What is the mechanism of the Clemmensen reduction? I have learned that we can reduce a ketone to an alkane using the Clemmensen reduction, where we treat the ketone with amalgamated zinc in the presence of $\ce{HCl}$.

My book didn't mention the mechanisms for this type of reaction and I couldn't find an answer by searching Google.

How does this reaction actually take place? Are there carbocation rearrangements in this reaction?

As with many complex mechanisms, many variants can and indeed have been proposed. The most widely accepted is the one proposed by Rosanti (J. Org. Chem. 1991, 56, 4269. DOI: 10.1021/jo00013a036).

The key steps of the mechanism are given below, crucially showing that the alcohol is not an intermediate (hence why secondary alcohols are not able to undergo a Clemmensen reduction to afford the corresponding alkane).

A single electron reduction of the ketone takes place to give a radical anion. The exact species responsible for this  has been debated, but is thought to either be the zinc chloride radical (as shown) or the mercury chloride radical (not shown)
The radical anion picks up a hydrogen from water (the reaction is carried out in aqueous hydrochloric acid), this generates a hydroxyl radical, but also means that the alkoxide derived from the starting ketone is now simply an anion rather than a radical anion.
Homolytic cleavage of the C–O bond, the oxygen has now been completely removed from the substrate, simply leaving behind a secondary radical which must gain a hydrogen (again from water) in order to provide the desired product..

As shown below, several other steps also operate in order to quench out the various radical intermediates produced during the course of the reaction.

  Proposed mechanism of the Clemmensen reduction

  Ref: Comprehensive Organic Named Reactions, Wang.

Here is the mechanism for reduction of aldehyde/ketone by Clemmenson reduction.


Conversion of 4-nitrotoluene to 2-bromobenzoic acid In the above conversion why doesn't bromine add to methyl group? Is the first reaction following free radical or polar reaction (Friedel-Crafts). If it's following free radical mechanism then Br should add to CH3, if it's F-C substitution then why there is no catalyst when NO2 is deactivating group?

What would be the change if I first do seconds step (i.e reduction to amine) and then bromination? Would it be wrong to do bromination at last?

Is the intermediate (2-bromo-4-nitrotoluene) the only major product of first step?

Sorry but I would like to answer your questions from last to first,

(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place

(2) -COOH  is deactivating meta directing group (due to presence of  >C=O) so friedel-crafts bromination would take place at meta position.

(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2

(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced  one of the -H at ortho (to -CH3) of benzene.

(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)


Which proton in cyclohexene is the most acidic? I know that $\mathrm{H^b}$ is most acidic due to conjugation (resonance).

But I am confused as to how to compare $\mathrm{H^a}$ and $\mathrm{H^c}$. I think $\mathrm{H^a}$ should be more acidic than $\mathrm{H^c}$ due to hyperconjugation.

When comparing acidities of protons, we need to find out what the corresponding conjugate bases are.

The conjugate base upon removing $\ce{H_\mathrm{b}}$ is an allyl anion. There is (and should be) no question that it is the most stable carbanion of the three we are going to consider.

The conjugate base upon removing $\ce{H_\mathrm{a}}$ is an $\mathrm{sp^3}$ configured carbanion while the conjugate base upon removing $\ce{H_\mathrm{c}}$ is an $\mathrm{sp^2}$ configured carbanion. This is nice, because it allows us to directly compare the two using their hybridisation.

In general, lone pairs (and that explicitly includes those coming from anions) want to be in the lowest-lying atomic or molecular orbital possible. Since an s orbital is lower in energy than a p orbital from the same shell, lone pairs prefer s orbitals. If there is no pure s orbital available — as would be the case in practically all carbon-centred lone pairs — an orbital with the highest possible s character is preferred instead. If we look at the configuration $\mathrm{sp^3}$ we can say that it has $25~\%$ s character while $\mathrm{sp^2}$ has $33~\%$ s character. Since $33~\% > 25~\%$, the $\mathrm{sp^2}$ configured carbanion is more stable — which is also supported by corresponding $\mathrm{p}K_\mathrm{a}$ values.

The method outlined above is a very crude approximation building on a high amount of simplifications but gives us the correct result in practically all carbon-centred anion cases.

Thus, the acidity order is:
$$\ce{H_\mathrm{b}} > \ce{H_\mathrm{c}} > \ce{H_\mathrm{a}}$$

Note: Ron noted in a comment above that the acidity of an $\mathrm{sp^2}$ hydrogen almost parallels that of an allylic hydrogen. So while most courses may teach this simple ordering, you should keep that in mind as additional information.

To check acidity see conjugate base by removing H+. So when you remove H+ from double bond you get a sp2 carbanion which is more unstable than sp3 carbanion. Now as weaker the conjugate base stronger is acid. hence hc>ha. Hope its clear.


Is there any mathematical or logical proof that carbon with 4 different groups would be chiral? In chemistry; chirality is generally defined in 2 ways.

Lord Kelvin's definition: "I call any geometrical figure, or group of points, chiral, and say it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself" (source: 1. chirality.org , 2. Biochemistry book by Lubert Stryer.)

"If a tetrahedral (sp3) carbon atom contains a different chemical group on each of its 4 hands, the central carbon atom is called chiral. Though there are some exceptions where we do not found any chiral carbon still some compounds can be chiral. Chiral molecules cannot be superposed with their 3-dimensional mirror image" (common classroom-definition for chirality).

But it seems the first definition is the basic definition; and the second definition seems to be a corollary or application of the first definition.
Now I want to know; is it possible to 'prove' or 'derive' the second definition (better to say second statement or second description) from the first (Lord Kelvin's) definition mathematically or geometrically or logically? and if possible then how to prove the second definition from the first definition?
This question is related but different from this question

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.

So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.

Chemistry

In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.

Geometry

Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.

Showing that the carbon is chiral

If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.

3D illustration

see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')

Is this trivial?

No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.

I don't think there is any mathematical way to prove it.It's only experimental.
See Chirality is defined for the compounds whose mirror image doesn't superimpose on each other.

-source : LG Wade Orgainc chemistry.
if you want more explanation then you can take help of LG Wade Or Solomons Organic Chemistry(both are available free on net)


Solubility, pressure and Henry's law Determine the solubility of $\ce{N2}$ in water exposed to air at $\pu{25^\circ C}$, if the atmospheric pressure is $\pu{1.2 atm}$, assume that the mole fraction of nitrogen is $0.78$ in air and the Henry's Law constant for nitrogen in water at this temperature is $\pu{6.1 \times 10^{-4} M/atm}$.

I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice.

What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$.

That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.

Henry's law is given by the formula $p=K_\mathrm h\chi$. Here $p$ is partial pressure of the gas $K_\mathrm h$ is Henry's constant for that gas and $\chi$ is mole fraction.Using this you can find it.


Will the following complex show optical isomerism So there is this question which asks whether the following complex

  $\ce{[Co(NH3)4Br2]+}$

will show optical  isomerism or not.

According to my knowledge it will show optical isomerism in cis form only and not in trans form so overall the compound shouldn’t be optically active.

This what I hav been taught. And moreover shouldn’t it be optically active in every scenario to be called so?

But the ans says that it shows optical isomerism.
I don’t understand why.. is my understanding wrong?

As already commented by Ivan Neretin, the cis form also does not show optical isomerism. Elaborating a bit further on the comments:

Complexes show optical isomerism when they are asymmetric (also known as chiral). They should not have any elements of symmetry, such as planes/axes/centres of symmetry. The trans form has a clearly visible plane of symmetry. It's a bit trickier to find the plane for the cis form. Here's a diagram showing the planes:

(source: Chemistry libretexts)


Why do we use pV term independent of U in the equation H = U + pV? I have read the following definition of enthalpy in my textbook:

A substance has to occupy some space in its surroundings depending upon its volume ($V$). It  does against the compressing influence of the atmospheric pressure ($p$). Due to this,the substance possesses an additional energy called $pV$ energy which is given by the product of pressure ($p$) and volume ($V$) of the system.
The sum of internal energy and $pV$ energy of any system, under given set of conditions, is called enthalpy. It is denoted by $H$ and is also called heat content of the system. Mathematically, it may be put as $$H = U + pV$$

Let's suppose we have some water in a test tube at room temperature open to atmosphere. It will have internal energy which consist of kinetic energy of  molecules, chemical energy, rotational energy and some other forms of energy. In order to occupy some space in its surroundings, the molecules of the water should vibrate in such a way that they cancel out the compressing influence of the air molecules present in the atmosphere.
So if we try to measure the internal energy of the system, isn't the energy required to make space for the surroundings already included in the internal energy of the system. Why do we add an extra term $pV$ along with internal energy $U$ to measure the heat content of the system?

The energy U of a system consisting of a single component and phase is

$ U = -PV + TS + \mu N$

As you rightly point out, U includes the energy required to create the volume by expanding against the surrounding (constant) pressure, namely the -PV term. Removing this contribution gives H.

Therefore

$ H = U + PV = TS + \mu N$

Equating TS with the "heat content" of the system leads to the statement in the textbook (this is my interpretation of this textbook's author's meaning). This kind of algebraic manipulation can lead to all sorts of confusion.

EDIT

Some people recoil (and downvote) when they see an expression such as

$ U = -PV + TS + \mu N$

as if this were somehow unphysical. Well, it does suggest that there is such a thing as an absolute value of U rather than only relative values, and it is hammered into our heads in school that there are only relative, not absolute, energies. But the point here is that there is an absolute value, relative to an empty system, and you can use N (the number of particles) as your simple scaling constant. The expression above captures the extensive property of U. You could equally write

$ \Delta U = -P \Delta V + T \Delta S + \mu \Delta N$

and no one will blink. Well, if you define

$\Delta U = \Delta N \times U_m$

$\Delta S = \Delta N \times S_m$

$\Delta V = \Delta N \times V_m$

with

$\Delta N = N-0$

you obtain an expression equal in spirit to the sacriligeous one above.

Usage of terminology such as "heat content" is potentially confusing, but we can use similar arguments to those used in explaining the PV term's meaning as work to create the system from scratch by increasing the particle number, while keeping intensive properties such as T and P constant, to suggest that the term TS is the heat that must be exchanged with the surroundings during creation of the system (from 0 to N particles). This is more of theoretical than experimental interest, in the same way we can hypothetically charge an object from 0 to a finite charge to compute its energy (relative to no charge).

Note that saying that H represents the heat content is confusing, since the application of the equation is not usually in the spirit I have just outlined, and if you look at

$ H = U + PV = TS + \mu N$

there is an additional term $\mu N$ term which cannot be easily associated with the concept of "heat content".

So, being generous, I would think the author of the book meant by the statement was, very loosely, that changes in H, that is $\Delta H$, are usually associated with the heat exchanged by a system at constant P (subject also to other restraints but more on that in other questions and answers referring specifically to the meaning of H).


How can I separate the vinegar from a 5% white vinegar to make a more concentrated solution. I was under the impression that if I put a semi full bottle of vinegar in the freezer the water would freeze but the vinegar would not until a lower temp.  Therefore I could remove the bottle after the water freezes and pour off the vinegar. Apparently the water and vinegar doesn't separate nicely before freezing so this didn't work so well.

I've seen boiling suggested but I'm afraid I'll lose some of the vinegar in the process.

What is the simplest way to concentrate the vinegar at home?

If you have access to a still you can distil the vinegar solution, however this can only go so far. You may find this section of the wikipedia article on azeotropy edifying - to summarise, adding ethyl acetate to a vinegar solution generates a water:ethyl acetate azetrope with a depressed boiling point that makes distillation more effective, however this may be beyond the pale for a home experiment and would obviously render the vinegar non-food-safe.

Ive tired this via distillation and afterwards it left a black substance so at the very least you need to distill it first to remove any solutes in the viniger.


Why does adding salt to water reduce the pH? I recently heard someone claim that adding table salt to vinegar caused HCl to form which helped them clean coins - clearly not the case - however I thought I'd disprove it by doing an experiment.

So I added NaCl to ethanoic acid and measured a 0.5 reduction (3 to 2.5) in pH using a pH probe. This seemed a bit weird so I repeated using tap water and the same happened (7 to 6.5).

The pH of water when NaCl is added should not change from 7!

The probe measures by detecting a change in voltage through the solution so it could be that, but you'd think that salts shouldn't have such an effect or be adjusted for. Can anyone repeat please?

What's going on?

This is probably an effect in which sodium ions from the salt are being weakly detected by the glass electrode, giving the appearance of a lower pH. Try add the same concentration of salt to samples of your reference buffers and recalibrate your pH meter to those before measuring the pH of the salt water. I would expect you would see pH 7. Also, keep in mind that the pH of unbuffered water is very sensitive to low concentrations of certain impurities -- from the air or from the added salt. For example, carbon dioxide in the air will lower the pH over time.

If you add NaCl to any neutral or acidic solution, the pH will decrease by a factor of at least 0.5. This is due to the fact that the pH is not the log of the concentration calculated with respect to the total volume, but the concentration should be calculated with respect to the volume of free water. In concentrated NaCl solutions, nearly all water molecules are adsorbed around Na+ and Cl- ions. The apparent concentration of H3O+ increases a lot for lack of water.


In which pH range are silica-based TLC plates stable? I am trying to understand why the experiment asks for raising the pH of a fairly acidic solution. The analytes should not be affected by it, so I thought the silica plate might be. Are silica-plates vulnerable in acidic conditions and, if yes, what happens?

Assuming "normal" TLC with silica as stationary phase, and a mix of ethyl acetate / hexane as eluent, charged compounds (i.e., salts) do not elute well; their adsorption is too tight to move forward.

Hence, if your reaction is performed under acidic conditions (sulfonation reaction, for example), drop a small sample of your reaction into a test tube or small vial, gently quench it against an aqeuous solution of $\ce{Na2CO3}$, extract it with some ether / MTBE / methylene chloride (well...) and spot the extract on the TLC plate.  With some training, this works easily around o.5 mL of total volume.

Conversely, if your reaction proceeds under basic conditions (like a Grignard reaction, for example), bring the sample back to neutral, for example with dilute HCl (or, with more fragile products like tertiary alcohols, with an aqueous solution of $\ce{NH4Cl}$), and then spot the organic extract on the plate.

I referred to "normal" TLC as there are other stationary phases than silica, too.  Alumina (basic, neutral, acidic) works with the same principle as silica.  These two contrast to applications of RP-TLC, were the stationary phase is coated with alkyl chains, and much more polar solvents are deployed as eluents, hence with much lesser issues of free salts / free acids. But the price tag for these plates is higher than for normal phase TLC.


Does ice increase the humidity of its surroundings? Does ice increase the humidity of its surroundings? I'm trying to use the BME280 for an Arduino project regarding locating ice, but I don't know whether the atmosphere is more humid around ice. I'm trying to see whether humidity increases near ice.

It depends on the current humidity.

Water vapor certainly does evaporate from ice, as you can see in a vapor pressure chart. If the humidity is 100%, and the air temperature is the same as that of the ice, then the evaporation and condensation are in equilibrium; i.e. water evaporates as fast as it condenses, so there is no net change in humidity.

However, if the ice is warmer than the air, or the air has less than 100% moisture content at that temperature, then ice will evaporate more than it condenses, raising the humidity.

Conversely, if the ice is colder than the atmosphere, it can lower the humidity.

No. It is the contrary. Water from the atmospheric humidity is condensed on objects whose temperature is 0°C.


Does a catalyst increase atom economy? Somebody I know insists that the use of a catalyst increases the atom economy. They did chemistry at school and were told that a catalyst increases the atom economy. He pointed me to several past exams where this was written as a valid answer.

However, I am convinced that this is false. After all, atom economy is:

$$\text{atom economy} = \frac{\text{molecular mass of desired product}}{\text{molecular mass of all reactants}} \times 100\%$$

I feel this has nothing to do with a catalyst.

Am I missing something?

There are a couple of ways to think about this. First, considering your atom economy equation. Typically a catalyst is used in place of a stoichiometric reagent. A catalyst is (strictly speaking) not a reagent, so its effect on the atom economy is neutral. The alternative stoichiometric reagent is a reactant, so it will reduce the atom economy. This leads to the conclusion that a catalytic method offers a better atom economy than an alternative, non-catalytic method.

From a slightly different point of view, atom economy is one aspect of measuring the amount of waste that comes out of a reaction. A more relevant quantity is really the ratio of the mass of the desired product(s) to the total mass of all of the reaction inputs (including reactants, catalysts, solvents, purification materials, etc). Since catalysts can be used in smaller amounts than stoichiometric reagents, there will physically be less mass put into the process, which increases that ratio.

Catalysts do have an effect on the atom economy, since catalysts have act sides, they eliminate  side products( waste), and they are not considered in the equation for atom economy, thus they will increase the atom economy.


Is there a simple field test for heavy water? Thought experiment

Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment?

Something like the way a piece of litmus paper (yes, it is laboratory equipment and easily available to the layman at that!) may be used to test whether a solution is an acid, or alkali.

There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.

If you have two glasses—knowing that one contains deuterated water, the other contains potable water; one might determine the contents by freezing both samples in a freezer at $\pu{0 ^\circ C}$. Since the melting point of the potable water is $\pu{0 ^\circ C}$ and the melting point of the deuterated water, $\ce{D2O}$, is $\pu{3.8 ^\circ C}$, the $\ce{D2O}$ will take longer to melt assuming they are side-by-side.

I would add, having done this,  practically speaking: this means you can hold the glass of frozen $\ce{D2O}$ in one hand and $\ce{H2O}$ in the other and while the regular water melts, the frozen $\ce{D2O}$ takes noticeably longer to melt.  And on a side note: imagine some of the toxicity in Eukaryotes is due to the subtle differences in thermal properties between these two water species.

Be aware: $\ce{D2O}$ must be in an enclosed container, otherwise it will exchange heavy hydrogen in the water for atmospheric light hydrogen in atmospheric moisture. If left uncovered, the experiment may work a couple times at first; however, it will become less effective (as the deuterium exchanges and evaporates).


Hydrazine as a reduction agent When using hydrazine as a reduction agent to reduce graphene oxide, which is the product hydrazine oxidises to?

I'm sorry for the wrong answer initially given; you might consider to unaccept this one.

I'm not sure whether this has ever been examined in detail. It is likely that nobody bothered to collect the (volatile) reaction products of hydrazine.

According to a procedure provided by the group of Rodney Ruoff (DOI), the reduction was performed in an open flask equipped with a reflux condenser. An aqueous suspension GO was first sonicated (to support exfoliation) until it became clear, then hydrazine hydrate was added and the mixture refluxed at 100 °C for 24 hours. During this time, the reduction product precipitated as a black solid.

No remarks on what actually happened to the hydrazine were given in this article.

The decomposition reaction of hydrazine ($\ce{N2H4}$) is as below:
$$\ce{N2H4 \rightarrow N2H2 + H2}$$

The forming $\ce{H2}$ will reduce the graphene oxide and hydrazine itself oxidises to $\ce{N2H2}$.


Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't? Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem.

TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot.

First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting:

The natural ionicity, $i_\ce{SF}$, of each $\ce{S-F}$ bond [in $\ce{SF6}$] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$, resulting in a sulfur center of charge $+2.69$. [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$. The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization.

If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$. So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram.
You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way:

The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation, although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak.
(An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910; see also refs. 12 and 13 in that article.)
Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$, then those "correct" resonance forms that we drew would involve $\ce{H-}$. I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable.
Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$, the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised.
In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$. Is there anything better? Well, there's the species $\ce{CH6^2+}$, which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$, and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258. While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$.
Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much. Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details.

Notice also how the Group 1 metals form peroxides much like hydrogen.
Why does potassium form peroxides but sodium does not?

Alkalai Metals bond with Sulfur/Chalcogens in a 2:1 ratio and Hydrogen is behaving like an Alkalai Metal.

Lithium Sulflide = Li2S

Sodium Sulfide = Na2S

Potassium Sulfide = K2S

Rubidium Sulfide = Ru2S

Cesium Sulfide = Ce2S

Francium Sulfide = Fr2S

.
.
.

Hydrogen Oxide = H2O

Hydrogen Sulfide = H2S

Hydrogen Selenide = H2Se

Hydrogen Telluride = H2Te

If Hydrogen were like a Halogen, it would form H4S, and H6S. However Hydrogen doesn't obey the octet rule like halogens as it only needs to get rid of one electron, and yet it shares 2 electrons in a bond with Sulfur. That is exactly what the alkali metals do: they need to get rid of one electron and they share 2 with sulfur.

  Na2S adopts the antifluorite structure, which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+. -https://en.wikipedia.org/wiki/Sodium_sulfide

Thus an atoms lacking or exceeding by 2 tends to form 2:1 ratios with atoms lacking or exceeding by 1.

Oxygen is really the problematic chalcogen: we are used to the Halogens behaving like hydrogen: See Dioxygen Diflouride and Hydrogen Peroxide.  And then Hydrogen goes out and acts like an alkali metal.


Difference between exothermic and exergonic In High School I learned that an exothermic reactions releases energy, while an endothermic reaction needs energy to occur. Now I learned that there is a separate, somewhat similar classification scheme of exergonic and endergonic reactions.

What is the difference between these two classification schemes? Are exothermic reactions always exergonic, and if not, can you give me an example?

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$.

If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}H < 0$, an exothermic reaction that releases heat to the surroundings (temperature increases)
$\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat
$\Delta{}_{\mathrm{R}}H > 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases)

In 1876, Thomson and Berthelot described this driving force in a principle regarding affinities of reactions. According to them, only exothermic reactions were possible.

Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, even during cold winter? Thanks to works by von Helmholtz,  van't Hoff, Boltzmann (and others) we may do. Entropy $S$, depending on the number of accessible realisations of the reactants ("describing the degree of order") necessarily is to be taken into account, too.

These two contribute to the maximum work a reaction may produce, described by the Gibbs free energy $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants ("degree or order") may change ($\Delta_\mathrm{R} S$ may be large).  For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}G < 0$, an exergonic reaction, "running voluntarily" from the left to the right side of the reaction equation (react is spontaneous as written)
$\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or
$\Delta{}_{\mathrm{R}}G > 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction)

Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction:

Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -> 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H < 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S > 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G < 0$).
Example, reaction of dioxygen to ozone, $\ce{3 O2 -> 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G > 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S < 0$) and simultaneously it is endothermic ($\Delta_{\mathrm{R}}H > 0$), too.
Water gas reaction, where water vapour is guided over solid carbon $\ce{H2O + C <=> CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S > \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic.
Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -> 2 H2O}$.  This is an exothermic reaction ($\Delta_{\mathrm{R}}H < 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S < 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| < |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction.  In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run.

After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction.

Yes, all exergonic reactions are exothermic. Consider a reaction occurring spontaneously we know that energy would be released i.e. '$\ce{\Delta H}$ is negative' (since a reaction or process absorbing energy makes it non-spontaneous) and according to the second law of thermodynamics, entropy(or disorder) of the system must increase.

Negative $\ce{\Delta H}$ and increasing, positive entropy together make $\ce{\Delta G}$ negative according to the equation: $\ce{\Delta G = \Delta H~ - ~T\Delta S}$ (where $\ce{\Delta }$ = change; G= Gibb's free energy; H= enthalpy; T= Thermodynamic temperature and S= entropy).
Therefore, if enthalpy change is negative and change in free energy is negative, they are both (respectively) exothermic and exergonic.
The same applies for endothermic and endergonic.


Is methanol more acidic than water? Methanol is slightly more acidic than water. Their $\mathrm{p}K_\mathrm{a}$ values, in water, are $15.5$ and $15.7$, respectively. All other aliphatic alcohols, however, are less acidic than water.

Is the following reasoning correct? This is my best rationalization; is there anything better or anything that can be added?

In both hydroxide and methoxide, we have an $\ce{O}$ bearing negative charge. In the former, we have a hydrogen attached to the $\ce{O}$. In the latter, we have a carbon attached to the $\ce{O}$. Carbon is more electronegative than hydrogen, therefore, carbon should be able to better withdraw electron density via induction from the $\ce{O}$. This is a stabilizing interaction.

We often think of methyl groups as inductively donating, but that's molecular profiling. They can inductively donate to a carbocation, partly because a carbocation is highly electronegative, so the carbocation pulls the electrons toward itself pretty well (so there's both pushing and pulling of electrons going on). When you have a methyl group attached to an $\ce{O}$, however, the methyl group is often inductively withdrawing. NMR data supports this.

Now the question is why aren't other aliphatic alcohols more acidic than water? In longer-chain aliphatic alkoxides, you don't just have a methyl group attached to the $\ce{O}$ bearing the negative charge - you have a bunch of methyl groups strung together. These lessen the amount of inductive withdrawal that the alpha carbon can do. Each$\ce{-CH2 -}$ unit attached to the alpha $\ce{-CH2 -}$ is somewhat inductive donating to the alpha $\ce{-CH2 -}$.

The relative acidities of these molecules in the aqueous solution$^{[1][2]}$ is:

$$\text{methanol > water > ethanol > isopropanol > t-butanol}$$

The relative acidities of these molecules in DMSO$^{[3]}$ is:

$$\text{methanol > ethanol > isopropanol > t-butanol > water}$$

The problem with only considering the acidities of these molecules in aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play.

The acidity trend in DMSO is explained by polarizability and the anionic hyperonjugation effect$^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol.

The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents.

  For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ($n$) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect.

  $\hspace{3.8cm}$

  Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$, and hence a greater stabilizing interaction$^{[5]}$.

Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules.

In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater affect on the alcohol's acidity than the dispersion of the negative charge, and overrides the the intrinsic acidity, inverting the order of acidity.

$^{[1]}$ Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[2]}$ Wikipedia, Ethanol
$^{[3]}$ Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[4]}$ ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO
$^{[5]}$ ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?

Methanol is more acidic than water because its conjugate base that is methoxide is weaket than waters conjugate base that is hydroxide and now we can say that weaker is the conjugate base more is its acidic character . or we can say that due to the presence of methyl group in CH3 bond between O and H become weak and it can easily release hydrogen where is in H2O bond between O and H is stronger so it will not easily release hydrogen and become a weak acid than methanol


Which chemical reactions don't involve bond breaking? I recently came across a question in which an option is 'Breaking a chemical bond is the first step in any chemical reaction', which lead me to think of reactions which do not involve bond breaking. The only reaction I can think of is dimer formation, but a Google search reveals no more. Surely this cannot be the only example?

The reaction of a Lewis acid with a Lewis base results only in bond formation. Your question gets at the essence of chemistry, though. A chemical reaction is the rearrangement of electrons between atoms. So generally, most reactions will involve breaking bonds. Even for monomers (e.g. alkenes) to polymerize, $\pi$-bonds within the monomers have to be broken to form $\sigma$-bonds in the polymer. You seldom have one without the other.

Per a source, to quote:

  Like ozone, the predominant oxidation reaction mechanism for chlorine dioxide proceeds through a process known as free radical electrophilic (i.e. electron-attracting) abstraction rather than by oxidative substitution or addition (as in chlorinating agents such as chlorine or hypochlorite).

So, the reaction:

$\ce{.ClO2 + e- -> ClO2-}$


Using combined gas law to calculate the volume of dry hydrogen gas Volume of wet hydrogen collected at room temperature: $\pu{40.48 mL}$, ambient pressure: $\pu{29.7 inHg}$ ($\pu{754 torr}$), ambient temperature: $\pu{295 K}$ ($\pu{22^\circ C}$), vapor pressure of water at $\pu{22^\circ C}$: $\pu{19.83 torr}$. Partial pressure of dry hydrogen: $\pu{9.87 inHg}$.
Find the volume of hydrogen under standard conditions.

The equation of combined gas law is $p_1V_1/T_1 = p_2V_2/T_2$.
Say if the volume of wet hydrogen gas is $\pu{40.48 mL}$, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?
One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of wet hydrogen if we are trying to find the moles of dry hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.

Consider the composition of the gas in your eudiometer.  Most of it is hydrogen, but some of it is water vapour.  In other words, not all the pressure is exerted by hydrogen, part of it is due to water.  You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.

You should have been given vapour pressure of water in a table.  You should also have taken down value of air pressure and ambient temperature.

WAIT! I got it;
the "volume of the wet hydrogen gas" is ALSO the volume of the DRY hydrogen gas. That's because Dalton's Law of partial pressure assumes/ is for CONSTANT volume. IE * at constant volume *: the total pressure of wet hydrogen at x mL = vapor pressure at x ML + dry hydrogen pressure at x mL. It's x mL for both.

Another way to think about it is IF the volume of the dry gas were less, then that would mean the pressure of the dry gas is more than the one we calculated for using Dalton's law of partial pressure * at constant volume * (Boyle's Law) and we are keeping the pressure that we calculated.


Stability of cis vs trans isomers? Recently, while reading about isomers I learned that cis isomers are more stable than trans isomers.

I referred many books as well as searched around the internet but I didn't find a suitable explanation.

Could someone please explain this to me?

As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration.
For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference between the two methyl groups either side of the double bond. However, there are exceptions to this rule which I couldn't find a quick explanation for. The rule applies to both cyclic and acyclic compounds although there are some obvious cyclic exceptions such as cyclic alkenes, where the cis isomer is more stable, especially for smaller rings, due to the large ring strain caused by the orientation of the trans isomer. This page has a good illustration of this:

cis is more stable than trans as long as no. of constituent C atoms is less than 11.
For more than 11 and for acyclic ,  trans is more stable....the reason is that like radicals and charges repel
however for less than11 the repulsion will be balanced best by cis


How Does a Precipitate Get a Colour? If you can use precipitate colours to determine an anion, what makes the precipitate have that particular colour?

Like for example what in sodium sulfate or barium chloride makes a white precipitate when these two combine?

The 'white colour' is due to the fact that the product does not absorb light in the visible part of the spectrum but forms as many fine crystals which in themselves scatter room lights. So the 'whiteness' is due to the scattering of light back towards your eyes after the light is repeatedly reflected of the many facets of the many crystals in the ppt. If you use red light to illuminate then the 'white colour' will become red. If the ppt is coloured then absorption of light occurs at some wavelengths as well as scattering at all wavelengths.

At the surface of a crystal some small fraction of light is reflected, but with millions of randomly oriented crystals numerous reflections occur. The amount of reflection at each surface depends on the difference in refractive index between the crystal and air for the wavelength of light used. If you could add a solvent say dichloromethane which has  a high refractive index and will probably not dissolve the ppt then the crystals can be made to almost disappear as scattering is now far less.  You can do the same thing with frosted glass beads which, when in air, scatter a lot but in water become virtually invisible.

The following answer applies to all questions regarding color.

Color is the result of a substance absorbing certain wavelengths of light and reflecting others.

Different molecules have different shapes (geometry). Different shapes absorb/reflect different wavelengths. In addition to the shapes, the other parameter regarding color is the temperature and pressure.

Closely packed molecules will absorb/reflect light different from loosely packed ones.

2 more things that can "produce colour" rather than reflection of photons hitting it, are:

photons generated from chemical reactions and
photons generated from radiation


What is the physical state of the gas adsorbed on a solid adsorbent? I am studying surface chemistry in high school I have this doubt and I am unable to find a answer your help is appreciated
 I searched for the answer on google I asked my chemistry teacher he said the question is beyond high school's curriculum and I even asked some of my friends but all in vain.

Well, it usually depends on the temperature: if temperature is too low, molecules cannot move so freely so they are localized on top of the solid without much movement.

However, in general cases the adsorbed gas behaves like a liquid in two dimensions. Near the surface, the gas molecules are 'trapped' in the force field of the solid, so they don't escape so easily. Then the freedom of movement occurs only in two dimensions (the molecules travel on the surface).

I made some time ago a movie calculating (using molecular dynamics) movement nitrogen molecules adsorbed on a graphene surface at 77 K. This can give you a clear image of what happens on the surface. As you can see, the molecules can travel on top of the graphene in two dimensions (we can call it a 2D phase). This movie is the case when still a monolayer of gas does not complete. There can be cases with less molecules, say, less pressure of nitrogen in the gas phase; or more molecules.

The physical states can be observed more clearly in phase diagrams of the adsorbed surfaces. Typical examples shows a plot of the adsorbed phase transitions in a P vs. 1/T plot. This topic is extensively covered in the nice monograph on Physical Adsorption by Bruch, Cole and Zaremba, for example.

Since solid means ordered structure and when gas gets adsorbed orders itself on the surface (at least partially) and is not free in volume, I would consider it a solid.


On the activity of liquids and solvents in equilibium constants This is a confusion I've had for quite sometimes, first when learning about equilibria (due to contradicting teachers), then when trying to teach it (due to contradicting sources, in books and on the Internet).

When facing an equilibrium (let's be original and take $\ce{a A + b B <=> c C + d D}$ as an example), the equilibrium constant is
$$K = \frac{a(\ce{C})^c \cdot a(\ce{D})^d}{a(\ce{A})^a \cdot a(\ce{B})^b}$$
where $a(\ce{X})$ is the activity of X.

For pure solids, the activity is 1; therefore solids "do not appear" in the expression of the constant.
For dilute solutions, the activity of the solute can be taken as its molar concentration. Thus, supposing A, B, C and D are not solids, the equilibrium constant can be written as

$$ K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

But what happens when facing liquids (e. g. water)? In that case, I have seen two things :

Just putting the "concentration" of the liquid into the equilibrium constant.
A pure (or nearly pure, such as the solvant in a dilute solution) liquid has an activity of (or nearly of) 1. Therefore, the liquid should not appear in the constant.

Let's apply both cases to acidity constants, for the reaction $\ce{HA + H2O <=> A- + H3O+}$.

As $[\ce{H2O}] \approx 55.55$ mol/l in dilute solutions,
$$K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}][\ce{H2O}]}
\textrm{ and }
K_a = \frac{K_c}{[\ce{H2O}]} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
As $a(\ce{H2O}) \approx 1$ in dilute solutions,
$$K_a = K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Which way is correct? I'd be more inclined to follow the option #2, but I've been stared at wide-eyed like a heretic when talking about that.

I concluded on this question after discussion with the physical chemistry professor at my home university, and reading some physical chemistry books.
Although many references include the concentration of water in the expression of Ka, this is incorrect: the activity of pure water is 1. In dilute solutions, $a(\ce{H2O}) \approx 1$.
Therefore, the second option is correct:
$$K_a=K_c=\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
This has a direct consequence on the values of pKa for the two couples of water: $pKa(\ce{H3O+}/\ce{H2O})=0$ and $pKa(\ce{H2O}/\ce{OH-})=14$, instead of the values $-1.7$ and $15.7$ that are often found in textbooks (which result from including $[\ce{H2O}]=55.55$ mol/l).
A good article on this matter can be found here: What is the pKa of water?.
References:

B. Leyh, Department of Chemistry, University of Liège, Belgium.
Atkins, P., & De Paula, J. (2006). Atkins’ physical chemistry (8th ed.). London, England: Oxford University Press.
Reid, P., & Engel, T. (2006). Physical Chemistry. Philadelphia, PA: Pearson Education.

You should go through the derivation of the equilibrium constant from thermodynamics and see exactly what is meant by activity, and the importance of defining a "standard state". The activity of water in a dilute solution is very nearly 1 because our standard state assumes pure water. We use the concentration of solutes because the standard state typically involves 1M solutions. Using the concentration of water is wrong unless you define your standard state with 1M water (in...something?).

If you want to be extra careful, you might write the activity of, for example, C in your first equilibrium as
$$\left( \frac{[C]}{[C]°} \right)^c$$
Where the little ° means "in the reference state".


Meaning of depiction of atomic orbitals There was a section about atomic orbitals in my organic chemistry textbook that I did not quite understand. First the author explained a bit about the Schrödinger equation, which I understood to be (vaguely, they did not go to any detail, even omitting the formula all together) a equation that describes the likeliness of positions of an electron in 3d space. $\Psi$ (i.e. the Schrödinger equation) can be solved for different quantum numbers that respectively represent the different orbitals.

This made me wonder what a picture (like below) of an orbital actually depicts, as it are chances that are described by $\Psi$, not boundaries. My best guess was that such picture represents an area that has a high chance, say 90% or more, of containing the electron (how they determine the shape, I don't know, but my best guess again here is that they only take those parts of the real orbital that add a significant percentage to the total chance of the electron actually being in the depicted orbital).

Are all these assumptions right?

Without going into a long discussion about what the wave function is, I will try to briefly answer your questions:

1) For atoms, the (approximate) solution to $\Psi$ is the product of two functions:

the radial components (distance from the nucleus)
the angular/spherical components (behavior going around the nucleus, like latitude and longitude)

2) Since the radial components of each component extend quite far, decaying to zero only at infinity, it would be very difficult to make a plot of the 3D object on a 2D surface that does not cover up the features that would be nearer the nucleus (further from eye), without using varying degrees of transparency.

The spherical component is usually solved analytically using complex numbers (i.e. a+b*i) but they can be transformed into real values by linear combinations. So, instead of two "channels" for real/imaginary, we can use only one "channel" (real) for each function, 1s, 3dxy, etc.  Often, these are colored so that if the sign of the real function is positive, it is e.g. red, otherwise, e.g. blue.

A second feature of these types of graphics is called the "isosurface"--which means, plot only if the value of the function is equal to a chosen value.  Sometimes the chosen value is taken to be "90% of the corresponding electron density contribution from this atomic orbital is enclosed" (which, b.t.w. is $\Psi^* \Psi$ or simply $\Psi^2$.  Though $\Psi$ may be real/complex/negative/positive valued (hence need all sorts of colors and "channels" to represent it) the electron density is positive and real.  This is very important, as complex quantities are not observable, but are rather mathematical constructs (vehicles) for arriving at things that are actually observable, such as the electron density.

But practically speaking, in order to see certain features like the doughnut in $dz^2$, people play a little fast and loose with precisely what the value of the isosurface is.

The shading on the surfaces that you have provided could be misinterpreted, because the colors do not correspond in any way to the value of the function being plotted, but rather seem to be an aid to the volumetric rendering of an isosurface that has already been set to either the absolute value or the square of the atomic orbital.

Finally, note that in your center drawing, which adds all of the individual atomic orbitals, that you are beginning to form a sphere.  It had better, as atoms are, in fact, spherical!${}^{\dagger}$  The formal name of the functions are spherical harmonics, and the set of e.g. $p_x$, $p_y$, $p_z$ or the five $d$'s taken together form a spherical shape.

${}^{\dagger}$ assuming the nuclear quadrupole is zero

What I understand of orbitals is simply that ELECTRONS can not be pin-pointed accurately, because they are in continous motion and that too not in a fixed circular path (as Bohr had depicted), so it is virtually impossible to know where an electron is in an atom. So scientist found a way to counter this problem by providing another model - THE QUANTUM MECHANICAL MODEL of the atom, where instead of electrons residing in orbits, it was stated that electrons have a possibility that it may reside in a particular region - what we call as orbitals. It is quite intresting to note that the probality of finding an electron never becomes 0, which means an atom in USA can have its electrons in Australia (atleast in theory).


Which element has a greater second electron affinity, sulfur or oxygen? I found a question asking the above which states that sulfur has a lesser second electron affinity than oxygen. But since the inter-electronic repulsion in sulfur is lesser, shouldn't it be willing to take up another electron more easily than an oxygen ion with a -1 charge?

The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$  orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$).

The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy.  This dependence with quantum number scales approximately $E\sim - n^{-2}$ so is very important [1].

The dependance of orbital energy also scales quadratically with effective nuclear charge, however effective nuclear charges falls significantly compared to the true nuclear charge due to shielding and repulsion from other electrons.  A common first order approximation for the effective nuclear charge is given by Slater's rules [2]

Combining these two approximations we can calculate very approximate orbital energies.

$$
\begin{align}E_{2p}(\ce{O^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(8-4.15)^2}{2^2} \\
& {} = - 3.7 \text{ Ry} \\
E_{3p}(\ce{S^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(16-11.25)^2}{3^2} \\
& {} = - 2.5 \text{ Ry}
\end{align}
$$

Where 1 Ry is the electronic energy of a hydrogen atom.

As you can see, to first order, the energy of the new electron is much lower when added to $\ce{O-}$ than when added to $\ce{S-}$, leading to a more favourable electron affinity for $\ce{O-}$ than $\ce{S-}$, driven by the fact that the orbital occupied has a lower principal quantum number $n$

[1] - See the energy of a hydrogenic ion - Wikipedia.

[2] - Slater's rules - Wikipedia

Sulfur  has 3 shells while oxygen has 2.This increase in shell reduces the nuclear charge and lessens the second I.P.Further Explanation:

Due to the extra shell, the electron in the outermost shell is held , but not as strongly as an oxygen atom has held its electron. So the second Ionization Enthalpy of sulfur will be less than that of Oxygen,because more energy is required to extract the electron from oxygen atom.


Isonitrile to nitrile conversion If we are given $\ce{CH3NC}$ and told to convert it into $\ce{CH3CN}$ by doing some chemical reactions.
My attempt
According to my textbook there are no chemical reaction of isocyanide conpounds.  I tried reduction of isocyanide giving a secondary amine. Then I don't know how to convert a secondary amine to cyanide compound.
Hints are too appreciated.
Edit:
I have got a solution but I don't know whether the reaction are spontaneous.

I have written these reaction on the basis of theoretical knowledge I don't know whether they are spontaneous or not. Correct me if I am wrong somewhere.

Just heat it up: Isonitriles can be thermally rearranged to nitriles.

$$\ce{R-NC ->[\Delta] R-CN}$$

Please have a look at some references, such as:

Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc., 1972, 94, 2704-2706 DOI

Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem., 1987, 52, 648-652
DOI

UPDATE

As far as the sequence outlined in the question is concerned:

The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life. Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth., 1961, 41, 13.
N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions.
Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts.
You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it!
For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter.

In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare.

It's too simple guys first you have to convert isocyanide into primary amine. It can be done by hydrolysis of isocyanide, you'll get primary amine and formic acid.
Then after getting primary Aimne, on reaction with Cu/500°C or Ni/500°C you will get Cyanide.


Aldol condensation in acidic medium Attempt:

I did aldol condensation in acidic medium followed by $\ce{E1cB}$ to get 1,3-diphenylbut-2-enone but the answer given is:

I am unable to devise a mechanism to reach the above product from the reactant. Could someone please suggest the mechanism?

I have drawn a three step mechanism as opposed to the concerted step reaction proposed in the other answer.

This mechanism has the added benefit of not leaving any room for formation of any side products as there is only one source carbon of acidic hydrogens in each compound.

The only mechanism I see is, you write the three molecules in enol form (acidic medium) and try to form a cyclic compound in a concerted step as you did in forming benzene from acetylene and then as it is acidic medium, you gotta eliminate water molecules from the new cyclohexane to end up at the product.


Is lithium is less reactive than sodium? Lithium has most negative electrode potential still its reaction with water is less vigorous than that of sodium

You can't just take the electrode potential as if that's going to correspond to the rate in which something is going to react.  For example, why do you think that aluminum is so stable in atmospheric conditions versus iron?  Aluminum forms a stable oxide layer that serves to protect it against further reaction, while iron oxide is weak, adheres poorly and is and porous.  There are numerous factors which affect a rate of reaction, and they vary in different situations.

Calculating rates of reaction is a very complicated thing, and it's usually easier / more accurate to just determine the Arrhenius parameters experimentally and use those.

Reactivity with water in the sequence Li (mp=180.5C), Na (mp=98C), K (mp=63.5 C) is dictated by melting point of the metal. Electrode potential dictates the equilibrium constant between M and M+. With voltages >1 the reverse reaction can be ignored.

Here is what happens when alkali metal reacts with water from phisical perspective. It contacts water, temperature rises to 100C (boiling point of water). Lithium remains solid, whereas Na and K turn into a liquid. This liquid spreads on the surface of water creating a large area of contact. Lithium remains solid. LiOH forms on it's surface and prevents further reaction. LiOH does dissolve in water (that is why Li react with water and Al doesn't), but it takes time. At the same time both sodium and potassium leak out of the protective shell of oxides/hydroxides.


What happens when a piece of copper is placed in 1M HCl? I know that the answer is that nothing happens, but the explanation is confusing me in some places. Firstly, we are given that:
\begin{align}
\ce{Cl2 + 2e- &-> 2Cl-} & (E &= \pu{1.36 V})\\
\ce{Cu^{2+} + 2e- &-> Cu(s)} & (E &= \pu{0.34 V})\\
\ce{2H+ + 2e- &-> H2} & (E &= \pu{0 V})
\end{align}

What's confusing me is that in the solution, we'll have $\ce{H+}$, $\ce{Cl-}$ and $\ce{Cu(s)}$, which are all products in the equations (except for $\ce{H+}$). I get that for the reverse equation (oxidation), the oxidation potential is given by the negative of the reduction potential. What I'm wondering is, if we look at copper and chlorine first, is the only option for the $\ce{Cl-}$ in solution to be oxidised (with $E = \pu{-1.36 V}$)? Also, can the copper also only be oxidised (with $E = \pu{-0.34 V}$)? And if both of these are true, then does that mean that these two can't react at all, not because of the total $E$ being negative, but because they can both only be oxidised?

Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.

Your question is: What happens when a piece of copper is placed in 1M HCl?

Do a thought experiment +  use some basic intuitive chemistry.

Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.

Now there are only two possibilities for each substance.

a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.

Now you have the problem well defined: Will Cu -> Cu$^{2+}$?

b) Follow the same reasoning: H$^+$, it can be oxidized to  H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$

c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to  Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$

Now you can ask only two questions:

Can copper metal reduce H$^+$ to H$_2$?

or

Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.

At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation
Ecell= E$_{cathode}$-E$_{anode}$

Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)

Ecell= 0.00- (+0.34) = - 0.34 V

The negative sign shows this not possible under these conditions.

Do the same for chloride ion.

Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell)
Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.

In short nothing will happen to copper in HCl.

P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.

OK, if you takes a piece of copper metal (I employ plumbing copper, high purity) in concentrated HCl the copper, in time, will be cleaned of its Cu2O coating. [EDIT] To quote a source:

  Copper has excellent atmospheric corrosion resistance. It becomes naturally covered with an oxide film, changing its color to dark brown/black in normal atmospheric conditions. Later, a green patina forms on copper outdoors with a varying intensity depending on where and how the surface is exposed.

So, perform an experiment on copper metal taking before and after pictures, and see if it is noticeably different.

Underlying chemistry, per Wikipedia, to quote:

  Copper(I) chloride...is a white solid sparingly soluble in water, but very soluble in concentrated hydrochloric acid. Impure samples appear green due to the presence of copper(II) chloride (CuCl2).

Also, forms a chloro-complex with concentrated HCl:

  It forms complexes with halide ions, for example forming H3O+ CuCl2− in concentrated hydrochloric acid.

So, expected reactions in concentrated HCl, a cleaning off of any Cu2O coating:

$\ce{HCl + H2O <=> H3O+ + Cl-}$

$\ce{Cu2O + 2 HCl -> 2 CuCl (s) + H2O}$

$\ce{CuCl (s) + H3O+ + Cl- <=> (H3O)CuCl2 (aq)}$

I have performed experiments where one adds much NaCl creating a soluble cuprous presence (in the form of NaCuCl2). Wikipedia also cites a preparation to Cupric oxychloride based on this solubility conversion of cuprous per Eq(6) and Eq(7) here, where:

$\ce{ Cu (s) + CuCl2 (aq) -> 2 CuCl (s)}$

$\ce{ 2 CuCl (s) + 2 NaCl (aq) -> 2 NaCuCl2 (aq)}$

There is no other apparent reaction occurring with the freshly cleaned Copper.


Can we really see the bonds? I was thinking is there really bond present at microscopic level or atoms/molecules are just nearby and are connected with force which is not visible(like gravitational force between earth and sun) and we make bonds just for understanding.

All credit to Zhang et al. "Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy" Science Vol. 342  no. 6158  pp. 611-614.

Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.

It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.

Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.

No I don't think that is possible.

It's all about the attractions (the wanting of a full octet) between the atoms, to be honest we don't even have the technology to view the actual atoms itself (if we do then we are at the initial stages). So these "bonds" that you are talking about are just imaginary connections between atoms so people don't get confused or overwhelmed.

Take an ionic bond, one of the strongest bonds out there, we know it is caused by strong electrostatic attractions between the atoms in which one completely takes an electron or gives it up. If there was like a connection between them it would look almost surreal. It would make more sense if they are beside each other.

Now for the covalent bond, in a covalent bond atoms are sharing electrons, in this case the electron clouds merge, this maybe hard to grasp but that is just how this works. That is why for most covalent bonds we look at the Lewis Structures.

Now take a look at an ionic "bond".

Take a look at the covalent "bond".

Check out the double and triple "bond". - Bad Image

By looking at the above picture you can see that a triple "bond" is just two more electrons.

So in reality to make these drawings of atoms easy, we have came up with a bond. There might be "bonds" but one has to note that they are attractions instead. Here I gave you a couple examples, mostly 2D but if you want to really get the idea you should take a look at the 3D structures that include everything. But you have to also note that we don't know how far our current depictions are true, I just gave you the most plausible ones, so don't believe everything as a fact.

To sum it up: You Will Not See The Bonds (Not the sticks and ball kind). Though you may see the attractions.


Why does Benzoin give positive Tollen's Test? Certain alpha hydroxy ketones which can tautomerise to form an aldehyde will give a positive Tollen's Test. I was pretty clear with this idea until I came across this question and the answer given is A,B,C. I can't figure out why Benzoin gives a positive Tollen's Test as it can't give an aldehyde by tautomerisation.

What exactly does happen to Benzoin over here and why does it give a positive test? Is my idea about tautomerisation of alpha hydroxy ketones to aldehydes wrong?

For benzoin, a similar mechanism of tautomerisation (of alpha hydroxy ketones tautomerising to aldehydes) happens , except for one step, where instead of the hydride ion (H-) migrating to the sp2 hydridized carbon, it is the Benzene carbanion (C6H5- or Ph-) that migrates to the sp2 hybridized carbon, since it is more stable than the hydride ion (an example of the stability of benzene carbanion over the hydride ion can be found in the pinacol-pinacalone rearrangement). Other than that, the rest of the mechanism is exactly the same.

Hope this cleared your doubt.


Why is the dipole moment of acetone higher than ethanol? It is true that acetone is less polar than ethanol.
I thought the dipole moment is proportional to polarity, and if so, the dipole moment of acetone should be lower than ethanol.
But the dipole moment of acetone is higher than ethanol.
Why is it higher?

Dipole moment is just one of aspects of solvent polarity, as is well known in HPLC separation  analytical technique where the following effects take part:

Dipole moment interaction, permanent and induced
The above for the whole molecule and for particular bonds
Electron pair donor
Electron pair acceptor (including $\ce{H+}$ donation)

Depending on solute molecules specifics, the solvent polarity order is variable, as various molecules have different responses to above interactions.

See also Using solvent triangle in HPLC

The bond dipole μ is given by:

$ \mu =\delta \,d$.

The bond dipole is modeled as +δ — δ- with a distance d between the partial charges +δ and δ-. It is a vector, parallel to the bond axis, pointing from minus to plus, as is conventional for electric dipole moment vectors.

In your case, acetone has a higher dipole moment than ethanol because it has a carbonyl group more towards the center and thus acetone has lesser boiling point than ethanol.

The definition is from Wikipedia, : https://en.m.wikipedia.org/wiki/Bond_dipole_moment

EDIT : Alcohol hydrogen bonding is the major effect here, as in the case of water. (by @Poutnik)

Thanks for the edit, @MathewMahindaratne

Think for yourself


Haloform reaction with limited bromine We were asked the reaction between 1 mole acetone and 1 mole $\ce{Br2}$ in basic medium. Since only a limited amount of bromine was given I got the final answer as 1-bromopropan-2-one (mono bromo acetone). However, in the given solution the entire haloform reaction was shown with $\ce{CHBr3}$ and acetate ion as the final product.

I am not sure how a single mole of $\ce{Br2}$ is sufficient for this(at least 3 moles are needed, I believe).

Let's take a look at the first product you propose: 1-bromopropan2-one.

The mechanism includes deprotonation by hydroxide to form an enolate. The resulting enolate is better stabilized with more electronegative atoms (Br in this case) and will form more readily than another molecule of acetone turning into an enolate ion. Thus, as the reaction proceeds, the product distribution will be determined by the most stable enolate anion. This will be the tribromoacetyl intermediate enolate ion and will drive the reaction forward. So a whole mol of bromine will be converted completely to the tribromoacetyl intermediate with a 1/3 mol of acetone. Notice how only a 1/3 will react to give the monosubtituted intermediate, than 1/3 mol of bromine will react to give disubstituted product, and lastly the last 1/3 mol of bromine will react to form the trisubstitued product. The last step is displacement by hydroxide via the nucleophilic acyl substitution mechanism (not SN2!) to form acetate and release bromoform. This reaction isn't really great if bromine isn't in excess.

$3$ $\ce{H}$ atoms out of total $6$ in a molecule of acetone is brominated.
So $\frac{3}{2}$ moles $\ce{Br}$ molecules is needed to react with 1 mol acetone.
Hence $1$ mol $\ce{Br}$ molecule will react with $2/3$ mol acetone.
Thus acetone is the limiting reagent and the reaction is completed.


How many valence electrons do elements in the d block have? I just noticed that on the periodic table it doesn't say how many valence electrons there are for each column in the d block.

How do I find out how many valence electrons elements in the d block have?

The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell.

But it is wrong to apply the concept to determine the valency of the element, because that varies with the compound it is present in. For example, iron can show two valencies, $2$ and $3$ in the compunds iron(II) sulphate ($\ce{FeSO4}$) and iron(III) sulphate ($\ce{Fe2(SO4)3}$) respectively. But from the above definition, iron has $8$ valence electrons.

As a matter of fact, this applies even to many elements outside d-block too, for example chlorine typically shows a valency of ($-1$) in compunds like $\ce{NaCl}$, but it can also show a different valency (more appropriately, oxidation state) in ions like perchlorate anion ($\ce{ClO4^-}$), where its valency is $\mathbf 7$ (while the oxidation state is $\mathbf{+7}$, notice the difference that there is no plus or minus sign in valency).

Elements in the d block are transition elements, and each posses one or two valence electrons in their respective s orbitals.  Most have two, with several notable exceptions: elements may "steal" an electron from the outermost s block and relocate it to the d block in order to reach a filled or half-filled ($5$ of $10$ electrons) state in the d block.  These elements must be taken on a case-by-case basis.


Why is oxygen an oxidizing agent? Why are diatomic oxygen molecules STILL reactive especially with metallic elements like sodium and copper even at room temperature?

You would think that since the two oxygen atoms already have the much needed 8 valence electrons when they bonded with each other that they wouldn't need to react with anything else but that doesn't seem to be the case.

To estimate reactivity solely based on the the octet rule is a crude, (semi-)empirical approach in the first place. Atoms and molecules don't know anything about their number of electrons in shells, but the formation of molecules are based on the minimization of (free-)energy of the system.
The octet rule gives a hint, for what kind of systems the potential energy may be small, but does not say anything about the relative energies of two systems, both fulfilling the octet rule.

To "exactly" predict the reactivity, you would have to solve the Schrödinger equation for both states of your systems, find the eigenvalue of the energies, consider entropic contributions. Furthermore you would need to "sample" all possible pathways of your reaction, to check if the energy barriers are small enough to be overcome at a given temperature.
(To be really exact, you still would need to estimate entropic contributions)

Yet still a chemists' intuition often can predict reactivity very well. Concerning your specific problem, one possible qualitative explanation would be, that the octet rule is not fulfilled for copper, but only for oxygen, before the reaction happens. Taking a look on the structure of Copper(II) oxide on the other hand, shows that it fulfils the octet rule for both, copper and oxygen.(Ignoring d-electrons for copper)

Furthermore, the reaction enriches the electron density in the strongly electronegative oxygen, while reducing the electron density in the "electropositive" copper, which also stabalizes the structure of copper oxide.

Oxygen (O2) generally exists as diradicals i.e. each oxygen bonded to each other through single bonds and the remaining two electrons remains on each oxygen atoms as radicals. So this structural feature makes oxygen act as a strong oxidizing agent. Even in our body oxygen coordinates through its radical and this structural arrangement makes oxygen can act as a strong oxidants in many places.


Na+ or Ca2+ . Which ion has the larger radius? Which ion has the larger radius? Some sites say $r(\ce{Na+}) > r(\ce{Ca^2+})$. Any explanation would be appreciated.

The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]:

\begin{array}{cc}
\hline
\text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \\
\hline
\ce{Na+} & 4 & 1.13 & 0.99 \\
& 5 & 1.14 & 1.00 \\
& 6 & 1.16 & 1.02 \\
& 7 & 1.26 & 1.12 \\
& 8 & 1.32 & 1.18 \\
& 9 & 1.38 & 1.24 \\
& 12 & 1.53 & 1.39 \\
\hline
\ce{Ca^2+} & 6 & 1.14 & 1.00 \\
& 7 & 1.20 & 1.06 \\
& 8 & 1.26 & 1.12 \\
& 9 & 1.32 & 1.18 \\
& 10 & 1.37 & 1.23 \\
& 12 & 1.48 & 1.34 \\
\hline
\end{array}

where C.N. - coordination number; C.R. - crystal radius; I.R. - ionic radius. The data is also available online for free. Comparing similar coordination environments, it's easier to notice that overall $\ce{Na+}$ is indeed larger than $\ce{Ca^2+}$:

Reference

Shannon, R. D. Acta Cryst A, 1976, 32 (5), 751–767 DOI: 10.1107/S0567739476001551.

According to the diagonal relation, $\ce{Na}$ and $\ce{Ca}$ have similar radii.

$\ce{Na+}$ and $\ce{Ca+}$ have thus similar radii as well but, when one more electron is removed from $\ce{Ca+}$, making it $\ce{Ca^{2+}}$, its effective nuclear charge increases, which will make its radius decrease.


Is 1,2-dihydronaphthalene "partially aromatic"? I've learned that if any one ring of a structure like the below is aromatic, the whole compound is aromatic.

But one of my friends suggested that such compounds cannot be aromatic and gave the following resonance structures as the reason:

I feel that the non-aromatic/anti-aromatic resonance structures wouldn't make significant contribution. Is my reasoning correct?

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.

What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.

But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.

So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.

For a compound to be aromatic or anti aromatic there should be conjugation throughout the peripheral , this compound lacks conjugation throughout the peripheral so it falls under non-aromatic category.


Why does sodium react with water? The reaction of sodium with water is quite infamous.

$\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$

But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$?

Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.

In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.

Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.

This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.

Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.

Sodium is more "electropositive" than hydrogen as it is down on the periodic table. This makes it more reactive to say elements like fluorine or oxygen. In this same way it is reactive to ions like hydroxide. Since water can turn into hydroxide and hydrogen ion the sodium ion is able to displace the hydrogen ion. This causes sodium to react to hydroxide to form sodium hydroxide and hydrogen. The two hydrogen atoms merge to form a hydrogen molecule.


Why is fluorine the best oxidising agent, but also the most basic? (Please note that all comparisons I have made below are with respect to halide ions only.)

According to what I have learnt, the fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density. This implies that it has the highest tendency of sharing/giving away its electron density.

Why isn't the same logic applicable when reducing strength is compared? Fluorine should have the highest tendency to lose electrons, and hence should be the best reducing agent.

Fluorine is the most electronegative element.
But due to its high effective nuclear charge it holds up its electron density very tightly, moreover its a non-metal so preferably accepts electron to form stable (F–) ion which has configuration of Ne.
F2 ------> F- + e- is very favorable therefore it gets reduced very easily (reduced form F- is very stable than F2) and so it oxidises other elements and self gets reduced.

... fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density.

A look at the Periodic table and the ionization energies is always a good idea to see what’s going on.

The highest energies, required to remove an electron from the atom, are needed for the Noble gases. That means they are most aggressively reclaiming a removed electron back.

The less the number of electrons around the nucleus of Noble gases, the higher the aggressiveness to retrieval a lost electron.
The elements in the column before the Noble gases ($\ce {Fl}, {CL}, {Br}, ...$) are doing the same in an attenuated way. And more than that, in bounds these elements try to complete their electron shell to the Noble gas configuration.

Take the Noble gas compounds. Fluorine is able to remove electrons from Xenon:

The tendency to fulfill the electron configuration is higher for Fluorine when the tendency for Xenon to preserve its electron configuration. The Octet rule works better for elements with high ionization energy when for elements with less ionization energies. (Perhaps instead of ionization energy in this case we have to remember the electron affinity, but this seems to me to be more a scholastic difference in definition.) According the above image Fluorine has a ionization energy of roughly 17 eV, Xenon has about 13 eV. That is why the octet rule in this case “works” for Fluorine and against Xenon.


Can I perform electrolysis without electrolyte? (with distilled water) When performing electrolysis at home, I usually used salt water as electrolyte. The problem is, that this method also produces chlorine and chlorine is aggressive to both humans and electrical equipment.

I was thinking of creating pure hydrogen and oxygen by using distilled water. I was taught that water is not conductive.

But since water has "positive" and "negative" part (which allows for it to dissolve polar bonds, doesn't it?). Couldn't I just tear the molecules apart if I use electrical field strong enough? If I can, how high the voltage must be?

If I can, how high the voltage must be?

Not possible at home.

For obtaining pure hydrogen and oxygen it is still required to use an electrolyte.  However, there is some amount of them that are not electrolyzed, allowing to electrolyze only water. I'd try sodium/potassium hydroxide or sulfate.

I've ran a 240v supply through a container of de-ionised water and even rain water to connect up to a light bulb. The light bulb never worked. I even placed my finger in the waters and that didn't get a shock or trip my rcd. So, imho, water without impurities does not conduct electricity and it's only the impurities that do (don't put your finger in water that has impurities!!) So if water isn't conductive how can it be broken down into the gases by electrolysis? Well, put simply, it can't, people always add an electrolyte to the water but it is the electrolyte that is being broken down, not the water. Hope that clarifies your question.


Does chemistry change under high voltage? Do chemical reactions change when you charge the entire reaction vessel plus or minus $\pu{1 MV}$ or more?  Is there a name for such chemistry?

I was looking at "electrochemistry" expecting to see a host of information on charged reactions (either statically or with a constant charge applied) but it all just appears to be focused on batteries, not so much on synthesizing novel materials.

Long story short, no, they don't.

Charging a vessel to 1 MV is not a big deal, if you look at it from the inside. To put things into perspective, imagine a vessel of about 10 cm across, which is kinda OK for a flask. Imagine it spherical to make the calculations easier.

Now, the capacitance of a sphere is $C=4\pi\varepsilon_0R$, which evaluates to 5.56 pF (yes, "p" stands for "pico").
So charging your vessel to 1 MV implies putting $5.56\;\mu\rm C$ of electric charge on it.
That's $3.48\cdot10^{13}$ extra electrons (or some negative or positive ions, that doesn't matter) ...
... which may seem big until you compare it to the amount of water that would fit into the vessel: $1.75\cdot10^{25}$ molecules.

So it is like 1 electron per quite a few billions of molecules. That would not make much of an influence. The molecules will just sit there as usual, feeling nothing; most of them have never seen an electron in their entire lives, and have no idea that their vessel is being charged.

Things will change a great deal if you touch the vessel from the outside, but that's just ordinary electrochemistry which you seem to put out of scope of your question.

I thought about this too but knew it will work only in space so that the different charges would not be involved into a developing field. If you study idle or colder gasses, they show that negative ions become more stable because of lesser collisions. And the same for positive ions but that is kind of obvious even with collisions. I find this challenging opportunities when studying elektrochemistry (and it's influence of temperature differences which are also more easy to install in space). Think of all molecules sharing electron; one could think of a shortage that could become quite different in space if we could add or diminish charge in processes. My interest has come from discovering the non-existence of electrical repelling. When millions of volts can be stable just being enough isolated; then how could one say that electrons are repelled when charging such spheres??? Eric Dollard proposes the same in his youtube story: the history of energy synthesis (47th minute). Repelling is a postulate and never investigated. The definition of an electrical field is OK but repelling is being "explained" outside this definition. Make two big plates (1 m2) charged equal and have a probe charge in between them (equal charged) and nothing will happen (untill you approach the borders of the plates when charges outside of the plates get their influence. The electrometer is a fraud since the quality of our earth to attract any kind of different charge is being perfected by the earthed shielding and so explains the attracting forces on the blades. Fieldlines don't go to likewise charged items because there is no voltage difference and so no attraction and this last argument is the one from Eric Dollard and his website. If people only would understand this extremely dumm and simple mistake it could revolutionize electrochemistry perspectives for our future. Also get aware of the fact that all elements are elektro negative, meaning they attract electrons (there is no such quality as "electro positiveness!!). This together with the fact that it costs at least around 13 ev to loosen an electron out of it's bonded state and the notion that our earth would be on zero is a total misrepresentation of the truth !!!!


Why is Cr(2+) a stronger reducing agent than Fe(2+) in water? According to me, $\ce{Fe^{2+}}$ should be a better reducing agent because $\ce{Fe^2+}$ - after being oxidized - will attain a stable $\ce{d^5}$ configuration, whereas $\ce{Cr^2+}$ will attain a $\ce{d^3}$ configuration. I think the half filled $\ce{d^5}$ configuration is more stable than the $\ce{d^3}$ configuration. Why is this not so?

The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept.

It is tricky to compare the "stability" of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the change in (free) energy that occurs from start to finish.

To put it another way, it may be that $\ce{Fe^3+}$ is more stable than $\ce{Cr^3+}$ on an absolute scale, but what we really care about is how much more stable $\ce{Cr^3+}$ is to $\ce{Cr^2+}$ compared to how much more stable $\ce{Fe^3+}$ is to $\ce{Fe^2+}$.

Let's examine all four using electron configuration as you have done:

$\ce{Fe^2+}$ is $\ce{d^6}$ or more probably $\ce{[Ar] 4s^1 3d^5}$ - two half-filled half shells!
$\ce{Fe^3+}$ is $\ce{d^5}$, which is $\ce{[Ar] 3d^5}$ - one filled half shell.

The difference between the two iron ions might not be that large.

$\ce{Cr^2+}$ is $\ce{d^4}$, which is $\ce{[Ar] 3d^4}$ or $\ce{[Ar] 4s^2 3d^2}$
$\ce{Cr^3+}$ is $\ce{d^3}$, which is $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^2 3d^1}$

The energy difference between chromium ions might be larger. Represented graphically, the reaction coordinate energy diagram for the two process might be:

  Now the number part.

We can go looking for some standard thermodynamic data to help make our case. The best data are for standard reduction potentials, because these data are for exactly what we want!

Taking the reduction potential data and writing the equations in the direction we care about, we have:

$$\begin{align}
&\ce{Fe^2+ -> Fe^3+} &&&E^\circ=\pu{-0.77 V}\\
&\ce{Cr^2+ -> Cr^3+} &&&E^\circ=\pu{+0.44 V}
\end{align}$$

Spontaneous reactions produce positive potential differences, so we can see right now that $\ce{Cr^2+}$ is a better reducing agent.

Let's go a step farther to free energy.

$$\Delta G^\circ =-nFE^\circ$$

However, to deal with free energy, we need a full reaction. Technically, comparing the two half-reactions in isolation is just as bad. However, the definition of the standard electrode potential and the standard free energy come with a common zero point in terms of half-reaction: $\ce{2H+ + 2e- -> H2}$.

The two full reactions (net ionic equations anyway) are:

$$\begin{align}
&\ce{2Fe^2+ + 2H+ -> 2Fe^3+ + H2} &&E^\circ=\pu{-0.77 V}&&&\Delta G^\circ=\pu{+74kJ/mol}\\
&\ce{2Cr^2+ + 2H+ -> 2Cr^3+ + H2} &&E^\circ=\pu{+0.44 V}&&&\Delta G^\circ = \pu{-42 kJ/mol}
\end{align}$$

And the winner is chromium, by a whopping $\pu{116 kJ/mol}$. In fact, the data might show that $\ce{Fe^2+}$ is more stable than $\ce{Fe^3+}$, while $\ce{Cr^2+}$ is less stable than $\ce{Cr^3+}$.

$\ce{Cr^2+}$ is a better reducing agent. As it attains a $\ce d^3$ configuration on loosing an electron while $\ce{Fe^2+}$ attains $\ce d^5$ configuration. In an aqueous medium $\ce d^3$ (half filled $\ce{t_{2g}}$ orbital) is more stable than $\ce d^5$.