hydroxide anions as lattice points with no water molecules in interstitial spaces.

Hydrated oxide is a compound of an oxide with water. And Hydroxide is a diatomic anion with chemical formula OH− ( a hydroxyl functional group). you can refer to this link for better understanding : http://www.memidex.com/hydrated-oxides. Also looking at the definition of hydration might be helpful.


What is the difference between distilled water and rainwater? I am studying different means of collecting water for consumption. From resources available on Internet, I understand that distilled water is not advised for regular consumption because of the lack of minerals and salts, while rainwater can be consumed if not overloaded with unwanted chemicals.

What confuses me is that I know water to be pouring down from clouds, which is water vapor turning into droplets through condensation. To me, this looks to be exactly the same as the process of distillation, i.e. evaporation and condensation of water.

So is my assumption about rainwater formation wrong? What is the difference between rainwater and distilled water? What causes this difference to happen?

Rain water passes through the atmosphere and picks up whatever is present there, including particulates such as dust, trace amounts of atmospheric gases, nitrates and nitrites, and so forth.

Rain water is relatively low in mineral content.  In practice, people drink rainwater from streams and wells, where it has had ample opportunity to pick up minerals from soil and rock.

https://www.google.com/search?q=rain+water+composition

Rain water contains very low amounts of salts and other nitrates but it takes in any gas present in air. That is why acid rain occurs when the air is polluted. The air contains oxygen so people recommend the rain to be consumable as it has lots of oxygen contents in them. But distilled water does not contain any minerals in them, so it is not recommended for consumption because of lack of minerals.


Why is the specific gravity of 0.9 % saline 1.0046? Medical Normal Saline (0.9% Saline) contains 9 g of salt dissolved in 1000ml of water.

Hence a litre of NS should weigh 1.009 kg.

Hence its specific gravity should be 1.009, but it’s only 1.0046.
(I've measured it, and verified it on-line. It is apparently, a fact.)
I thought I understood SG, it’s a simple concept but I am missing something.

EDIT: after more reading, I learned that dissolving NaCl in water results in a reduction in the original volume. So my question now becomes:
We're dissolving NaCl in pure water at a temperature where 1.000kg mass of water has a volume of 1.000 litres. By my (erroneous) reasoning:
Say the volume decreases by 1 ml, giving a final volume of 0.999 litres.
So the density of the solution would be 1.009 kg/ 0.999 l = 1.010 kg/l
Giving an SG of 1.010 (because water has a density of 1.000 kg/l, as above) If the volume didn't change, it would be 1.009, but it increases because of the reduction in volume.
BUT: The known, published SG of 0.9% w/v NaCl solution is 1.0046.
Can anyone explain why, please? or at least, why the SG is below 1.009, when it should be greater?

0.9% medical saline is defined to contain $\pu{0.90 g}$ sodium chloride per $\pu{1000 ml}$ solution at $\pu{22 °C}$. Its density is $\pu{1.0046 g/cm^3}$ at $\pu{22 °C}$.

The density of water is $\pu{0.9978 g/cm^3}$ at $\pu{22 °C}$.

If the volume would stay constant when you add $\pu{0.90 g}$ sodium chloride to $\pu{1000 ml}$ water, the resulting density would be

$$\pu{0.9978 g/cm^3} + \pu{0.0090 g/cm^3} = \pu{1.0068 g/cm^3}$$

The density of the 0.9% saline shows us that $\pu{1000 ml}$ of it contain

$$\pu{1004.6 g} - \pu{9.0 g} = \pu{995.6 g}$$

or

$$\frac{\pu{995.6 g}}{\pu{0.9978 g/cm^3}} = \pu{997.8 ml}$$

water.

This means that the volume of the liquid phase increases when we add sodium chloride to water, what explains the difference in density from $\pu{1.0068 g/cm^3}$ to $\pu{1.0046 g/cm^3}$.

Density of NaCl is indicated at 2.16g/cc. Given this 9.0g of NaCl added occupies 4.16667ml of the flask. This leaves 995.833333ml of water in the flask. Therefore:

995.833333ml water * 1g/ml = 995.833333g water + 9.0g NaCl in the flask = 1004.833333g in 1000ml = 1.00483g/ml


What is meant by viscous liquid? I think all the liquids are viscous. So I was wondering what is meant by a viscous liquid as it is written in the first line of page on Oil at Wikipedia.

  An oil is any neutral, nonpolar chemical substance that is a viscous liquid at ambient temperatures and is both hydrophobic (immiscible with water, literally "water fearing") and lipophilic (miscible with other oils, literally "fat loving"). ...

This is somewhat an elementary discussion, but viscosity is typically measured in some sort of viscometer. A typical one is the Oswald viscometer shown below.

Fluid is poured into the apparatus through tube at $f$ to fill neck and bulb at the bottom. The fluid is then sucked up above line $c$. The fluid is then released and drains down the tube to line $c$ at which time the timing for the flow rate starts. When the fluid falls to the line at $d$ is reached then a flow rate can be measured. By using samples of known viscosity, the flow time can be calibrated to the viscosity of the sample.

Now such measurements are in units of pascal-seconds. For reference a short list will be provided.

     Substance           mPa-s
     acetone                0.0306
     water                  0.0894
     sulfuric acid          0.242
     olive oil             81.
     motor oil SAE 40     319.
     glycerol           1,200
     corn syrup         1,380.6
     pitch              2.3×10^8

Wikipedia also lists tables for Viscosity of selected substances

Now let's defined three ranges of viscosity:

Low Viscosity -    0.1 mPa-s >= sample
Medium Viscosity - 0.1 mPa-s <= sample <=1.0 mPa-s
High Viscosity - sample > 1.0 mPa-s

Of course the "exact" ranges are arbitrary. So a viscous liquid is a High Viscosity liquid on our scale.

Generally viscosity is an ability to resist flow. More viscous liquid will flow slower. Oil and honey are viscous liquids i.e. liquids with high viscosity.


Why does a certain number of moles of gas occupy a set volume? In class, I was taught that 1 mole of an ideal gas occupies a set volume eg 22.4 L at STP. Why does the gas only occupy a certain volume and not just keep expanding forever, having limitless volume?

Ideal gases neglect inter-molecular forces so shouldn't they just keep expanding?

The reason the gas doesn't expand to infinity is because it is in a vessel.

Or, to be more precise, the ideal gas equation defines the relationship between temperature, pressure and volume for a fixed amount of gas.

What is normally done to understand the relationship is to put the gas into a vessel of fixed volume and measure the pressure it exerts at a given temperature. The pressure changes with temperature according to the formula (real gases may deviate slightly).

Alternatively, we can use a more complex apparatus that allows us to measure the volume occupied by the gas as we vary the temperature while maintaining a constant pressure (say atmospheric pressure). Same equation, different constraint.

Your misunderstanding is that you have not thought through the conditions under which you are applying the rules. The thought experiment you are doing is releasing the gas in the vacuum of space where there is no external pressure. In those circumstances the gas will expand to infinity but this doesn't violate the ideal gas equation as there is no constraining volume (you could say volume is infinite).

When applying the rules you have three variables (you usually keep the amount of gas, the number of moles, fixed) and you have to constrain two of them. If you don't, are an infinite range of possible combinations of the variables (e.g. constrain temperature and there is still nothing that tells you what the volume and pressure are as they can vary together over any values that satisfy the equation).

That is based on the equation pV=nRT for ideal gases, where p is pressure, V is volume, n is amount, R is a constant, and T is the absolute temperature in kelvin.

The derivation of this equation can be found from many sources, so I am not going to include it here.


How to transform a tertiary alcohol to a carboxyl? I am trying to transform a lipid with a tertiary alcohol into a carboxylic acid.

More specifically, I want to convert to diplopterol (see below) into something homologous to cholesteryl hemisuccinate (see below).

Any suggestions?

Diplopterol:

Cholesteryl hemisuccinate:

I am not sure that what you claim to want to do is what you actually want to do.

Cholesterol hemisuccinate is prepared from cholesterol, but not by converting the alcohol functional group into a carboxyl group. The alcohol oxygen becomes one of the oxygen atoms in the ester, but not the carboxyl oxygen. Cholesterol hemisuccinate can be prepared in the lab probably from cholesterol and succinic anhydride in the presence of a non-nucleophilic base (like pyridine or triethyl amine), which can act as a proton shuttle. An acidic workup will be needed to remove the base and protonate the carboxylic acid. The scheme below shows this reaction (sans acidic workup). The alcohol oxygen is highlighted in red.

An identical reaction could be done with your molecule, diplopterol. Dissolve both (1 eq) diplopterol and succinic anhydride (1.1 eq.) in dichloromethane or diethyl ether (or any low polarity aprotric solvent you like). Cool on an ice bath. With stirring, add pyridine (1.1 eq.) dropwise. Let the reaction warm to room temperature, probably about 2 hours. The product will likely precipitate as the pyridinium salt. Check the progress by TLC or some other means. The product will probably be more polar than the reactant. Work up by adding dilute HCl (probably 0.1 M). Extract the aqueous layer with your organic solvent of choice. Combine the organic layers. Wash with brine. Dry over magnesium or sodium sulfate. Remove the solvent by evaporation (rotary being preferred), and purify by chromatography if necessary.

Alternatively, if the product does precipitate at the pyridium salt, collect the salt by filtration, wash the precipitate with ether and then with cold ethanol (will wash away the reactants). Then, add the precipitate to the 0.1 M HCl and extract with ether if necessary. Further purification is probably not needed.

You can try reaction with sodium acetate, anhydride acetic at 120°C for 6h for acetilation product. Because of tertiary alcohol, more hard condicions it´s needed.