Bond angles for the hydrides I noticed the fact that all the hydrides of the elements belonging to group IV has bond angle 109.5° while in group V it varies from 107.3° for $\ce{NH3}$ to 91.3° for $\ce{SbH3}$. Similarly we see in group VI that water has highest bond angle then it significantly decreases to 89.5° for $\ce{H2Te}$. Is there any specific reason for hydrides of group IV metals to have remarkably equal bond angles, and other groups to have varying angles?

The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides.

According to VSEPR, electron domains (i.e. bonds and lone pairs) are arranged around the central atom in geometries that minimize electrostatic repulsion between the electron domains. These geometries also happen to be solutions to the Schrödinger equation for the set of molecular orbitals as linear combinations of atomic orbitals, and in the molecular orbital case, they arise not from electrostatic repulsion, but strictly from the mathematics of the linear combinations producing a minimum energy self-consistent set of orbitals.

VSEPR is qualitatively simpler for small molecules. All of the binary hydrides in Group IV, V, and VI have steric number four around the central atom. This means that the number of lone pairs and the number of bonding domains add up to four. These species can also be described using the $\ce{AXE}$ method, where $\ce{A}$ denotes the central atom, $\ce{X}$ denotes an atom bonded to $\ce{A}$ and $\ce{E}$ denotes a lone pair of electrons.

The Group IV hydrides all have four bonds and no lone pairs. We describe this arrangement as $\ce{AX4}$.
The Group V hydrides all have three bonds and one lone pairs. We describe this arrangement as $\ce{AX3E}$.
The Group VI hydrides all have two bonds and two lone pairs. We describe this arrangement as $\ce{AX2E2}$.

There are several factors that affect the degree of repulsion:

Lone pairs - Lone pair domains are more diffuse and have higher repulsion than bonding domains.
The nature of $\ce{X}$ - Larger and/or more electropositive atoms have higher repulsion.
Bond order - Bonds with higher bond orders have higher repulsion than bonds with lower bond orders.

Since $\ce{X=H}$ in all cases, we only need to worry about the first factor: lone pairs.

The arrangements $\ce{AX4}$, $\ce{AX3E}$, and $\ce{AX2E2}$ all have four electron domains, and thus the base electron domain geometry group is tetrahedral.

The differences in bond angle between $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2O}$ originate from the increasing number of lone pairs:

$\ce{CH4}$ has no lone pairs: the bond angles are the ideal tetrahedral bond angles of $109.5^\circ$.
$\ce{NH3}$ has one lone pair: the bond angles are contracted a little because of the grater repulsion of the lone pair to $107.8^\circ$.
$\ce{H2O}$ has two lone pairs: the bond angle is contracted further due to the repulsion of two lone pairs to $104.5^\circ$.

  All of the Group IV hydrides will have perfect tetrahedral geometry due to having four bonds to the same atom and no lone pairs. Thus $\ce{CH4}$, $\ce{SiH4}$, $\ce{GeH4}$, $\ce{SnH4}$, and $\ce{PbH4}$ all have bond angles of $109.5^\circ$.

For Group V, the bond angles are:

$$\begin{array}{c|c}
\ce{AH3}&\text{angle}\\ \hline
\ce{NH3}&107.8^\circ \\
\ce{PH3}&93.5^\circ \\
\ce{AsH3}&91.8^\circ \\
\ce{SbH3}&91.7^\circ \\
\ce{BiH3}&90.5^\circ \\ \end{array}$$

Notice that there is a large decrease from $\ce{NH3}$ to $\ce{PH3}$ and then very small changes for $\ce{AsH3}$, $\ce{SbH3}$, and $\ce{BiH3}$.

You could make the argument that the larger orbitals of the heavier elements will have high repulsion than the smaller orbitals of nitrogen. This is where VSEPR breaks down. VSEPR works well for some compounds because there is significant $s-p$ mixing when constructing the molecular orbitals. This mixing need not happen for the heavier elements. One molecular orbital solution for $\ce{PH3}$, for example, indicates very little contribution from the $\ce{P}\ 3s$ orbital. We can describe the $\ce{P-H}$ bonds as predominantly $\ce{P}_{3p}\ce{-H}_{1s}$ bonds. Thus the geometry around $\ce{P}$ is determined mostly by the orientation of the $p$ orbitals.

If we look at the Group VI hydrides, we see the same trend: a large decrease from $\ce{H2O}$ to $\ce{H2S}$ and then very little change. The same argument can be made for primarily $\ce{A}_{p}\ce{-H}_{1s}$ bonding in the heavier Group VI hydrides as for the heavier Group V hydrides.

$$\begin{array}{c|c}
\ce{AH2}&\text{angle}\\ \hline
\ce{H2O}&104.5^\circ \\
\ce{H2S}&92.1^\circ \\
\ce{H2Se}&91.0^\circ \\
\ce{H2Te}&90.0^\circ \\
 \end{array}$$


Barron states that 4.18×10⁸ joules equal 1 kcal, is this correct? Here is a paragraph in Barron's SAT chemistry prep book:

  Work itself is measured in Joules, and so is energy. In some problems, however, energy may be expressed in kilocalories. The relationship between these two units is that $4.18\times10^8$ Joules equals 1 kilocalorie (kcal).

Is this correct? Isn't 1 kilocalorie 4180 Joules?

This should never have happened in the scientific community, but:

  The energy values of food are measured by the Calorie, with a capital C. A Calorie is equal to 1,000 calories or 1 kilocalorie (kcal).

Source: here

It is possible that Barron's got confused between Calorie and calorie. Even so, they still would be off by a factor of 100.

According to Wikipedia:

  The SI unit of energy is the joule, with symbol "J"; one small calorie is now defined as exactly 4.184 J, and one large calorie is 4184 J. However, the two units are still used occasionally in technical work, and the large calorie is still widely used in nutrition.

So you are correct in saying that 1 "small" kilocalorie is 4180 joules. 1 "large" kilocalorie is $\pu{4.18e6 J}$, and maybe Barron's is now suggesting an "extra large" calorie.

More likely it is just a typo.


Catalytic energy profiles (is this Wikipedia image misleading?) I am looking at the image below from the Wikipedia page for energy profiles in chemistry. I do not understand why the "$E_a$ (with catalyst)" is labeled as it is. I would argue that is not the activation barrier with the catalyst. Shouldn't all of the individual steps need to be considered to get the rate of production of product (or, if one step has an especially high barrier, then that step can be assumed to be rate-limiting and that step's $E_a$ is the barrier with the catalyst)? I suppose in other words, if I do $r = A\exp(-E_a/k_{B}T)$ where $E_{a}$ is the activation energy with catalyst, that will not get me the rate, right?

For the record, I have taken grad-level courses in chemistry, but I got myself into a position of questioning everything I once learned and here I am.

Of course, in a strict sense the reaction path using a catalyst should be split into four sub-reactions (all going by the exact image) and each of those sub-reactions should get their own activation energy assigned. Two of the four steps would be endothermic, two exothermic. (Note that it is not necessary for the endothermic steps to occur first; it could be the other way around, mixed or anything.)

However, there is still some truth in labelling ‘$E_\mathrm{a}$ with catalyst’ as such. Even though you have much smaller steps you will still need to add that labelled energy in net terms to overcome the barrier. The principal difference is that it need not be added all at once but can be added stepwise. You could even say that the total activation energy (i.e. adding up all the individual $E_\mathrm{a}$) be even higher, but you regain some of that from the intermediate reactions.

If you decided to supply only enough energy so that a molecule could jump the barrier from intermediate 2 to intermediate 3, that would not make the reaction go. For one, the barrier to get to intermediate to is larger by itself. For two, you would only get that much above the baseline which might be enough to cross the barrier to intermediate one but no further.


What is the reason of explosion of batteries? In social sites we see many pictures showing awareness of not using smartphone while charging because of the reason that battery explode while charging, is really batteries explode and if yes than what is the reason behind it.

The most common cause for a battery to explode is the contents of it heating to a point where it exerts a force on the battery too large for it to handle. Basically, if the battery feels very hot, it could be under large pressure and the object that it is powering should be powered off if possible. The recent reports of smartphones catching fire are due to a manufacturing error by Samsung on the Galaxy Note 7. A portion of these phones were not insulated correctly and it caused a short circuit that ended up burning holes through the phones. This was not an explosion, but the phones did catch on fire. In conclusion, you can use your phone while charging if it's not a Galaxy Note 7 (which has been recalled).


Where do the percents in empirical formula questions come from? A typical empirical formula problem reads something like this:

  Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur. source

How would those percentages be determined in real life? Spectroscopy?

A percent composition problem would obviously yield percentages, but those start with compound formulas--there'd be no need to determine the empirical formula from the percentages.

How would those percentages be determined in real life? Spectroscopy?

You are right, this analysis comes in the domain of analytical chemistry. The fundamental requirement for formula determination is that your sample must be extremely pure.

There is a whole field. So there are two broad classes of approaches. One is called classical analysis where one would dissolve the sample in an "appropriate" acid or base mixture, and this is followed by target element specific titration or precipitation. This technique results in very reliable data but it is very time consuming

Today most people would use an instrument called the inductively coupled plasma optical emission spectroscopy (ICP-OES), you "srpay" the dissolve sample into a state of matter (plasma) which is as hot as the surface of the Sun. Under those those conditions, compounds break down into their constituent atoms and emit characteristic wavelengths.
For example, if we are interested in knowing how much aluminium is there, we would construct a calibration curve and interpolate the amount of Al in your unknown sample.
The same thing (although less effectively) can be done for sulfur.

Sometimes you need both classical or instrumental analysis.


Is it possible to define the coordinates of an atom using other coordinates while optimizing in gaussian 16? I know it is possible to fix the value of composite expressions using generalized internal coordinates (gic) in gaussian 16. But is it possible to assign that value to the coordinate of another atom.

For example if I have the following coordinate file

H x1 y1 z1
H x2 y2 z2
O x3 y3 z3

Is it possible to assign x3 = (x1 + x2)/2 . When optimizing this structure I would like that the x coordinate of oxygen always be the average of the x coordinates of the two hydrogen atoms.

I just figured out that there is a way to do this. Though its not that clean I think. I can define the initial structure where x3 = (x1+x2)/2 and then freeze the following expression using gic.

x3 - (x1+x2)/2


Container to supercool water The internet is full of videos like this showing water bottles for drinking used to supercool water. These bottles, not unexpectedly, appear a bit finicky, since they have lots of edges to serve as nucleation sites for ice crystals. What container would work best for this sort of demonstration?

Is the most important quality just to eliminate edges or does the material have an important effect? It's important in lab to add boiling stones as nucleation sites when boiling water in glassware. Could I use this trait for freezing too? Is the important quality of the glass the high surface energy or the smoothness of the surface? Alternatively, it seems like something like PTFE coating with super low surface energy might be effective?

One way to approach this problem is think about what kind of material would best promote crystallization, or ice-nucleation, and then essentially use a material that does not have those characteristics.

Consider cloud seeding for example. The whole premise of this is that there is a dearth of ice-freezing-nuclei in the atmosphere, so in the presence of supercooled water droplets, you introduce a compound with a similar crystal structure to ice, like silver iodide for example. Even crystalline compounds with different structures like sodium chloride and dry ice are also effective. So, since we're trying to do the opposite of ice-nucleation, we want opposite characteristics of the surfaces in contact with the supercooled water; nothing crystal-like in nature, something as smooth and clean as possible on a very small scale.

Teflon is a highly hydrophobic synthetic polymer, meaning it's similarity to both liquid and solid water is minimal. A smooth, clean, unscratched Teflon surface is probably the ideal choice. A cheaper option using the same concept would be other organic polymers like polyethylene (probably what the bottles in your video link were made of), polypropylene, HDPE, etc.

Another option that is not crystal-like by definition is any type of consumer glass container. Again, any nano-scale defects or contamination will degrade it's ability to inhibit ice formation from supercooled water.


How is the bonding in the [Au6C(PPh3)6]2+ cluster explained? How can carbon form six bonds in such a compound as $\ce{[Au6C(PPh3)6]^2+}?$ I understand how carbon with five bonds can be formed $(\ce{CH5+},$ for instance), but it shouldn't have enough electrons to form six bonds.

This is by no means a trivial case.
You are dealing with a so-called template or interstitial atom placement within a cluster shell.
There is a thorough review available [1, pp. 18–20] (reference numbers updated):

  An alternative way of satisfying the closed shell electronic requirements and retaining a positive charge on the cluster is possible if an interstitial atom is introduced.
  For an octahedral gold cluster an additional gold atom is too large to satisfy the steric requirements of the interstitial cavity, but a small main group atom, e.g. C, N or O, is able to satisfy these steric requirements.
  Figure 11 gives the relevant interaction diagram for octahedral $\ce{[Au6C(PPh3)6]^2+}$ and an interstitial carbon atom, which contributes just the right number of electrons to lead to a $\mathrm{[S^σ]^2[P^σ]^6}$ pseudo-spherical ground state conﬁguration.
  It is noteworthy that this simple theoretical analysis was ﬁrst published in 1976 [2], and the existence of $\ce{[Au6C(PPh3)6]^2+}$ was conﬁrmed by Schmidbaur and his coworkers in 1989 [3–5].
  This compound had been isolated and structurally characterised previously, but the interstitial carbon had not been identiﬁed.
  The introduction of the interstitial atom strengthens the radial interactions signiﬁcantly as a result of effective overlaps between the carbon $\mathrm{2s}$ and $\mathrm{2p}$ orbitals and the matching $\mathrm{S^σ}$ and $\mathrm{P^σ}$ cluster molecular orbitals.
  In broad brush terms the stabilisation of the valence orbitals of the central atom are stabilised by $\mathrm{nβ^σ}$/degeneracy of the molecular orbitals.
  If $\mathrm{β^σ(s)} = \mathrm{β^σ(p)} = \mathrm{β^σ(d)},$ the relative stabilisations are $\mathrm{6β^σ(s)},$ $\mathrm{2β^σ(p)},$ and $\mathrm{3β^σ(d)}.$
  Therefore, the greatest stabilisation involves the $\mathrm{s}$ orbitals of the central
  atom and increases as the number of metal atoms, $n,$ increases.
  For filled shells the stabilisation energies are independent of geometry as long as the ligand polyhedron approximates to a sphere.
  It follows that gold clusters with main group interstitial atoms are characterised by a pec of $12n + 8$ (sec 8) valence electrons, since each $\ce{AuPPh3}$ fragment is associated with a filled $\mathrm{d}$ shell containing $10$ electrons and a bonding $\ce{Au–P}$ bonding molecular orbital.

  Fig. 11 Molecular orbital interaction diagram for $\ce{[Au6C(PPh3)6]^2+}.$
  Similar analyses may be constructed for trigonal bipyramidal $\ce{[Au5N(PPh3)5]^2+}$ and tetrahedral $\ce{[Au4O(PPh3)4]^2+}$

References

Gold Clusters, Colloids and Nanoparticles II; Mingos, D. M. P., Ed.; Structure and Bonding; Springer International Publishing: Cham, 2014; Vol. 162. DOI: 10.1007/978-3-319-07845-8.
Mingos, D. M. P. Molecular-Orbital Calculations on Cluster Compounds of Gold. J. Chem. Soc., Dalton Trans. 1976, No. 13, 1163–1169. DOI: 10.1039/DT9760001163.
Schmidbaur, H., Grohmann, A., Olmos, M. E.; Organogold chemistry. In: Schmidbaur, H. (ed) Gold progress in chemistry, biochemistry and technology. Wiley, Chichester, 1999. pp 648–731.
Steigelmann, O.; Bissinger, P.; Schmidbaur, H. Assembly of the $\ce{[CAu6]^{2⊕}}$ Cluster with a Tailor-Made Diphosphane Spanning the Octahedral Edges. Angewandte Chemie International Edition in English 1990, 29 (12), 1399–1400. DOI: 10.1002/anie.199013991.
Schmidbaur, H.; Brachthäuser, B.; Steigelmann, O. Direct Observation of the Central Atom in $\ce{[C\{Au[(C6H5)2(PC6H4NMe2)]\}6](BF4)2}$ by $\ce{^{13}C}$ NMR Spectroscopy. Angewandte Chemie International Edition in English 1991, 30 (11), 1488–1490. DOI: 10.1002/anie.199114881.


What is the product after treating ethene successively with chlorine and water, sulfuric acid, and potassium hydroxide? Question:

  Ethene reacts with $\ce{Cl2}$ in $\ce{H2O}$ to form A. A on reaction with $\ce{H2SO4}$ followed by heating gives B. B on reaction with alcoholic $\ce{KOH}$ gives C. Find C.

My answer is:

A: $\ce{CH2(OH)CH2Cl}$
B: $\ce{CH2=CHCl}$
C: ethyne.

Explanation:

Ethene on reaction with chlorine water undergoes addition reaction to give A. A on reaction with $\ce{H2SO4}$ followed by heating undergoes dehydration to give B. B on reaction with alcoholic $\ce{KOH}$, undergoes elimination to give ethyne.

But the answer given is $\ce{CH2=CH-O-CH=CH2}$.

Where I am going wrong and what are the products?

The first reaction is a hydrochlorination reaction:

The second reaction is an acid catalyzed ether synthesis. The the solution is full of other 2-chloroethanol, which is capable of acting as a nucleophile in an $\mathrm{S_N2}$ reaction to eject $\ce{H2O}$. This is reaction will have a much lower energy barrier than the dehydration you proposed, which requires the extraction of an aliphatic hydrogen. The ether synthesis mechanism can be seen here:

The final step is a series of two $\mathrm{E2}$ reactions (shown condensed here):

$\hspace{4cm}$


Reaction of metal carbonate with acids It is said that acids react with metal carbonate to form salt carbon dioxide and water.

Sodium carbonate reacts with hydrochloric acid to form sodium chloride , carbon dioxide and water
But if we ionise them then sodium carbonate dissociates to form  two sodium ions and carbonate ion and hydrochloric acid disassociates to form two hydrogen ion and two chlorine Ion so the sodium Ion will combine with the chloride Ion to form NaCl and the hydrogen ions will combine with carbonate ion and form carbonic acid but this doesn't happen and they form carbon dioxide and water.
WHY?

Hydrochloric acid is already dissociated to $\ce{H+}$ and $\ce{Cl-}$

Sodium carbonate as ionic compound dissolves as

$$\ce{Na2CO3(s) <=> 2 Na+(aq) + CO3^2-(aq)}$$

Chloride and sodium ions are in this reaction called the spectator/bystander ions, not being involved in the reaction.

The ongoing reactions are these :

$$\ce{ 2 H+ + CO3^2- <=>> H2CO3 <=>> H2O + CO2(aq)}$$

followed by the dissolution equilibrium:

$$\ce{ CO2(aq) <=> CO2(g)}$$

If solution is lead to crystallization, the spectator ions then meet to form ionic crystals, keeping their charges.

$$\ce{Na+(aq) +  Cl-(aq) <=> NaCl(s)}$$


Applying Raoult's Law to a condensed gas mixture So, here's the question which I'm having trouble solving :

Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.

We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively.  We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)

This gave us:
 $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$

$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$
Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$

Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.

This gave us
$$x_{benz} =p_{benz} /P_{total} = 6.35 / 8.25= 0.770$$

$$x_{tol} =p_{tol} /P_{total} = 1.90 / 8.25 = 0.230$$

By the way, here's the answer to my question :

$$p_{tol} = x_{tol} * p^°_{tol} = 0,770 × 3,79  = 2,91 \text{ kPa}$$

$$p_{benz} = x_{benz} * p^°_{tol} = 0,230 × 12,7  = 2,92 \text{ kPa}$$

So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you

You have confused the vapor-phase mole fraction with the liquid-phase mole fraction.  The liquid-phase is mole fraction in usually denoted as $x_i$ for compound $i$.  The vapor-phase mole fraction of $i$ is $y_i$.

Raoult's law is that the partial pressure over an ideal solution is equal to the pure-component vapor pressure times the liquid mole fraction, $p_i=x_i*p^°_i$.

Separate from Raoult's law we have the ideal gas law. It says that that says that partial pressures are proportional to mole fraction in the vapor phase, i.e. that $p_i=y_iP$.  So when you took the ratios of $p_{benz}/P$ and $p_{tol}/P$, you were solving for the vapor-phase mole fractions, i.e. for $y_{benz}$ and $y_{tol}$.

Your mistake comes in taking those $y_i$ values and treating them as $x_i$ values.  That's a mis-formulation of Raoult's law.

You can read more about Raoult's law and the ideal gas law for binary ideal mixtures in my answer to a previous chem.se question.


How do steric effects affect inductive effect in compounds with tetravalent nitrogen? I want to compare the electron density on methyl group between tri‐tert‐butyl(methyl)azanium (1) and triisopropyl(methyl)azanium (2), i.e. essentially the inductive effects of $\ce{-\overset{+}{N}R3}$ groups.
$$\underset{\large\textbf{1}}{\ce{Me-\overset{+}{N}(t-Bu)3}}\qquad\underset{\large\textbf{2}}{\ce{Me-\overset{+}{N}(i-Pr)3}}$$
The only difference I see in these compounds is that due to steric effects the bond angle would be greater in compound 1 than in 2. Based on this info alone I am not able to judge which one would have greater electronegativity on $\ce{\overset{+}{N}},$ or so as to say greater inductive effect due to the $\ce{-\overset{+}{N}R3}$ group.
How should I go about comparing the electron density on methyl group in both of these species?

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).
$\ce{Me-\overset{+}{N}(^iPr)3}$
We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.4367\\
\ce{H} & \;\;\, 0.1954\\
\ce{H} & \;\;\, 0.1957\\
\ce{H} & \;\;\, 0.1966\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.151$.
$\ce{Me-\overset{+}{N}(^tBu)3}$
Again, we only need to look at the Me group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.3087\\
\ce{H} & \;\;\, 0.1436\\
\ce{H} & \;\;\, 0.1430\\
\ce{H} & \;\;\, 0.1414\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.1193$
So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.
Explanation?
I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *

Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—"That methyl groups attached to carbon atoms are electron donors must not be generally assumed."


Cis/Trans Isomerism in Cycloalkenes How do I tell which one is cis/trans? I know the definitions: cis is same side; trans is across; but

1) What am I comparing here.

2) How do I compare/visualize?

In simple molecules like what you've drawn above, see if the hydrogens are on the same side of the double bond or not.  If they are on the same side, then the compound is cis; if the hydrogens are on opposite sides, then the compound is trans.  In more complex cases, prioritize the four substituents on the double just as you've been doing with "R\S" nomenclature.  If the two highest priority groups are on the same side then the compound is cis, if they are on opposite sides, then the compound is trans.  The terms "cis" and "trans" are still used, but especially in the more complex cases they have been replaced with the first letters of the German words "entgegen" (E, opposite) and "zusammen" (Z, together).  So cis-2-butene could also be called Z-2-butene.


change of pH in urine I'm looking to lower the pH of a urine solution from 12 to 7. I'm planning to do it in 2 ways. First is to use a strong acid such as HCl. Secondly, using vinegar, a weak acid. However, I am having troubles calculating the pH in both methods as it other chemicals in urine such as ammonia may react to it. Since it's a theoretical experiment, can I just assume that the acid would only react to water? But how will I do it either way?

Not knowing what concentrations of the other chemicals that are present, it is impossible to do a theoretical calculation. The only way to do this would be experimentally.


Ammonium is a weak acid but isn't it considered a strong conjugate acid as a salt's ion? I'm learning about salt hydrolysis in school and this just jumped out at me. How can $\ce{NH_4^+}$ be both a weak acid and a strong conjugate acid? Since $\ce{NH_4^+}$ is a weak acid it has a strong conjugate base $\ce{NH_3}$. However, as part of a dissolved salt, since $\ce{NH_4OH}$ is a weak base, then $\ce{NH_4^+}$ is considered a "strong" conjugate acid. How can we reconcile these facts? Are the definitions slightly different for a salt ion and a normal acid?

Just because something is a weak base does not mean its conjugate acid has to be a strong acid. Likewise, just because something is a weak acid does not mean that its conjugate base is a strong base.

The only connection between acidity of an acid and basicity of a conjugate base, henceforth labelled $\ce{HA}$ and $\ce{A-}$, is that they are related by the ion product of water, $K_\mathrm{w}$ by the following equations:

$$\begin{align}\ce{HA &<=> H+ + A-}&\quad K_\mathrm{a} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{1}\\[0.8em]
\ce{A- + H2O &<=> HA + OH-}&\quad K_\mathrm{b} &= \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\tag{2}\\[0.8em]
K_\mathrm{a} \times K_\mathrm{b} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\\[0.6em]
&= \frac{[\ce{H+}][\ce{A-}][\ce{OH-}][\ce{HA}]}{[\ce{HA}][\ce{A-}]}\\[0.6em]
&= [\ce{H+}][\ce{OH-}] = K_\mathrm{w}\tag{3}\\[1.2em]
\Longrightarrow \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} &= 14\tag{4}\end{align}$$

Therefore, the sum of the acid and base constants must equal $14$. But if $\mathrm{p}K_\mathrm{a} = 6$, then $\mathrm{p}K_\mathrm{b} = 8$ — meaning the conjugate base of a weak acid is a weak base. This basically mirrors ammonia’s case.


Balancing a chemical equation I'm in chemistry and I still have a hard time comprehending how to balance a chemical equation.
What are the main steps and general things to know?
thank you.

Basically, balanced chemical reactions must 1) have the same number of each element on each side, and 2) have the same net charge on each side. Consider the reaction $H_2 + O_2 → H_2O$. It is unbalanced because there are two oxygen atoms on the left side but only one on the right. If we assign variable coefficients to each reactant and product, $a H_2 + b O_2 → c H_2O$, we can set up a system of equations by multiplying the variables by the number of atoms in each species. For oxygen, $2b = c$ and for hydrogen, $2a = 2c$. While the system has infinitely many solutions, we take the lowest integer solution. Thus, the balanced equation is $2 H_2 + O_2 → 2 H_2O$.

Chemical reactions involving charges are a bit more complicated. A good explanation can be found this tutorial.


Fluorine isotopes and isomers I was looking up fluorine isotopes out of curiosity. The Wikipedia entry gives info on them but adds there are also two isomers which are $\mathrm{^{18m}F}$ and $\mathrm{^{26m}F}$. What is an isomer of an atomic nucleus please someone?

Just like electrons nuclei can exist in a number of quantised states. We see evidence of this when, during radioactive decay, a gamma ray is produced. This is a nucleus in an excited state decaying to the ground state, and the excess energy being radiated away as a gamma ray photon. This is precisely analogous to an electron in an atom falling from a high energy state to a lower energy one, and thus producing radiation which we can see in an atomic spectrum. So an isomer of a nucleus is simply that nucleus in a state which is not the ground (lowest energy) state.

Most isomers are very short lived, and thus the gamma radiation appears to be more or less coincident with any other radiation that is produced during a radioactive decay. For instance the $\ce{^{18m}F}$ isomer you mention has a lifetime of 162ns according to https://en.wikipedia.org/wiki/Isotopes_of_fluorine . However the odd one is much longer lived, and for a remarkable case of one that is observationally stable take a look at $\ce{^{180m}Ta}$. 0.012% of natural Tantalum consists of this isomer, and it has never been observed to decay to the ground state - as is usual in such cases symmetry forbids the transition. And what I find even more remarkable is that the ground state itself is unstable with respect to both electron capture and beta emission, decaying to isotopes of Hafnium and Tungsten with a half life of just over 8 hours! See https://en.wikipedia.org/wiki/Isotopes_of_tantalum for details.

Sorry - slightly belated edit, but it occurred to me that isomers do have one very direct use in Chemistry, namely Mossabuer spectroscopy, see https://en.wikipedia.org/wiki/M%C3%B6ssbauer_spectroscopy. In essence very small changes in the energy of the gamma ray that is emitted when the isomer decays to the ground state can be detected by this technique, and can be related to changes in the chemical environment within which the atom finds itself.


Why do hydroxyl groups only produce a singlet in proton NMR? Take, for example, methanol ($\ce{CH3OH}$). The hydroxyl proton ($\ce{-O$\color{red}{\ce{H}}$}$) is three bonds away from three more protons ($\ce{-C$\color{blue}{\ce{H}}$_3}$), which aren't equivalent to it, so why would this not produce a quartet using the $n+1$ rule?

You are both right and wrong.

At low temperatures, the NMR spectrum will indeed be as you predicted: the $\ce{H}$ from $\ce{OH}$ will produce a quartet and likewise the $\ce{H}$ from $\ce{CH_3}$ will produce a doublet. Note that at low temperatures the formation of $\ce{H}$ bonds is favored, hence leading to more stable components and significantly decreasing proton exchange.

At higher temperatures however, the following spectrum will be observed:

Due to the increase in temperature, the $\ce{H}$ bonds are weakened, and hence shields its signal. Most importantly, the intermolecular proton exchange, non negligible at those temperatures, will render the interactions between the two groups of $\ce{H}$ undetectable via NMR. Hence why you get two singlets.


Ordering the reduction potentials for Cr3+, Mn3+ and Fe3+? I have been asked to order, with explanation, the standard reduction potentials for the following:

$\ce{Cr^3+ + e- -> Cr^2+}$
$\ce{Mn^3+ + e- -> Mn^2+}$
$\ce{Fe^3+ + e- -> Fe^2+}$

(all ions in aqueous conditions)

I am really uncertain what factors to consider; I am even struggling to make this a general born-haber process. I know that $\mathrm{3^{rd}}$ ionization enthalpies will come into play, and that $\ce{Fe}$ will be expected to have the highest one. Will hydration enthalpies also be relevant? Will there be a ligand exchange component (with the solvent, water) that I have to consider also?

Firstly, I'd like to say the explanations we propose here are merely our efforts trying to explain what we observed experimentally in the first place. Let's get some data:

$${\begin{array}{|c|c|} \hline \ce{Cr^3+/ Cr^2+} & \ce{Mn^3+ / Mn^2+} & \ce{Fe^3+ / Fe^2+} \\ \hline \pu{-0.41 V} & \pu{+1.57 V} & \pu{+1.97 V} \\ \hline \end{array}}$$

And while we're at it, let's ponder on electronic configurations as well:

$${\begin{array}{|c|c|} \hline \ce{Mn^2+} & \pu{[Ar] 3d5 4s0} \\ \hline \ce{Fe^3+} & \pu{[Ar] 3d5 4s0} \\ \hline \end{array}}$$

You were correct in thinking that $\ce{Fe^3+}$ will have the highest reduction potential due to the third ionization enthalpy coming into play, all we've got to do now is to think for manganese and chromium. Why don't we put it differently:

  Why is $\ce{Cr^2+}$ reducing and $\ce{Mn^3+}$ oxidizing when they both have the same $\pu{d4}$ configuration?

It's because if manganese acts as an oxidizer, it'll get converted into $\ce{Mn^2+}$, which is stable owing to the $\pu{d5}$ configuration. Chromium, on the other hand is notorious for showing the $+3$ oxidation state in its compounds. So, that's a win-win kind of situation we're dealing with.

The subtle role played by the solvent molecules when they're acting as ligands comes into limelight on asking yet another relevant question:

  Which is a stronger reducing agent: $\ce{Cr^2+}$ or $\ce{Fe^2+}$?

Answer: $\ce{Cr^2+}$. This is because of more CFSE stabilization of $\ce{Cr^3+}$ as compared to $\ce{Fe^3+}$ when water molecules behave as a ligand for our complex.


Spectroscopy: What is an instrument response function? Many papers involving ultrafast spectroscopy measure the instrument response function and model it as a Gaussian, but the sources on any of this are never cited and I have no intuition for what this could mean. What is the instrument response function intuitively and why is it represented as a convolution?

Let me provide the intuitive idea so that you take the lead from here and understand the notion behind "instrument response" function. Any analytical/physical chemist, who uses an intrument, should have some basic idea of the instrument response. After that you should look up formal concepts of convolution on Youtube.
Whenever you collect experimental data using an instrument, you would like to have no distortions induced by the instrument in your data. For example, if your eyesight is weak, and you are trying to read a road sign, but it appears blurred to you, your eyes and brain have "convoluted" the true image of the road sign. Therefore, your eye/brain response (=instrument response) is to blurr the image and this is what you do not want. At least, one should be aware of this fact that the true image is not a blurred picture.
Hence people who wish to study lineshapes, linewidths in a spectrum they would like to know how their spectrometer set-up is distorting the true experimental result. In an ideal case there should be no distortion but a real instrument can introduce its own effects which we call as convolution.
Let us see a example to illustrate what Gaussian broadening does to a "true" signal shape, using an light absorbance detector (of a chromatograph). I did this just to understand the response function of that instrument.
Imagine that a light source is turning on and off periodically. It should look like a square wave to a light detector (signal is up when the light is on and it is down when the light of off). However, if you have increasing Gaussian broadening done by the instrument, you begin to loose the true shape of the square wave. If you have too much of Gaussian broadening, the square wave, appears to be completely distorted. Look at the top and the bottom figure. The top shows the relatively true signal shape, and the bottom figure shows the effect of Gaussian broadening on the same signal. All true information is lost.
Hope that highlights the importance of instrument response function in analytical chemistry/instrumental analysis.


Does water really 'go bad' after a couple of days? Among my friends it is a sort of 'common wisdom' that you should throw away water after a couple of days if it was taken from the tap and stored in a bottle outside the fridge, because it has 'gone bad'.
First of all, the couple of days is not very well defined, which already makes me a bit suspicious. Second, I cannot think of anything in tap water that would make the water undrinkable after a couple of days already.

Can someone clarify this issue for me? Does tap water really 'go bad' after a couple of days outside the fridge? Why?

First of all, it depends on how the tap water was treated before it was piped to your house. In most cases, the water was chlorinated to remove microorganisms. By the time the water arrives at your house, there is very little (if any) chlorine left in the water. When you fill you container, there is likely to be some microorganisms present (either in the container or in the water).
In a nutrient rich environment, you can see colonies within 3 days. For tap water, it will probably take 2 to 3 weeks. But that doesn't mean that the small amount of growth doesn't produce bad tasting compounds (acetic acid, urea, etc.).

BTW Nicolau Saker Neto, cold water dissolves more gas than hot water. Watch when you heat water on your stove. Before it boils, you will see gas bubbles that form on the bottom and go to the surface (dissolved gases) and bubbles that disappear while rising to the surface (water vapor).


H-NMR with 10H integration? I have the following two spectra (IR and 1H-NMR) see below.

The chemical formula is: C16H16O2

Thus DBE = 9

My approach:

From the IR, I can see that there's no broad peak around 2500-3300, thus I can rule out any carboxylic acids and alcohols.

Then I'll look at the sharp peak at 1734 which makes me suspect an aldehyde.

From the H-NMR I see the following:

This is where I run into some issues.

I see peak a and c as -CH2- groups next to another R-CH2-, but in two different chemical environments.

And I suspect that peak d might be caused by two benzene rings (2x5H=10H) in a symmetric molecule, like; Ar-R-Ar but I can't stitch these informations together to something useful.

I hope that someone can help me out.

I will ignore the IR as I always do and work only with the NMR.

The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents.
The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either side of these is either an oxygen or a quarternary carbon as there are no additional couplings.
The final singlet points to a $\ce{-CH2-{}}$ group with the same constraints as above.
We still have $\ce{CO2}$ and one double-bond equivalent to take care of. One functional group particularly suggests itself. Then, we have to play connect the fragments.

   $\ce{CO2}$ is likely an ester functionality since we do not have a carboxylic hydrogen to spare.

   One of the two triplet $\ce{CH2}$ is shifted downfield rather strongly. This suggests that it is directly bound to the ester group oxygen, giving $\ce{-CH2-CH2-OC(=O)-{}}$.

   Both of the remaining $\ce{CH2}$ groups will be benzylic but the singlet is more deshielded than the triplet. That suggests that the singlet $\ce{CH2}$ is not only benzylic but also α to the ester.

   Therefore, the final result is $\ce{C6H5-CH2-CH2-OC(=O)-CH2-C6H5}$.


SARS-CoV - relative size of the spike protein I am taking an online course in image analysis and have been asked to use the attached image to determine the size of the SARS-S protein of the SARS-CoV relative to the total of its proteins. However, I have been given no explanation of the image, and with a physics background, I simply do not understand it. Can someone please explain the image (specifically the SARS-CoV column).

Although this is a homework question, I also think OP deserves a help as orthocresol suggested. I'm not going to go into details but leave it at orthocresol's suggestion of this being a Western Blot of protein analysis, which is absolutely correct. Hence, OP should start reading of electrophoresis and western blots techniques since, I assume, s/he taking molecular biology class.
In brief, the given image is a X-ray photograph of the nitrocellulose membrane after transferring the proteins from the SDS-PAGE gel (after electrophoresis). I'm suggesting that because the image looks like visualized by radioactivity (I suggest OP to read about these techniques in details to understand). In SDS-PAGE gel electrophoresis, the proteins separated mainly by molar mass of each protein while heaviest protein go first since electric current run from top to bottom as sepicted in following diagram:

As marked in the diagram, SARS-CoV consists of SARS-M ($\pu{20 kD}$, SARS-N ($\pu{54 kD}$, and SARS-S (~$\pu{175 kD}$ proteins. I may give you your answer but I strongly suggest to read about the subject for better understanding. I do not support online courses, but there are no choice you can made during a bad pandemic era.


Is there such a thing as an acid without a hydrogen? I'm in AP Chemistry and we are learning about the Brønsted-Lowry model and my teacher mentioned that "for the most part" acids have hydrogen, could there be such a thing as an acid that has no hydrogen in it?

It depends on which definition of acids and bases you are using.

According to the Arrhenius theory, acids are defined as a compound or element that releases hydrogen (H+) ions into the solution. Therefore, there are no Arrhenius acids without a hydrogen atom.

According to the Brønsted–Lowry acid–base theory, an acid is any substance that can donate a proton and a base as any substance that can accept a proton. Hence, there are no acids without a hydrogen atom according to this theory.

But according to the Lewis theory of acids and bases, an acid is any substance that can accept a pair of nonbonding electrons. In other words, a Lewis acid accepts a lone pair of electrons. According to this theory, acids without a hydrogen atom can exist. (A coordinate bond is formed between the Lewis acid and base. The compound formed by the Lewis acid and base is called a Lewis adduct)

A great example of this would be $\ce{BF3}$. It is neither an Arrhenius acid nor a Brønsted–Lowry acid but it is a Lewis acid. The boron atom accepts a pair of nonbonding electrons from another atom or ion to complete its octet. Here $\ce{BF3}$ is a Lewis acid as it accepts a pair of nonbonding electrons.

The fluorine ion here is a Lewis base as it donates a pair of electrons.

If you want to get more rigorous about the definition of Lewis acids: "a Lewis acid is a type of chemical that reacts with a Lewis base to form a Lewis adduct".

More about Lewis acids and bases here.


Unable to find sources for uncommon specific heat I'm calculating the energy, temperature, and gas volume outputs of gunpowder. To solve this I need to find the specific heat of Potassium sulfate (K2SO4) and Potassium carbonate (K2CO3). If anyone can refer me to any directory that can list the values for these it would be very helpful.

Thank you for your time.

For potassium sulfate see here.

For potassium carbonate see here.

Generally, search in the NIST Chemistry WebBook and when you find the compound of interest, either click on "gas phase thermochemistry data" or "condensed phase thermochemistry data" as appropriate.


Electronegativity in krypton and xenon? Why do krypton and xenon have high electronegativity? Noble gases are supposed to be "happy" with the amount of electrons they have, because they have 8 valence electrons (thus, most noble gases have no electronegativity). So why do krypton and xenon have electronegativity? Why do they tend to "want" one more electron? It makes no sense... their outer shells are already full!

I'm going to do a research paper on this question, so I would really like some good resources that could help me answer this question for myself. Know any?

Why do Krypton and Xenon have high electronegativity?

As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family.  They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron.  However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules.  In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.

$$\ce{Xe + 3F2 → XeF6}~~(1)$$
$$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$

The most common definition of electronegativity is based on Pauling's work and is given by the equation

$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$

where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.

Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$.  From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.

In essence, for any element that can react and form a molecule, we can calculate its electronegativity.  Since Kr and Xe do react to form molecules, we can use the above methodology to calculate its electronegativity.

Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity.  Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2).  We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.


What is the actual reason behind PCl₅ existing as an ionic solid? Why does $\ce{PCl_5}$ exists in the solid state as $\ce{[PCl_6]^{-}[PCl_4]^{+}}$? What is the problem with it that it doesn't exist in its original form in solid state? Like why isn't it just $\ce{PCl_5}$ in solid state as well?

In the gas phase ions cannot be stabilized through solvation or lattice forces so ions are destabilized relative to non-ionic entities in the gas phase.  $\ce{NaCl}$ is a case in point, in the gas phase it exists as an equilibrium of monomers and dimers.  Yet in the solid phase it exists as a highly ionic crystal.  This example demonstrates that going from a gas to a solid can dramatically affect the relative stabilities of ionic and covalent structures.

The same is true in the case of $\ce{PCl_5}$.  In the gas phase it exists as a covalent compound with a trigonal bipyramid structure.  However, in the solid phase, it turns out that enthalpic affects favor the ionic tetrahedral/octahedral structure $\ce{[PCl4]^{+}[PCl6]^{-}}$.  Simply said, the ionic structure turns out to be thermodynamically more stable in the solid phase where positive and negative charges can be arranged to stabilize the entire lattice; whereas in the gas phase the positive and negative ions cannot be stabilized so the covalent form is favored since it is now lower in energy than the ionic form.

Another reason why the ionic form is preferred in the solid phase is that packing regular geometric shapes such as tetrahedra and octahedra together allows for a more stable arrangement since they can be packed closer together and in a more ordered arrangement than non-regular (kind of an odd shape) trigonal bipyramids can be arranged.

Edit: Response to OP's comment on ion stabilization

If we generate an ion in the gas phase, it will have a certain energy associated with it. If we put that ion in a solution its energy will be much less.  We say the ion has been stabilized by placing it in solution.  Some solvents will stabilize the ion more than other solvents.  This relates to the dielectric constant of the solvent which is a measure of the solvent's polarity.  The higher the dielectric constant for a solvent, the more polar the solvent and the better its ability to stabilize and lower the energy of an ion dissolved in it.

Here is a picture showing a sodium ion dissolved in water.  Note how the negative ends of the dipoles in the water (the negatively polarized oxygen end of the water molecule) molecules congregate and orient themselves around the positive ion.  This orientation of the negative ends of multiple water molecules around a positive sodium ion is what stabilizes the ion in solution.  There is no such stabilization in the gas phase.  In the gas phase it is just a "naked" ion without any external stabilization, so in the gas phase an ion will have a higher energy.

image source see the "Solvation od Ions" section

Here is another picture showing showing (on the right) sodium chloride in a crystal lattice.  The sodium and chloride ions pack around each other to perform the same function as a solvent.  The opposite charges arrange themselves around each other to minimize electrostatic repulsion between like ions and maximize the stabilizing electrostatic attraction between oppositely charged ions.  Through these types of arrangements, the "lattice" can stabilize ions much like a solvent can stabilize ions.  Consequently, both solvents and crystal lattices stabilize ions and lower their energy compared to the ion's energy in the gas phase where no such stabilization exists.

image source


What is a water blank? What is a water blank?

I am doing some research and I am suppose to prepare a solution. I should mix 10g of soil with 95ml of water blank in order to create a soil suspension.

My problem is I am not English and I have no idea what a water blank is. I have a feeling the type of water is pretty important here though. (can't find what it is on the internet).

Help please. A simple definition/description will do.

Generally a blank means no sample and just solution. So it would not contain the 10g of soil in this case. I think for soil analysis it's used as a control so you can have something to compare to.

Here's some links I found by doing an internet search.
http://asl.irri.org/lims/index.jsp?page=about&feature=soilanalysis

http://www.soiltest.vt.edu/PDF/lab-procedures.pdf


Why does an OH bond absorb a higher frequency than a CH bond? I'm wondering why a bond between an O and an H is going to absorb a higher frequency than a bond between a C and an H (any hybridization).

O is heavier than C in terms of atomic masses, and I'm assuming the O-H bond will be weaker because of the electronegativities, so why would it absorb a higher frequency?

Thank you.

First, let me just put down the approximate $\ce{X-H}$ stretching frequencies (where, $\ce{X = O, C}$) just as a reference.

$$\begin{array}{c}
 & \ce{X-H}\ \text{stretch}\ \mathrm{(cm^{-1})}  \\ \hline
\ce{X = O} & 3600  \\ \hline
\ce{X = C} & 3000 \\ \hline
\end{array}$$

Now, as a simplification I will first discuss a diatomic molecule, thought of as a two balls of some mass $\mathrm{m}$ linked together with a spring. Solving, Schrodinger's equation for this system (called a harmonic oscillator), yields the following relation:

$\mathrm{\bar{\nu}} = \frac{1}{2\pi c}\sqrt{\frac{k}{m}}$

For, dissimilar masses, (say, $\mathrm{m_1}$ and $\mathrm{m_2}$), we can switch to the centre of mass coordinates, and the $\mathrm{m}$ in the preceding equation is replaced with the reduced mass ($\mu$) defined as,
$\frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} $

Consequently, $\ce{C-H,O-H}$ bonds in general have much higher stretching frequencies than do corresponding bonds to heavier atoms other than $\ce{H}$.The mass effect becomes evident when deuterated isotopes are examined. The stretching frequency of a free $\ce{O-H}$ bond is ca. $3600\ \mathrm{cm^{-1}}$, but the $\ce{O-D}$ equivalent is lowered to ca. $2600\ \mathrm{cm^{-1}}$. Since deuterium has a mass twice that of hydrogen, the mass term in the equation changes from 1 to 1/2, and the frequency is reduced by the square root of 2.

Naively, one would expect the frequency of the $\ce{C-H}$ bond to be higher, based on the preceding discussion, but there is another factor at play, the force constant $k$. This, is different for $\ce{O-H}$ and $\ce{C-H}$ bonds, and depends on the bond strength (stiffness of the spring)

Bond strengths can be estimated from bond disassociation energies (BDEs), which is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolytic cleavage at $\mathrm{0\ K}$. This values are frequently tabulated, and can be easily found online (here for example; caveat the values here are at $\mathrm{298\ K}$)

$$\begin{array}{c}
 & \ce{X-H}\ \text{BDE}\ \mathrm{(kJ\ mol^{-1})}  \\ \hline
\ce{X = O} & 428.0\ (21)  \\ \hline
\ce{X = H} & 337.2\ (8) \\ \hline
\end{array}$$

Anyway, this tells you that the $\ce{O-H}$ bond is stronger; Why? Well, simply look at the approximate atomic radii of Oxygen and Carbon. The Oxygen is about $\mathrm{48\ pm}$ while carbon is about $\mathrm{67\ pm}$. The longer atomic radii implies a longer $\ce{C-H}$ bond length, ergo a weaker bond.

Stronger bond, implies larger $k$, which in turn means greater frequency of vibration.

So, the $\ce{O-H}$ bond is NOT weaker. Remember, when we define bond disassociation energy, we are looking at homolytic cleavage. I really don't understand the point you try to make about electronegativities.


A solution is cloudy at lower concentration and becomes clear at higher concentration. Why? This might be just an "everyday life science" question. I am a university professor but not in Chemistry so please educate me. Today I was adding some floor cleaner formula to water as instructed by the label. In the beginning, the solution became cloudy. However, as I added more formula, the solution became clear. I am curious why this can be and whether there is a name for such a phenomenon in Chemistry.
More details and pictures:
Just a little bit of formula added to the water; the solution was cloudy:

Then I added more formula; the solution becomes clear:

The cleaner is called "HOOVER FloorMate Grout & Tile Plus".
Out of curiosity, I checked its data sheet which states that it is a :

Blend of proprietary non-hazardous surfactants and polymers.

At very dilute concentrations, the solvating power of water overcomes the natural tendency of surfactant molecules to agglomerate into a separate phase. The shear numerical excess of water molecules just about totally dissociates the bulk surfactant and dissolves it as separate molecules. So a mixture of surfactant and water can be a true solution - clear and homogeneous.
However, as the concentration of surfactant increases, aggregates form (micelles) which can absorb oleophilic molecules (e.g., the polymers in the formula). With sufficient surfactant, the polymer is dispersed enough and into small enough micelles to appear clear. In the second graph, the knee of the curve is at the CMC; at this point, the aqueous phase is saturated with single molecules and some are beginning to stick together, starting to be able to contain polymer.

To get to the point: Your floor cleaner needs to be concentrated enough to work well. This suggests that there is enough surfactant to exert a solvent effect on the polymer, dispersing the chains, separating the molecules, making the smallest dimension significantly less than a wavelength of light. The original floor cleaner product could appear clear to the eye. Addition of a little of the product to a lot of water reduces the micelle formation of the surfactants by favoring dissolution of single surfactant molecules: the polymer chains then have fewer surfactant molecules to surround them. Polymer molecules escape, agglomerate and become visible - but only a little, since you can reverse the process.
A more delicate detection method is to use the Tyndall effect: in a darkened room, shine a narrow beam of light through a liquid (say from a single LED flashlight). The beam may be visible, much like a flashlight beam in a smoky room, and can detect some materials at levels in the ppm range (e.g., milk in water).

I suspect your floor cleaner would be represented by the middle test tube in the series of five. Small micelles are present but the liquid seems clear, and will seem to have a small or negligible Tyndall effect. The polymers will be dissolved in the many micelles, which are too small yet to be very visible. When you dilute, you go to the left one step, the surfactant becomes more dispersed into single molecules, exerting less solvent power on the polymer. Either the micelles  grow so large as to become visible, or actually collapse, allowing the polymer molecules to agglomerate.
Adding formula back to the mixture increases the surfactant (the solvent power), redisperses the polymer molecules into tiny micelles, and the liquid becomes clear again. The key (hypothesis) is that the pure polymer dispersion would be cloudy by virtue of its large particle size (~1 micron?), but a high enough surfactant concentration makes the formula essentially clear by generating micelles much smaller than the wavelength of light.


Why is Ag+ a soft acid? A soft species is typically large, polarizable and has a low charge (magnitude). The silver cation is not highly charged (this supports softness) however, it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. Consequently, shouldn't it also be less polarizable?

It does have a tendency to form bonds that are significantly covalent in nature which backs up the assertion that $\ce{Ag^+}$ is soft; however, I don't understand why.

According to HSAB theory, Soft Acids have:

large atomic/ionic radius;

low or zero oxidation state bonding;

high polarizability;

low electronegativity.

The ion $\ce{Ag+}$ satisfies all the aforementioned properties.
I don't agree with you when you say it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. The effective nuclear charge has nothing to do here.

According to http://www.webelements.com/silver/atom_sizes.html: The radius of this ion is $114\ \mathrm{pm}$ when it's in 4-coordinate, tetrahedral geometry and it's $116\ \mathrm{pm}$ when  it's 4-coordinate, square-planar geometry. Please just compare these radii with the radius of  $\ce{Al^3+}$ ion : $53.5\ \mathrm{pm}$ which is considered as a hard acid.
I hope it's clear now!


Shape of the container and the number of possible energy states A question came up when studying the formula for calculating the total number of possible energy states of a particle within a container.

Why is the number of possible energy states independent of the shape of the container and merely dependent on the number of states in only 3 perpendicular container dimensions ($x$, $y$ and $z$), while a momentum vector can be directed at any direction within the container? Shouldn't the length of the container in that same direction as the momentum vector also determine the number of energy states in that direction?

For a sufficiently small container, the size and shape of the system is significant, e.g. some energy states are prohibited. For example, CdTe quantum dots behave differently from the bulk material.

However, the kinetic theory is based on the assumption that, "The number of molecules is so large that statistical treatment can be applied." The momentum is treated as a vector having x, y and z components and does not depend on the shape of the container (though a cube is convenient for illustration of three orthogonal directions). The equations can be derived using spherical coordinates, and they still must integrate collisions in all directions, so are equivalent.


How to form an acetal from an enol ether? I am having trouble understanding how the protecting group (ethylene glycol) is applied to the alkene next to the ether of highlighted step above. I thought maybe TsOH protonates the alkene as seen in the mechanism below so that it forms a carbonyl but it doesn’t seem right as a carbocation should be found on the tertiary carbon. I have no idea if I am going about this correctly, any suggestions?

Steps (Source: from comment made by Zhe)

Tosic acid protonates alkene to form oxocarbenium through resonance.
One of the alcohols from ethylene glycol attacks the activated carbonyl and toluenesulfonate deprotonates the oxygen.
Methoxy group is protonated by the regenerated torsic acid.
Methanol is eliminated via SN2 with the other hydroxyl group of the ethylene glycol.
Tosic acid is regenerated by deprotonating the 1,3-dioxolane group.


Identifying organic compounds in Infrared spectrum What further information would allow you to make a positive identification of an organic compound from an infrared spectrum?

If you're analyzing an IR spectrum to find out what compound you have, you'll first want to look past the 1500/cm region to see what functional groups you have. For example, a 1700/cm peak will tell you you have a carbonyl, alcohols will have a peak around 3000/cm, etc. From there, pull up spectra of well known compounds with those functional groups, and compare your compound's 1500/cm to 500/cm region to them. This region is called the "fingerprint region" because it's practically unique for all molecules, so if the spectra match there, they're likely the same compound.


Do conjugated polymers absorb very low energy photons? I was reading that in conjugated compounds, the extent of conjugation decides what wavelength photon it can absorb. Longer the conjugation "chain", lesser the photon energy required. So, molecules like lycopene

with 11 conjugated bonds is able to absorb in the blue range, appearing red. If this is the case, consider the polymer poly-acetylene

which has a conjugated "chain" spanning "thousands". Does this mean, this molecule would absorb infra red and lower energy light?

In principle yes, longer conjugated regions absorb at longer wavelengths. The simplest model when thinking about this is to consider a particle in a box, or on a ring if considering cyclic conjugation as in the chlorophyll's. However, in very long polymers, such as polyacetylene, the conjugation length is in fact rather small, approx 10 to 20 double bonds. Nanometer thick  films appear brown and absorb below 800 nm ish whereas if the conjugation length were very large, as per the chemical structure you draw, the molecules would absorb to longer wavelengths and appear black.

This limited conjugation is due to two things, first the molecules are  twisted and bent in forming the polymer and these distortions alter the electronic properties just enough to shorten the conjugation length. Second even when this twisting etc. were removed there are two ways of arranging the double bonds depending how you start. At their junction these two forms form what is called a 'solition' or 'topological kink'  and this is  a non-bonding electron between two regions of conjugation. This also limits conjugation.


Formal charges on polyatomic ions Why is it that sometimes a polyatomic ion's Lewis structure is adjusted to remove the formal charge on the central ion, but sometimes it is not?

For example, in the $\ce{I3-}$ ion (triiodide), the Lewis structure is the following:

This gives a single formal negative charge to the central iodine. In my textbook, it says to remove formal charges from the central atom whenever possible, so couldn't we form a double bond with one iodine and therefore move the formal charge off of the central atom like this?

Comparatively, in $\ce{IO4-}$ (periodate), the Lewis structure is:

There, the single formal negative charge is on one of the oxygens instead of on the central ion, making it a resonance structure.

So basically what I'm asking is why is there a formal charge on the triiodide ion's central atom but not on other ions' central atoms? Like why is it okay that the triiodide ion has a formal charge on the central atom when it could be moved off to create a resonance structure, such as in $\ce{IO4-}$?

First off, in neither triiodide nor periodate have the structures you drew. They all assume octet expansion on iodine which has been disproven. It was traditionally explained by the ‘participation’ of iodine’s empty 5d orbitals but these are energetically too far removed and do not take part in bonding to any noticable extent.

Second off, it is wrong to give a formal charge to the central iodine of triiodide. In fact, it is the two outer iodines that carry the negative charge. The structure should be given as follows with a four-electron-three-centre bond between the two iodines:

$$\ce{I-I\bond{...}\overset{-}{I} <-> \overset{-}{I}\bond{...}I-I}\tag{1}$$

Because iodine adheres to the octet rule, there is no contribution by any double-bonded structure. That would create a superoctet structure.

In periodate, there is no question that oxygen is far more electronegative than iodine and that iodine’s oxidation state of $\mathrm{+VII}$ does not go well with a formal or partial negative charge. Thus, it is only logical for the outer oxygen atoms to bear the negative charge. A better way to draw it would be in adherence to the octet rule, i.e. four $\ce{O-}$ substituents and a central iodine with a $3+$ formal charge.

There is nothing inherently wrong with a formal charge on the central atom, though. Take for example tetrahydridoborate $\ce{BH4-}$, the addition product of borane and hydride. Here, boron should only have three valence electrons but if bonded to four hydrogens it must have four. Thus, it carries a formal negative charge. (In reality, the hydrogens carry the negative charge, though, since they are again more electronegative than boron.)


Relative acidity of acetic acid derivatives with strongly withdrawing substituents What would be the order of $\mathrm{p}K_\mathrm{a}$ values of the following:

$\ce{O2N-CH2-CO2H},$
$\ce{Me3\overset{+}{N}-CH2-CO2H},$
$\ce{NC-CH2-CO2H},$
$\ce{HO-CH2-CO2H}?$

Since only the groups attached to $\ce{-CH2-CO2H}$ are changing, we must look at the point of difference in each group and I was able to figure out that since there is no conjugation, we cannot compare these groups on the basis of the mesomeric effect.
So, we must compare them on the basis of negative inductive effect as the more stable the conjugate base will be, the more acidic our compound is and hence the lesser is its $\mathrm{p}K_\mathrm{a}$ value. But I am not able to come to a final conclusion which matches with what's given in my book.
Is there some other factor playing role in this as well? Because if we just look at it on the basis of inductive effect, the order for increasing values of $pK_a$ should be
$$\ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{O2N-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
which seems to be wrong.

The order for the increasing $pK_a$ values given in my book is:
$$\ce{O2N-CH2-CO2H} < \ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
Note that the order given in my book may be subject to errors.

EDIT:
After pondering over this for a lot of time and with no convincing answers or explanations, I finally came to a potential explanation for this. Could Hydrogen bonding be the reason for the compound with $-NO_2$ to be more acidic than the one with $-Me_3\overset{+}{N}$ group ?

EDIT-2:
Thanks to Buck, it's now clear that my book's answer is right. Could someone please give a possible explanation to this from organic point of view?

Some cheminformatic descriptors for the puzzling compounds can be found on the PubChem website:

SMILES    N+(=O)O ;
Nitroacetic acid; 625-75-2; alfa-Nitro
acetic acid; 2-nitroacetic acid; ALFA-NITROACETICACID;
CID: 43581;
MF: C2H3NO4;
MW: 105.05g/mol;
InChIKey: RGHXWDVNBYKJQH-UHFFFAOYSA-N   ;
IUPAC Name: 2-nitroacetic acid   ;

and

SMILES N+(C)(C)C;
TRIMETHYL GLYCINE;
N,N,N-trimethylglycinium; Cystadane;
CHEBI:41139;
carboxy-N,N,N-trimethylmethanaminium;
CID: 248   ;
MF: C5H12NO2+   ;
MW: 118.15g/mol;
InChIKey: KWIUHFFTVRNATP-UHFFFAOYSA-O  ;
IUPAC Name: carboxymethyl(trimethyl)azanium

With glycine betaine it is usually meant the chloride salt, which would result in chloride counterions. For dilute solutions presumably those chloride ions don't matter, but they would at higher concentration.
Nitroacetic acid even has its own Wikipedia page. The acidity reported there is pKa 1.68, which is similar to that found in other sites such as in ChEMBL, pKa 1.75.
Glycine betaine or trimethylglycine also has Wikipedia and ChEMBL entries. From Wikipedia, acidity (pKa) is 1.84. From ChEMBL, 2.26.
So from this data alone it would certainly seem like the book is correct. Still, those pKa values are very very similar.


Why is MnSO4 colourless? I get it that transition metal salts have colour due to d-d transition from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$ orbitals in octahedral and the opposite in tetrahedral but why would $\ce{Mn^2+}$ not have any characteristic colour? Is it so because the orbitals are degenerate in $\ce{MnSO4}$? or because d-d transition only takes place when there is completely empty d orbital?

Manganese(II) sulphate is not colourless. It is a faint pink. Compare this image taken from Wikipedia, where a full list of authors is available:

However, it is true that the colour is rather faint, and you need a white background to properly see it. That’s because manganese(II) is a $\mathrm{d^5}$ system and a high-spin one of those meaning that all five d-orbitals are occupied by a single electron:

Exciting an electron directly is forbidden due to the spin-retaining rule: The excited electron must occupy an already populated orbital, which is only possible if it flips its spin, which is forbidden by selection rules. Special mechanisms are needed to account for the (obviously possible but faint) transition here; the page Orthocresol linked in the comments names spin-orbit coupling.


Can a drink with such composition have 0 caloric value (whether manufacturer lied)? I bought a drink which have next statements:

  The nutritional value, g/100 ml:
      no proteins, fats, carbohydrates.
  Energy value (calorific value):
      kJ / 100 kcal to mg: absent.

Can a drink with such composition have 0 nutritional/calorific value ?

Composition:

prepared drinking water,
flavoring base "Cola"

dye: caramel,
acidity regulator: phosphoric acid,
caffeine - no more than 150 ml/g,
natural flavor enhancer,

sweetener

aspartame,
acesulfame potassium,
sodium saccharin,
sodium cyclamate,

preservative sodium benzoate.

It contains a source of phenylalanine.

You have to keep in mind the bane of all introductory chemistry students: excess precision and rounding. Every measurement has error bars, even if those error bars are implicit. It's senseless to report a value that has (much) more precision than your error bars.

A value that's measured as zero isn't necessarily exactly zero. ... it's only "mostly zero".

The particular rounding rules that are used in nutritional labels depend on the specific regulatory agency, but in the US, the FDA provides rounding rules for nutritional labels. It counts anything less than 5 (dietary) Calories per serving as being zero. Likewise, anything with less than 0.5 g of protein, fat or carbohydrate can be listed as having zero grams.

This is how Tic-Tac breath mints can be advertised as being "zero calorie" despite being made almost entirely of sucrose (table sugar): a serving contains less than 5 Calories, and so can be rounded to zero on the label.

In your case, there's a few things which could contain calories. Carmel color is made from sugar, and may contain residual saccarides. Aspartame is a methyl ester of a dipeptide, which is converted in the body to its constituent amino acids, providing a small amount of calories. (Gram for gram, aspartame would contain almost the same amount of calorific energy as sugar does. It's about 200 times sweeter than sugar, though, so on an equivalent sweetening basis it has much fewer calories.) Depending on what the "natural flavor enhancer" is, that might also contain carbohydrates, proteins or fats. (Natural flavors are often fats or fat-soluble, so flavorings often contain trace amounts of dietary fats.)

For a single serving (and for 100 ml) there's only a small amount of carbohydrate, only a scant amount of amino acids, a negligible amount of fat, and an insignificant number of calories. They're a rounding error. But it is true that if you had enough of the drink, you could indeed have an appreciable number of calories, carbohydrates, protein, etc. It's just that it would come along with enough water to make actually consuming that amount physically unfeasible.


Gold's atomic structure and bonding in relation to the separation techniques used How is Gold's atomic structure and bonding related to the separation techniques used to distinguish it from other substances?

I'm just confused on what I'm actually searching for, I have read so many blogs on Gold extraction, refinement, occurences and more topics, and either I'm skipping the important parts or the answer's just not there. What can I search to find what I'm looking for? The separation technique I chose is flotation, can someone lead me in the correct path and/or give me a few hints on what to start of with? Thanks in advance, much appreciated!

There are differences in gold extraction for different mines - depending on the chemical nature of the raw ore, the impurities present, the availability of resources etc.

A good example of gold extraction comes from the Super Pit mine in Kalgoorlie, Western Australia; their gold extraction process is outlined in their website Mineral Processing, which is outlined below:

Crushing and grinding: this is a physical process where the mined raw ore is crushed, ground and water added to make a slurry.

Flotation: chemicals are added to the slurry to remove everything other than gold and pyrite.  Water is also removed at this stage.  The slurry is agitated forming air bubbles, which then rise through the slurry, carrying mineral grains.  From Introduction to Mineral Processing:

  The ability of a mineral to float depends upon its surface properties. Chemical modification of these properties enables the mineral particles to attach to an air bubble in the flotation cell. The air bubble and mineral particle rise through the pulp to the surface of the froth or foam that is present on the flotation cell.

Roasting: water is re-added and the concentrated slurry is roasted to remove $\ce{SO2}$

CIL: or carbon in leach, chemicals are added to dissolve the gold and have close to pure gold precipitate on the carbon.

Elution: A mixture of caustic soda and cyanide is used to liberate the gold from the carbon.

For your specific request for resources about flotation, the example process below is for a sulphide copper ore containing iron bearing pyrite:

(Image Source)

Suggested additional resources about flotation:

Froth Flotation – Fundamental Principles

A great resource is the book Handbook of Flotation Reagents: Chemistry, Theory and Practice (chapters are behind a paywall).


Why is HBr colorless while Br2 isn't? Bromine ($\ce{Br2}$) has a dark reddish color. If it reacts with something like benzene, this results in the formation of Bromobenzene and $\ce{HBr}$:
$$\ce{Br2 + C6H6 ->[{Cat.}]C6H5Br + HBr}$$
It can be observed that during this reaction, the color of bromine disappears.
One can thus conclude that $\ce{HBr}$ is colorless. Why is this the case? If we look at the molecules, it seems to me that they are basically the same (speaking in how many electrons there are in each orbital, for example), except for that in $\ce{HBr}$, there is a hydrogen atom instead of another $\ce{Br}$.
So what makes the $\ce{Br2}$ bond "special", so that this molecule can absorb light in the visible spectrum but $\ce{HBr}$ can't?

I think that I have a basic understanding of the quantum mechanic model of an atom (e.g. atomic orbitals and how their energy levels correspond to different wavelengths of light being absorbed). Here are some of my ideas trying to explain the above described question:

In $\ce{Br2}$, both atoms have the same electronegativity, while in $\ce{HBr}$, there is a difference of $\ce{\Delta EN=0.76}$. The only consequence I could imagine from that is that in $\ce{HBr}$, the probability of an electron in the bond to be located closer to the $\ce{Br}$ atom is higher than being close to $\ce{H}$, while in $\ce{Br2}$, the electrons (or rather their probability) is evenly distributed

$\ce{H}$ has an electron configuration $\ce{[1s^1]}$, $\ce{Br}$ has an electron configuration $\ce{[1s^2]\,[2s^2]\,[2p^6]\,[3s^2]\,[3p^6]\,[3d^{10}]\,[4s^2]\,[4p^5]}$. So I would assume that the bond in $\ce{HBr}$ is formed by the overlapping of a $\ce{1s}$ and a $\ce{4p}$ orbital, while in $\ce{Br2}$, two $\ce{4p}$ orbitals overlap. The $\ce{4p}$ orbital has a higher energy level than $\ce{1s}$, so this might be the cause for the different absorption spectrum. However, I am not sure if this is true and how the energy levels behave when two orbitals overlap to form a new one.

The colour of molecules come from electronic transitions that are very sensitive to the nature of the molecule
To simplify a little, molecules absorb light when the energy difference between electrons in an occupied orbital and an unoccupied orbital match the energy of a photon of light (and that light will have a specific colour). This much you already understand.
But the energy levels of the electrons that matter for colour are usually the bonds in the molecule that join one atom to another, not the other electrons in the atom that don't get involved in the bonding. So atoms are rarely thought of as having an inherent colour that is independent of what they are attached to (some isolated atoms do have colour: sodium vapour has narrow absorption bands in the yellow, for example, but this isn't important in sodium compounds).
The assumption you make that is false is that "If we look at the molecules, it seems to me that they are basically the same". If this were true there would be very little interesting in the whole subject of chemistry. When bromine reacts with benzene a Br-Br bond is replaced by a Br-C bond and a Br-H bond. Those are very, very different bonds. A Br-Br molecule has a readily accessible electron transition that results in a dark red colour. That this particular transition is very sensitive to the exact nature of the molecule is obvious as other dihalogens (with very similar electronic structures) have very different colours (iodine is dark purple, chlorine is light yellow). But the orbitals available in other Br-X bonds will be very different as will their energy. Often the corresponding colour of light will be in the UV and there will be no visible colour.
The essential point is that colour comes from electronic transitions that vary a great deal with the atoms involved in the bond. And in many colourful compounds those transitions involve molecular orbitals including multiple atoms across the whole molecule. There is no simple relationship between atoms and the colours of molecules. Every bond is (very) different.


pKaH of pyridine versus other imines Why is pyridine so much more acidic than other similar imines (5.5 versus ~9, see picture)? Wouldn't delocalization of the positive charge into the aromatic system stabilize it and hence decrease acidity? Also the lone pair of pyridine is not delocalized into the perpendicular $\pi$-system and hence not any less available than the one of the projected imine.

(Source: Organic Chemistry by Clayden et al. p. 1149. )

Why is pyridine so much more acidic than other similar imines?

When we discuss the acidity of a compound, we are really discussing where the equilibrium between the acid and its conjugate base lies.  Therefore, when we compare acidities we are really comparing equilibria, the relative stabilities of the reactant (acid) and product (conjugate base).

The following diagram illustrates the equilibrium between the iminium ion and the corresponding imine, it's conjugate base.

We can't draw any good resonance structures for the imine, but we can draw 2 resonance structures to describe the iminium ion, A and B in the figure. In the pyridinium ion - pyridine case we can draw 2 resonance structures for pyridine (just like in benzene), but we can't draw any resonance structures for the pyridinium ion that delocalize the positive charge without destroying the aromaticity of the ring.

So, in the iminium ion - imine case

we can stabilize the iminium ion through resonance - this stabilizes the iminium ion and shifts the equilibrium to the left making the iminium ion less acidic.

Whereas in the pyridinium ion - pyridine case

we can't draw good resonance structures that stabilize the positive charge in the pyridinium ion - this makes pyridinium ion more acidic.

Both factors operate in the same direction and tend to make the pyridinium ion ($\mathrm{p}K_\mathrm{a}=5.5$) more acidic than the iminium ion ($\mathrm{p}K_\mathrm{a}$ $\sim$ 9).


Why is sugar used in low-cost homemade rockets? I was thinking about making a "rocket candy". It's relatively inexpensive rocket based on sugar and some oxidiser, mostly $\ce{KNO3}$.

The fuel is usually formed by heating the mix of both compounds so that the sugar sticks onto the potassium nitrate. They can also stick together by being dissolved and dried.

My question is, what makes sugar the best fuel? Paraffin or resin can be melted more easily and the oxidizer could be poured on them while they are hot. To me, that seems like a much better idea.

But there were many people before me and they have surely tried many things. So, what is the reason the sugar is more advised than paraffin or resin?

the best fuel is actually sorbitol. It is a bit costlier than sugar, but still cheap. It melts easily without decomposition and can dissolve some amount of potassium nitrate. Common sugar decomposes before melting.

Paraffin has one bad habit: it quite easily vaporizes. It is bad, because it makes mixture much harder to ignite and it becomes prone to self-extinguishing because of it. Many polymers also easily decompose producing much gases. Paraffin is used in some specific formulations with more active oxidizers or as a non-fuel additive.

Black powder is actually quite common, but it has very high combustion rate and is prone to explosive combustion, so it is not recommended for beginners. It also requires ball mill to make, and constructing one is time consuming.

Added: The main advantages of rocket candy are readily available components, simplicity of preparation (no special equipment required), stable burn and safety in comparison with many other compositions.


Ferric Chloride solution weight When I notice that ferric chloride is sold as $40\%$ solution w/v, does it mean weight / volume?

Is it a matter of for example adding $40$g of ferric chloride crystals to a water solution where the total volume will then be $100$ml?

Or do I have to workout via the molar mass of $\ce{FeCl3}$ to water? How do I calculate the weight of crystals necessary to create a $40\%$ solution of $\ce{FeCl3}$?

Thanks

EDIT:

The system seems to think this is a homework question (whatever that is). If I divide $100$ml of water (so $100$g) by $40\%$, I get $40$g of $\ce{FeCl3}$. If I dissolve $40$g of $\ce{FeCl3}$ in $100$ml of water, will I get a $40\%$ solution of $\ce{FeCl3}$? And a total volume of $114$ml? ($\ce{FeCl3}$ is $2.9\frac{\mathrm g}{\mathrm{cm^3}}$, so $40$g is approx $13.8$ml)?

You're right in that 40% weight/volume means that you take 40 grams of ferric chloride and dissolve it in 100 mL of water. This is the definition of w/v(%) and there is no need to work out the molar concentration - the main benefit of this method of stating concentration.

The total volume will not be (volume of liquid + volume of solid) in most cases of dissolution. In most cases, you will observe very little change in the volume of the resulting solution as the solid molecules separate and fit into the spaces between the liquid molecules. You may even occasionally observe a decrease in volume as shown in this experiment.


Crystal field splitting of other orbitals According to crystal field theory (CFT), $\mathrm{d}$-orbitals of central metal atom splits due to non uniform repulsion of ligands around the metal atom. Why isn't $\mathrm{p}$-orbital splitting discussed?

Argument: As in case of square planar complexes $\mathrm{(dsp^2)},$ $\mathrm{p}$-orbitals are involved which are facing different conditions of repulsion. So, $\mathrm{p}_{xy}$ must have higher energy than $\mathrm{p}_{xz}$ or $\mathrm{p}_{yz}.$ This happens in case of trigonal pyramidal structure too.

According to Crystal Field Theory, d-orbitals of central metal atom splits due to non uniform repulsion of ligands around the metal atom. Why isn't p-orbital splitting discussed?

To be blunt, the reason nobody mentions p-orbital splitting or non-splitting is that neither the teacher nor the learner has ever understood the crystal field theory properly in its true depth (including myself).  They just propagate what they saw in standard textbooks, repeat the same in class, the students pass never to see CFT again. The story ends. If that student ever becomes a teacher, the story continues.

You are mixing apples and oranges. Crystal field theory has nothing to do with hybridization. The original work of Hans Bethe, the man behind crystal field theory, wrote a 72-paged highly theoretical paper in Annalen der Physik in German. The translations are available and I quote the abstract where he does clarify what happens to p-orbitals. How many can claim that that they read this paper completely before teaching CFT. I cannot. The abstract clarifies your misconception.

  The influence of an electric field of prescribed symmetry (crystalline field) on an atom is treated wave - mechanically . The terms of the atom split up in a way that depends on the symmetry of the field and on the angular momentum I (or J) of the atom . No splitting of s terms occurs , and p terms are not split up in fields of cubic symmetry . For the case in which the individual electrons of the atom can he treated separately ( interaction inside the atom turned off) the elgenfunctions of zeroth approximation are stated for every term in the crystal; from these there follows a concentration of the electron density along the symmetry axes of the crystal which is characteristic of the term. - The magnitude of the term splitting is of the order of some hundreds of cm-1. - For tetragonal symmetry, a quantitative measure of the departure from cubic symmetry can be defined, which determines uniquely the most stable arrangement of electrons In the crystal.


White organic solid that forms CO2 with NaHCO3, forms a diazo dye withe beta-naphthol, and reacts with bromine water A white crystalline solid "X" gives the following chemical tests:

It liberates $\ce{CO2}$ with $\ce{NaHCO3}$
It forms a coloured dye in diazotisation and coupling with $\beta$-naphthol
With $\ce{Br2}$/$\ce{H2O}$, it forms a white precipitate of 2,4,6-tribomoaniline.

"X" can be identified as:

I thought that (d) could be the answer as COOH releases $\ce{CO2}$ from $\ce{NaHCO3}$, but can sulphonyl group release $\ce{CO2}$ from $\ce{NaHCO3}$?

(i) indicates presence of 'strong' acidic group: -(SO3H) and -(CO2H) since both are strongly acidic to protonate NaHCO3 and yield CO2.

(ii) indicates presence of Primary Amine which can fully undergo diazotization reaction and then with beta-naphthol form azo dye (coloured dye).

(iii) excess of bromine water forming s-tribromoaniline directly indicates that Option 2(b) is correct.

An article on point (iii): journal_iisc


Why is acetone and diethyl ether miscible, but not water and diethyl ether? Acetone and Water are miscible, and water and diethyl ether are not. So why are acetone and diethyl ether miscible if water and acetone seem to have similar polarities?

Acetone's dipole moment is 2.91D while that of ether is only 1.15D.  Water is a very polar substance, so acetone will dissolve in it while ether will not.

Two substances are miscible when their intermolecular forces (IMFs) are similar enough such that the forces of attraction between molecules of different substances are similar in strength to the forces of attraction between molecules of the same substance.  If we look at the miscibility of A and B as an example, if A attracts B about as strongly as A attracts A and B attracts B the substances should be miscible.  If A and A have much stronger attraction than A and B the substances are less likely to be miscible.

In your example it's not that some molecules of ether will not dissolves in water and vice versa, it's just that if the two liquids are in contact the water will stay with water and the ether with ether for the most part.  This is because water's strongest IMF is hydrogen bonding while ether's is dispersion with some dipole-dipole forces.


What's unique about the most commonly used hetero-cyclic elements? I have question about hetero-cycles: Why do we use (the most common hetero-cyclic elements) nitrogen, oxygen, sulfur as hetero compounds most of time?

A possible reason for this is that heterocycles with O, N or S are often easier to obtain and more stable than the corresponding compounds which contain other main group elements as heteroatoms. Phosphole, for example, is less stable than furan, pyrrole and thiophene because of its lower aromaticity. Borole, the structural analog with the lighter heteroatom B, has not yet been isolated in unsubstituted form. Ring strain and the stability of X-C and X-H bonds (X = heteroatom) also need to be considered. Bond lengths tend to increase with heavier heteroatoms. With decreasing electronegativity of X in relation to C, the X-C bond becomes less covalent and is more prone to be cleaved by hydrolysis, for example.


Lennard-Jones potential and vibrational energy level diagram explanation I don't understand how to interpret graphs like the following:

Forget the difference between the Harmonic and Morse (I think I get that), It's more what the horizontal lines mean and what the boundary line means. I know that vibrational energy is quantized and I can rationalize the shape of the Lennard-Jones potential but I don't understand how they're combined in this diagram. Please could you walk me through what's going on?

This diagram is conceptually horrible, because it is a mix of quantum and pseudo-classical physics.

Forget for a while the vibrations. If you calculate single point energies (within Born-Oppenheimer approximation) for different internuclear values, you'll get something like blue curve.

Now, for the $r_e$ distance between nucleus (again, within Born-Oppenheimer approximation) you can calculate the "vibrational (normally harmonic) frequency" in a pseudo-classic fashion. For this frequency, you can assign energies (of a quantum harmonic oscillator having this frequency). This energy values correspond to the difference between energy values for the horizontal blue lines, and the well (blue).

The widths of the horizontal lines have not physical meaning.

For the sake of completeness, the $D_e$ would be the energy of equilibrium state if no vibrations are considered. $D_0$ when the fundamental vibrational state energy is add. The former energy is not measurable because this is not a real energy for the system.


Determining the voltage of a cell Setup: A magnesium electrode dipped in a magnesium sulfate solution and an iron electrode dipped into iron (III) sulfate solution

Am I right in thinking that you need to first identify the redox reactions to determine to determine overall voltage of the cell at standard conditions?

I have identified Mg to be oxidized to Mg(2+) at the anode, but I am not sure which is reduced at the cathode. Would it be the sulfate ions?

If it is indeed the sulfate ions, I have calculated the standard electrode potential to be 2.54 volts (is this correct)?

[...] I am not sure which is reduced at the cathode. Would it be the sulfate ions?

orthocresol has already pointed out in his comment that there's more in $\ce{Fe2(SO4)3}$ than just the sulfate.

Take a look at the cation:

$$\begin{align}
\ce{Fe^3+& + e- &<=>&Fe^2+\quad &+0.77V} \\
\ce{Fe^3+& + 3e- &<=>&Fe(s)\quad &-0.04V}
\end{align}$$

(Standard electrode potentials are cited from wikipedia.)


Non-Newtonian substance - Tahini Raw Tahini seems to me like a non-Newtonian substance. When I mix it fast, it gets tougher. When I mix it slowly, it gets softer.
Is there any truth to that?

Yes, tahini (and cornstarch-water mixtures) are dilatant, non-Newtonian fluids, with viscosity increasing with shear strain. See http://en.wikipedia.org/wiki/Dilatant for definition and https://www.youtube.com/watch?v=D-wxnID2q4A, https://www.youtube.com/watch?v=BleCJJAKkgw for demo.

Tahini is much better in hummus than in a swimming pool, though. ;-)


What is the mechanism of addition of glycine into 5-chloro-N-methylisatoic anhydride? The following transformation is involved in a step in the synthesis of diazepam:

(†)

What is the mechanism of the transformation from the second to the third intermediate (N-methylisatoic anhydride to 1 via glycine in triethylamine)?

There are a few possible electrophiles and nucleophiles that could be involved here. Either the aniline could attack the glycine carboxyllic acid, or the glycine could attack either of the anhydride carbonyls. The paper, however, says that the glycinate is formed, so I don't think the former option is viable. Based on the pKa's involved, what might be the most reasonable proposal?

(†) J. Org. Chem., Vol. 45, No. 9, 1980, 1679

I propose the following mechanism. Contrary to the answer provided by DHMO, I do not believe that the hemicarbamate intermediate is stable. Instead, I assume that it breaks down into carbon dioxide and an amido anion immediately upon formation or very shortly afterwards. This gives us a nucleophilic amido residue which can attack glycine’s carboxylate function. If need be, we can supply catalytic protons to the carboxylic group before the amido nitrogen’s attack but I have omitted that in the scheme.

Scheme 1: Proposed reaction pathway.

I consider the amino group of glycine to be the first attacker since it is most nucleophilic, and I consider it attacking the ester group since that should be more susceptible than a substituted hemicarbamate. Upon breakdown of the tetrahedric intermediate to give the first amide group, the free hemicarbamate anion is liberated which should immediately liberate carbon dioxide as stated above. The formation of the final amide linkage is thermodynamically favourable, even in spite of the unfavourable ring size.