Is there a shortcut method to calculate the Steric Number? The steric number is equal to the number of $\sigma$-bonds + the number of lone pairs of electrons on the central atom. It gives us the number of hybridised orbitals.

It is pretty straight-forward to calculate it, but the problem here is that one must always draw the Lewis Structure before one can actually get to calculating the steric number, and then the number and types of hybrid orbitals. Even that is quite simple for a smaller compound, even like XeF$_6$, but when it comes to complex hydrocarbons, it's a little difficult.

My question is that is there any well-known (or not so well-known, but working) shortcut to doing this, so as to save time? It would be great if anyone could share their ideas and help me out.

Thanks in advance.

Short answer: no.

The steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups. For hydrocarbons and other organic compounds, you need to consider isomerism. Given the capability of carbon to form complicated bonding patterns, even simple formulas can produce a fair number of isomers with different bonding patterns and steric numbers.

  Let's look at some examples.

$\ce{C4H10}$

This formula corresponds to two compounds with the structures shown:

In this case, both compounds have all four carbon atoms with steric number of 4.

it is not always true that a set of hydrocarbon isomers will always have the same steric number for all carbon atoms or even the same set of steric numbers.

$\ce{C4H8}$

This formula corresponds to six isomers:

Note that four of these structures have two carbon atoms with steric number 4 and two carbon atoms with steric number 3. The other two have all four carbon atoms with steric number 4.

Any method to calculate steric number for carbon atoms in an organic compound using just the formula will fail. You must examine the structure.


Wouldn't radiolabelled phosphorus in DNA break it apart as it disintegrates? The Hershey-Chase experiment was designed to prove that DNA is the genetic material in organisms. In this experiment, two batches of viruses were grown in two separate media A and B, with A having an abundance of nutrients involving $\ce{^32_15P}$, and B having a lot of $\ce{^35_16S}$.

Leaving solution B aside, we have $\ce{^32_15P}$ undergo a beta decay:

$$\ce{^32_15P -> ^32_16S + e- + \bar{\nu_e}}$$

This process releases $\pu{1.709 MeV}$ of energy, which is quite sufficient to break apart the phosphodiester linkages.

With all the phosphorus in the viral DNA being radiolabelled, this would very well break apart all the phosphodiester linkages. and give us a soup of sulfur-based nucleosides. In such a case, how come this hasn't affected the virus' mechanism to transfer the DNA to a host cell?

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway.

The same applies to any other case of radiolabeling, in biochemistry or elsewhere, and you seem to be missing the very point of it: All this is not important.

Say, we have some $\ce{A + B -> C + D}$, and we label some atom in $\ce{A}$ so as to find out whether it will go to $\ce{C}$ or to $\ce{D}$. Then we run the reaction as usual. Now, radioactive atoms do not decay all at once; some decay earlier and some later. Those which decay earlier (that is, before and during the reaction) are lost to us, and we don't care about them one bit. Then we separate $\ce{C}$ from $\ce{D}$ and measure the radioactivity of each, and it is those atoms which decay now that matter.

Oh, and it is not like we label all atoms in $\ce{A}$. No, a very tiny fraction would suffice. Those viruses would probably get but a few radioactive atoms each. Some will be lucky enough to reproduce. Some will be killed by an early decay and remembered as martyrs of science.

So it goes.


How is o-Fluorophenol more acidic than phenol even after having Hydrogen bonding? Source: Concepts of Organic Chemistry by Dr OP Tandon, Himanshu Pandey, Dr AK Virmani
Page no: 241

Perhaps the words in the picture are explained at greater length in the text, or perhaps actual numbers are given for the dissociation constants (https://www.quora.com/Why-is-o-flurophenol-is-more-acidic-than-p-flurophenol).
As it turns out, the dissociation constants are 8.7, 9.3. 9.9 and 10.0 for ortho, meta, para and unsubstituted phenol. Ortho-fluorophenol is the most acidic - perhaps in spite of the possible hydrogen bonding. The inductive effect of the fluoro group is surely greatest at the ortho position.
In the series of nitrophenols (Acidity order of nitrophenols), the para (p$K_a$ = 7.16) is slightly more acidic than the ortho (p$K_a$ = 7.2), so it could be said that the ortho substituted compound, which would be expected to be more acidic (if you assume that the resonance effects at ortho and para positions would be similar), but isn't, we have to look for some other explanation, and hydrogen bonding could stabilize the neutral molecule.
Well, even if the actual effect is stronger than 0.04 p$K_a$ units, the hydrogen bonding effect seems to be rather small, and is overcome in the fluorophenols by the $-I$ effect of fluorine. Just scratch out the words "H-bond decreases acidity". (Lightly.)


Mechanism of Halogenation of Alkene In the above reaction, we use a non-polar compound (such as CCl$_4$) as a solvent. Then, why does the Br$_2$ polarise on attack by the alkene, even though a non-polar solvent does not encourage polarisation of a compound. Is the mutual polarisation enough for the reaction to occur.

Further, in different books the mechanism is either explained by the above mechanism or by the Molecular Orbital Theory (the alkene's filled $\pi$ orbital (HOMO) will interact with bromine's empty $\sigma$* to give the product).

Which one of these is the more popular and/or the officially accepted explanation.

Alkene double bond is an area of high electron density and hence high (partial) negative charge. Bromine molecule, approaching double bond, shifts its electrons of covalent bond towards farther Br atom, it becomes polarized and an induced dipole.

That's the simplest explanation in terms of electron movement I would use.


Absolute configuration in stereochemistry: R/S nomenclature I have all the rules of Fischer projections figured out.

But, when the problem asks for finding the configuration of a molecule with 1 chiral carbon and in a Fischer projection form, do we need first to make the manovrations that are allowed ( like fliping 180 degrees, or changing the places of 2-2 substituents )  or do we directly find the priorities ?

Just to remember the rules to assign an R/S configuration:

In order to assign the configuration as R or S:

Identify each of the chirality centers (most commonly an sp3 C with 4 different groups attached) Then at each chirality center....
Assign the priority (high = 1 to low = 4) to each group attached to the chirality center based on atomic number.
Reposition the molecule so that the lowest priority group is away from you as if you were looking along the C-(4) σ bond. If you are using a model, grasp the lowest priority group in your fist.
Determine the relative direction of the priority order of the three higher priority groups (1 to 2 to 3)
If this is clockwise then it is the R-stereoisomer (Latin; rectus = right handed)
If this is counter-clockwise then it is the S-stereoisomer (Latin; sinister = left handed)
If there is more than one stereocenter, then the location needs to be included with the locant, e.g. (2R)-

You can identify these previous rules on this example:

Source


How to determine interaction of photon with a hydrocarbon Bixin Bixin is an apocarotenoid found in annatto with 25 carbons with  carboxylic acids at each end of the chain. It is an insoluble orange red pigment.

I am interested to study photon interaction with this molecule. Please, give me some head start on what to read. It is a red pigment, therefore it absorbs photons below 670 nm or so. How can I determine if this molecule allows luminescence in any degree? Electronic transitions?  or just any refraction or scattering.

In a first step I would widen the available search criteria.  While accessing the English entry of wikipedia, collection of the chemical name, CAS number, smiles code (copy/paste), pubchem and chemspider ID are presented.  These identifiers may become helpful in the subsequent searches in other databases.  Nor would I hesitate (as long as my web browser allows proper display .and. there are at least some words barely understood in reading) to access the other versions (Spanish, Portuguese, ...), too.

In the public libraries of US' NIH, just the keyword "bixin" yields 150 (pmc) or 116 (pubmed) hits, that after some narrowing yield entries like this abstract in Spectrochimica Acta or this about triplet states of bixin, published in J. Agric. Food Chem.

While the British chemspider does not yield spectra directly, the literature references listed by this database include even surprising finds like doi 10.1039/C4CS00309H (2015ChemSocRev3244) about Vegetable-based dye-sensitized solar cells.

Do not omit collections like researchgate, as here, or figshare. If your institute has subscriber access, chemical databases like Reaxys (Elsevier) or Scifinder Scholar (ACS) may indicate primary literature including spectroscopic data, too.  No longer chemistry/physics centred, a search in Thomson Reuters' Web of Science, and more engineering-oriented Scopus (by Elsevier) may round the picture.


Why is the melting of ice a spontaneous reaction? So here's my question: we know that melting of ice is an spontaneous reaction but is it spontaneous? We have to supply energy in the form heat for the reaction to take place.

First of all,

  what exactly is spontaneous reaction?

In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction.

Now let us consider melting of ice example,

Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice.

This process took place without any intervention from your side. Heat absorbed by the ice from surrounding is due to set of conditions not due to your intervention.

Thus we can consider melting of ice as a spontaneous process(or reaction).


Why can't we use fusion to make ourselves rich? As we know on fusing two hydrogen atoms we can convert it to helium and by breaking them we get two $\ce{H}$ atoms. So why can't we convert mercury to gold by removing 1 proton per atom?

While it is possible to synthesize gold from mercury-196 (exists as 0.15% of natural mercury) the resultant gold will be radioactive and will not be gold for long!

There is actually the spallation neutron source which creates gold from liquid mercury.

Wikipedia


Is the electric conductivity of chlorine ions influenced by its oxidation state? I'm a high school student researching the conductivity of electrolyte solutions. In this excerpt of the CRC Handbook, I noticed that different ions containing chlorine have different limiting molar conductivity, and there seems to be a trend between molar conductivity and the oxidation state of chlorine:

Ion
Oxidation state of chlorine
$\Lambda$ in $\pu{e-4 m2 S mol-1}$

$\ce{ClO2-}$
+3
52

$\ce{ClO3-}$
+5
64.6

$\ce{ClO4-}$
+7
67.3

Does the trend really exist? If so, why did it happen? I found previous researches on the conductivity of $\ce{Fe^2+}$ and $\ce{Fe^3+}$, and the difference seemed to be a result of electrons spinning.

It is the similar effect as for molar conductivity of cations of alkali metals, which grows from $\ce{Li+}$ toward $\ce{Cs+}$, in spite of increasing ionic radius.
The point is, the effective radius of the hydrated ions has the opposite trend as the "naked" ion radius trend. $\ce{Li+}$ ion with the smallest radius has the strongest electrostatic field and the largest hydration sphere and has therefore the lowest molar conductivity.
The similar effect I expect for chlorine oxyacid anions. In this case, there is also decreasing degree of delocalization of the negative charge in trend $\ce{ClO4-} \gt \ce{ClO3-} \gt \ce{ClO2-}$.


Reduction half-reaction for the reduction of elemental oxygen Write the reduction half-reaction for the reduction of elemental oxygen in the presence of water, producing hydroxide ions

This does not make sense to me - a half reaction (from what I understand) is when you have an "element + some number of electrons -> element charged" or "charged element -> element + some number of electrons"

Reaction:

$\ce{HOH + O -> 2OH}$

Oxidation States: They all seem to be neutral to me - but I may be missing something.

So, how is a half reaction being introduced into this and what would it look like? Also, how can i know which element/compound is reduced/oxidized and is the reducing agent/oxidizing agent?

I am really sorry if this seems simple but I am a chem noob and I was out for a month and my teacher is behind schedule so he is giving us worksheets without teaching and i do not understand a lot. Thank you for your time

O2 + 2HO+ 4e- —> 4OH-
That is the correct answer


Relative stabilities of tropylium and cyclopropenyl carbocations How can the stabilities of the tropylium and cyclopropenyl carbocations be compared?

Both are aromatic according to Hückel's rule. I think that to compare the stabilities, we must either count the number of resonance structures, or perhaps we could say that cyclopropenyl is less stable due to angle strain. Am I on the right track?

Both ions are indeed strongly stabilized by aromatic coupling, and salts containing both ions are known.  But, of course, aromaticity is not the only thing that determines stability.  Tropylium ion, with its lesser ring strain and more atoms over which to distribute the positive charge, forms ionic salts with more anions (e.g., ionic $\ce{C_7H_7Br}$ versis covalent $\ce{C_3H_3Br}$) and in that sense is more stable.
The ionic structure of some tropylium salts is documented in Wikipedia[1][2][3][4]

In 1891 G. Merling obtained a water-soluble bromine containing compound from a reaction of cycloheptatriene and bromine.[1] Unlike most hydrocarbyl bromides, this compound, later named tropylium bromide, is a water-soluble solid and is insoluble in hydrocarbons, chloroform, and ether. It is purified by crystallization from hot ethanol. Reaction with aqueous silver nitrate immediately gave a precipitate of silver bromide. The structure of tropylium bromide was deduced to be a salt, C7H7+Br–, by Doering and Knox in 1954 by analysis of its infrared and ultraviolet spectra.[2][3] The ionic structures of tropylium perchlorate (C7H7+ClO4–) and tropylium iodide (C7H7+I–) in the solid state have been confirmed by X-ray crystallography.[4] The bond length of the carbon-carbon bonds were found to be longer (147 pm) than those of benzene (140 pm) but still shorter than those of a typical single-bonded species like ethane (154 pm).

References:

Merling, G. Ueber Tropin. Ber. Dtsch. Chem. Ges. 1891, 24 (2), 3108–3126. DOI: 10.1002/cber.189102402151.

von E.doering, W.; Knox, L. H. The Cycloheptatrienylium (Tropylium) Ion. J. Am. Chem. Soc. 1954, 76 (12), 3203–3206. DOI: 10.1021/ja01641a027.

Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Aromaticity as a Cornerstone of Heterocyclic Chemistry. Chem. Rev. 2004, 104 (5), 2777–2812. DOI: 10.1021/cr0306790.

Kitaigorodskii, A. I.; Struchkov, Y. T.; Khotsyanova, T. L.; Vol'pin, M. E.; Kursanov, D. N. Crystal structures of tropylium perchlorate and iodide. Russ Chem Bull 1960, 9 (1), 32–36. DOI: 10.1007/BF01178699.


Volume calculation perchloric acid 70% HClO4= 100.5, d HClO4=1.8, (ACO)2O=102, d (ACO)2O=1.1, H2O=18, d H2O=1

A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution. below is my answer, but what I'm not sure is that the density the problem has given is for 100% pure acid or 70%. below I calculated the density for 70% assuming 1.8 is for 100%.

d_70%=  100/(70 x 1/1.8+30 x 1/1)=1.452
M=10αd/MW=10x70x1.452/100.5=10.11 M

M_1 V_1= M_2 V_2→10.11 x V_1=0.1 x 100→ V_1≅1ml

B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid. Okay about this one. So all I know is that I need to remove all water in the acid by adding acetic anhydride, and the ratio is 1/1:

I thought about how to calculate the volumes of both acid and acetic anhydride, and maybe I can use the density, or tried to create some equation, but none actually reached a final answer. So I tried this, I'm not sure this actually makes sense:

70/100 x 1/1.8=0.39 ml

M_pure=  ( 0.7/100.5 x1000)/0.39=17.86 M

M_1 V_1= M_2 V_2→17.86 V_1=0.1 x 100→V_(pure acid)=0.56ml

1-0.56=0.44 ml water

0.44ml x 1g/1ml x 1mol/18g x  (1 mol (ACO)_2 O)/(1 mol water)  x 102g/1mol x 1ml/1.1g=2.27ml acetic anhydrade

So 1 ml perchloric acid and 2.27 ml acetic anhydride, up to 100 with acetic acid

Given:

HClO4= 100.5, d HClO4=1.8
(ACO)2O=102, d (ACO)2O=1.1
H2O=18, d H2O=1

A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution.

  I had to do some poking around to understand the density number. Without a temperature the density of liquids is nebulous. Wikipedia lists a density of 1.768 g/ml which seems to be for the 72.5% azeotrope.

So just blundering with the information given assume density is for 70% by weight perchloric acid.

Molarity $\ce{HClO4}$

$$\dfrac{0.70\times\pu{1.8 g/ml} \times \pu{1000 ml/L}}{\pu{100.5 g/mol}} = 12.5 M$$

Volume of conc $\ce{HClO4}$ for 100 mls of 0.1 molar

$$\dfrac{\pu{0.1 L}\times \pu{0.1 M}}{\pu{12.5 M}}= \pu{0.0008 L} = \pu{0.8 ml}$$

B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid.

This is a bit gnarly.

Volumes are not typically additive. So typically an excess of the acetic acid would be made and this would be used to dilute the 0.8 ml of perchloric acid to 100 ml. However in this case the 0.8 ml of perchloric acid contains water too. Also no density is provided for glacial acid acid, nor for the 0.1 Molar perchloric acid in glacial acetic acid. So the problem isn't really solvable as is.

The other wrinkle here is that mixing exact stoichiometric proportions is impossible. If an anhydrous solution was wanted, then an excess of acetic anhydrous would be used or some other means to dry the solution would be necessary.

Never the less, let's blunder through some sort of solution, as poor as the problem is...

Assume that the volumes are additive.

  So we'll get the right proportions of the three components but only somewhere near 100 ml of solution, not 100 ml exactly.

$$V_{solution} = V_{anhydride} + V_{water} + V_{\pu{conc acid}}$$

$$V_{\pu{conc acid}} = \pu{0.8 ml}$$

$$\therefore V_{anhydride} + V_{water} = \pu{99.2 ml}$$

However as mentioned before, the concentrated perchloric acid contains 30% water by weight.

$$\pu{g water in acid} = 0.3 \times \pu{0.8 ml}\times \pu{1.8 g/ml} =\pu{0.4 g}$$

Since water has a density of 1 g/ml that is also 0.4 ml.

We were given that (ACO)2O=102, d (ACO)2O=1.1 so for the anhydride the moles per ml is:

$$\dfrac{\pu{1.1 g/ml}}{\pu{102 g/mol}} = \pu{0.0108 mol/ml}$$

We were given that H2O=18, d H2O=1 so for the water the moles per ml is:

$$\dfrac{\pu{1 g/ml}}{\pu{18 g/mol}} = \pu{0.0556 mol/ml}$$

Now since the perchloric acid contained 0.4 ml of water we need to figure out how many ml of acetic anhydride we need to neutralize that water.

$$\dfrac{\pu{0.0556 mol/ml}\times\pu{0.4 ml}}{\pu{0.0108 mol/ml}} = \pu{2.1 ml}$$

So we now need 99.2 - 2.1 = 97.1 ml of water/acetic anhydride mixture in a 1:1 molar ratio. For 1 ml of acetic anhydride the ml of water needed is

$$\dfrac{\pu{0.0108 mol/ml}}{\pu{0.0556 mol/ml}}= \pu{0.194 ml water}$$

Let $x$ be the multiplicative factor

$$x = \dfrac{97.1}{1.194}= 81.3$$

So:
0.8 ml conc perchloric acid
83.4 ml acetic anhydride
15.8 ml water

I loathe that the problem is inferring that volumes are additive.

Looking up some data...

\begin{array}{|c|c|c|c|} \hline
 & Mol. Wt. & Density\ (g/ml) & mol/ml \\ \hline
\pu{Water} & 18.015 & \pu{0.9950 @ 25 °C} & 0.05523\\ \hline
\pu{Acetic anhydride} & 102.09 & \pu{1.082 @ 20 °C} & 0.01052 \\ \hline
\pu{Acetic acid} & 60.05 & \pu{1.0446 @ 25 °C} & 0.01740 \\ \hline
\end{array}

  The density data is out of Pubchem. I looked for a while and couldn't find density values at same temperature, but the values are close enough to make the point...

assume 1.0000 ml of acetic anhydride and appropriate water to give 1:1 molar ratio

\begin{array}{|c|c|c|} \hline
 &  ml & g \\ \hline
\pu{Water} & 0.1904  & 0.1894\\ \hline
\pu{Acetic anhydride} & 1.0000 & \pu{1.082 g} \\ \hline
\pu{Totals} & 1.1904 & 1.271 \\ \hline
\end{array}

So density of acetic acid by combining volumes is
$$\dfrac{1.271}{1.1904} = 1.067$$
which gives an error of
$$\dfrac{1.067-1.0446}{1.0446}\times 100\% = +2.1\% $$

Thus the total solution as calculated above would be closer to 98 ml than 100.


Approximating the concentration of ions when a salt dissociates In my textbook written that:

  Given that $K_\mathrm{sp}(\ce{Ag2CrO4})=9.0\times 10^{-12}$, consider the solubility of $\ce{Ag2CrO4}$ in a $0.100\:\mathrm{M}$ solution of $\ce{AgNO3}$.

  Initial concentrations (before any $\ce{Ag2CrO4}$ dissolves) are,
  $[\ce{Ag+}]_0=0.100\:\mathrm{M}$ (from $\ce{AgNO3})$ and $[\ce{CrO4^{2-}}]_0=0\:\mathrm{M}$

  The system comes to equilibrium as $\ce{Ag2CrO4}$ dissolves according to the reaction:
  $$\ce{Ag2CrO4(s)\rightleftharpoons 2Ag+(aq) + CrO4^{2-}(aq)}$$
  $$K_\mathrm{sp}=[\ce{Ag+}]^{2}[\ce{CrO4^{2-}}]=9.0\times 10^{-12}$$

  We assume that $x\:\mathrm{M}$ of $\ce{Ag2CrO4}$ dissolves to reach equilibrium, which means that:
  $$x\:\mathrm{M\:\ce{Ag2CrO4}(s)}\rightarrow 2x\:\mathrm{M\:\ce{Ag+}(aq)}+x\:\mathrm{M\:\ce{CrO4^{2-}}\:(aq)}$$

  Now we can specify the equilibrium concentrations in terms of $x$:
  $$[\ce{Ag+}]=[\ce{Ag+}]_{0}+\Delta[\ce{Ag+}]=(0.100+2x)\:\mathrm{M}$$
  $$[\ce{CrO4^2-}]=[\ce{CrO4^2-}]_{0}+\Delta[\ce{CrO4^2-}]=(0+x)\:\mathrm{M}$$

  Giving:
  $$9.0\times 10^{-12}=[0.100+2x]^{2} \cdot [x]$$

Also, here is line I want to discuss (again, taken from the textbook):

  Since the $K_{\mathrm{sp}}$ value for $\ce{Ag2CrO4}$ is small (the position of the equilibrium lies far to the left), $x$ is expected to be small compared with $0.100\:\mathrm{M}$. Therefore:
  $${9.0\times 10^{-12}=[0.100+2x]^{2}[x] \approx [0.100}]^{2}[x]$$

When can this approximation be made and when can it not be made?

Really, your question seems to be a math question.

You need to solve the cubic equation:

$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$

Cubic equations are exactly solvable; however, it is common to use an approximate method.

Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.

Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.


Using osmotic pressure to find the molar mass of a compound Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms $(\pu{12.5E-6 g})$ of what you hope is pure saxitoxin (a non-electrolyte) from a poisonous (and quite cross) puffer fish. You dissolve this sample in $\pu{3.10 mL}$ of water and determine that the osmotic pressure of
  the resulting solution is $\pu{0.236 Torr}$ at $\pu{19 °C}$ $(\pu{760 Torr} = \pu{1.00 atm}).$ What is the molar mass of the compound?

I am having trouble reaching the correct answer of $\pu{310.8 g/mol}$ for this question. I first found the pressure to be $\pu{0.0003105 atm}$ and plugged numbers in for the formula $\Pi = MRT.$ I got $\pu{2.04E-5 mol L-1}.$ I used this to find the amount of substance and got $\pu{6.3E-8 mol}.$

Lastly, I divided $\pu{12.5E-6 g}$ by $\pu{6.3E-8 mol}$ and got $\pu{1.98E-14 g/mol}$ which is not the right answer.

I feel like I have the right process so if you can help show me where I messed up, I would appreciate it.

I couldn't resist. The osmotic pressure equation is

$$Π = cRT,$$

where $Π$ is osmotic pressure in atmospheres, $c$ is molar concentration, $R = \pu{0.082057366080960 L atm mol-1 K-1}$ (exactly) and $T$ is temperature in Kelvin.

In this problem, $Π = \pu{0.236 torr}$ times exactly $\pu{1 atm}/\pu{760 torr}$ and $T = \pu{(273.15 + 19.0) K}.$ So $c = \pu{1.2953E-5 M}$. But $\pu{12.5E-6 g}$ of saxitoxin was dissolved in $\pu{3.10 mL}$ of water, which is $\pu{4.032E-3 g/L}$. Hence $\pu{1 mol}$ of saxitoxin has a mass of $\pu{311 g}.$


Magic Potion Trick For Kids I am currently trying to make a magic potion with chemistry.

I want to turn black liquid clear to reveal a hidden message.

I am currently using water, blue and red food colour and bleach. This works but I want something more dramatic and safe for kids.

Any Ideas :)?

regards Daniel

One possibility:  Starch/iodine solution + sodium thiosulfate.  The starch/iodine solution goes from dark blue/black to clear when sodium thiosulfate reduces the iodine.


Which one of the two is the stronger base: methoxide ion or hydroxide ion? If we consider the experimentally determined pka values, it looks like hydroxide ion should be the right answer but in many books it has been mentioned that all alcohols are less acidic than water implying that all alkoxide ions are stronger bases than hydroxide ion. This has created a lot of confusion.

We have $\ce{MeO-}$ and $\ce{HO-}$
$\ce{-CH3}$ has a +I effect. Therefore it destabilizes the methoxide ion. $\ce{-H}$ has a comparatively lesser +I
This leads to the conclusion that $\ce{OH-}$ is more acidic than $\ce{CH3O-}$.
This seems like a valid argument, but the order varies in gaseous/aqueous phases, because we consider polarizabilty and solvation effect, rather than inductive effect there


How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid? In the following reaction, $K_c=4.0$:
  $$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$

  A reaction is allowed to occur in a mixture of $\pu{17.2 g}$ $\ce{C2H5OH}$, $\pu{23.8 g}$ $\ce{CH3COOH}$, $\pu{48.6 g}$ $\ce{CH3COOC2H5}$, and $\pu{71.2 g}$ $\ce{H2O}$

  (a) In what direction will a net change occur?

  (b) How many grams of each substance will be present at equilibrium?

This is what I did for part (a):
$$Q_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{C2H5OH}][\ce{CH3COOH}]}$$

First, get the concentration of $\ce{C2H5OH}$, $\ce{CH3COOH}$, $\ce{CH3COOC2H5}$, and $\ce{H2O}$ using the given amount of each substance.

$$\begin{align}
[\ce{C2H5OH}]
  &= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} &
  &= \pu{0.373 mol} \\
[\ce{CH3COOH}]
  &= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} &
  &= \pu{0.396mol} \\
[\ce{CH3COOC2H5}]
  &= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} &
  &= \pu{0.552mol} \\
[\ce{H2O}]
  &= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} &
  &= \pu{3.96 mol} \\
\end{align}$$

Assuming the overall volume is $\pu{1 L}$, then
$$Q_c = \frac{(\pu{0.552 M})(\pu{3.96 M})}{(\pu{0.373 M})(\pu{0.396 M})}
      = 14.8 $$

Since $K_c$ is $4.0$, and $Q_c$ is $14.84$, $Q_c$ is larger than $K_c$ and the direction of the reaction is to the left, towards the reactants.

I know how to solve for (a) but I have no idea how to get the amount of each substance at equilibrium for (b). The answer for (b) should be $\pu{68 g}$. I tried doing the ICE chart, but I did not get $\ce{68 g}$. Do I use the ICE chart for this part of the question or which method should I be using?

Knowing that $K_c=4.0$, equilibrium is achieved when $Q_c = K_c$.

Let's set up an ICE table and write down our equilibrium expression like you did:

Reactants:

$$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & \pu{0.373 mol} & \pu{0.396 mol} & \pu{0.552 mol} & \pu{3.96 mol}\\
\hline
\text{Change conc.} & -x & -x & +x & +x\\
\hline
\text{End conc.} & \pu{0.373 mol} - x & \pu{0.396 mol} - x & \pu{0.552 mol} + x & \pu{3.96 mol} + x \\
\hline
\end{array}

Assuming that our reaction is at equilibrium:

$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$

We plug in our values for our change in equilibrium:

$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$

Solving for $x$, which is the change in concentration:

\begin{align}
4 &= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\
0 &= 3x^2-7.588 x-1.59509
\end{align}

Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for $x$:

\begin{align}
x_1 &= 2.72449 \\
x_2 &= -0.195154
\end{align}

The negative zero is chosen, since the reaction will evolve to the left (being $Q_c>K_c$), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:

\begin{align}
\ce{[CH3COOH]}
  &= \pu{0.373 mol} - (\pu{-0.195154 mol}) &
  &= 0.568154\\
\ce{[C2H5OH]}
  &= \pu{0.396 mol} - (\pu{-0.195154 mol}) &
  &= 0.591154\\
\ce{[CH3COOC2H5]}
  &= \pu{0.552 mol} + (\pu{-0.195154 mol}) &
  &= 0.356846\\
\ce{[H2O]}
  &= \pu{3.96 mol} + (\pu{-0.195154 mol}) &
  &= 3.764846
\end{align}

Therefore:

\begin{align}
\pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &= \pu{34.1 g}\\
\pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &= \pu{27.3 g}\\
\pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &= \pu{31.4 g}\\
\pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &= \pu{67.8 g}
\end{align}

Initial product sum is $\pu{160.8 g}$.

Equilibrium product sum is $\pu{160.6 g}$. $\pu{0.2 g}$ is likely due to truncation error in calculations.

This problem is really cruel... whoever assigned it... :/


Why are alcohols with longer chains less polar? In my self-study, I recently came across the following question:

"Choose the solute of each pair that would be more soluble in hexane ($\ce{C6H14}$). Explain your answer.

(a) $\ce{CH3(CH2)10OH}$ or $\ce{CH3(CH2)2OH}$ ..."

Undecanol is more soluble in hexane because it is apparently less polar than propanol. Further Internet searches revealed that alcohols decrease in polarity as the chain length increases (assuming a basis in an alkane; don't know if this rule generalizes which I guess is a sub-question) but it is still not clear why. None of the sources I found explained this. The way I imagine things, the carbon-hydrogen bonds should add a bunch of zero vectors to the the O–H vector that both molecules share, giving the same polarity in both cases. But this is apparently wrong.

So why is undecanol less polar than propanol then? Does it have something to do with a more advanced bonding theory?

If you're looking at the overall polarity of the solvent, you're not only consider the value of the individual dipole moment on a single molecule, but also their “density”, i.e. the number of these dipoles per unit of volume. You are right that the dipole moment of an individual linear-chain alcohol will, in first approximation, be independent of the chain length. However, in a given volume, you'll fit fewer of those long molecules, and the polarity of the overall solvent is thus lower.


On the detection of molecules in electrospray ionisation (ESI) Of the ESI spectra that I have encountered, positive & negative mode, the detected species always have charge (positive or negative) corresponding to the charge mode that the machine is in. Is it therefore the case that the ESI detector can only detect charged species, or that the overwhelming majority of species are charged?

I feel you have a misconception. ESI is an ionization technique not a detection technique. The sole purpose of the electrospray ionization is to impart a charge on a molecule in the gas phase. The molecule could be neutral, anionic or cationic in the solution phase. The inventor of ESI used to say that ESI makes the elephants fly, i.e., very large molecules could also be ionized in the gas phase.
The important point that you must learn is that if you have an ionized molecule in the solution, it is not necessary that when you bring the molecule in the gas phase, it will have a electrostatic charge. It may become a neutral molecule.
In order to "sort" molecules according to their mass to charge ratios in the gas phase, it is best to ionize them, i.e., remove an electron or add a proton so that the molecule of interest acquires an electrostatic charge in the gas phase. Note that I am emphasizing gas phase again and again. Forget what is the state in the solution. Almost 100 years ago, physicists had worked out how "charges" will move in an electric or magnetic field predictably. If the molecule is electrically neutral it will not interact effectively either with electric or magnetic field. Hence you would not be able to sort them according to their mass/charge rations. This is the job of the mass spectrometer.
Once your molecule has a charge, it is very easy to direct them here and there in the mass spectrometer and even focus all the ions at place  (one of the major advantage of ionized molecules), before exposing them electric fields or magnetic fields.


Can we make a salt using metal oxide? Get a metal like sodium and then you burn it to make sodium oxide because burning something is adding oxygen. Then you add the sodium oxide to an acid like hydrochloric acid. That should make a salt (sodium chloride) and water. Evaporate or filter the water away and you are just left with the salt itself.

Is this possible to do or am I just completely wrong?

Your description is almost correct - note that when sodium metal burns in air, it forms primarily sodium peroxide in addition to sodium oxide. Sodium peroxide will react with hydrochloric acid to give sodium chloride and hydrogen peroxide. Heating will decompose that to water plus oxygen. You should still be able to get the sodium chloride product that you expected.


Is menthol an especially volatile molecule? I have been studying a bit about odors and plant essential oils. Several sources claim that the fragrant terpenes found in these oils, such as menthol, eugenol, limonene, linalool, etc. are volatile compounds.

I understand that all molecules are technically volatile, but since these molecules all have boiling points of over $170~\mathrm{^\circ C}$, and their vapor pressures are not particularly high, I don't see why they are considered 'volatile compounds'.

I have two guesses:

a) There is some fundamental aspect of chemistry that I am ignoring, and these compounds do have a set of properties that makes them especially volatile. This to me sounds very interesting, and I would really like to learn what is that makes these compounds especially well-made for being vaporized.

b) The compounds are not especially volatile. They are synthesized in large amounts, concentrated to high concentrations in the plant, and our nose is very sensitive to them. They are called 'volatile' compounds because we happen to notice their smell when they are in the air.

For my question I've chosen to focus on menthol. Is there anything about menthol that makes it especially volatile?

The term "volatile oils" is a traditional term.

From the 1916 book A TREATISE ON PHARMACY FOR STUDENTS AND PHARMACISTS 5th edition at page 217:

  Although the boiling points of volatile oils are considerably above that of water, the oils pass over rapidly with the vapor of boiling water, and in the leading establishments in this country [USA] and Europe volatile oils are now distilled by passing steam under pressure into stills which contain the material on a series of perforated trays...

So it means volatile to the degree that it could be steam distilled.


What is the oxidation state of Mn in HMn(CO)5? What is the oxidation state of $\ce{Mn}$ in $\ce{HMn(CO)5}$? If $\ce{H}$ has an oxidation state of $+1$, then $\ce{Mn}$ should have oxidation state of $-1$, which I'm not sure is possible.

On the other hand, if $\ce{H}$ has an oxidation state of $-1$, then $\ce{Mn}$ has an oxidation state of $+1$. However, Wikipedia says:

  The pKa of $\ce{HMn(CO)5}$ in water is $7.1$.

This means that the $\ce{Mn(CO)5}^-$ complex exists, implying a $-1$ oxidation state for $\ce{Mn}$. Again, I am not sure that a metal can adopt a negative oxidation state.

On negative oxidation states, in general

Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive oxidation states, but good authors will be careful to add a disclaimer along the lines of "only 'common' oxidation states are listed".

The key is in the ligand that the metal is paired with. In metal ions with positive oxidation states, the ligands that they pick up tend to be Lewis bases, like $\ce{H2O}$, $\ce{Cl-}$, $\ce{NH3}$, ... This should "make sense" intuitively, because the positively charged metal ion wants to gain electron density, and these ligands are happy to give electron density to it.

In $\ce{Mn(CO)5-}$, the ligand is carbon monoxide: a very poor Lewis base, but it does possess low-lying π* orbitals. In the parlance of coordination chemistry, CO is a poor σ-donor, but a great π-acceptor. This means it tends to seek out metal centres which are relatively electron-rich instead of electron-poor, and thus it makes sense that metals with a negative oxidation state are its preferred partners. [I'll defer a full explanation of the $\ce{M-CO}$ bonding to a proper inorganic textbook.]

It turns out that it's actually quite difficult to find examples of carbonyl complexes with positive oxidation states.[2] Nearly all carbonyl complexes have the metal in oxidation states of zero or lower: see, e.g. $\ce{[Cr(CO)6]}$, $\ce{[Fe(CO)4]^2-}$, $\ce{Ni(CO)4}$, and of course, $\ce{[Mn(CO)5]-}$. Another common feature is that all of these complexes obey the 18-electron rule.

On [HMn(CO)5]

So, back to the question of the oxidation state in $\ce{[HMn(CO)5]}$... IUPAC has the final say on oxidation states;[3,4] see also this writeup by Martin. Quoting from ref. 4,

  Oxidation state equals the charge of an atom after its homonuclear bonds have been divided equally and heteronuclear bonds assigned to the bond partners according to Allen electronegativity [...]

Following this, we have to assign the electrons in the $\ce{Mn-H}$ bond to the hydrogen, since hydrogen is more electronegative than manganese on the Allen scale (3.30 and 2.75 respectively). There is also evidence that in metal carbonyl hydrides like these, the hydrogen bears a partial negative charge, although this argument is not without its caveats.[5] Therefore, the oxidation state of $\ce{Mn}$ should be +1. However, IUPAC offers a way out of this:

  [...] except when the electronegative atom is bonded reversibly as a Lewis-acid ligand, in which case it does not obtain that bond’s electrons.

According to this, it is possible to have exceptions to the general algorithm for determining oxidation state in the previous quotation, and there is a case to be made for this exception being valid here: the electronegative atom ($\ce{H}$) is indeed bonded reversibly ($\mathrm pK_\mathrm a = 7.1$) as a Lewis acid ($\ce{H+}$). If we accept this line of argument, then $\ce{H}$ does not "obtain" the two electrons in the $\ce{Mn-H}$ bond, and the oxidation state of $\ce{Mn}$ returns to −1, which is perhaps more in line with our chemical intuition.

I leave it to the reader to draw a conclusion for themselves; but this is an excellent illustration of the most important lesson here: oxidation states have their limits, and should not always be taken as meaningful descriptors of the chemical bonding in a species, especially when ambiguity is present, since this makes their assignment essentially arbitrary. If we assign an oxidation state of −1 to hydrogen here, it doesn't magically make the complex any less acidic than it was before. And IUPAC are aware of it (ref. 4):

  The applications of OS in chemistry are wide and deal with a cornucopia of chemical compounds and materials. It is therefore not surprising that, for some compounds, one value does not fit all uses, or that dedicated measurements or computations are needed to ascertain the actual OS. In those rare cases when the most convenient OS becomes a matter of choice, this fact must be clearly stated.

Refs. 3 and 4 make for excellent (and fairly accessible) reading on the topic, for those wishing to find out more about these edge cases. Both can be accessed without a subscription.

Notes and references

Negative oxidation states for non-transition metals exist (e.g. alkalides), but these are probably more "exotic" than organometallic complexes.
They do exist, for example $\ce{[Ir(CO)6]^3+}$, although the bonding situation is markedly different (as evidenced by, e.g., IR spectroscopy).
Karen, P.; McArdle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081. DOI: 10.1515/pac-2013-0505.
Karen, P.; McArdle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839. DOI: 10.1515/pac-2015-1204.
Sweany, R. L.; Owens, J. W. Evidence for a negative charge on hydrogen for some metal carbonyl hydrides. J. Organomet. Chem. 1983, 255 (3), 327–334. DOI: 10.1016/S0022-328X(00)99326-4.


Why is cyclopropane more acidic than propane Though I found links about ring strain and other things, I actually fail to understand the exact reason for this difference in acidity.

Indeed, according to the Evans pKa table the cyclopropane $\ce{C-H}$ bond (pKa ~ 46) is more acidic than the $\ce{C-H}$ bond on the central carbon in propane (pKa ~ 51).

We know that, in general, the acidity of $\ce{C-H}$ bonds follows the order $\mathrm{sp > sp^2 > sp^3}$.  This is because the more s character in an orbital, the more stable (lower energy) electrons are in the orbital.

The $\ce{C-H}$ bonds around the central carbon in propane are close to being $\mathrm{sp^3}$ hybridized, whereas the $\ce{C-H}$ bonds in cyclopropane are roughly $\mathrm{sp^{2.46}}$ (see this earlier answer to see how the hybridization in cyclopropane was determined).

Now, when we compare the hybridization of the $\ce{C-H}$ bonds in cyclopropane and propane we see that there is more s character in the cyclopropane $\ce{C-H}$ bonds, this explains the increased acidity.


Why do seemingly reliable sources give NFPA 704 Health ratings ranging from 1-4 for elemental mercury? Recently a user posted the question Do Mercury have a NFPA 704 classification?.

Why the present question is not a duplicate:
Although there is no accepted answer to the above question, I think it's a fairly obvious "yes", and that my question rather is regarding how there could be such a diversity of classifications arising from this single, widely used standard.

Background:
According to the US National Fire Protection Association (NFPA), NFPA 704: Standard System for the Identification of the Hazards of Materials for Emergency Response:

  This standard shall address the health, flammability, instability, and related hazards that are presented by short-term, acute exposure to a material under conditions of fire, spill, or similar emergencies.

These are typically displayed in the widely recognizable "fire diamond" (generic image from Wikipedia, not for mercury):

Here, blue indicates health hazard, red indicates flammability, and yellow indicates chemical reactivity.

Data:
The following are some values given from different sources for the "health" rating (scale of 0-4, where 0 is the most benign and 4 the most hazardous) for elemental mercury:

The National Institute for Occupational Safety and Health (NIOSH): Health = 1
MSDS from NIST: Health = 2
MSDS from Fischer Scientific: Health = 3
MSDS from Sigma-Aldrich: Health = 4 (link from comment on first link above by MaxW)
SDS from Sigma-Aldrich: Health =2

Question:
My question is simply: why is there such disparity in the NFPA health ratings given for elemental mercury?

I believe that the radical rage of possible hazard ratings for elemental mercury arise from different assumptions about how the mercury is handled. I can't prove this, but it makes sense when you consider the possible hazards arising from mercury use.

Properly handled liquid mercury is not particularly harmful. There are no acute risks and a competent person could drink it and handle it with little risk. In a chemical laboratory, for example, it is common to use mercury manometers to provide limited gas overpressure for supplies of nitrogen or argon. This involves bubbling gas through mercury. As long as the vapour is safely vented to a scrubber, this is not a hazard to people in the lab.

Equally, chemists can use mercury safely with few precautions as long as they clean up any spills thoroughly. The hazard primarily comes from vapour buildup which, if it is allowed to happen, will create serious long term chronic problems.

One example of potential danger is if non-experts come into contact with mercury. Some old gas-flow meters for domestic gas supplies used mercury which could be discharged from the meter in certain rare situations, leaving droplets of mercury around the floor near the meter. This can (and has) created a great deal of hazard in a domestic environment. This hazard is magnified enormously, for example, if the householder tries to clean the mercury spill by vacuuming it up. The result is a fine spray of mercury droplets inside a warm vacuum cleaner, which vastly increases the vapour emission and may spread fine droplets widely through the house. If badly ventilated, this can be a serious chronic issue.

The two situations span a wide range of hazard. But the difference is the context: in experienced hands mercury is not very hazardous; in other hands it is potentially very dangerous.

In short, the different rating may reflect different assumptions about who is using the mercury: someone who knows what they re doing or someone who does not.


Molecular Imaging - Any surprises to be had? Molecular imaging, using STM and AFM technologies, appear so far to visually prove most of what we already know about chemical structures, such as VSEPR theory. For instance, here are the ideal structures and AFM of a couple of cyclization reactions:

We've also taken single-atom imaging to pretty scary levels. Behold a field emission electron microscope image of the s and p orbitals of carbon, long relegated to conceptual mathematical images in textbooks:

The question is, have there been any surprises to be had from these images, where something we thought was structured one way was in fact radically different? You can see in the AFM image that some of the bond angles deviate from the ideal regular shapes a little, and it may also be that the electrons in the pi bonds, instead of being regularly distributed, seem to prefer the "far" end of the phenyl rings, but the predicted structures are uncannily accurate. Again, with the orbital imaging, the s orbital isn't perfectly spherical and we don't see the full p-orbital shell, just two "lobes", but again, they're powerful evidence that our theories are right so far.

Has this ever not been the case? Have any of these recent images turned chemistry on its head by showing something totally unexpected?

I don't think this is going to put basic quantum chemistry upside down. Because chemistry is complicated, and these tools are "inaccurate". Let me illustrate : it was thanks to spectroscopy of atoms that people discovered the spin of the electron. It then explained a lot of fact that puzzled chemists. A molecule is something awfully complicated, that we can only solve with approximations and powerful computers. So to my mind, molecular observables are unlikely to contradict quantum theories. Now you need the LHC to split particles in order to test the theories.

This guy will tell you that there are surprises to be had about large molecules, the ones from natural sources (or plain coal) that are a pain in the ass to study by NMR,  or enzymes (like this one, studied with single molecule diffraction).

There will be always surprises. But if we could predict them, would they still be surprises ? (sorry that was too long for a 'comment')


In chemistry what do they mean by disorder when in the context of entropy? Entropy is often the amount of disorder is present in a system, but when they refer it to the degree of disorder in a system what do they mean. It just seems the term is quite vague and not very descriptive. Does it refer to the number of choices a molecule possesses in interacting with other molecules in a system or does it mean something else?

What you have said about entropy is generally correct. For a given energy it is related to the number of accessible states the system can be in. If a system has 100 energetically accessible configurations then we would say it has more entropy than a system with only 10. You can call this disorder because you have no knowledge of which of the 100 (or 10) states the system is actually in.

That said, entropy has a very well define mathematical meaning. I'll start with the punchline and then breakdown what it means. Entropy ($S$) is equal to the natural log of the multiplicity function ($W$, or $g$ sometimes) times boltzmann's constant ($k_b$):
\begin{equation}
S = k_b \ln(W)
\end{equation}

The multiplicity function $W$ is the number of states in the system of a given energy. So what does this mean? Let's imagine we have a magnet with three unpaired electrons which can either be pointed up or down. (e.g. $\uparrow \uparrow \uparrow $,  $\uparrow \downarrow \uparrow$, etc. ) Now let's say that there's a magnetic field applied so that up and down have different energies. Each $\uparrow$ adds $\epsilon$ to the energy while each $\downarrow$ subtracts $\epsilon$ from  the energy. This system can now exist in states ranging from +3$\epsilon$ to -3$\epsilon$.

Let's look at the E = +3$\epsilon$ and -3$\epsilon$ states first. In both cases, W($\pm 3\epsilon$) =1 becasue there is only one configuration for each energy, $\uparrow \uparrow \uparrow $ and $\downarrow \downarrow \downarrow$  respectively. Thus, using the formula, S = $k_b \ln(1) = 0$. Thinking of entropy as disorder, this makes sense. If the energy is $\pm3\epsilon$ then we know exactly which state the system is in and there is no chance whatsoever.

Now let's look at the case of E = $\pm \epsilon$. Here, we have multiple options for each state. In the E = +$\epsilon$ there are three possible states: $\uparrow \uparrow \downarrow$, $\uparrow \downarrow \uparrow$, and $\downarrow \uparrow \uparrow$. Similarly, for the E = -$\epsilon$ case there are also three states: $\uparrow \downarrow \downarrow$, $\downarrow \uparrow \downarrow$, and $\downarrow \downarrow \uparrow$. Thus, $W(\pm \epsilon) = 3$, and the entropy is $S = k_b \ln(3)$

$k_b$ is called The Boltzmann Constant and is just a scaling constant. It doesn't contain any physics and exists only because we happen to use different units for temperature and energy. In SI it's equal to ~$1.38×10^{−23}$ J/K, but you can equivalently measure temperature in energy units and then $k_b =1$ (with no units).

The physical insight is that entropy entirely a consequence system being allowed in multiple states.


How to find the potential of a mixed Iron/ Cer solution? Given the following half-reactions:
   \begin{align}
    \ce{Ce^{4+} + e^{−} &-> Ce^{3+}} & E° &= \pu{1.72 V}\\
    \ce{Fe^{3+} + e^{−} &-> Fe^{2+}} & E° &= \pu{0.771 V}\\
 \end{align}
   A solution is prepared by mixing $\pu{3.0 mL}$ of $\pu{0.30 M}$ $\ce{Fe^{2+}}$ with $\pu{2.0 mL}$ of $\pu{0.12 M}$ $\ce{Ce^{4+}}$.

  What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)? Calculate $\ce{[Ce^{4+}]}$ in the solution.

My attempt at solving the problem (which was incorrect):
First to find the potential (relative to SHE):
\begin{align}
E&=E^\circ-0.05916\cdot\log(\ce{Fe^{2+}/Fe^{3+}})\\
0.771&=E^\circ-0.05916\cdot\log((3\cdot0.3)/(3\cdot0.3 + 2\cdot0.12))\\
E^\circ&=\pu{0.7649 V}\\
\end{align}
I don't think this is correct.

Then:
I am not sure how to find $E^\circ$ cerium or how to find $\ce{[Ce^{4+}]}$.

First it helps to have an idea of what will happen. Since 1.72 V is a
lot more than 0.771 V (a difference of 1V in chemistry goes a
long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much
better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of
enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$.
Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$
will be reduced to $\ce{Ce^{3+}}$.

In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$
left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look
at the number of moles of each ion that went into the reaction at the
moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts
with one unit of $\ce{Ce^{4+}}$.

So now you've figured out which is the limiting reagent (the one that gets
consumed almost entirely). Let's suppose for the sake of argument it's
the $\ce{Fe^{2+}}$. Because you know how many
moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know
how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can
find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the
equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$,
because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's
all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.

If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same
logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and
$\ce{Ce^{3+}}$ in the equilibrium mix.

So now you have 3 concentrations out of 4.  Finding the 4th requires
knowing the equilibrium constant, which you can calculate from the
difference in potentials. (Write the combined full reaction based on the half-reactions.)

You'll find that the 4th concentration is
very small, justifying the claim that one of the reagents
was almost entirely consumed.

If you're with me so far, please update the question with these
results, and if you still need help with finding the potential at that
point, we can take it from there...


Temperatures Below Absolute Zero This seems like a farce...

  Quantum gas goes below absolute zero
  Ultracold atoms pave way for negative-Kelvin materials.
  by Zeeya Merali, Nature News 03 January 2013, DOI: 10.1038/nature.2013.12146

I have seen several recent articles like this one  and I am extremely skeptical. Perhaps my skepticism is based upon an ignorance of physical principles, but I am less offended by apparent violations of the speed of light than by considering that temperatures below absolute zero could exist. I have studied under scientists who were using advanced techniques and vast amounts of energy to get small samples cooled down to a few nanokelvin. The physics seems to imply that if a sample were to reach zero, even the nuclear forces cohering the particles would cease. This was hypothesized in hushed tones, in awe and reverence. Now this article (which seems akin to scientific blasphemy) is causing an existential crisis. What is the deal with negative absolute temperature?

It is a quirk in how temperature would be measured using quantum mechanics. In general thermodynamics is about "bulk" properties and ignores quantum mechanics. Resolving these sort of quirks is why it would be so wonderful to have a theory of everything that would tie all the forces together.


Why is lithium the most reducing alkali metal, and not caesium? Caesium has a larger size, and the effective nuclear charge that the valence electron experiences will be far less compared to that of lithium's, right? But lithium is still considered the strongest reducing agent among all the alkali metals, and this is evidenced by its large and negative reduction potential. Why is this so?

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated.

$$\begin{array}{cc}
\hline
\ce{M} & E^\circ(\ce{M+}/\ce{M}) \\
\hline
\ce{Li} & -3.045 \\
\ce{Na} & -2.714 \\
\ce{K} & -2.925 \\
\ce{Rb} & -2.925 \\
\ce{Cs} & -2.923 \\
\hline
\end{array}$$
Source: Chemistry of the Elements 2nd ed., Greenwood & Earnshaw, p 75

However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward.

The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction

$$\ce{M(s) -> M+(aq) + e-}$$

by the equation

$$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$

where $K$ is the absolute standard Gibbs free energy for the reaction

$$\ce{H+ + e- -> 1/2 H2}$$

and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on.

$$\require{AMScd}
\begin{CD}
\ce{M (s)} @>{\large \Delta_\mathrm{r}H}>> \ce{M+(aq) + e-} \\
@V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\
\ce{M (g)} @>>{\large IE_1(\ce{M})}> \ce{M+ (g) + e-}
\end{CD}$$

We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$:

$$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$

(the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone, this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$.

However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$.

$$\begin{array}{ccccc}
\hline
\ce{M} & \Delta_\mathrm{atom}H & IE_1 & \Delta_\mathrm{hyd}H & \text{Sum} \\
\hline
\ce{Li} & 161 & 520 & \mathbf{-520} & 161 \\
\ce{Cs} & 79 & 376 & \mathbf{-264} & 211 \\
\hline
\end{array}$$
Source: Inorganic Chemistry 6th ed., Shriver et al., p 160

This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) > E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion.

Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood & Earnshaw, p 75).


Melting point-general chemistry This one has stumped me for a few days.  What is the specific name of the general chemistry principle that is responsible for the difference in melting points of pure and impure solids.  I have exhausted all other resources available and a precise principle name eludes me.

This is discussed pretty thoroughly in materials engineering textbooks when melting and freezing of (mostly) metals in general is considered. A good reference is Gaskell’s “Introduction to thermodynamics of materials”. But you will find helpful explnations to fill in the gaps of what uou know by reading up on undercooling in Wikipedia etc. Feezing and melting are related.

And if you read my answer to highest melting point materials, I mention some related issues.

One issue is that you cannot consider materials as “pure” or “impure”: a pure material of any amount large enough to test is in practice almost impossible to make. In fact any real world material is impure once you consider a large enough scale: which is on the order of anything bigger than (say) millions of atoms. You cannot actually make a “pure” material of appreciable size, but can only approach it once you get larger than a relatively small size of sample. In addition, a lot of materials have only been made as thin coatings, or as individual small crystals. And their behaviour is not the same as larger amounts of what you might think is the same material. A single crystal behaves very differently from a polycrystalline mass, and the surface of any material is very different from its interior, so a thin film is dominated by surface effects, as are very fine particles.

So you are looking at degrees of impurity, in terms of orders of magnitude. And the general effect of purity level is not linear but more like exponential in terms of its effect on material behaviour.

A related concept is to look at the theoretical strength of a material. From a theoretical point, a pure and defect free material would be incredibly strong, but in practice this is impossible to achieve, you can only approach it.


Why is there a need of polar coordinates to solve the Schrödinger wave equation for the hydrogen atom? Why is there a need of polar coordinate to solve the Schrödinger wave equation for the hydrogen atom?
I went through some standard text books but I am feeling rather confused about the explanation.

Tl;DR

When solving the Schrödinger equation for the hydrogen atom, its potential energy is dependent on $r$. Using Cartesian co-ordinates $r$ is expressed as $\sqrt{x^2+y^2+z^2}$ which greatly complicates the process of solving the equation. Therefore we use polar co-ordinates as the variables are$~r,\theta,\psi$ which means that there is no nasty square root sign as we don't have to split $r$ into terms of $x,y,z$.

Long Version

This is because to solve the Schrödinger wave equation for the hydrogen atom you require to use the technique known as, separation of variables. Separation of variables is a method for solving partial differential equations. For a partial differential equation in a function $\Phi~(x, y~...)$ and variables $x, y,~...$ we assume that it can be factored into functions which is function of only one variable:
$$\Phi~(x, y,\dots) \equiv X(x)~Y(y)~\dots$$

By breaking the resulting equation into a set of independent ordinary differential equations, we are able to easily solve for $X~(x), Y~(y), ~...$ and then plug them back into the equation to get our original function, $\Phi~(x, y~...)$. For example, lets consider this differential equation of
$$\frac{\partial^2 U(x,y)}{\partial x^2} - \frac{\partial U(x,y)}{\partial y} = 0$$

To solve this partial differential equation we use the separation of variables method. We assume that $U(x,y)$ can be written as $X(x)~Y(y)$. Therefore the equation becomes:
$$\frac{\partial^2 (XY)}{\partial x^2} = \frac{\partial (XY)}{\partial y} $$

Now, the beauty of this is that $X$ and $Y$ are independent of each other meaning that $X$ is not a function of $Y$ and $Y$ is not a function of $X$. Therefore we can rearrange that equation to give:
$$\frac{Y~\partial^2 X}{\partial x^2} = \frac{X~\partial Y}{\partial y} $$

The equation can be rearranged to give:
$$\frac{X''}{X} = \frac{Y'}{Y}$$

Since each side of the equation is a constant with respect to the other since one is a function of x and the other is a function of y, we can write:
\begin{align}\frac{X''}{X} &= C, &\frac{Y'}{Y} &= C\end{align}

Now each differential equation can be solved independently of other to obtain $X(x)$ and $Y(y)$. From that, we can easily find $U(x,y)$.

So as you can see, for the method of separation of variables, it is crucial that you are able to split the variables apart so that you can solve each differential equation independently of other variables. This same technique can be applied to solve the Schrödinger equation for the particle in the $3$D box. However when we try to do this for the hydrogen atom we get a problem. The Schrödinger equation for the hydrogen atom using Cartesian co-ordinates looks like this:
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)\psi + \frac{2m}{\hbar^2}\left(\frac{e^2}{r}+E\right)\psi = 0 $$

This is a partial differential equation in three variables and we suspect that we need use separation of variables technique to solve it. So lets assume that $\psi(x,y,z) \equiv X(x)~Y(y)~Z(z)$. Now all we have to do is substitute $\psi$ with that and then separate the $x$, $y$ and $z$ variables so that they are together then solve them independently.

But there is one problem. In the equation, $r$ which is the radius is equal to $\sqrt{x^2+y^2+z^2}$ That is a problem because due to the square root sign there is no way we can separate the $x$, $y$ and $z$ variables.

Therefore to solve this equation more easily, we use polar co-ordinates. By using polar co-ordinates, we take advantage of its spherical symmetry which is convenient as the entire system also has a spherical symmetry. So we now transform the Cartesian Schrödinger equation to polar co-ordinates, so that it is dependent on the variables, $r, \theta, \phi$. The Schrödinger equation using polar co-ordinates is:
$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial}{\partial r}\psi\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial}{\partial \theta}\psi\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial \phi^2}\psi
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)\psi = 0
$$

This may seem really complicated at first, however with some manipulation, we will be able to separate out all three variables, $r, \theta, \phi$.

We assume that $\psi ~(r, \theta, \phi) \equiv R(r)~\Theta(\theta)~\Phi(\phi)$. For simplicity we will write them as $R, \Theta, \Phi$ without showing the functionality. The Schrödinger equation becomes:

$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ \Theta\ \Phi}{\partial r}\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial R\ \Theta\ \Phi}{\partial \theta}\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 R\ \Theta\ \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
This simplifies to:
$$
  \frac{\Theta\ \Phi}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{R\ \Phi}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{R\ \Theta}{r^2\sin^2\theta} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
Divide both sides by $R\ \Theta\ \Phi$ and multiply by $r^2\sin^2\theta$:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Inspection shows that no other terms except for the $3$ terms has a functional dependence on $\phi$. Therefore with respect to the other variables, the third term can be treated as a constant, which we will call $-n^2$ for convenience. So we obtain: $$\frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2} = -n^2$$ The general equation to this can be easily solved. So we have separated one of the three variables. Now the equation can be written in the form:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + -n^2
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Dividing by $\sin^2\theta$ and rearranging:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right)
   + \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta}
    = 0
$$
Now by inspection, the first two terms depend on $r$ while the last two terms depend on $\theta$. As before, we will set the two terms equal to a constant, $\beta$, so that:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right) = -\beta
$$
$$
     \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta} = -\beta
$$
Now we have successfully separated all 3 variables into 3 separate single variable differential equations that can be solved relatively easily.


What do the brackets mean in this inline structure representation? Based on the way the molecular formula is written — $\ce{Br(CH2)4Br}$ — I'm assuming this means that there are $4\ \ce{-CH2}$ groups attached to $\ce{Br}$.  Am I correct? I've always thought that $\ce{Br}$ can't accept so many groups though.

When you see a formula like the one in the question in organic chemistry, the person writing it down either already assumes a lot of chemical knowledge on the reader’s side or wants to teach them how to interpret this. So to a practising organic chemist, one inline formula may mean one thing while a similar one (with brackets in the same positions etc) may mean something completely different. Sometimes even, inline formulae are (and must be, due to their nature) potentially ambiguous. Therefore, organic chemistry relies a lot on actually drawing out the structures, whether it be each atom explicitly or in the sum formula formalism.

In the case of $\ce{Br(CH2)4Br}$, organic chemists will know that bromine will typically only form one single bond. Additionally, each $\ce{CH2}$ unit is known to be able to form two bonds — either two single bonds or one double bond. Therefore, an organic chemist will automatically expand the formula to $\ce{Br-CH2-CH2-CH2-CH2-Br}$ or 1,4-dibromobutane.

However, you might also sooner or later encounter inline formulae such as $\ce{BrC(CH3)2CH2Br}$ or $\ce{(CH3)3COH}$. Here, the brackets do actually mean repeated addition of said group to a single common atom. so the first of these two formulae describes 1,2-dibromo-2-methylpropane and could also be written $\ce{H3C-C(CH3)(Br)-CH2Br}$ while the second one represents tert-butanol or 2-methylpropan-2-ol.

Immediately realising which type of brackets is used requires practice and experience.


Why is it not possible to determine the rate constant this way? The following equation can be used to determine the rate:
$Rate = \frac{-\Delta [A]}{\Delta t}$

This equation can also be used:
$Rate = k[A]$

By simple substitution, $\frac{-\Delta [A]}{\Delta t} = k[A]$ should be true. Therefore, $k$ can be represented as:
$\frac{-\Delta [A]}{[A]\Delta t}$

However, calculating the rate constant for a given reaction is obviously not that simple. Why does this equation not work then, if the mathematics of it check out? I have tried using this equation for some problems, and as I expected, they do not yield the correct answer as opposed to using the integrated rate equation.

The average rate $\Delta[A]/\Delta t$ over time interval $\Delta t$ is only approximately equal to the true instantaneous rate $\rm d[A]/dt$ at time $t$. The difference between the two can be significant if the instantaneous rate changes over the chosen time interval.

Here's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.

To see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:

The larger the time interval, the larger the difference between $\Delta[A]/\Delta t$ and $\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.

So why not just use very small time intervals? Well, there's a limit to how small you can make $\Delta t$ if you're computing $\Delta[A]/\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals.


Determining reaction formula from optical absorption graphs The rate of a reaction between A + B was followed spectroscopically at three different wavelengths, and the percent absorption at each wavelength was plotted as a function of time.  Which of the following mechanisms is consistent with the spectral data?

  (a) $\ce{A + B -> C}$
  (b) $\ce{A + B -> D, D -> E}$
  (c) $\ce{A + B -> F, A + B -> G}$
  (d) $\ce{A + B -> H, H + A -> I}$
  (e) none of the above

The correct answer is (b), which I can believe, but not sure why.

So, it can't be (a), because in $\lambda_2$, the curve went up then down, indicating the presence of an intermediate. However, I'm not sure why this can't be (d) or (c) as well. As long as F, G, and H are active at $\lambda_2$, then wouldn't all of the reactions show a similar curve?

I'm not sure what to do with $\lambda_1$ and $\lambda_3$, as I think they just tell me that A + B is consumed, and that the product and/or intermediates aren't active at either $\lambda_1$ or $\lambda_3$.

Let's see. You are right in that it can't be (a). Now consider (c): A and B are consumed, while F and G accumulate. Where is that intermediate that goes up, then down?

As for (d), I believe there is no way we can tell that from (b) based on the picture alone, without looking at the numbers. See, suppose $\lambda_1$ actually measures B, and A is in huge excess; what would be the qualitative difference between the two? I can't see any.


Determining solubility of silver sulfate in its saturated solution Question:

  Calculate the molar sulfate concentration in a solution that is $0.200$ M in $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$.

My attempt:

Looking through old notes and googling lead me to this webpage. Example #2 looked similar to my problem, and my notes from general chemistry involved something called an ICE table to deal with stoichiometry and concentrations of ions. This is what I did:

Write the dissociation equation  $\ce{Ag2SO4 + <=> 2Ag+ + SO4^2-}$
Write the solubility constant formula $$K_\mathrm{sp} = \frac{\text{products}}{\text{reactants}}$$

I was given the value of $K_\mathrm{sp}$ as $1.6 \times 10^{-5}$ and I assume these values are listed in an appendix in my text or online.

Also $\ce{Ag2SO4}$ is supposed to be a solid, I think? Because the question states the solution is "saturated" with $\ce{Ag2SO4}$ and does "saturation" not imply that the $\ce{Ag2SO4}$ is going in and out of solution as a solid and as an ion? And therefore since it is a solid we do not consider it for the equation (not sure why, but this is similar to what I have in my notes)?

$$1.6 \times 10^{-5} = \text{products}$$

$$1.6 \times 10^{-5} = \ce{[Ag+]^2 * [SO4^2- ]}$$

Here is an image if that is acceptable of the ICE table I used (note: $\text{I+C=E}$).

$$\begin{array}{|c|c|c|}\hline
&\ce{Ag+}&\ce{SO4^2-}\\\hline
\text{I}&\pu{0.200M}&x\\\hline
\text{C}&\frac x2&0\\\hline
\text{E}&\pu{0.200M}+\frac x2&x\\\hline
\end{array}$$

Now I substitute the values I obtained for concentrations from the ICE table into my formula:

$$1.6 \times 10^{-5} = [0.200~\text{M} + \frac{1}{2}x]^2 \times [x]$$

Now since $K_\mathrm{sp}$ is less than $1 \times 10^{-5}$ we can ignore the $\frac{1}{2}x$? I guess because it is very small and therefore negligible?

$$1.6 \times 10^{-5} = [0.200~\text{M}]^2 \times [x]$$

$$ \frac{1.6 \times 10^{-5}}{[0.200~\text{M}]^2}  = x = 0.0004$$

$$ x = 4.0 \times 10^{-4}~\text{M}$$

Therefore the concentration of $\ce{SO4^2-}$ is $ 4.0 \times 10^{-4}~\text{M}$?

Solubility constants
You correctly state that they are defined as:
$$K_\mathrm{sp} = \frac{[\text{products}]}{[\text{rectants}]}$$
And in your case ($\ce{Ag2SO4}$) this would be:
$$K_\mathrm{sp} = \frac{[\ce{Ag+}]^2 \cdot [\ce{SO4^2-}]}{[\ce{Ag2SO4}]}$$
And concerning the reactand, the ‘concentration’ of a solid is always defined as $1$, so it is removed from the equation (technically this is an activity not a concentration but that doesn’t matter in this case).
The problem itself
What do you have? Let’s take another look at it:

Calculate the molar sulfate concentration in a solution that is $0.200\,\mathrm{M}$ in:
a) $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$

So we have dissolved $\ce{Ag+}$ with a concentration of $[\ce{Ag+}] = 0.200\,\mathrm{M}$ in aquaeous solution and also a solid $\ce{Ag2SO4}$ precipitate. You also state you are given the value of $K_\mathrm{sp} = 1.6 \times 10^{-5}$. And you are supposed to calculate $[\ce{SO4^2-}]$.
You have all the information you need and all you need to do is solve a single equation.


How to sloppy freeze a dihedral angle within a range in a Gaussian modredundant optimization? I would like to perform an optimization in Gaussian 09 of my molecule with two dihedral angles being frozen. I would like that the values of these dihedral angles are not fixed, but can relax within a range of ±5°. In Gaussian 03 it seems to be possible using Opt=ModRedundant keyword and the following syntax:

[Type] N1 [N2 [N3 [N4]]] [[+=]value] [A | F] [[min] max]]

I have tried using the same syntax in Gaussian 09, but it doesn't work. Is it possible to perform such calculation in Gaussian 09 or 16 without using the scan approach?

The calculation fails with the following error:

The following ModRedundant input section has been read:
Invalid extra data found.

TL;DR: It is impossible (=not documented if it were) with Gaussian 09 Rev. E.01 and older, it might be possible with Gaussian 16, but in all cases not via opt(modredundant).

I am unaware of any sloppy freezing possibility in pre-Gaussian 16, and for the latter one has to dig much deeper in the manual and start constructing it; but at least it works in theory.

Gaussian 03

Unfortunately, you have misinterpreted the commands you found in the manual. You can find the full description for the opt keyword in via The Internet Archive, but I'll share the important parts here for convenience.

The construct

[Type] N1 [N2 [N3 [N4]]] [[+=]value] [A | F] [[min] max]]

indeed exists for a opt(modredundant) input section, but it behaves not as a sloppy freeze, but as a range exclusion criterion, which you can find a little lower in the manual:

  [...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max then define a range (or maximum value if min is not given) for coordinate specifications containing wildcards. The action specified by the action code is taken only if the value of the coordinate is in the range.
  [...]

For example, the input

D * * * * F 175.0

freezes all defined dihedral angles that are smaller than 175.0°.

The input

D * 4  5 * F 160.0 175.0

freezes all defined dihedral angles around the bond between atom 4 and 5 in the interval 160.0 - 175.0°.

The input

D 3 4  5 6 F 160.0 175.0

is a syntax error because it does not contain a wildcard.

Gaussian 09

The option modredundant to the opt keyword still exists, but has a slightly modified input. Specifically the [[+=]value]] part has been removed. Again you can look up the full page on The Internet Archive.
The examples above should, however, still behave the same.

  [...]
  Lines in a ModRedundant input section use the following syntax:
      [Type] N1 [N2 [N3 [N4]]] [A | F] [[min] max]]
  [...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max only apply to coordinate specifications containing wildcards. The action then specified by the action code is taken only if the value of the coordinate is in the min–max range (or below maximum value if min is not given).
  [...]

Therefore you also still get the error "Invalid extra data found".

Gaussian 16

In this version, while opt(modredundant) is still maintained, a new feature is rolled out. As far as I can see, there are no changes to Gaussian 09 regarding the modredundant input section. You can find it live on the Gaussian homepage or a copy on The Internet Archive.

According to this, the above examples should still produce the same results.

The new feature is Generalized Internal Coordinates or GIC. See a description on the Gaussian homepage or on The Internet Archive. You can see the GIC in action in my answer on Constrain bond angle In Gaussian molecular structure optimization and on Gaussian: Relaxed scan with modredundant optimization and dummy atoms.

In theory you should be able to freeze a GIC if it meets certain criteria for example from the manual:

  The following example removes an angle coordinate generated by default if ≥179.9°, substituting a linear bend:

A(1,2,3) Remove Min=179.9        !  Remove angle coordinate if too large.
L(1,2,3,0,-1) Add IfNot A(1,2,3) !  Add linear bends only if
L(1,2,3,0,-2) Add IfNot A(1,2,3) !+ the angle coordinate not active.

I have prepared a small example, but I am really not sure if it worked as intended. For cis-butadien the dihedral distortions should be small, so there isn't much visible. There are no syntax errors, but I am not really sure whether the section actually did what it should. Try it yourself and investigate further. (! recognises comments)

#p PM6                ! Semiempirical method for fast testing
opt                   ! Optimise
geom(AddGIC)          ! Use GIC, can also be specified with opt(AddGIC)
scf(xqc,MaxCycle=250) ! Switch to quadratic conv. if necessary, more cycles
symmetry(none)        ! Do not use symmetry

distorted cis-Butadiene (semi-constrained)
optimisation run
[Free title card can have multiple lines]

0 1
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.089000
 H     1.026720     0.000000    -0.362996
 H    -0.513358    -0.889166    -0.363001
 C    -0.754249     1.306394    -0.533333
 H    -0.738680     2.222526     0.055206
 C    -1.531373     1.266770    -1.931370
 H    -2.044734     2.155933    -2.294370
 C    -1.615181    -0.098364    -2.761662
 H    -2.540288    -0.117779    -3.335874
 H    -0.765588    -0.163525    -3.439797
 H    -1.596709    -0.942933    -2.074433

sloppyfix=D(1,5,7,9)               ! Define an alias for the dihedral
sloppyfix freeze min=5.0 max=355.0 ! freeze in intervall

! Empty line above is important.

That the setup somewhat works can be seen with the following example of dihydrogen. First the atoms are separated 400 pm (=4.00 Å), and the value should be frozen at 250 pm (=2.50 Å), the PM6 optimised value is 76 pm (=0.76 Å).

#p PM6 opt(AddGIC,MaxStep=5) scf(xqc,MaxCycle=250) symmetry(none)

Free title card

0 1
 H    -2.000000     0.000000     0.000000
 H     2.000000     0.000000     0.000000

theHHbond=R(1,2)
theHHbond freeze max=2.5

! Empty line above is important.

Note that it will be frozen after the condition triggers, so it will not be able to the value that triggers it. Here are the values and messages you'll find in the optimisation run:

                           ----------------------------
                           !    Initial Parameters    !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              4.0            estimate D2E/DX2                !
 --------------------------------------------------------------------------------

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     4.97283  -0.03796   0.00000  -1.00000  -1.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.037962     0.000450     NO
 RMS     Force            0.037962     0.000300     NO
 Maximum Displacement     0.500000     0.001800     NO
 RMS     Displacement     0.707107     0.001200     NO
 Predicted change in Energy=-4.653046D-02
 Lowest energy point so far.  Saving SCF results.
 Internal coordinate with a value of  3.9728 a.u. has been conditionally frozen
 because it is in Min/Max range.
 The coordinate's label:
 theHHbond
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     3.97283  -0.07253   0.00000   0.00000   0.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.000000     0.000450     YES
 RMS     Force            0.000000     0.000300     YES
 Maximum Displacement     0.000000     0.001800     YES
 RMS     Displacement     0.000000     0.001200     YES
 Predicted change in Energy=-0.000000D+00
 Optimization completed.
    -- Stationary point found.
                           ----------------------------
                           !   Optimized Parameters   !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              2.1023         -DE/DX =   -0.0725              !
 --------------------------------------------------------------------------------
 Lowest energy point so far.  Saving SCF results.
 Largest change from initial coordinates is atom    1       0.949 Angstoms.
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

Note that in this example there is only one coordinate, ergo the optimisation finishes when the freeze condition is triggered. In this case there is quite a significant deviation from the condition max=2.5 and the actual value when it is frozen 2.1023. I would expect it to behave better with larger systems.

I would assume that the variable remains frozen for the remainder of the optimisation, with no more changes possible.
Standard disclaimer: Geometry optimisations with frozen parameters are critical for production purposes. Always check if you have a well converged molecular structure (stationary point) with no imaginary modes.


Sublimation of Iodine Why does iodine sublime? I have researched it myself and I have got the same answer; it sublimes because it directly converts from solid to gas. But why don't bromine or chlorine also sublime?

Iodine sublimes for the same reasons that all solids do: because it has some equilibrium vapor pressure an normal conditions. Now, the value of that pressure varies greatly in different solids. For many of them, it is so extremely low that for all practical reasons it can be considered non-existent. (Say, if you'd leave a sample of NaCl for a thousand years, I don't think you will be able to detect the weight loss due to sublimation.) For iodine, the vapor pressure is significant, so a sample left in the open will vanish completely in a matter of hours.

Some sources may specify that iodine goes directly from solid to gas when heated, never becoming a liquid. That's an urban legend; iodine does have the melting point and the boiling point at standard pressure, and between the two it will be liquid. It is not easy to demonstrate, though; if you heat it in the open, it will probably sublime away even before reaching the melting point, and if you do it in a closed container, the vapors get so thick you can't see a thing.

That being said, there are indeed some compounds that would never turn liquid at ambient pressure. That's because their triple point is above that ambient pressure. Carbon dioxide is one example; water on Mars is another.

As for chlorine and bromine, I don't really see the point of your question. At normal conditions they are not solid, so they can't undergo sublimation simply by the definition of it. If you cool them down enough to be solid, they will.


What is the structure of N₅P₃? What is the structure of the molecule $\ce{N5P3}$? $\ce{N5P3}$ is not documented online, so could anyone please comment on the structure of the molecule:

How many ligands does each phosphorous atom have?
How many ligands does each nitrogen atom have?

$\ce{N5P3}$ is more commonly written as $\ce{P3N5}$, and known as triphosphorus pentanitride. It's a crystalline solid at ambient conditions and not a molecular compound. From the first publication that reported the pure compound and its structure [1]:

In the solid a three-dimensional cross-linked network structure of corner sharing $\ce{PN4}$ tetrahedra has been identified with 2/5 of the nitrogen atoms bonded to three P atoms and 3/5 of the nitrogen atoms bonded to two P atoms.

Only the combination of spectroscopical and diffraction methods, especially electron microscopy for structural analysis, enables a detailed structural and crystallographic characterization of $\ce{P3N5}$ as no conventional single-crystal data for structure determination have been available for this compound.
A representative for an unique binary structure type was found for the first time which combines a tetrahedral structure with the very rare molar ratio of 3:5 for the constituent elements.

References

Schnick, W.; Lücke, J.; Krumeich, F. Chem. Mater. 1996, 8 (1), 281–286 DOI: 10.1021/cm950385y.


Why is it necessary to correct the pressure of a gas that is collected over water? Basically, what are some mistakes that you can make while using the downward displacement of water method and how to fix them? This is actually a homework question and I'm not too sure what it means either. The original question:

  State and explain Dalton’s Law of Partial Pressures. Why is it necessary to correct the pressure of a gas that is collected over water? {Background: Gases formed in a chemical reaction are collected using a technique called the downward displacement of water, feel free to research this technique before you respond to this question}

Generally, gases which are insoluble in water are collected by putting them into an inverted vessel via a tube, so that the gas rises to the top and 'displaces' the water downward. Take a look at the following picture—

Now, the problem is that water has a certain vapour pressure at any temperature. In other words, whenever water is exposed to a gaseous phase, it will start to evaporate until the partial pressure of the water vapour in the gas phase reaches a certain value (this value depends on the temperature). This value is known as the 'vapou pressure of water'.

In this case, the gas you collected by displacing the water will be contaminated by water vapour. Therefore, the pressure of the gas which you measure will be the sum of the partial pressure of the product gas, and partial pressure of water vapour.
$$\begin{align*}
Measured\;pressure &= Partial\;pressure\;of\;product\;+\;partial\;pressure\;of\;water\;vapour \\
&=Actual\;pressure + Vapour\;Pressure
\end{align*}$$ This means that you must correct the pressure, by subtracting the vapour pressure of water from the measured value to get the actual pressure of the produced gas.

Note: There are standard tables which list the value of vapour pressure of water at different temperatures. These tables are generally used for the correction.


How do dibromopyrazolones undergo base-promoted decomposition? I was not sure but seeing the base as NaOH I used the fact that it will take most acidic hydrogen that is N-H and electron pair will come over it and it will resonate from two sides. I could make upto this intermediate but next I am not really sure the ring will break up and give the product as option D.

Although a mechanism was not reported in Carpino's paper of 1958, there is a more concise mechanism than the one offered by @Soumik Das. There is no need to dissemble structure 1 to a tricarbonyl compound. Vinylogous elimination of the elements of HBr from dibromide 1 leads to species 2. [Alternatively, attack of hydroxide at the carbonyl of 1 would also suffice.] Attack of hydroxide at the carbonyl group of 2 produces 3, which via a Grob-like fragmentation affords the propargylic carboxylate 4 of the acid 5. ADDENDUM: The ring opening of 3 should not be construed as concerted. Just saving space.

Carpino, L. A., J. Am. Chem. Soc., 1958, 80, 599.


How do I write many cyclic voltammetry curves in Nova? We are making an electrochemical research on supercapacitors that involves checking stability of our material. That requires repeated writing of cyclic voltammetry curves. Our goal is to write at least 2000 of them, but there is a problem with Nova software (version 1.8, to be specific): if we want to write even 50 cycles, Nova complains that "only last 30000 point will be written". We can adjust how many points are plotted, but when it comes to 2000 cycles, it's impossible to extract any information from them (15 points per one curve is just ludicrous).
Has anyone dealt with such problem? Is there a simple solution?

Try to use command "repeat n-times", that will make n-data files and each can have 30 000 points.


Radical Bromination of a 1-isopropyl-4-methylcyclohexane I'm considering a problem whereby 1-isopropyl-4-methylcyclohexane with the indicated stereochemistry shown undergoes radical bromination at the indicated position. I understand there are other products produced. However, if the radical initiator (not shown) extracts the hydrogen at the 4 position, then the bromine radical attacks, there will be two distinct products. However, I am having trouble determining if these molecules are chiral? Are they diastereomers to each other?

Also, as a follow up, how would you name these (reagent + two products) to indicate their stereochemistry?

They are not chiral because $\ce{C}-4$ is only considered to be bonded to 3 unique groups because the ring is completely symmetrical through $\ce{C}-1$ and $\ce{C}-4$.

As for naming these compounds, the reagent is simple: cis-1-isopropyl-4-methylcyclohexane (or cis-1-(1-dimethylethyl)-4-methylcyclohexane if you're a stickler for the rules).

The products are a little tricker. Let's start with the basics: it is 1-bromo-4-isopropyl-1-methylcylohexane. Now to account for the steric relations of these substituents, label the reference substituent (labeled r) followed by a hyphen and the substituent locator number and name. The relative configuration of other substituents are then reported as cis (c) or trans (t) to the reference substituent.

This means that in the order that you present the products, their names are r-1-bromo-t-4-isopropyl-1-methylcylohexane and r-1-bromo-c-4-isopropyl-1-methylcylohexane.

[Source - (page 1060)]