text:
The dissolution of ammonium chloride
The dissolution of ammonium chloride is used to cool a container of water placed in the solution. It's an endothermic process. What absorbs the heat and what loses it?

Related: What is the heat of dissolution of sodium acetate trihydate?

When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.
$$\Delta G = \Delta H - T \Delta S\tag{1}$$
There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):
$$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$
Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.
$$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$
Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).
$$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$
All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table 1.[1]
$$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$
So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.
Thus, the answer to your question is:

The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)

Reference:
[1]: H. D. B. Jenkins, D. F. C. Morris, Mol. Phys. 1976, 32, 231. DOI: 10.1080/00268977600101741.

To keep it simple, the "heat" is lost by the water and the "ammonium chloride" has a net gain.
The ammonium chloride starts off as a solid crystal. The positive aluminum ions and the negative chloride ions attract, so it takes energy (as "heat") to break the crystal apart. The solvation of the anions and cations releases energy, but in this case it isn't enough to offset the energy spent breaking the crystal apart. So overall solution cools.
The thing to remember here is that for salts the "molecules" don't exist per sey in solution. Thus we could make a solution which is 1.00 molar in potassium chloride and 1.00 molar in sodium nitrate and we'd get the same solution as 1.00 molar potassium nitrate and 1.00 molar sodium chloride.


text:
How can we combine all the three gas laws into a single ideal gas equation?
In all texts that I have read, it has been stated that the combined gas law for ideal gases was derived from the individual gas laws proposed by Boyle, Charles and Avogadro.
My confusion is this is that, in each individual law, some variables of the system's state are to be kept constant. However, none of these conditions are considered in the combined gas equation. Also, how is it possible for us to suggest that for two states with variables $P$, $V$ and $T$:
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Suppose you have two states which have nothing in common (that is, nothing is kept constant): $P_1,T_1,V_1$, and $P_2,T_2,V_2$. Well, you may just move from the first state to $P_1,T_2$ (that would be a constant pressure process which you know how to deal with), then from there to $p_2,T_2$ (that's a constant temperature process).

http://physics.stackexchange.com/questions/99347/deriving-combined-gas-law-from-boyles-and-charles-laws

I think the historical answer is that it was messy. We like to think of science as pushing back the darkness in a defined way but that just isn't what happens. It is usually more like "blind men describing an elephant." Then someone with sight comes along and explains that one has the tail, one has a leg, one has the trunk and one is rubbing the side. Then the pieces fit.

It is indeed the case that the rearrangement of the different gas laws is not very satisfactory as it employs the different variables without any constraints. This issue has been addressed by S. Levine in J. Chem. Educ., 1985, 62 (5), p 399, but in the case you do not have access to this reference, I'll shortly repeat his arguments here.
As $P$, $V$ and $T$ are thermodynamic sate functions, we may write the total differential of $T$ as
$$\mathrm{d}T=\left (\frac{\partial T}{\partial V} \right )_P \mathrm{d}V + \left (\frac{\partial T}{\partial P} \right )_V\mathrm{d}P \equiv X\mathrm{d}V+ Y\mathrm{d}P$$
since this differential must be exact, the following relation must hold
$$\left (\frac{\partial X}{\partial P} \right )_V=\left (\frac{\partial Y}{\partial V} \right )_P$$
We can find simple functions satisfying this condition
$$X=cP\text{ and }Y=cV$$
where $c$ is a constant that we may choose to be $c=1/R$ for later convenience.  Substitution of these functions in the total differential gives
$$\mathrm{d}T=\frac{1}{R}\left (P\mathrm{d}V +  V\mathrm{d}P\right )$$
and integrating on both sides from 0K to T yields
$$T-T_0=\frac{1}{R}\left [(PV)_T-(PV)_{T_0} \right ]$$
and since $T_0=0$ and $PV=0$ at $\pu{0K}$ we arrive at the ideal gas law.

An easier way to combine the simple gas laws is the following. We take Boyle's, Charles' and Avogadro's laws:
$$PV=k_1(n,T),\quad \frac{V}{T}=k_2(n,P),\quad \frac{V}{n}=k_3(T,P),$$
where $k_1,k_2,k_3$ are constants that depends on the fixed value of $n$, $T$, or $P$.
Now, from Boyle's and Charles's laws we have
$$V=\frac{k_1(n,T)}{P}=Tk_2(n,P)\iff \frac{k_1(n,T)}{T}=Pk_2(n,P)$$
In the left hand side of the last equation, all the terms are function of $n$ and $T$ only, and in the right hand side all the terms are function of $n$ are $P$ only. Thus, both terms have to be equal to a constant $C_1(n)$:
$$ C_1(n)=\frac{k_1(n,T)}{T}\iff k_1(n,T)=C_1(n)T\implies V=\frac{C_1(n)T}{P}$$
With the same argument, we can combine this last equation for $V$ with Avogadro's law:
$$V=\frac{C_1(n)T}{P}=nk_3(T,P)\iff \frac{C_1(n)}{n}=\frac{P}{T}k_3(T,P)$$
and both sides of the equation have to be equal to another constant $C_2$:
$$\frac{C_1(n)}{n}=C_2\iff C_1(n)=nC_2\implies V=\frac{nC_2T}{P} $$
Changing the name of $C_2$, we have the magnificent:
$$PV=nRT$$

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text:
Reduction of copper oxide by hydrogen
In the concluding part of my previous question, wouldn't there be an explosion near the hole in the tube from where the excess hydrogen is coming out instead of a steadily burning flame?

Please refer to the previous question.

Pure hydrogen does not explode. When mixed with air, it does. Don't let them mix.

But if it is pure hydrogen, and no air, how does it even burn with a flame.

@NisargBhavsar It may be questionable to bump old questions by minor updates.

@Poutnik I am sorry but I didn't understand what you meant. Can you please elaborate?

@Poutnik Nisarg wanted to link the previous question which was not done before. Such edits are welcomed. It may even attract answers after being left unanswered for years.

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text:
Electrophilic substitution in phenyl alkyl ether

Phenylalkyl ethers undergo usual halogenation in the benzene ring, e.g., anisole undergoes bromination with bromine in ethanoic acid even in the absence of iron (III) bromide catalyst.

Why do we use ethanoic acid as a solvent? Why don't we use CS2 as a solvent? Does ethanoic acid has something to do in activating phenyl?

Fairly sure you can do it in nearly any solvent you like. And acetic acid is a ton cheaper than carbon disulfide.

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text:
Are imine or amine bonds similar to ethereal bonds, from the perspective of an enzyme?
I'm interested in enzymes that can cleave (hydrolyse or however) an ethereal bond ($\ce{C-O-C}$). Enzymes can evolve to act on chemicals/bonds that are similar to what they previously acted on, so enzymes with similar functions are also interesting.
I've been told by a couple of chemists that carbon-nitrogen-carbon bonds are similar to ethereal, so I was wondering if, and in what way that is true. Are the bonds in an imine group ($\ce{C-N=C}$) or a amine group ($\ce{C-NH-C}$), more similar to those in an ethereal group compared to a $\ce{C-C}$ pair, for example?
I guess from the perspective of an enzyme, if you were approaching the imine from the direction of the lone pair, that would feel pretty similar to an oxygen maybe?

See, some enzymes are awfully picky about their substrates, some are less so.

I was asking about two different nitrogen groups, but I wasn't sure of the names. I think the question now says what I meant

@IvanNeretin Yeah, I guess my question was how unpicky would an enzyme need to be to act on both, or switch substrate. I've noticed different isochorismatases that act on a nitrogen or oxygen, but they don't have much sequence similarity so that might not mean anything.

This does depend a lot on what the enzyme does and what it expects. Take the following three structures, one from an ether (e.g. diethyl ether), one from an amine (diethyl amine) and one from an imine (N-ethylethanimine).

Scheme 1: Ether linkage, amine linkage and imine linkage (from left to right) including hydrogen atoms on adjacent carbons.
When looking for a ether, an enzyme may well attempt to form a hydrogen bond to the ethereal oxygen. This might work in the case of an amine — there is a spare lone pair on nitrogen that can accept hydrogen bonds, but the additional hydrogen present can present ordering and space problems —, but would in the case of an imide.
However, note that the carbons next to ether and amine bonds have four atoms attached to them (two hydrogens and another carbon each), while the imine has only one hydrogen. The $\ce{C=N}$ bond is planar, meaning that the configuration of the neighbouring atoms is also relatively fixed. Thus, the imine may have a pretty different shape — and may not fit into the enzyme’s active pocket. This might actually also mean that the imine cannot accept the hydrogen bond offered by the enzyme because it cannot orient itself favourably.
As I said above, it depends on the enzyme whether it cares about hydrogen bonding more or whether the overall shape is more important. In general, I would consider an amine to be closer to an ether structurally, and therefore more likely to be accepted. Possibly, the amine hydrogen can point away, allowing for the amine to accept a hydrogen bond nonetheless. In that case, the amine would be the clear winner.
All these heteroatom bonds are clearly different from $\ce{C-C}$ bonds due to the bonds’ inherent polarity (electronegativity of $\ce{O}$ and $\ce{N}$).


text:
pH(H3O+ ion concentration) of a mixture of 2 weak acids
I need to find out the $\ce{H3O+}$ ion concentration in a mixture of 2 weak acids $\ce{HA1}$ and $\ce{HA2}$.
They will be ionised as
$$\ce{HA1 + H2O <=> A1- + H3O+}$$
And
$$\ce{HA2 + H2O <=> A2- + H3O+}$$
Then, if $c_1$ and $c_2$ are the concentration of the acids and their dissociation constants are $K_{1}$ and $K_{2}$,then how much will be the value of $[\ce{H3O+}]$ ?
The textbook states without proof  that it is approximately equal to $\sqrt{(K_{1}c_{1} +K_{2}c_{2})}$.
Any help on how to approach the result is appreciated.

If they are weak acids then how come they ionize (not dissociate) irreversibly?

@user34388 sorry for the error. They ionize only like any other weak acid.pls check, I have edited the question


text:
1,1-diformyl propanal

The correct IUPAC name of the following compound is:

(A) 1,1-diformyl propanal
  (B) 3-formyl butanedial
  (C) 2-formyl butanedial
  (D) 1,1,3-ethane tricarbaldehyde

The structure of 1,1-diformyl propanal has been given. And the IUPAC name has been written. But I can't get the same IUPAC name because I'm a beginner. Also I don't know how to get structure here in the question, so I didn't do it.
Help me with the IUPAC naming of the compound with step-by-step explanation . Please.

@JamilAhmed The problem is that the first carbon atom in propanal is already almost full, so it cannot have two more functional groups.

@JamilAhmed Does your exercise look like Exercise I, Q.4 on page 7 of this pdf document? http://tekoclasses.com/IITJEE%20CHEMISTRY%20%20ENGLISH/NOMANCLATURE%20TYPE%201.pdf

@Loong, Q.4 exactly.

@JamilAhmed Then the answer is not A. It is not 1,1-diformyl propanal. Problem solved?

The proper and straightforward way to solve nomenclature problems is to construct the systematic name for the given structure according to the current IUPAC recommendations. However, for such multiple-choice questions, it may be easier to check the given answers and try to draw a structure for each proposed name as follows – especially when the particular nomenclature rules that are considered in the exercise are not exactly known.
The name 1,1-diformyl propanal (A) can be ruled out since the atom ‘1’ (i.e. the $\ce{C}$ atom of the aldehyde group) of propanal has only one $\ce{H}$ atom that could be substituted; therefore it is impossible to place two formyl groups at atom ‘1’. Furthermore, after substitution of the $\ce{H}$ atom of the aldehyde group, the compound would no longer be an aldehyde.

The name 1,1,3-ethane tricarbaldehyde (D) does not make sense since ethane has only two $\ce{C}$ atoms; therefore it is impossible to place a carbaldehyde group at atom ‘3’ of ethane.

It is possible to draw structures for the names 3-formyl butanedial (B) and 2-formyl butanedial (C) based on the structure of butanedial:

Note that the resulting structures for 3-formyl butanedial and 2-formyl butanedial are actually identical, and both structures correspond to the structure that is given in the question. The correct choice is 2-formyl butanedial (C) rather than 3-formyl butanedial (B) since the locant ‘2’ is lower than ‘3’. (Note that the correct spelling of the name is actually 2-formylbutanedial, i.e. without space.)

Anyway, even the name 2-formylbutanedial is actually not in accordance with current IUPAC recommendations. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), aldehydes can be systematically named in various ways:

P-66.6.1 Systematic names of aldehydes
Aldehydes are systematically named in three ways:
(1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;
(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;
(3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

Therefore, the preferred IUPAC name for the compound given in the question is ethane-1,1,2-tricarbaldehyde.


text:
C-C bond dissociation energy
Have the various $\ce{C-C}$ bonds in straight chain hydrocarbons, have the same dissociation energies? Or does it depend on the location of the carbon in the chain?
In other words, are the terminal bonds weaker or stronger than the internal bonds?

The terminal bonds are stronger than the internal bonds.
In simple hydrocarbons where strain or electronegative substituents are absent, the bond strengths will correlate with the stability of the resultant radicals that would be formed when the bond breaks.
Because radicals are electron deficient species (they are one electron short of an octet) they are stabilized by electron donating substituents (alkyl groups for example).  Hence, a tertiary radical is more stable than a secondary radical which is more stable than a primary radical which is more stable than a methyl radical.
Using this information we can see that
$$\ce{R-CH2. + .CH2-R}$$
is more stable than
$$\ce{CH3. + .CH2-R}$$
Hence the internal bond will have a lower bond dissociation energy and break more readily.

then another question arises: Which of the R-CH2 radicals is more stable? (R=C2 or R=C3)? or (R=C3 or R=C4)? and so on

Does R=C2 mean that R is an ethyl group and R=C3 is an isopropyl group and R=C4 is a t-butyl group? If so, then this earlier answer may provide some guidance.  Inductive effects are likely similar for R=C2, C3 and C4.  Resonance effects would be different due to hyperconjugation and they would favor R=C2 as explained in the earlier answer.  However, the effects are very small in the earlier answer where a positive charge is being developed.  They would be even smaller here since we

are dealing with a radical rather that a carbocation.

R=C2, R=C3 and R=C4 means ethyl, n-propyl and n-butyl group, respectively.

C2 might be a bit more stable than C3, no significant difference between C3 and C4

If so, is the following conclusion true? ethyl is more stable than methyl. also, ethyl is more stable than n-propyl and n-propyl is a bit more stable than n-butyl and there is no significant difference in stability of longer alkyl groups? I will be very thankful if you introduce a reference.

According to here:

There are three places where a butane molecule ($\ce{CH3-CH2-CH2-CH3}$) might be split. Each has a distinct likelihood:
48%: break at the $\ce{CH3-CH2}$ bond.
$\ce{CH3* / *CH2-CH2-CH3}$
38%: break at a $\ce{CH2-CH2}$ bond.
$\ce{CH3-CH2* / *CH2-CH3}$

This seems to suggest that internal bonds are weaker than terminal bonds (there are two terminal bonds, so we need to divide 48% by 2).

From data that I gather:

The enthalpy of formation of methyl radical is 146 kJ/mol (source)
The enthalpy of formation of ethyl radical is 119 kJ/mol (source)
The enthalpy of formation of propyl radical is 100 kJ/mol (source)

Therefore:
$$\ce{CH3*+*CH2CH2CH3 -> CH3CH2*+*CH2CH3 (\Delta H= -8kJ/mol)}$$
Again showing that the internal bond is weaker.

Surely this only applies to butane.


text:
Why H2O2's molecular geometry is what it is...?
I came up with this question when I was trying to determine whether H2O2 is polar or not.
H2O2 looks like this when looked parallel to oxygen-oxygen bound:(source: Pubchem)

Each hydrogen is in one of the three positions available around each oxygen(the two nonbonding electron pairs occupy the other two); even the oxygen-oxygen bound can rotate to set hydrogens in any position relative to each other.  Therefore I don't see the problem in following nonpolar structure(pretty neat!)

What concept am I missing while determining molecular geometry?
Is a H2O2 molecule sometimes polar and sometimes nonpolar?
Thanks for your help!

The answer is not quite easy, you can find the answer of your question there https://deepblue.lib.umich.edu/bitstream/handle/2027.42/71115/JCPSA6-42-6-1931-1.pdf?sequence=2 but I guess it would be better if someone could explain it too you. I have not time right now, but there is difference of energy in comparaison with ethene about the rotation on the single bond. It is an atropisomer https://en.wikipedia.org/wiki/Atropisomer

The conformation of hydrogen peroxide (H2O2) is dominated by the lone pairs rather than the hydrogen atoms. Instead of the expected anti conformation in the free state the dihedral angle is 94°. The conformation of hydrogen peroxide is therefore neither eclipsed nor staggered but an intermediate structure. When in the solid state hydrogen bonding will cause the shape and angles to change.

I suppose you could reason with orbitals/VSEPR that for hydrogen on one oxygen to line up with a lone pair on the other oxygen as a kind of hydrogen bonding, the planar version would require the oxygens to be sp2 hybridized rather than sp3 hybridized, which is unfavored due to electron pair repulsion favoring sp3.

Maybe useful: http://chemistry.stackexchange.com/a/69587/40029

As stated in the link provided by @Shadock, hydrogen peroxide, like many more other molecules, is subject to hindered internal rotation. When you rotate the two OH groups with respect to each other around the O-O bond, this requires a certain energy depending on the relative angle. In the eclipsed configuration (OH groups pointing in the same direction), it's easy to see that there must be a maximum in the potentiall due to the Paulli repulsion. In the anti-eclipsed configuration (both OH groups pointing in opposite directions) there is another maximum as you anti align the two dipoles of the groups. In between these maxima, you'll find a minimum and the potential can be expanded as a Fourier series in the relative bond angle
$$V(\gamma)=\frac{V_2}{2}\cos(2\gamma)+\frac{V_4}{2}\cos(4\gamma)+\ldots$$
where $V_2$ and $V_4$ are related to the heights of the two barriers. If the barriers were infinitely high, the OH groups would act as a quantum harmonic oscillator in one of the potential minima. On the other hand, if the barrier were very small, the OH groups would rotate freely around each other. In the case of hydrogen peroxide, the eclipsed barrier is so high that the OH groups can hardly go over/through it. The lower barrier is a different story. It is high enough to support a few vibrational levels, but not high enough to prevent the OH groups from tunnelling through the barrier and as a consequence, these energy levels are split in two. (In fact, these two levels are again split by the tunnelling through the higher barrier, but since this is so high the splitting is very small). Mathematically, the Schroedinger equation for this periodic potential is equivalent to the Mathieu equation (or more general Hill's equation).
In the case of H$_2$O$_2$ the tunnelling is so fast, that the molecule vibrates around the lower barrier with a period of ~100 fs. As a consequence, to determine the polarity, you have to average over this motion, resulting in a net polarity.
Note that, if the barrier would be so high that tunnelling would be sufficiently slow, H$_2$O$_2$ would be a chiral molecule!

I'm sorry but i don't get the second maximum. Why anti aligning the two dipole groups increases potential?

An intuitive example is the alignment of two bar magnets; if you align the poles in the same direction you have an attractive force whereas you'll experience a repulsive force when you anti align them, i.e., put the two north or south poles together.


text:
Calculation of proportion of zwitterion from this please.
This is the question. I tried to post it in the title section but it was rejected because it was considered lengthy. Question: Valine is one of the 20 natural amino acids and has an acid -COOH (pka1=2.6) and an amine-NH2(pka2=9.6). If we consider a valine solution with pH=10.6, what is the proportion (%) of the zwitterion form of the solution? A) 2.4 B) 3.6 C) 5.7 D) 9.1 E) 10.2.
Sorry if this might be too elementary but I don't know how to solve this question. I need help, thanks.

Since this seems to be a homework question, let me lead you along the solution. What is a Zwitter ion? Consider the molecule as (H2N-R-CO-OH).

This is not a homework question. It'

It's a past medical entrance exam question from my country under the chemistry section in  2013. I came across the question while revising for the exam and didn't know how to appreciate it. Please help, thanks.

Ok for a zwitter ion the acid is deprotonated and the amine group is protonated, thus the molecule has no net charge. It is given that the pKa of the acid is 2.6 and the pH of interest is 10.6. That is so alkaline that virtually no molecules will have the acid group protonated. So the problem boils down to figuring of what percentage of the amine group molecules are protonated.

Hi. You can refer to the homework policy here. The term "homework" is used more loosely here, as the link describes.

(I shouldn't answer this because it is a homework question with no effort, but since someone posted a wrong answer I'm answering anyway)
Only the pK of 9.6 is relevant because the pH is 10.6; therefore, think of the situation as having just two species, zwitterion (acid) and negative (conjugate base).
Because pH is one unit above the pK, the ratio of acid to conjugate base is 1 to 10.
Ratio of acid to total is 1 to 11.
1/11 = 0.0909

Proportion of zwitterion = pka amine : pka acid
                                           =9.6 : 2.6
                                           =3.6923


text:
How can I separate benzidine from a food colouring (FD&C red 40)?
I'm doing a research paper on the food colouring red $40$ (also known as Allura red) and the fact that it contains a carcinogen: benzidine.
I am having trouble finding a way to separate the benzidine $(\mathrm{C}_{12}\mathrm{H}_{12}\mathrm{N}_2)$ from red $40$ $(\mathrm{C}_{18}\mathrm{H}_{14}\mathrm{N}_2\mathrm{Na}_{2}\mathrm{O}_{8}\mathrm{S}_2)$ to determine the concentration of benzidine in the colour since they are both organic.
Paper chromotopgraphy wouldn't work because I need to quantify the amount of benzidine as well, which is the aim of my project.
I am a highschool student and I have very limited resources available to me and therefore the method of separation shouldn't be very complicated but instead, rather basic. Does anyone have any suggestions.

You could possibly try paper chromatography.

Paper chromatography would help me to separate the two, however I wouldn't be able to quantify the amount of benzidine, which is really what my research paper is all about.


text:
Shape of nickel tetracarbonyl
Is nickel tetracarbonyl a square planar complex or a tetrahedral complex?
According to the crystal field theory, $\ce{CO}$ is a strong field ligand and should hence cause the $t_{2g}$ electrons of nickel to pair up, hence leaving 1 $d$ orbital free, which must pair up with empty $s$ and $2p$ orbitals forming 4 $dsp_2$ orbitals yielding a square planar complex. But my textbook says its tetrahedral, with the electrons not pairing up and the outer d orbitals being used.

the Ni atom in the molecule is d10.

So you're saying that the 4s2 electrons are demoted to the t2g orbitals, pairing them up and hence resulting in the formation of the tetrahedral complex?

Yes! In the formation of a metal complex the 4s electrons are always "demoted" to the 3d orbitals. I would not quite put it that way, but in effect, that is what happens. d-electron count always includes the electrons that were originally from the 4s orbitals (unless they were ionised, of course). I've written an answer on it before: http://chemistry.stackexchange.com/questions/41962

And now, with a $\mathrm{d^{10}}$ transition metal, no matter how strong-field the ligand is, you won't be able to derive any ligand-field stabilisation energy. So, electronic effects can't dictate the geometry here. What other factors might favour a tetrahedral complex over a square planar complex?

nitpick: $t_{2}$ not $t_{2g}$ for tetrahedral molecules

According to the work by Ladell et al., the structure of $\ce{Ni(CO)4}$ in the solid state ($\pu{-55 \pm 5 ^\circ{}C}$, ambient pressure) is tetrahedral with refined $\ce{Ni-C}$ distances of $\pu{(1.84 \pm 0.03) Å}$, and $\ce{C-O}$ distances of $\pu{(1.15 \pm 0.03) Å}$.  You find the corresponding .cif file for example in the COD database filed there as entry COD 2310876:

Note this primary reference has been cited by (at present) 103 other publications.
Reference:
Ladell, J.; Post, B.; Fankuchen, I. The Crystal Structure of Nickel Carbonyl, $\ce{Ni(CO)4}$. Acta Cryst. 1952, 5, 795-800; doi: 10.1107/S0365110X52002148.

Technical detail:
Initially, the reference card of the entry displays only the unit cell and the crystallographic motif.  To view the packing in the unit cell as shown in the illustration above, put your mouse over the interactive display provided by JSmol, and click once with the right-hand mouse button to open a new pull down menu.  From here, choose entry Symmetry, followed by Reload {1 1 1} to display any molecule with at least one atom within the unit cell.  (In the present case, the multiplicity equates to eight molecules per unit cell.)  You may zoom-in and out with a mouse wheel, or walk around the model by click-and-drag with the left-hand mouse button.


text:
Isopropyl alcohol to clean Nylon plastic
I have a hardwood wooden frame, possibly lacquered, containing some very dirty nylon plastic pieces, a bit like the teeth of a comb, where the frame goes along top and bottom.  I know it's nylon because the manufacturers say so, ditto they mention lacquer.
I would normally just scrub the nylon section with soapy water, but the wooden frame MUST NOT warp!
I have some isopropyl alcohol, 99 or 97%, i can't remember, and propose to wipe the nylon clean.  It's 120cm long and has lots of slots in it, so this may take quite a lot of propanol.
My question is:  Is solid nylon 'plastic' affected by isopropyl alcohol of this %?
If I am better off diluting it a bit, would it then warp the wood?  I know it will take the lacquer off - it's not good with furniture finishes generally!
I do know to do this job outside :-)
Thank you

I don't care too much about the lacquer, but the wood must not warp or it will be useless for it's intended function.  thank you

Your nylon plastic should be thoroughly resistant to concentrated isopropyl alcohol.
According to this encyclopedia.com entry, nearly all of the nylon produced (in the US anyway) is one of two types: type 66 and type 64. Each of these is stated in the article to be insoluble in water and most organic solvents, but soluble in strong acids.
As some anecdotal evidence of nylon's resistance to common solvents, this Quora discussion referenced a frustrated biker's need to find a suitable solvent for removing some nylon that had melted onto a motorbike exhaust pipe. There was really no good consumer-type of solvent listed (and nothing remotely chemically similar to isopropanol), and it actually included a link to a patent for a compound specially designed for dissolving nylon-based materials, just because they are so chemically resistant.
That all said, it wouldn't be a bad idea to do a "spot test" of the isopropyl alcohol on a small, inconspicuous as possible piece of the plastic. There's always an outside chance that it's not exactly what you think it is or that for some reason your particular nylon-based plastic is not as immune to a solvent like isopropanol as the above arguments imply that it should be.


text:
Consequences of a neutron decay (beta radiation) and reasoning for an electron capture

I came about the following doubts (extremely comprehensive in nature) in a logical sequence, so I have made it a point to explain either in full detail and then proceed to the interrelation between the two. I have also shared my hypothesis for the for the observed events.

In the following reaction for radioactive dating:
$$\ce{C^{14}_6 \longrightarrow N^{14}_7 + e_{-1}^0  + \overline{\nu_e}}$$
The neutral carbon-14 atom initially had $6e^-$, $6p$ and $8n$ in which a neutron decayed into a proton and an electron leaving a nitrogen atom with $7p$, $6e^-$and $7n$ and the high energy electron from the neutron decay escaped from the atom :
$$n^1_0\longrightarrow p^{1}_1 + e_{-1}^0  + \overline{\nu_e}$$
$\hskip2in$

Now it is evident that the nitrogen atom has a positive charge and in totality, it is a positively charged gas (which could be a matter of great concern since a lot of charge at one place could ionize the surroundings and could even lead to the dielectric breakdown of the atmosphere, provided there is a lot of charge.)
The above predicted results are not normally observed partly because it has a half life of about 5700 years (corresponding to extremely low concentrations) and that the nitrogen atom "might" take an electron from the surroundings, or the electron might just not escape from the gas... (This is unclear.)
Moreover, in the following reaction : (with a half life of about $11s$)
$$\ce{Mg^{23}_{12}(g) \longrightarrow {Na^{23}_{11}}^-(g) + e^{0}_{+1} + \nu}$$
The magnesium-23 has $12p$, $12e^-$ and $11n$ to start with and after the $\beta^+$ decay, sodium-23 atom with $12n$, $11p$ and $12e^-$ is left behind, which is a negatively charged ion. The $\beta^+$ radiation proceeds according to the reaction:
$$p^1_1\longrightarrow n^{1}_0 + e_{+1}^0  + \nu_e$$
With a high concentration of Mg-23 atoms we could(or should) achieve a negatively charged sodium gas in less time. Now the same doubt arises that whether the ions will remain so, or if lose an electron to the surrounding (which would still not solve the problem), or capture the positron (which is also unlikely due to the high speed).
Now, to neutralize the charge on the atom, it should lose/gain an $e^-$. Here comes the k capture reaction in which an electron and a proton combine to form a neutron according to the following reaction:
$$p^1_1+ e_{-1}^0\longrightarrow n^{1}_0   + \nu_e$$
Accordingly when an electron is "captured" by a charged species to neutralize the charge, that electron orbits the nucleus rather than go through all the orbitals and collide with the nucleus (which is made clear by the following observed reactions).
$$\ce{{Na^{22}_{11}}^+(g) + e_{-1}^0 \longrightarrow Na^{22}_{11}(g)}$$
where the positively charges sodium atom takes an electron (which revolves around the nucleus) to neutralize its net charge and the reaction isn't known to follow as:
$$\ce{{Na^{22}_{11}}^+(g) + e_{-1}^0 \longrightarrow Na^{22}_{10}(g)}$$
wherein we assume it to be a k capture reaction and algebraically add the subscripts and the superscripts. Since it isn't a nuclear reaction, the electron wouldn't really go through towards the nucleus.
This way I don't really see why an $e^-$ should ever come on the left hand side in a nuclear reaction (while it does in the k capture reaction).
But the k-capture or electron capture is found to occur in nature, so how can an electron really collide with the nucleus? Even if the innermost electron is allowed to be "sucked in" by the nuclear force, the angular momentum would still be conserved and the electron would revolve with faster and faster tangential velocities and much lower radial velocity (relativistic effects might have to be considered). This way, wouldn't it take infinite time for it to happen, with the radial speeds becoming infinitesimal?
Also, when the electron and a proton are really close to each other, the binding energy is really high, and the point when the electron and the proton "collide", the whole system would have an undefined potential energy (most probably $-\infty$ eV, since in the preceding case, the potential energy was decreasing). Wouldn't it result in the emission of huge amount of energy (after obviously subtracting the kinetic energy)?
P.S: It is slightly (rather wildly) off topic, but since we are considering the case when the electron is revolving at speeds near $c$ while coming towards the nucleus, do we take the increased mass into the consideration of the mass of the atom altogether, or is it frame dependent (even galaxies move at speeds beyond $c$ with respect to us)?

I wish more people were like you, in the sense of making their formatting right on their first question. Welcome to Chem.SE.

I'm still not sure what the question is after reading this twice through, other than you have confusion on nuclear physics. As for you P.S., you have an incorrect view of electrons in orbitals - they do not revolve around the nucleus, they occupy (are) orbital wave functions.

Yeah, I understand your point @JonCuster, I am aware of the orbitals (as also mentioned somewhere in the question, and was used just for the ease of asking). The question is in two parts - 1. What happens to the positively charged nitrogen ion (to support the doubt, I have also cited a faster reaction in the question). 2. Assuming it takes an electron from the surroundings(by still not really solving the issue), I have a doubt in the k capture reaction. With knowing this, reading the question once more might explain my question exactly. If not, please let me know. :)

It seems that the gist of the question is about the distance in the force equation for Coulombic attraction.
$$ F = k*\frac{q_1 * q_2}{r^2}$$
Naively, as the distance between the charges goes to zero, the force should go to infinity. That obviously doesn't happen. The same problem exists for black holes; the gravity should be at a singularity which doesn't really exist. The point is that the traditional formulas are "good enough" under most circumstances, but fail under extreme conditions. For instance the theory of relativity predicts that time slows for a body which is motion. GPS satellites actually need such a correction.

I'd rather underline that Coulomb's law still works at these distances; it is just that an electron is not a dot.

@MaxW I understand, but when the electron is really near the nucleus and not far, we would have a really low potential energy, but the speed might exceed in such configuration which is why we don't really have anything below the 1s orbital. So when you lower your potential energy, you gain kinetic energy, but when they "collide" would they not realease some energy, understanding that the kinetic energy will approach infinity (due to relativistic corrections), so the mass would also increase, so isn't this really going weird?

The excess energy gets emitted as an electron neutrino. See https://en.wikipedia.org/wiki/Electron_capture // Reread my post. As the distance approaches zero the energy doesn't go to infinity. The proton and the electron have real volumes as particles. Also at the distance that they merge Coulomb's Law doesn't apply. Rather Coulomb's law works in the other direction. As the electron from a beta particle gets 100 atomic diameters away from the nucleus it is at essentially an infinite distance since we can measure the energy loss to 4 significant figures (I'd guess).

I should have pointed out that the simple Coulomb's law also breaks down for the inner orbitals of heavy elements. Ab inito calculation for the inner orbitals (ie 1s) have to be corrected for relativity. So some of the gained energy appears to be converted to mass.


text:
NMR Splitting for Adjacent, Equivalent Hydrogens?
Why don't chemically equivalent hydrogens on adjacent carbons split the 1H NMR signal? There appears to be an answer here, but I'm having trouble understanding it because I cannot rationalize what happens with the magnetic field.
Let's say we're analyzing acetylene. When we subject the molecule to an external B-field, both (equivalent) protons have 2 possible spin states. For each proton, I would expect that the other proton's magnetic moment may either augment or oppose B0, causing a doublet. When we look at the signal for the other proton, it's the same story--I would expect 2 peaks identical to the 2 for the first proton (effectively doubling the integration).
Where have I gone wrong in extending spin-spin coupling to identical hydrogens? Why does the magnetic moment of adjacent, equivalent protons not affect the transition energy?
Edit: I asked my organic chemistry professor this question, and his response was that there does exist spin-spin coupling of identical protons, but the coupling constant is equal to 0. The reasoning involved quantum mechanics that went over my head.

The previous answer explains that the energy levels are there and are split as you surmise, but due to selection rules  rf transitions  into and out of these levels are 'forbidden', thus no nmr signal can be observed.  This is what your prof. probably meant by 'zero coupling constant' in the $A_2$ case. The transitions that are allowed have the same energy hence no line splitting is observed.

There are two ways to look at this.
a) All transitions you can see are between a state where the spins are parallel and a state where they are antiparallel.
$aa <-->  ab$  or  $ab <--> bb$
Their distances are identical, as are the states $ab$ and $ba$, so, as porphyrin said in his comment, you only see one line.
b) If the two protons were chemically different, you would see two doublets.
If the difference in CS were small, you'd see the "roof effect": An AB spin system. The two inner lines come closer and grow at the cost of the outer lines.
But there is no difference in CS, and so the outer lines vanish, and the two inner lines coalesce into one singlet.


text:
Mechanism of toluene oxidation with CrO3
What is the mechanism for the following oxidation of toluene with $\ce{CrO3}$ in acetic anhydride?

Use of chromic oxide ($\ce{CrO3}$): Toluene or substituted toluene is converted to benzylidene diacetate on treating with chromic oxide in acetic anhydride. The benzylidene diacetate can be hydrloyzed to corresponding benzaldehyde with aqueuous acid.

This appears to be the Étard reaction, which proceeds via a 2,3 Sigmatropic rearrangment.  You can read about it here.

In etard reaction Chromylchloride is used and if it were like etard reaction then methyl of toluene will not get disubsituted.

I don't see why CrO2Cl2 would behave any differently than chromic oxide.  Can you explain why you would expect a different outcome?

Please see this http://chemistry.stackexchange.com/questions/58381/etard-reaction-chromium-complex

https://en.wikipedia.org/wiki/%C3%89tard_reaction

wikipedia mechanism is based on papers of 1950-60s .. pubs.acs.org/doi/abs/10.1021/ic200877f this is a recent mechanism proposed

What is the evidence that the Etard reaction is [2,3]-sigmatropic rearrangement? Read this: https://chemistry.stackexchange.com/questions/140039/%c3%89tard-reaction-of-ethylbenzene/140153#140153

The following mechanism is based on the findings made in The Mechanism of the Etard Reaction[1].

Reference:
(1) Necsoiu, I.; Balaban, A. T.; Pascaru, I.; Sliam, E.; Elian, M.; Nenitzescu, C. D. The Mechanism of the Etard Reaction. Tetrahedron 1963, 19 (7), 1133–1142.


text:
How to make copper plating solution?
I intend to make the following composition of copper plating solution:
\begin{array}{c:c}
\rm material   & \rm concentration \\\hline
\rm Copper sulfate & \pu{200 \frac{g}{L}}  \\
\rm sulfuric~acid  & \pu{100 \frac{g}{L}}  \\
\ce{Cl-}       & \pu{50 ppm}   \\
\end{array}
Do I use pure $\ce{H2SO4}$ ($95$–$98~\%$) and pure $\ce{HCl}$ ($37~\%$)?
Please check if I am correct with calculation.
first, copper sulfate: $$\pu{200 g} \times 5 = \pu{1000g}$$
then ($98~\%$) $\ce{H2SO4}$: $$\pu{100 \frac{g}{L}} \times \frac{\pu{5 L}}{\rm 1.84 g.ml^{-1}~ (\rho~of~H2SO4)} = \pu{271 \frac{ml}{L}}$$
then use ($37~\%$) $\ce{HCl}$: $$\pu{0.05 g/L} (= \pu{50ppm}) \times \pu{5L} \times \pu{1.19 g/mL}\ (\rho~ \text{of}\ \ce{HCl}) = \pu{0.21 ml/L}$$
so $$\pu{5000ml} - \pu{1000ml} - \pu{271ml} - \pu{0.21ml} = \pu{3728.79 ml}$$ — fill up this much water at final

I don't know what you are plating (nor do I care...) but here are some general observations.
(1) Plating is messy and exacting analytical measurements are overkill. Your calculation of 3728.79 ml water for instance is wild. I'm not sure what you intend to use for a container, but just diluting to 5 l +/- 100 ml would be more than enough precision.
(2) It is unclear if 200 g/l for copper sulfate is anhydrous copper sulfate or the pentahydrate. You're near the saturation limit for dissolving anhydrous copper sulfate at room temperatures.
(3) Your sulfuric acid calculation has the right number but wrong units. You need 271 ml per 5 liters.
SAFETY: You will require eye protection and appropriate clothing.  When diluting sulfuric acid always add the sulfuric acid to the water. The reaction is very exothermic. (If you try to add water to the acid it will boil violently and splatter sulfuric acid which is more than extremely dangerous.)
You should also not work alone, and you should have some sort of plan as to how you would handle an accident. What are you going to do if you get acid in your eye, or what happens if you drop the bottle of sulfuric acid? It turns out that concentrated sulfuric acid is very slippery...
(4) Your hydrochloric acid calculation is wrong. You have the right density, but HCl acid is only 37% HCl, the rest is water. Also HCl is only 97.3 % Cl, that pesky H is the rest. Your calculation is also again for 5 liters not 1 so the units that you show are wrong. I'm not sure about copper sulfate, but it wouldn't surprise me if there were some chloride in the sulfate.  That little hydrochloric acid isn't going to change the acidity of the solution. I'd be tempted to use copper (II) chloride as the source of chloride. It stores a lot easier, and would be easily measured. (Hydrochloric acid releases HCl fumes unless absolutely sealed - like in a glass ampule.)


text:
Shouldn't 1 mole of any ideal gas at open space occupy more than 22.4 L volume at S.T.P. due to diffusion?
So if diffusion occurs,1 mole of any ideal gas should occupy a much greater volume in open space at STP.

Well, we are assuming we are putting the gas in a container (with variable volume). Otherwise it's meaningless to talk about the volume of a gas, right?

Standard pressure was redefined in 1982. Now the IUPAC definition of STP is such that a mole of gas has a volume of 22.7 liters at STP. https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

Here is the calculation for the volume occupied by one mole ($n$) of an ideal gas at STP (temperature $T = 273.15\ \mathrm K$, pressure $P = 1\ \mathrm{bar}$):
gas constant: $R = 0.08314\ \mathrm{L\ bar\ mol^{-1}\ K^{-1}}$
ideal gas law: $PV = nRT$
$$\begin{align}
V &= \frac{nRT}{P}\\[6pt]
&= \mathrm{\frac{1\ mol\ \times\ 0.08314\ L\ bar\ mol^{-1}\ K^{-1} \times\ 273.15\ K}{1\ bar}}\\[6pt]
&= 22.71\ \mathrm L
\end{align}$$
Another way of looking at this of course is that any 22.71 liter chunk of particulate-free air, of arbitrary shape, will contain about 1 mole of air. Molecules of the different gases composing the air will diffuse in to and out of this region, but it will always contain about 1 mol of air at STP.

The ideal gas equation defines the relationship between temperature, pressure, volume and quantity. The situation it models is how some of those variables change when others are altered.
For a fixed quantity (1 mole) of gas at STP all the variables are fixed except volume. But the situation you describe does not have a fixed volume.
What you should imagine doing with the equation is this. Put 1 mole of gas into a vessel where you can measure the pressure and volume while keeping the temperature constant. The equation describes how the pressure and volume will vary as you change one or the other within the vessel. When the pressure is 1atm, the volume will be 22.7 litres. If you now open the vessel, you no longer have a fixed volume and the equation is irrelevant. The gas will diffuse into the room and eventually will have mixed with the atmosphere.
You could still use the equation to measure the density of the atmosphere. The atmosphere has a fixed pressure and, assuming the mix of gases behaves in an ideal way (a good approximation) you can tell that the number of moles of gas in a given volume should conform to the ideal gas equation (i.e. 22.7L at STP). If you know the mix of compounds making up the atmosphere the mass will be the weighted average of the molar masses of the constituents.

If diffusion occurs, either the pressure or temperature (or both) of the system is affected such that the system is not at STP.
Simply rearrange $PV = n\mathrm RT$ into $V = n\mathrm R \times \frac TP$.
Assuming constant moles $n$, the pressure or temperature must change.


text:
Hydrolysis of benzalchloride
What should be the mechanism for hydrolysis of benzalchloride to benzaldehyde?

Here
B denotes Brønsted base
and HA denotes acid

What is the product of second step C6H5CCl ? The valency of C is not satisfied.

Its a carbene sort of a structure

How carbon gets double bonded oxygen in last step?

check the answer now

Where goes the proton of water in first step of second picture?

It doesnt matter ,its there somewhere in the solution

Could you please add a little bit clear pic?

sure will do after sometime

I don’t believe this unless you back it with a mechanism-researching paper.

@Jan is that wrong?

No it isnt.
It is based on research work    .
THE MECHANISM OF THE HYDROLYSIS
OF BENZAL CHLORIDE
By
Kozo TANABE and Tadaatsu 100*)
(Received December 20, 1964)

@PrakharSankrityayan That’s not even a citation. Citations include authors (first names abbreviated to initials), Journal, year, potentially volume and page number.


text:
Gattermann-Koch reaction - Role of cuprous chloride
I am unable to figure out what the role of $\ce{CuCl}$ in Gatterman-Koch reaction is. Does it provide $\ce{Cl-}$ ions by dissociating into $\ce{CuCl2}$ and $\ce{2Cl-}$, but that can also be achieved by using more $\ce{HCl}$? So why do we add $\ce{CuCl}$ in the reaction?

According to this site:

The formylation of aromatic compound can be accomplished by treatment with $\ce{CO}$ and $\ce{HCl}$ in the presence of Lewis acid. The reaction is performed either under pressure or using copper(I) chloride or anhydrous aluminum chloride whose role may be to aid the reaction between $\ce{CO}$ and $\ce{HCl}$ via the complex which it forms with $\ce{CO}$.

As you can see, it does not involve in the mechanism (provide $\ce{Cl-}$ via dissociation of $\ce{CuCl}$) but act as a promoter to speed up the reaction.


text:
Mechanism for an Étard reaction (chromium complex)
In the following reaction, a chromium complex is formed during the Étard reaction:

I have the following questions:

How is this complex formed?
How does its hydrolysis take place? (What's the mechanism?)

Wikipedia gives a different mechanism for the same reaction. I am in doubt which is correct?

@Jan is this the correct mechanism?

Wikipedia’s mechanism is basically the same.

How ? There is disubstitution in book and mono substitution on Wikipedia. I am unable to figure out how after disubstitution hydrolysis takes place? I understand hydrolysis when monosubstituted.

Ooooh, now I see it, the two escaped me. But you can think of $\mathrm{S_N2}$ type attack of a water molecule to chromium, displacement of one alcoholate which then rearranges to displace the second one. I’m not in a place where I can draw compounds now though (and won’t be until in a few days).

http://pubs.acs.org/doi/abs/10.1021/ic200877f . The proposed mechanism is radical, though it is yet to be proven experimentally. However, the mixture of products is in agreement with this proposal. 2) The mechanism would be extremely hard to prove, but I suppose it should be an addition/dissociation type mechanism, similar to nucleophilic substitution at $sp^2$  carbon atoms.

@Jan I am uploading the picture of hydrolysis please check it in the answers column.

Pretty much what I would assume, yes.

The ene reaction mechanism for the Etard reaction is questionable given this data. https://chemistry.stackexchange.com/questions/140039/%c3%89tard-reaction-of-ethylbenzene

Does this answer your question? Mechanism of toluene oxidation with CrO3


text:
Why does water have no latent heat?
When we continue to heat water at its boiling point it evaporates and some latent heat is stored in the steam formed, but when we heat ice at its melting point then why doesn't any latent heat get stored in the water?
Why does water have no latent heat like ice and steam?

Who said that there is no latent heat stored in water during melting?

Water's latent heat of melting, in fact, is greater than those of most other compounds.

Is it the latent heat present in water called "latent heat of fusion" or something else?what is it's value?

@navneet the heat of fusion of water is about 80 cal/g.

@OscarLanzi Both "heat of fusion" and "latent heat of fusion" are correct.

Water does have a heat of fusion.  It's less than the heat of vaporization, like most materials, because fusion is a much less radical rearrangement of the molecules than vaporization.
Even so, the heat of fusion of water (about 80 cal/g) is actually quite large compared with most other materials when they melt.  See the comments.

Both "heat of fusion" and "latent heat of fusion" are correct.


text:
Liquid floating on top of water due to density or insolubility?
When a liquid is floating on top of water, how can one tell whether it is because of its insolubility in water or because it has lower density than water?

If it is insoluble and more dense then it will settle down

Is a crocodile long or green?

Both are contributing factors.
The insolubility ensures that it does not mix with water.
The lower density of the liquid on top ensures that it stays on top of the water instead of below the water.

Sand is insoluble with water and when you mix it with water you see sand particles in the bottom of the water container, while when we  put hot water in cold water "water is soluble in water obviously" hot water float at the top of cold water due to the difference in density. So to make things clear, insoluble liquids which have more density than water stay at the bottom and form a heterogeneous mixture of the container while insoluble liquids which have less density than water float on top of the water and form a heterogeneous mixture and soluble substances mix with the water to form a homogeneous mixture


text:
Omission of locants for functional groups in IUPAC names
I'm having a bit of conflict while naming molecules as per IUPAC the nomenclature rules. If we take the following compound for example,

IUPAC would say that that correct name is 2-phenylacetaldehyde. But then for this molecule,

IUPAC designate it the name is chloroacetaldehyde.
Why does the aldehyde with a benzene ring as a functional group receive a '2' while the aldehyde with a halide as a functional group does not? I figured the molecules mentioned above wouldn't require a number to designate their functional groups given they're both substituents to an ethane parent chain.
Also, would it be acceptable to drop 'aldehyde'  and replace it with 'ethanal'. I've always been taught to add '-al' with a '-CHO' functional group?

$\hspace{6.2cm}$
There is only one possible phenylacetaldehyde, so its IUPAC name is in fact phenylacetaldehyde. 2-Phenylacetaldehyde is its systematic IUPAC name, but many chemicals' official and systematic IUPAC names differ, like propan-2-one for example. The reasoning for this is often historical and to reflect what is most commonly used in the scientific community.
$\hspace{4.9cm}$
Chloroacetaldehyde is quite obviously an aldehyde, but  switching the position of the hydrogen and the chlorine as you suggested results in a completely different class of molecule known as an acyl chloride. There is therefore only one possible chloroacetaldehyde and no need to specify the chlorine's location, though its systematic IUPAC name is indeed 2-chloroethanal.

The answer is correct; however, I should like to make it clear that the preferred IUPAC names are indeed “phenylacetaldehyde”, “chloroacetaldehyde”, and “acetyl chloride”, respectively.


text:
Reaction of fumaric acid and performic acid
Performic acid ($\ce{CH2O3}$) has strong oxidizing properties. So when it reacts with fumaric acid, is this the product (2,3-dihydroxybutanedioic acid) that I should expect?

My first thought would be epoxidation, but I don't suppose dihydroxylation is out of the question.

Peracids usually epxydize compounds with C-C double bonds

Wiki claims that performic acid is capable of "epoxidation, hydroxylation and oxidation reactions".

Do the three double bonds in fumaric acid help to stabilize the fumaric acid by interactions of the π-bonds? If so, does this affect which product is preferred?


text:
What is the difference between hydroxyl, hydroxide and hydroxy?
Is there any difference between hydroxide, hydroxyl, and hydroxy used in chemistry or are they all the same?

Hydroxide - Hydroxide is a diatomic anion with chemical formula $\ce{OH-}$. It consists of a hydrogen and an oxygen atom held together by a covalent bond, and carries a negative electric charge.
Hydroxyl radical- {hydro- + oxy(gen) + -yl + radical} - The hydroxyl radical, •OH, is the neutral form of the hydroxide ion (OH-1). It is the monovalent group -OH which contains hydrogen and oxygen. The suffix -yl indicates it's a radical just like carbonyl.
Hydroxy or hydroxy group: A hydroxy or hydroxy group is the entity with the formula $\ce{-OH}$. It contains oxygen bonded to hydrogen. Hydroxy is also prefix which means containing hydroxyl group.


text:
How to use zinc sulfide to produce hydrogen (via photolysis)
According to wikipedia, zinc sulfide added to water and then brought into the presence of (sun?) light will split the water into hydrogen and oxygen. Is this practical (for demonstration purposes) and are any other materials/apparatus needed? I cannot seem to find much useful information which answers this question elsewhere.

Your source did not mention oxygen gas?

The reference in your source states that "$\ce{ZnS}$ is one of the most widely investigated photocatalysts because it rapidly generates electron-hole pairs under photoexcitation, and exhibits a relatively high activity for $\ce{H2}$ production under UV light."

It further links to this site which contained the set-up.

@user34388, why don't you sum up your comments and write the answer to the question?


text:
How to select the parent chain in mechlorethamine?
Here is the structure of mechlorethamine:

Although the longest carbon chain is ethyl, why does Wikipedia name the compound as bis(2-chloroethyl)methylamine (or N,N-di(2-chloroethyl)methanamine)?
It seems that that is the preferred name, rather than 2-chloro-N-(2-chloroethyl)-N-methylethanamine which does give a result on ChemSpider.
I understand that the longest carbon chain does not necessarily yield the parent chain, as with cyclic compounds (where the chain lies within the ring) and compounds with multiple bonds (where the longest chain with multiple bonds is preferred over a longer one with no multiple bonds). I am guessing that perhaps the rule of more substituents is used to determine the parent chain here, but the name on Wikipedia seems to have two substituents compared to 2-chloro-N-(2-chloroethyl)-N-methylethanamine which has three. What is happening here?

Please note that many “IUPAC names” given on Wikipedia are not in accordance with current IUPAC recommendations or simply wrong. In this case, the preferred IUPAC name is 2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine.

Also note that “the longest chain with multiple bonds is preferred over a longer one with no multiple bonds” is no longer true. This rule has been changed in the 2013 IUPAC recommendations.

Related: http://chemistry.stackexchange.com/q/47777/7951

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), secondary and tertiary amines can be systematically named in various ways similar to the methods for primary amines:

P-62.2.2 Secondary and tertiary amines
P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.
(1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;
(2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
(3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2) are set off by parentheses to distinguish them from these deprecated names.

Therefore, the preferred IUPAC name for the tertiary amine given in the question is generated according to method (1) using the suffix ‘amine’, the name of a parent hydride, and further N-substitution. (Other methods may be used in general nomenclature.)
Concerning the choice of the principal chain, the rules for amines state:

P-62.2.2.2 Selection of the principal chain or senior ring system in secondary and tertiary amines.
Names of amines formed substitutively by using the retained name aniline or the suffix ‘amine’ are based on a principal chain and a senior ring system (see P-44.1). When a choice for parent hydride is possible between a ring and a chain, the ring is preferred. In names using ‘amine’ as a parent hydride, substituent groups expressed as prefixes are cited in alphanumerical order; the prefix(es) immediately preceding the term ‘amine’ is (are) enclosed in parentheses.

Thus, the rules essentially refer to the usual seniority order for parent structures that is stipulated in P-44.1 and the following subsections. The relevant criteria for the compound given in the question are

P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (…) and the seniority of suffixes (…).

and

P-44.3.2 The principal chain has the greater number of skeletal atoms (…)

In case of the compound given in the question, both methane (methylamine) and ethane (ethylamine) are parent structures that have the maximum number of substitutents corrsponding to the principal characteristic group (i.e. the suffix ‘amine’) since the compound contains only one amine group. The principal chain for the compound given in the question is ethane rather than methane since it has the greater number of skeletal atoms (i.e. the longest chain).
In accordance with the above-mentioned method (1), using the name of the parent hydride ‘ethane’ and the suffix ‘amine’ with the appropriate locant yields ‘ethane-1-amine’. The chloro substitutent is placed at atom ‘2’ of the parent hydride ‘ethane’, which leads to
‘2-chloroethan-1-amine’. Further N-substitution with the substituents ‘methyl’ and ‘2-chloroethyl’ results in the preferred IUPAC name
2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine.
Note that the locant ‘1’ is omitted in ‘ethanamine’; however all locants must be cited in ‘2-chloroethan-1-amine’ and ‘2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine’.

The alternative name bis(2-chloroethyl)(methyl)amine can be generated according to method (3). This is not the preferred IUPAC name, but it is a correct systematic name that may be used in general nomenclature.
Note that, according to method (3), the parent hydride in this case is ‘amine’ (i.e. neither ethane nor methane). The names of the substituent group(s) R, R' or R" are added as prefixes in alphabetical order, for example: ‘ethyl(methyl)(propyl)amine’. The multiplicative prefixes ‘di’ or ‘tri’ are added as usual, for example: ‘diethyl(methyl)amine’. However, the multiplicative prefixes ‘bis’ and ‘tris’ are used for substituted substituents (such as ‘2-chloroethyl’) and other compound or complex entities, which results in the name bis(2-chloroethyl)(methyl)amine for the compound given in the question.
Note the correct use of parentheses according to Subsections P-62.2.2.1 (3) and P-62.2.2.2.

The name N,N-di(2-chloroethyl)methanamine, which has been proposed in the question, is simply wrong. It violates multiple nomenclature rules.


text:
What is the region of this wavelength (462 nm)?
My lecturer said 462 nm is infrared region. I think the wavelength is visible light. Which region is the correct one? I need confirmation about this one.

Light with a wavelength of 462nm is in the visible light region (blue light visible by the human eye).
For certain, your lecturer was wrong that the light is in the infrared region, this occurs at much higher wavelengths closer to 700nm. More possibly he meant to say ultraviolet, which occurs somewhere below 400nm (shorter wavelengths), though even this isn't technically correct.
The Wikipedia article gives a fairly good overview of the various regions, along with a schematic describing where they come:

I would like to say that there is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected.For example, SLF radio waves at 60 Hz may be received and studied by astronomers, or may be ducted along wires as electric power, although the latter is, in the strict sense, not electromagnetic radiation at all we can just loosely divide them based on their wavelength but without any strict range.
As far as you question is concerned 462nm can never be infrared because viable light has roughly range from 400nm to 700nm as the name suggests infrared ie.inferior to red that is it comes below red in usual em spectrum diagram.(Note that red light is at higher wavelength region in visible spectrum) so you lecturer had make a mistake (slip of tongue) may be wanted to say UV rays but that is also wrong.

Why downvote? Is there anything wrong?

You couldn't say it's UV - margin is not that big. Also definition with production exists only for gamma AFAIK.

@Mithoron I have corrected that for UV. But the definition for production exists for other waves too Wikipedia says:For example, SLF radio waves at 60 Hz may be received and studied by astronomers, or may be ducted along wires as electric power, although the latter is, in the strict sense, not electromagnetic radiation at all

@Mithoron Also I am not saying that all types of waves depend on source they just depend on mode of production and detection. Yes it is true that X rays and gamma rays are differentiated based on their source.


text:
Why there is no one scale universal in measuring temperature?
a) Why we use Fahrenheit scale for measuring body temperature, Celsius scale for whether report, and for no use in daily life of kelvin scale it has importance in scientific calculations or in my course books.
b) Why kelvin can not be negative... If there is solid even then we can not reach negative of it... Why?

(a) We don't use Fahrenheit; it's always been an enigma to me why you do. (b) Temperature is energy; when you take away all energy, and all molecular motion ceases, you can't go below that.

@IvanNeretin Temperature is not energy.

Should this be migrated to physics.SE?

@user34388. When you heat something up, what are you doing to it? You're transferring energy from one thing to another.

@NotWoodward That's a simplified model of temperature.

(a) The US and its various territories is the only country that still uses Fahrenheit. There was an effort years ago to move the US to the metric system that flopped. Sooner or later we'll have to bite the bullet. In the meantime we still sell stuff by the pound in the grocery store, gasoline by the gallon, and 1/4 inch socket sets. :-(
(b) The truth is that absolute zero on the kelvin scale is an abstraction. The wonder is exactly what would happen if you could get to that temperature. Thus a great deal of effort has been spent trying to get to low temperatures to test the scientific theories about the interaction between matter and temperature. At that sort of low temperature weird quantum mechanical effects are observed, for instance Bose–Einstein condensates. See the Wikipedia article on Absolute Zero for more information.
Also because of the way our scientific theories are axiomized you can have negative Kelvin temperatures. Negative temperatures are strictly a quantum phenomenon.


text:
Why is the spin magnetic moment for nickel(II) given as 1.7 Bohr magnetons in some references?
I read some theoretical articles concerning $\ce{Ni-O}$ system, where $\ce{Ni}$ sits at the center of the octahedron formed by $\ce{O}$. The calculated magnetic moment on the $\ce{Ni}$ sites is about $1.7$. And these articles said that this is the typical magnetic moment of $\mathrm d^8\ \ce{Ni^2+}$. I think according to $\mu=\sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons, $\mathrm d^8\ \ce{Ni^2+}$ should have 2 unpaired electrons and $\mu=2.8$.

Can you link the article? My guess is that one has to think about something like the effective number of unpaired electrons or something weird like that.


text:
What is the need for so many theories of acids and bases?
What is the need for the  Brønsted-Lowry theory and the Lewis theory of acids and bases?Is there any problem with the Arrhenius theory?

First came the Arrhenius theory. It defines an acid as a species that dissociates to produce $\ce{H+}$ in solution, and a base as a species that dissociates to produce $\ce{OH-}$ in solution. (Examples: $\ce{HCl}$, $\ce{NaOH}$)
Next came the Brønsted-Lowry theory, which expands the notion of acid and base. It defines an acid as a species that loses $\ce{H+}$ in a reaction, and a base as a species that gains $\ce{H+}$ in a reaction. All Arrhenius acids and bases are Brønsted-Lowry acids and bases; hence the Brønsted-Lowry theory generalizes the concepts of Arrhenius acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{NH4+}$, $\ce{CH3COO-}$)
Finally came the Lewis theory, which further expands the notion of acid and base. It defines an acid as a species that accepts an electron pair, and a base as a species that donates an electron pair. All Brønsted-Lowry acids and bases are Lewis acids and bases; hence the Lewis theory generalizes the concepts of Brønsted-Lowry acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{Li+}$, $\ce{CO}$)

Remark: Hopefully the repetition and structure within my answer has been suggestive. The benefits of generalization are obvious: the theory can explain more, and more things can be explained; this is the natural progression of science.

Nice post, but I'd point out that by working up from the Arrhenius theory through the Brønsted-Lowry theory to the Lewis theory one develops a good knowledge of the subtleties of chemistry. Like in math you don't need a  10 pound theorem to prove a two pound problem.

Shouldn't "solution" in Arrhenius theory be qualified by "aqueous" or is it that the dissociation can take place in any solution?

@MrAP, yes, an aqueous solution is implied.


text:
Baldwin's rules exceptions: cations
Why do cations violate Baldwin's rules?
For example, the following acetalization mechanism is plausible even though it contains a 5-endo-trig step:

Image taken from: Clayden, J. Organic Chemistry; Oxford University Press: Oxford, 2013

The acetal formation in question is actually an example of a 5-endo-trig cyclisation, which, according to Baldwin's rules should be unfavourable.
The first thing to realise about Baldwin's rules is that they are largely empirical, that is, based on observation of data rather than theoretical proof (though they can be rationalised based upon orbital overlap arguments).
The second thing to realise is that, as with any rule, there are many exceptions.

From Wikipedia: Baldwin's rules discuss the relative rates of ring closures of these various types. These terms are not meant to describe the absolute probability that a reaction will or will not take place, rather they are used in a relative sense. A reaction that is disfavoured (slow) does not have a rate that is able to compete effectively with an alternative reaction that is favoured (fast). However, the disfavoured product may be observed, if no alternate reactions are more favoured.

The 5-endo-trig acetal formation is actually one of the most well known exceptions to Baldwin's rules (Baldwin himself acknowledged this).
In an 1981 JOC paper (J. Org. Chem. 1981, 46, 2196), Mike Perkins and Jack Barrett attempted to understand why it was that these acetalizations were so 'favourable'.
Using very basic computational methods (MNDO methods), they studied the energy needed in order to get the molecule into the correct conformation for the hydroxyl to attack the oxonium (this is the ring closing step, and the one which you might expect to he the disfavoured step).
The conclusions of the paper are that the barrier is actually much smaller than expected, making formation of the acetal acceptable, in spite of baldwin suggesting otherwise.

Is there any suggestion why the activation energy is much smaller than expected? Cations violate the rules frequently, some intrinsic property has to account for that.

Read the paper and see what you think.   And well, oxoniums are activated species. They're intrinsically more reactive which will overcome some of the geometric issues

Okay, I see they made a correlation  between torsion barrier of the double bond and the feasibility of the 5-endo-trig process. Could you add in your last sentence "torsion" before "barrier"?


text:
Why does a distinction between homogenous and heterogenous mixtures exist?
Take a known homogeneous mixture like air (oxygen and nitrogen) for example. The two have different atomic masses, so the molecular weight is different. If I let the solution sit there for a long time, eventually they should separate due to their different mass.
It seems like this logic can be generalized to every other solution, so why are there categorizations like homogeneous mixtures and heterogeneous mixtures, when they just seem to be more extreme versions of another?

The effect of Brownian motion will prevent sedimentation for small (on the order of $\mu$m) species.

Your premises are wrong, the gases would never separate.

@a-cyclohexane-molecule There will be a net force greater for some particles than others depending on their atomic mass. So, even with brownian motion the net change should result in sedimentation.

@IvanNeretin Theoretically they should separate after a significant amount of time.

@IvanNeretin If I'm wrong, you mind pointing out the flaw in my logic and not the obvious fact that they wouldn't separate in a small amount of time in the laboratory?

The gravitational force is counteracted by the force resulting from the concentration gradient generated in solution, resulting in dynamic equilibrium.

This is actually a very good question if you think about it!

You’re actually asking a very interesting question and many people put little thought into it. It is true, the words heterogenous and homogenous when applied to mixtures may be considered misleading. For example, gases will actually separate to a certain extent (not perfectly, but noticeably) if they are not disturbed. All of us will have heard that ‘carbon dioxide is heavier than air and thus drops to the bottom of a container’. Any many of us will have seen an experiment with a number of candles at different heights in a container where the lower candles are extinguished first due to lack of oxygen.
In the atmosphere, however, gases mix much better due to winds. They whirl around the gases to re-mix them. Hence why even though nitrogen is lighter than oxygen both gases are found in relatively equal proportions at different heights.
Similarly, a solution of $\ce{NaCl}$ in water is often thought to be fully homogenous. However, there are actually separation techniques that make use of a $\ce{NaCl}$ concentration gradient in centrifugation: the more dense part of the solution with a relatively higher $\ce{NaCl}$ content drops to the bottom and different biomolecules are held up at exactly the concentration that matches their density.
Therefore, it seems reasonable to say ‘there really only is one type of mixture which will separate itself given enough influence of gravity’. I’m not sure whether this fully holds true, though.
First and foremost, it makes a difference whether a mixture consists of macroscopicly separated domains with a high relative concentration of one component or whether the components are randomly distributed at a molecular level. The latter is the typical description of a homogenous solution, the former is a description of two substances that don’t mix even though they are finely dispersed (e.g. fat and proteins in milk). A well-mixed sodium chloride solution in a closed vessel will not separate at any reasonable timescale while even homogenised milk will, given enough time, separate into its components because it is entropically favourable.
Secondly, there are actual physical differences between the two. For example, heterogenous solutions such as colloids can be detected by shining a focussed light beam through them. A homogenous solution will not scatter light but in a heterogenous solution you can follow the path of the light through it due to scattering. The low-level physical explanation is that of microscopically different refraction indices between the colloidal particles and the solution around it. A ‘phase’ (and a homogenous solution is defined as a single phase) is defined such that all intrinsic parameters (density, refraction index, …) are constant throughout the entire solution. Thus, while the solution looks homogenous, it is distinct from a homogenous solution at a molecular level.
Therefore also, the carbon dioxide that is collected at the bottom of the tub extinguishing candles as mentioned above has actually just not been mixed properly with the surrounding air. If there is a fan, whirling the air in the tub around, the candles are expected to be extinguished at approximately the same time. And if one hears about ‘more dense sea water’ that travels along the ocean ground, that is simply due to incomplete mixing of the different water levels.

Conclusion: While indeed it may seem that the distinction between homogenous and heterogenous is arbitrary and examples can be found that seem to support the view, there is a well-defined physical difference between the two and the examples used are actually examples of not properly homogenous solutions.

The separation of gases won't occur in a laboratory scale experiment, but does occur on a planetary scale. So at sea level the atmosphere will have slightly richer oxygen content than at the outer reaches of the atmosphere. The effect is most pronounced for helium. Helium released into the atmosphere migrates upward in the atmosphere until it reaches the fringe of outer space. Then an occasional bump away from earth will be with enough force to enable the helium molecule (an atom) to reach escape velocity and the helium "boils off" the planet. Thus the atmosphere retains no helium.

For a "homogeneous mixture" no matter what sample i take the sampless will all have the same composition. This would be like a solution of sodium chloride.
For a "heterogeneous mixture" I could, theoretically at least, take samples that would differ in composition. Milk could be a good example. The fat globules are very small, but theoretically at least,  I could get an "aqueous" sample and a "fat" sample by judicious sampling.

Then take a planet the size of earth and fill it with sodium chloride solution. Shouldn't it, like you said, separate on a large scale? In that case taking samples will not all have the same composition.

On earth even sea water has a different composition on the surface than it does in the deep ocean. The ocean is a dynamic system not a static one. You won't see any separation of a salt solution from the top of a standard one liter beaker of solution compared to the bottom layer of that solution.

Then if I take a sample on top, the mixture will be different from the sample on the bottom. Thus it isn't a solution.

Yes it is a solution, as real as can be, just with slightly different composition. Separation is what happens between, say, water and oil; this is not going to happen here.


text:
Inter electronic repulsion in s and d orbitals
This was written in a lecture slide:

Two electrons present in the same d-orbital repel each other more strongly than do two electrons in the same s-orbital.

Why is there a difference in the amount of inter electronic repulsion in each orbital? Does it have to do with the shape of the orbital?

It is related to the shape of these orbitals, but importantly, also with spin-orbit coupling. Two electrons in one orbital must have spin directions opposite to one another. There is no spin-orbit coupling in s-orbital, so these electrons repel only because of their charge. The electrons in d-orbital share a common direction of magnetic field to align relative to. Given that their preferred positions relative to the non-spherical shape of d-orbital and each other are near opposite ends along the direction of the magnetic field, their spin magnetic moments being opposite contributes to repulsion.

There are multiple reasons for the above mentioned phenomenon but mainly due to their shape and size.
In detail:

           -[D-orbitals,Source:Google]

1.Shielding: D-orbitals with double-dumbbell shape occupies larger volume of space then the small spherical s-orbitals. This leads to less shielding in d-orbitals.
2.Exchange energy: According to Pauli's exclusion principle,electrons occupying same orbital must have opposite spin states.This leads to a repulsive interplay called Exchange energy which is high in d-orbital due to its complex orientation.Hence,stronger repulsion between electrons.
3.Electron-electron repulsion: Due to the presence of nodal plane and multiple lobes,electrons are more spreaded which leads to more interaction and so, repulsion.


text:
Free-Radical chlorination: Find all unique mono-chlorinated products

I'm having trouble with this problem. I suspect that there is a typo, as the methylcyclohexene described in the wording does not match the methylcyclopentene shown in the image(someone please confirm if my suspicion is correct or if I made a mistake in interpreting nomenclature, I am very new to Organic Chemistry). If it does appear to be a typo, I would like help in solving for the methylcyclopentene(the molecule described by the image).
I'm not quite sure how to tackle this problem, as this is actually the first time I've had to mono-chlorinate something with a double bond. I'm confused about how stereochemistry factors in, since there's no symmetry in the molecule, and the double bond onto one carbon suggests a sp2 triplanar configuration where there is only one H position that a radical chlorine can take. This problem isn't like anything I've seen so far, so I would appreciate any help from some veteran organic chemists.

Well of course it's incorrect. Also you need to remember that it's two step - first alkenyl radical then halogenide

@Mithoron halogenide is the German word for halide

Nice first question!

Hint for the stereochemistry part of the question: as soon as you have any cyclic $\mathrm{sp^3}$ carbon with two different substituents attached, stereochemistry comes into play.

Can someone check these for me:

At the end of the methyl group, there is only one monochlorinated product because every hydrogen in the CH3 is identical to each other?                  2. The cyclic carbon that is double bonded and is attached to the methyl group cannot be chlorinated                                                                                3. The cyclic carbon that is double bonded and is not attached to the methyl group can be chlorinated, but because of its sp2 configuration, there is only 1 chlorinated product

the 3 other cyclic carbons that are all single bonded CH2 structures will have a sp3 configuration when chlorinated, and will become stereoisomer centers, so they have 2 monochlorinated products each. Total number of monochlorinated products for methylcycloPENTENE(adding up each carbon's corresponding numbers) is thus 1+0+1+2+2+2= 8 monochlorinated products?


text:
All resonance structures of the hydrogen molecule
This answer written by Jan says that "a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures."
Please reproduce those resonance structures of the hydrogen molecule.
If nobody is able to come up with that many, the answer with the most resonance structures would be accepted.

Note to close-voters: if you think this is off-topic, I can edit it.

I'm not sure if Jan's anecdote may have been a throwaway comment.  Given the two electrons present its hard to imagine 4 combinations let alone 80. In any case. I like the challenge. May the odds be ever in your favour @Jan

@NotWoodward considering the three regions created by a hydrogen molecule there are already 6: 2+0+0, 0+2+0 (the usual one), 0+0+2, 1+1+0, 1+0+1, 0+1+1.

@Mithoron I did not mean to challenge Jan's professor?

For reference: The professor in question was Professor Klapötke from the LMU Munich.

This is a reference kick-starter contestant solution.
There are 6 resonance structures:

$\ce{H:H}$ (the usual one)
$\ce{:H- H+}$
$\ce{H+ H:^-}$
$\ce{*H*H}$
$\ce{H*H*}$
$\ce{*H H*}$

Don't forget this one: https://xkcd.com/1442/

Come on, people. What's that even supposed to mean? Well, if we are just playing with dots, here are a few more:  $\rm{\dot H\dot H}$, $\rm{H\ddot H}$,  $\rm{\ddot HH}$,  $\rm{\dot H\mathop{H}\limits_{.}}$.


text:
Why does ethene react explosively with fluorine?
In this website it is claimed that "[e]thene reacts explosively with fluorine to give carbon and hydrogen fluoride gas", via this equation:
$$\ce{CH2=CH2 + 2F2 -> 2C + 4HF}$$
Why does the fluorine hog the hydrogen atom instead of attacking the $\pi$-bond?
Additionally, may I ask for the mechanism of this reaction?

It's seems wrong because fluorine reacts very vigorously to form not only fluorinated but also several oxidation products. tetrafluoromethane also formed during the reaction. The exact mechanism is unknow

@N.ALodhi Would I be correct in saying that the tetrafluoromethane may have formed from the C in the product above? $\ce{C +2F2 -> CF4}$

yes, you are right, here is the link http://link.springer.com/article/10.1023/B:RJAC.0000024584.69292.6d

@N.ALodhi However I do suspect that there could be a $\ce{CH2F2}$ intermediate, meaning that the carbon atom is not completely separated before the tetrafluoromethane starts forming.

There is no mechanism, in the sense that there are too many things happening all at once. Name just about any remotely sensible particle composed of C, H, F, and it would be there.

@IvanNeretin Like this or this for $\ce{CH4 + 2O2 -> CO2 + 2H2O}$?

Yeah, like that.

The equation gives only the reactants and final products and tells us nothing about the mechanism; stoichiometric equations rarely do.This is an explosive reaction so it will proceed by many steps usually involving radicals. To get feedback, and so explosion, one step must be of the branched chain form $$\ce{R_1^. +  M -> R_2^. + R_3^.}$$ where $\ce{R^.}$ is a radical and $\ce{M}$ a reactant species (e.g. $\ce{CH2=CH2}$). As two radicals are produced for one reactant radical a chain reaction ensues, the reaction rate accelerates and so bang!
Any explosive reaction has initiation, propagation, branching and termination steps, some also have an inhibition step. The initiation step in your reaction probably forms the radical $\ce{FCH2-CH2^.}$ although I don't know the mechanism.

Is there a typo in your equation?

I did not downvote.

ok, one can't tell who does this. I hope that you found the answer helpful. You can always up vote!

My answer is, in principle, correct so why down vote? It does not make sense to do this and no, no typo. Two radicals are produced, one can be the same and one different, this is a generic way of illustrating that two radicals are  produced from one initial one. If you feel happier the right hand side can be written as R2⋅+R3⋅ which I have changed in my answer.

I'd make one other point. To be "explosive" the reaction is certainly producing heat. So the rate of reaction starts off slow and as the system heats up the rate of reaction increasing dramatically.


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Why must the forward reaction and the backward reaction follow the same pathway?
It is often claimed that a (normal) catalyst catalyzes the forward and the backward reaction to the same extent, so the equilibrium position is not shifted.
I would like to ask why must the forward reaction and the backward reaction follow the same pathway?
I could imagine a mechanism for this reversible reaction from $A$ to $D$:
$$\ce{A <=> D}$$
$$\ce{A -> B<=> D}$$
$$\ce{A <=> C <- D}$$
So that the forward reaction and the backward reaction must follow different pathways, as one step of both pathways are irreversible.
Is this scenario impossible in reality? If so, why? If not, is there any example in real life?

So you think if you remove the catalyst for the B-D step, you'd have an endless energy source? Fat chance. ;-) There is no such thing as an "irreversible" reaction, especially not two that are in an equillibrium.

A catalyst cannot change the equillibrium of a two component reaction, because that'd always violate the principle of conservation of energy.
Also there is no such thing as an irreversible reaction. If the backward reaction is very unfavourable energy-wise, the catalyst won't change that, but you cannot have two highly exothermic reactions in your funny circle anyway. Where should the energy come from?

Thank you. I did not know that there is no such thing as irreversible reaction until I saw this page.

A reaction could proceed by many different pathways, for example an extreme way would be splitting the reactants into atoms and combining them to form products, but this is essentially impossible, the energy required would be huge and how would the atoms reassemble just into the product required?
In any reaction the molecules follow the lowest energy pathway, which is to say that the lowest activation barrier between reactants and products will be crossed in preference to any other pathway. The same argument applies to a catalysed as to a 'normal' reaction even though a catalysed reaction proceeds by a different mechanism. The lowest energy pathway also ensures in some general way that close to minimum number of bonds are broken in the reaction.
The reason for this is that the activation energy (barrier height) is usually far higher than average thermal energy and many collisions happen between reactants before by chance there is enough energy to cross the barrier. The probability $p$ of having sufficient energy $E$ is given by the Boltzmann exponential distribution, $p \approx exp(-E/(RT))$. This is a rapidly falling distribution as $E$ increases thus the minimum energy pathway is generally hugely favoured over others.
In biochemical reactions it is possible for the same molecules to be synthesised by different pathways, involving several steps, but the same arguments apply to each stage in the reaction within the enzymes used.Synthetic chemists will have many examples of the same products being formed from the same starting molecules and progressing by different sequences of pathways in a series of $separate$ reactions.  However, the yields will be generally be different; one of the aims of synthetic chemistry is often to optimise yields, but I take the question to ask that the reaction is not that of a separate sequence of steps but in one process.


text:
Why do higher fluorine oxyanions not exist?
I asked this question from my teacher and he said that's because fluorine has a high electronegativity and does not form dative bond with oxygen. That's why only hypofluorite $\ce{FO-}$ exists and not $\ce{FO_n-}$ ($n = 2,3,4$).
But then I was wondering why $\ce{BF4-}$ exists. And according to this Wikipedia article, fluorine does form dative bond.
so why the anions in the title do not exist?
Thanks for your help!

I can't find the word "dative" in your linked article. Could you provide a quote?

I thought this reaction involves dative bonds: "Three equivalents of HF react to give the intermediate boron trifluoride and the fourth gives fluoroboric acid:
B(OH)3 + 4 HF → H3O+ + BF−
4 + 2 H2O"

I would say that it is because the electronegativity difference in $\ce{F-O}$ is too small, but I would wait for people much more knowledgeable than me.

There are two factors at play here:

Fluorine is more electronegative than oxygen. In oxyanions, the central atom carries a substantial positive charge due to the oxygen atom pulling away electron density. In oxyfluorides, oxygen actually donates electron density to fluorine and oxygen would logically be the central atom. There are theoretical studies into $\ce{OF3+}$ but no successful syntheses yet.
The fluorine atom is tiny, even by period 2 standards. There's simply not enough space around the proximity of the atom to fit three or more other atoms.

In the hypothetical fluorine- and oxygen- containing ions, F is the central atom. However, to have 2 or more O atoms bonded to it means that F will have more than 8 valence electrons. This is impossible as F is a period 2 element and has no energetically accessible orbitals for octet expansion.
This is because F is more electronegative than O - so you're right about F not donating electron pairs to O, hence F forms only covalent bonds here.
As for $\ce{BF_4^-}$ , B in $\ce{BF_3}$ is electron-deficient with 6 valence electrons and will gladly accept an electron pair from the nucleophile $\ce{F-}$ .
Hope this answers your question!

About the first part I think dative bond is not the same as valence shell expansion. About the second part; you mean BF3 is more electron-deficient than oxygen?

Yes you're right - to clarify, F does not form dative bonds with O, or expand its octet by forming covalent bonds with O.

For your second question - yes, as B is bonded to 3 electronegative F atoms. But the point was that the octet structure of O and B affects the bonding. O has electrons available for covalent bonding unlike B in BF3 which has an empty orbital to accept a lone pair from F to form a dative bond :)