How does absence of nd orbital in Oxygen affect its valency? The original question was 'Oxygen exhibits Oxidation state -2 to +2 but other elements of grp 16 exhibit only +2,+4,+6. Explain Why'. In the solution given i can't understand meaning of this line 'It doesn't have nd-orbitals due to which its valency doesn't increase'. Please explain how?
PS:I am a chemistry noob. So please provide as much details as possible :)

Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.

The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.

Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.

You need to review the concepts surrounding atomic shell structure and quantum numbers, and how they relate to the organization of the periodic table, I think.  Briefly, since oxygen is in the second row of the periodic table, we know that its valence electrons occupy the second orbital shell, corresponding to orbitals with principal quantum number $n=2$.  Recall also that for a given principal quantum number $n$, only sub-shells with orbital angular quantum number $l$ up to $n-1$ are allowed.  Therefore, the valence shell of osygen only includes s ($l=0$) and p ($l=1$) orbitals.  On the other hand, the other members of group 16 will by definition have higher principal quantum numbers, which means that their valence shells include d ($l=2$) orbitals.  It is these orbitals that can be used to rationalize the existence of higher oxidation states for those elements, for example by the $sp^3d$ and $sp^3d^2$ hybridization schemes in VSEPR (valence shell electron pair repulsion) theory.

Now, that explains why heavier elements in group 16 can have higher (i.e. more positive) oxidation numbers than +2.  Your question as stated appears to have an error however, since it says that those elements "exhibit only +2,+4,+6" ... that's not technically correct .. for example there are compounds where sulfur has an oxidation state of -2 (e.g. $H_2S$).  However I don't think it matters for the answer to your question here.


Are these alternate names for long chain hydrocarbons with complex substituents accepted? Let's consider the following compound:

$\ce{CH3-CH2-CH(CH3)-CH2-CH2-CH(CH(CH3)-CH(CH3)-CH2-CH(CH3)3)}$

Its IUPAC name is 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane

$\ce{CH3-CH2-CH(C2H5)-C(CH3)(C(CH3)3)-CH2-CH2-CH(CH3)-CH2-CH2-CH3}$

Its IUPAC name is 3-ethyl-4,7-dimethyl-4-(1,1-dimethyl-ethyl)-decane

I wish to ask if the following names are acceptable (respectively):

4-(1-methyl-propyl)-2,2,7-trimethyl-nonane
4,7-dimethyl-4-(1,1-dimethyl-ethyl)-3-ethyl-decane

Do complex substances follow a rule regarding where they should be placed in the IUPAC name or can they be placed anywhere?
Do prefixes like bis, tris, tetrakis, etc. need to be considered while naming?

Okay so here is the situation: You have two subsitutions, methyl and 1-methylpropyl as the two substituents here. You wish to know in which order the substituents must be written in the name. You knew that substituents are arranged alphabetically, and both methyl and 1-methylpropyl start with the same letter m.

Consider a dictionary. It has over thousands of words that start with the letter C. But we can still arrange the words in order by looking at the second, third, fourth, fifth, etc. letters of the names. Similarly, if the first letters are the same, look at the second, third and fourth letters.

Here's where we arrive at the conclusion. methylpropyl will be ranked below methyl as per dictionary order. Thus the name of the compound would, in fact, be 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane and the other name is not accepted.

I believe you can continue from here to name the second compound.

Important notes:

The inclusion of di, tri in the name must only be applied for complex subsituents such as in the case of 5-(dimethylethyl)decane. In cases such as 4-ethyl-5,5-dimethyldecane, do not use d as the starting letter. This is becase d is not the part of the name of the subsitituent (methyl) and only indicates how many times this subsitituent occurs in the compound.
Do not add hyphens between methyl and decane such as "methyl-decane". Use "methyldecane" Hyphens must only be used between numbers and subsituents for which the number represents its position in the name.

Substituents are always written in alphabetical order (without considering the numerical prefixes such as di, tri, etc). Also, in case of trimethyl and methyl propyl, having same first alphabet (i.e. m), generally the substituent with shotter name is written first. But as per my knowledge, there is no such official rule. So the name that you have suggested for the first case is acceptable. But your second name is incorrect as alphabetical order is not considered.


Which ions determine ionic conductivity of a salt solution? Let's take $\ce{ZnSO4}$ salt solution in water for example. So the ions inside my solution will be $\ce{H+}$, $\ce{OH-}$, $\ce{Zn^2+}$ and $\ce{SO4^2-}$.

My questions are:

Is the ionic conductivity measured for the solution gives an idea of mobility of zinc ions and sulfate, or is it a result of all four ions combined?
When we change the salt concentration of zinc sulfate, do we change the ionic conductivity of zinc and sulfate ions only, or all four ions?

Also, any undergrad university level source to understand this concept will be helpful.

Di water H2O has very poor conductivity. Because H+ and OH- only driven by the ionic product.

2 H2O <=> H3O+ +OH-

If any other ions will be add the condictivity rises . So that can be for example by acids (H+), alcaline  solutions (OH-) and salts (Cations and anions). The conductivity of the salts  >>>>>> as of the water. So mainly in your case zinc and sulphate drive it. But to calculate correctly all 4 ions participate.


why are amines more water soluble than alcohol for similiar molecular weight? kindly someone please explain the reason behind this behaviour despite the fact that oxygen is more electronegative and forms stronger hydrogen bond.

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong.


Why does an acid exist when dissolved in water? Acid is a substance that releases $\ce{H+}$ ions when mixed with water But if it releases $\ce{H+}$ ions the  acid will ultimately not exist. For e.g we take the acid $\ce{H2SO4}$. If we add water then it becomes $$\ce{H2SO4 + H2O -> H+ + SO4^{-2} + H2O}$$ So the bond between hydrogen and sulphate will be broken therefore the acid will not exist, but we see that diluted acids are possible. Why?

Your doubts are legitimates, but there are some aspects that you should take in account:
1-Not all acid are fully dissociated
Not all acids fully dissociate, so if you take acetic acid you have in fact an equilibrium with some acetic acid molecules not dissociated. So your question holds only for strong acid like sulfuric acid you mentioned.
2-You are using pseudo-Arrhenius definition, why don't you use Lewis acid definition?
If you think the acid as a 'substance that that releases $H^{+}$' this is still an acid even if it dissociate, and even here you should notice that acidity is linked with $H^{+}$. However if you want to dig deeper there are many definitions of acid if you use Lewis acid theory and so you think the acid as a specie that accepts electrons you can easily see that in fact that the acidity is linked mainly with $H^{+}$!

Yes, exactly; only ions can't not exist. Water molecule has dipole nature. When acid is mixed with water, then because of ion-dipole interaction which is stronger than dipole -dipole interaction b/w water molecules and ion-ion interaction b/w acid molecules, opposite ends of water dipoles surround the ions of acid ( means the dipole getting delta + charge will attract negative ion of acid and vica versa). So, now ions get hydrated as process of hydration takes place. They are still free entities, but surrounded by water dipoles. So, the solution of acid and water exists.


What exactly does a "system" include in chemistry? If we have reactants $\ce A$ and $\ce B$ and the reaction(s) between them haven't started yet, are those reactants our "system"?

After the reaction takes place, are products then also included in the "system"?

Basically, is it the case that the term "system" refers to all the chemicals we have in a reaction?

Note: This was not originally written by me. This was first written by @airhuff. I just thought it was relevant in this question as well.
According to this Wikipedia article:

...a physical system is a portion of the physical universe chosen for analysis. Everything outside the system is known as the environment. The environment is ignored except for its effects on the system.
The split between system and environment is the analyst's choice, generally made to simplify the analysis. For example, the water in a lake, the water in half of a lake, or an individual molecule of water in the lake can each be considered a physical system.

Every system has some relationship to it’s surroundings. As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.

Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.

Open system: Both energy and matter can pass into and out of the system.


Why is acetic acid more acidic than phenol? Acetic acid (ethanoic acid, $\mathrm pK_\mathrm a \approx 5$) is more acidic than phenol ($\mathrm pK_\mathrm a \approx 10$), which is reflected in their reactivity with a weak base such as sodium carbonate (adding acetic acid leads to effervescence of carbon dioxide).

However, if we look at the structures of acetate and phenoxide ions, the delocalization of negative charge should be greater in the phenoxide ion as the charge moves inside the benzene ring. On the other hand, in the acetate ion, the negative charge is delocalized between two oxygen atoms only.

How is it possible that acetic acid is still a stronger acid, even when its conjugate base has fewer resonance structures?

There are actually several factors that affect the stabilization of negative charge. The first is what kind of atom centers the charge can be delocalized on and the second factor is how many atom centers the charge can be delocalized on. The first factor is more important than the second. Just think it like quality is more important than quantity. Symmetry is another factor but not very important in your case.
Let's look at the ethanoic acid vs phenol case. In the acetate ion, the negative charge is delocalized on two oxygen atom centers, while in the phenoxide ion, the charge is delocalized on one oxygen and three carbon atom centers. Because oxygen is much more electronegative than carbon, the delocalization of negative charge over two oxygens is better than delocalization over one oxygen and three carbons. Quality beats quantity here.

If you have 2,4,6-trichlorophenol, because the strong election withdrawing chlorine atoms is helping carbon center to stabilize negative charge, it is a much stronger acid than phenol itself. This is an example of the inductive effect at work, as opposed to resonance.

The acetic acid is more acidic then the phenol because in resonating structure the acetic acid forms the equivalent structure. There is rule that the acidic character of equivalent structure is more then that of non equivalent structure.

Equivalent structure are those structures in which same charge remains on same atom in resonating structure.

Non equivalent structures are those structures in which same charge resides on different atoms.

Here in acetic acid if we draw resonating structure then same negative charge resides on oxygen atom but if we draw the resonating structure of phenol the negative charge move on from oxygen to carbon and it becomes a non equivalent structure.

Hence acetic acid is more acidic than phenol.


Aromatic Substitution in acidic medium Find product A.
I tried solving but why isn't the answer this
The answer is given this

Three parts to this answer:

You have two electro-philes, the tertiary carbo-cation formed by protonating the alcohol and breaking off a water molecule and then the bromine.  Looks like you got this part.

Assume the two electro-philes attack consecutively.  The methoxy group is more strongly activating and ortho/para directing than the alk-yl group for the second attack.  So the bromine and the newly formed carbon-carbon bond will be meta to each other.

The sterically favored positions for each are as shown in the given answer.  The carbo-cation favors the position para to the methoxy group.  The bromine picks the position ortho to the methoxy group that's away from the alk-yl function.

Please excuse my use of hyphens.  Device auto-correct was putting in wrong words and I had to defeat it.

The group $\ce{O-Me}$ is an o/p orientator for the second substitution not meta as you answer


Time dependence of integrated rate equations Is it possible for an integrated rate equation to not be dependent on time? What law or principle says so in chemical kinetics?

What if an integrated rate equation obtained only depends on the concentrations of the reactants? Is it possible? Can you refer me to books and references?

By definition, an integrated rate equation shows the change in concentration as a function of the progress of the reaction (time). So I don't understand why professor insists that an integrated rate equation can exist without having the variable time in it.

From a mathematical standpoint: If the integrated rate law is independent of time, then the rate of change of the concentration must equal zero (by the fundamental theorem of calculus). So you're looking for a reaction in which the concentration of the species you want in the rate law (presumably the reactant) does not change appreciably.

As others pointed out, this is a common case for intermediates under the steady-state approximation (rate of production = rate of consumption, net rate zero). But that's just for intermediates. For a substrate, you would need the substrate concentration to be essentially constant. That's probably not a "reaction" that will be interesting to study. A trivial solution, if you will.

Ok, so if we got the reaction

$A\rightleftharpoons B$

with $k_1$ for $A \rightarrow B$ and with $k_2$ for $B \rightarrow A$ then

$\frac{d[A]}{dt}=-k_1 [A]+ k_2 [B]$

if you integrate that you end up with

$k_1 [A]=k_2 [B]$

which is time independent


Terminology for "proton stripping"? This is going to probably be an easy question for an organic chemist, but I am looking for the proper terminology to describe the following reaction mechanism.

In particular, what is a more precise way than "hydrogen stripping" to describe the process where a photogenerated hole on an oxygen can pick off the hydrogen and its associated electron to create a radical that reforms to make formaldehyde?

Further, would I say something like "an HCO radical attacks an adsorbed methoxy to create methyl formate" or is there a more precise way to talk about it? The proper reaction names would be ideal.  Many thanks from a physical chemist who took orgo way too long ago!

EDIT: Thanks Orthocresol for the hint about hydrogen abstraction. Upon looking into things more, hydrogen abstraction aka hydrogen atom transfer (HAT), can be viewed as a subset of proton coupled electron transfer reactions (PCET). The difference is that HAT occurs with minimal solvent reorganization and between similar donors whereas PCET is more general where there can be heat transfer (i.e. it is diabatic versus adiabatic.) So I believe that PCET may be the proper name compared to "hydrogen/proton stripping".

One last question: does anyone know what the reaction between the HCO radical and the methoxy would be called?  Thanks again!

Photocatalysis - reactions caused by light. The term seems to be predominately related to free radicals.

This is obviously a misnomer. The photon is consumed which isn't true for a "real catalyst."

I know that lasers have been used to create excited molecular states which then react. I investigated this for reactions in gas phase of different atomic isotopes of a chemical.

I think if you get any more specific than that then it degenerates into "commonly accepted names" for certain reactions. Sort of like you just have to know what acetic acid is. There is no way to "figure out" what the name means if you don't just know.


Is nucleus at the centre of the atom? If so why? In postulates of Bohr's model of the atom, it is said that the nucleus is at the centre of the atom. In Rutherford’s α scattering experiment he said that out of 20,000 alpha particles only 1 particle is deviated by $180^\circ$. If the particles in the atom are moving continuously (like an electron), why is that nucleus at the centre of the atom? If that is so, then when we focus all the α particles at the nucleus (the point where an α particle got deviated by $180^\circ$) all should get deviated by $180^\circ$ right?

What I mean is that why is the nucleus at the centre of the atom?

Why is the nucleus the centre of the atom?

Definition. Whatever the electron does, chemists define the position of the nucleus to be the centre of the atom meaning that electrons can be polarised away.
Quantum mechanics. When studying an isolated atom, the only thing that influences electron behaviour is the nucleus. You can calculate the ‘path’ of an atom‡ relative to the position of the nucleus with quantum mechanics and your result will be that quite a few electrons will occupy a perfect sphere whose centre is the nucleus.

The second point may require elaboration. But it really is quite simple. You have a very lightweight, very fast particle with a charge of $-1$ (your electron) and you have a very heavy, very slow particle with a charge of $+1$ (your proton). By the time the proton has moved an inch, the electron has moved a mile so we can approximate that only the electron really moves. (This has the fancy name Born-Oppenheimer approximation.) Now remember that both are attracted to each other. The proton doesn’t move, but the electron feels the ‘desire’ to ‘move towards’ the proton. (Inverted commas because this is a macroscopic view that only gets us so far when dealing with quantum effects.) If you average this over time, you will find that the electron will have spent the same amount of time on one side of the nucleus as on the other and the mean electron position will be in the nucleus. Spherically outwards from the proton, the chance of locating the atom decreases monotonously with distance; the angle does not matter. We have a perfect sphere with the proton being the centre.

There are so many different approaches to this question, I am sure. But, if you look at it strictly from a very simplistic perspective, the nucleus has a net positive charge and the electron cloud has a net negative charge. The electron cloud gets as close as it can to the nucleus, but not too close due to inter-electron repulsion. That makes the most probable distribution of the electron cloud to be a uniform distribution with the nucleus at the center. Because the position of the nucleus is only a probability, not all focused alpha particles would back-scatter by 180 degrees experimentally.


Is sodium ion an acid? As sodium ion has a 1+ charge and it can form the ionic compound NaOH with OH1-, hence it is a lone pair accepter (is this correct?). Therefore sodium ion should be an acid?

According to the bronsted lowry definition of an acid, the sodium ion is not an acid. This is because the bronsted lowry definition of an acid states that an acid is a proton donor. The sodium ion is not a proton donor and so according to this definition it is not classed as an acid.

I hope this answers your question :)


Hybridization of nitrogen in trisilylamine, (SiH3)3N? I want to know the hybridization of the central atom in $\ce{(SiH3)3N}$.

I think it should be $\mathrm{sp^3}$, because $\ce{N}$ is attached to three silicon atoms and one lone pair. But actually it is supposedly $\mathrm{sp^2}$.

How is this so?

Ordinarily and according to Bent’s rule, we would expect nitrogen’s lone pair to be in an $\mathrm s$ orbital and nitrogen using its three $\mathrm p$ orbitals to form three bonds to the three silicon atoms. This configuration would allow for the greatest stabilisation.
However, due to nitrogen’s small size, this perfect world already falls apart for ammonia ($\ce{NH3}$), where nitrogen is bound to three otherwise tiny hydrogen atoms. Because an electronically perfect angle of $90^\circ$ would generate much too much steric strain between the hydrogen atoms, $\mathrm s$ contribution is mixed into the bonding $\mathrm p$ orbitals to a certain extent; for ammonia, this extent happens to be almost perfect $\mathrm{sp^3}$ — results for other amines will vary. This electronic situation is not ideal, however it is clearly better than having $\mathrm{sp^2}$ hybridisation and the lone pair in a $\mathrm p$ type orbital. An $\mathrm{sp^2}$ hybridisation of nitrogen in ammonia can be reached, but only as the transition state of nitrogen inversion.
Carrying on to the compound $\ce{N(SiH3)3}$, we would be inclined to again assume a hybridisation of $\mathrm{sp^3}$ in line with the previous paragraph. However, Beagley and Conrad performed electron diffraction studies on $\ce{N(SiH3)3}$ and found the molecule to be practically planar within experimental error.[1,2] A planar molecule without doubt means that nitrogen is $\mathrm{sp^2}$-configured in $\ce{N(SiH3)3}$.
The question remains why. There must be some kind of stabilising interaction of nitrogen’s remaining $\mathrm p$ orbital with something else to keep that molecule planar. Beagley and Conrad suggest — in line with what was thought at the time — that this be due to π bonds with silicon’s remote $\mathrm d$ orbitals.[1]. Numerous evidence, much of which is collected on this site, speaks the opposite (namely that $\mathrm d$ orbitals do not play any role in the bonding situation of main group metals).
Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:
$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$
In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Note and reference:
[1]: B. Beagley, A. R. Conrad, Trans. Faraday Soc. 1970, 66, 2740–2744. DOI: 10.1039/TF9706602740.
[2]: Actually, $\angle(\ce{Si-N-Si}) \approx 119.5^\circ < 120^\circ$. The authors state:[1]

The apparant slight deviation from planarity is associated with a shrinkage effect11 on the $\ce{Si\dots Si}$ distance of about $\require{mediawiki-texvc}\pu{0.007 \AA}$ (see [table]). Spectroscopic results12 are entirely in agreement that the molecule is planar.
11 A. Allmenningen, O. Bastiansen and T. Munthe-Kaas, Acta Chem. Scand., 1956, 10, 261. [sic!]
12 E. A. V. Ebsworth, J. R. Hall, M. J. Mackillop, D. C. McKean, N. Sheppard and L. A. Woodward, Spectrochim Acta, 1958, 13, 202. [sic!]

We call this weird thing as back bonding. The lone pair kinda delocalises or seeks refuge in the empty d orbital of Si, basically providing each N-Si, bond, on an average, a one third of an extra bond.


Symmetry lost in orbitals? I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.

I'll explain via an example:

Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).

Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.

This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they themselves wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)

So, I'd like a clear explanation of how/why the symmetry breaks.

What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.

The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like

$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$

and placing an electron in this orbital does not break rotational symmetry.

The following is to clarify the degeneracy issue debated in the comments.

Hypothesis: Electron orbitals are quantum wave functions (the truth value is left to reader).

The text below is an example adabted from Liboff: introductory quantum mechanics, pages 119 - 121.

The system is in degenerate state of

$\Psi = \frac{3\phi_2+4\phi_9}{\sqrt{25}}     (5.17)$

Probability of finding energy $E_n$ is:

$P(E_n) = \langle\phi_n|\phi_n\rangle$

$P(E_2) = \langle\phi_2|\phi_2\rangle = \frac{9}{25}$

$P(E_9) = \langle\phi_9|\phi_9\rangle = \frac{16}{25}$

$P(E_(2+9)) = \langle\phi_2|\phi_9\rangle = 0$

$P(E_(a+b)) = \langle\phi_a|\phi_b\rangle = 0$, if $a \neq b$

Following is a direct quotation: "In an ensemble of 2500 identical one-dimensional boxes, each containing an identical particle in the same state $\phi(x,0)$ given by (5.17), measurement of $E$ at $t=0$ finds about 900 particles to have energy $E_2 = 4 E_1$ and 1600 particles to have energy $E_9 = 81E_1$."

So, the above states that the particle can be only in one state, not in many states and not in any linear combination of states. The ensemble average can have a linear combination of states.

Now, adapt this to orbitals. Electron can be only in one orbital. On average, electrons of bulk material are on linear combination of orbitals.


Trans vs. cis, or Z vs. E I am new to organic chemistry and am confused regarding the following question:

Does the geometrical isomerism of 3-methyl-2-heptene fall into the category of trans vs. cis, or the category of Z vs. E? And what is the rule that determines whether it falls into the first or the second category?

My textbook is not clear on this.

Many thanks!

(Z) is the same like cis

(E) is the same like trans

BUT cis and trans is usually only for double bonds with 2 different substituents. Confusion can arise when, for instance, you have 2-halobut-2-enes, because some take the highest priority when determining if this is cis or trans whereas others may take the identical groups.

You can use (Z) and (E) for more than 2 substituents, for those with higher priority. Like in your example.

In general it is better to use (Z) and (E) nomenclature to avoid these ambiguities.


Why is carboxylic acid hard to enolize? Why do we need PBr to help enolize a carboxylic acid. My textbook doesn't give a reason for this at all. Does it have to do something with the acidity of carboxylic acid. Thanks.

Think about the pKa of the carboxylic acid protons compared to the pKa of the enolisable proton (hint: the acid is more acidic).

Once the carboxylic acid is turned into the carboxylate, enolisation is not possible, as the carbonyl cannot simultaneously conjugate with the carbon and the oxygen.

The use of phosphorus tribromide is known as the Hell-Vollhard-Zelinski reaction, it allows carboxylic acids to be directly brominated at the alpha position without initial protection of the carboxylic acid via in situ formation of the acyl bromide which hydrolyses upon workup.

-OH group in carboxylic acids have a much better resonance than an acid chloride due to difference in the size of halogen atoms vs carbon; this difference even increases with Br.
This lack of resonance from LP of Br onto carbonyl group in acid chloride makes Enolization possible, which is very difficult to be in -COOH.

Using PBr3 to convert an RCOOH in RCOBr brings in Enolization, which can give Substitution with Br-Br.


What is the expected product in the reaction of hydrogen chloride with allylbenzene? Shouldn't it be 4th option since since the carbon attached to the ring is secondary as well as resonance stabilized?

I've searched SciFinder for this and related reactions (allyl benzene + H-X). In the results, there are no examples of the halide at the benzylic position in the product, and there are many examples of the halide at the secondary postion arising from the more substituted end of the alkene. One such example is below.

It is certainly warranted to consider both the direct addition and the product from rearrangement (since a resonance stabilized, benzylic carbocation would be formed as an intermediate), and if one were performing the experiment in the lab, it would be a good idea to look for both products. It seems that in practice, the addition of halide to the secondary carbocation is faster than the hydride shift. Predictions about kinetics like this are very difficult, and exam questions that require them could be considered unfair unless there is an overwhelming and predictable rationale for one product over another.

I believe the fourth option would require a removal of a hydrogen.

As well, the double bond is electron donating so it acts as a nucleophile to pick up the proton.

Double bonds like protons/hydrogen more than a Carbon with four single bonds to other carbon and hydrogen. Its happy with a full octet and (nearly) equally electronegative buddies, while double bonds could sure go for a few more protons to fill up.


Effect of dilution on pH for HCl Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?

For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately?
Normally by using $SV=constant$ I get pH will change by 1 unit (without taking common ion effect).

How will common ion effect affect the answer? Help.

Yes, you will definitely need to account for common ion effect.

However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.

as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.

You should get a result around 6.79=pH.


Primitive cells and their variations Is there any intuitive way(without involving higher maths) to explain possible variations in seven primitive types of (unit)cells? I would like explanation in context of why end centred is possible only for orthorhombic and monoclinic but not in other unit cells.

I will not consider hexagonal cell in this answer.

Unit cell is generally parallelepiped, which has three sides $a$, $b$ and $c$ and three angles between them $\alpha$, $\beta$ and $\gamma$. The sides can be f.e. of equal length and angles can be same and equal to $90^\circ$, this way you will have cubic unit cell. You can try to vary sides and angles and you will discover, there is only those seven types of unit cells.

When I way learning this I found useful third chapter in "Foundations of material science and engineering". At least there is nice table and images to show those cells. You can find it by typing "Crystal structures and crystal geometry" into Google.

Well, now answer last question. Let's say we have cubic base-centered (more correct than end-centered, I think) cell. From top it would look like this:

I draw only two cubic base-centered cells (I can add more if it will be more clear) next to each other. However, as you can see, there is one smaller cell (red) inside of our base-centered lattice (green) and that is simple tetragonal cell (If you draw it in three dimension, you will see, that the height is $\sqrt{2}$ times greater). As it has lower volume, this is the primitive cell, not that base-centered.