How should elements be colored? And how many bonds does an atom have? I'm working on my own 3D molecule viewer for my AP Chem class. So far everything's going together nicely. There's a couple of things I'm not sure about though...

Is there a certain rule of thumb or standard for what colors each element inside a molecule should be? For example, Jmol renders oxygen red and hydrogen white:

For my app they are currently grey:

Also, how can I check if there are single bonds, double bonds, etc.? As you can see here, I haven't programmed that yet into my app. Tylenol:

I look up coordinates for each molecule in an NCI/CADD group server online database. This database gives me the number of elements, and the X, Y, and Z coordinates for where I should put each element. For example, water:

3
H2O
O   -0.0000   -0.0589   -0.0000
H   -0.8110    0.4677    0.0000
H    0.8110    0.4677    0.0000

It doesn't tell me anything about where bonds go. More info about .xyz file format here.

For the coloring part I would suggest you look at the CPK coloring convention.

As for the decision whether a bond is a single or a double bond: You could make this decision based on the bond length between two atoms. But you will need to find typical values for single, double and triple bonds for every combination of elements which seems a bit too much trouble. On the other hand, the depiction of double and triple bonds is used mostly for bonds between carbon, oxygen, sulfur, phosphorus and nitrogen (for other elements like metals the bond order is not readily identifiable, so I would leave those as single bonds). So, the amount of data to be collected might not be that large.

A very naive way to make the decision how many bonds an atom has could be the following (though this might not be very efficient):

calculate the distances between the atoms
for each element set a maximum distance another atom can be away from an atom of the element in question to be considered a bonding partner
for each element set a maximum number of bonds $N_{\mathrm{max}}$ an atom of this element can have, e.g. 4 for carbon, 4 for nitrogen, 3 for oxygen, etc. (for metals this might be a bit difficult to decide)
for each atom in the system order the list of atoms within bonding distance from lowest to highest distance. From this list choose only the first to $N_{\mathrm{max}}$th atoms as bonding partners. Then look whether some of the bonding partners qualify for double or triple bonds - if so you have to kick out some of the bonding members from the end of your bond candidates so that you don't exceed your maximum bond number.

There will be better ways out there to do this but this one worked for me in the process of creating "static" pictures of molecules.

Update:

You asked what size you should use for your atoms. Again, there is no definite answer to this - use whatever looks best for you. Nevertheless, I found that the sizes calculated according to the following formula give rather good looking atom size ratios for the different elements (maybe you have to multiply the resultant values by a constant to give good default sizes):

\begin{equation}
  \text{size} = \frac{1}{2 \sqrt{2}} \cdot \sqrt[6 \,]{\text{atomic number}} \ .
\end{equation}

Furthermore, here is a nice source for some typical bond distances. Other good resources would be the CRC Handbook of Chemistry and Physics or the book by Holleman and Wiberg if you can lay your hands on them. Those might come in handy if you try to adapt my method and look for values for the maximum bond distances. Another advice concerning the maximum bond distances would be that typical values for the bond lengths of bonds containing (transition) metals are between 200 and 300 pm, so a value of 350 or 400 pm might be a good choice for the maximum bond distance for those species.

Concerning the maximum bond number each element can have: here, the typical rules for drawing Lewis structures apply. You can find good resources for this on this site, e.g. here, and in every introductory general chemistry text. But you should be mindful of some special cases. You should not forget the cationic forms of the elements, when thinking about their maximum bond number. For example, carbon usually has 4 bonds, but there are very rare cases where it comes with 5 bonds ($\ce{CH5^{+}}$) although those are practically not important and some not so rare cases where it has 3 bonds (carbenes, carbocations, carbanions), so you might consider a maximum bond number of 5 for carbon, although for nearly all practical purposes 4 might do. Another example is oxygen. It usually has 2 bonds but when it is protonated or in Meerwein's reagent ($\ce{Me3O^{+}BF4^{-}}$) it has 3 bonds. Transition metals are very flexible when it comes to the number of bonds they have - it can be everything from 1 to 6 sometimes even more, so you should choose a high number, e.g. 8, for them.


Find the initial volume of the weak acid given a titration graph Question:

  Below is a picture of a titration between a weak acid, $\ce{HA}$, and a 0.150 M $\ce{NaOH}$ solution. Find the initial volume of the weak acid in milliliters in this titration.

My work so far:
From the graph, I think that the two most important points are (0,3.0) and (20.00,8.0) - I don't know the reason, but it looks important.
At (0,3.0), there is no $\ce{NaOH}$. This equilibrium takes place:
$$ \ce{HA_{(aq)} + H2O_{(l)} <=> A_{(aq)}- + H3O+_{(aq)} } $$

The constant $K_\text{a}$ can be found from:
$$
 K_\text{a} = \frac{[ \ce{H_3O^+}][\ce{A^-}]}{[\ce{HA}]}
$$
with $[\ce{H_3O^+}] = 10^{-3}$ since $\text{pH}=-\log_{10}[\ce{H_3O^+}]=3$
This is a dead end for me.
At (20.00, 8.0), $[\ce{H_3O^+}] =10^{-8}$, this number has decreased because some $\ce{H_3O+}$ have reacted with the newly dissolved $\ce{OH^-}$, although how much I can't tell. I have reached another dead end
Other facts I do know:

0.150 M of $\ce{NaOH}$ means for every 1 liter of the solution, there will be exactly 0.150 moles of $\ce{NaOH}$.
$\ce{NaOH}$ dissolves completely as a strong base. So, for every 1 mole of $\ce{NaOH}$ there will be 1 mole of $\ce{OH^-}$

Outline

There are three main points of interest in the titration graphic, and two you have found yourself. The three points are

The start at $V_\text{titr}=0$
Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$
The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$

Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.

Now, armed with this information, here is how to proceed:

Find out total amount of acid
Find out total concentration of acid
Calculate the initial volume from the data gathered in steps 1 and 2

1  Total amount of acid

The total amount of acid can be gleaned from information point 3, the equivalence point.

$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$

2  Total concentration of acid

The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning:
$$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$

Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species:
$$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$

The total initial concentration of all acid is then trivial:
$$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$

Note that we here actually calculated the activities and not concentrations. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration:
$$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$

3  Calculation of the initial volume

The initial volume is now calculated following the simple concentration formula:
$$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$

Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values:
$$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$


Why is a set of p-functions added to the minimum basis sets of Li and Be? The minimum basis set consists of the smallest number of functions necessary to contain the electrons of the neutral system. In the book Introduction to Computational Chemistry, Second Edition, by Frank Jensen, it is stated that "a set of p-functions" usually is added to the minimum basis sets of Li and Be atoms, despite them containing 3 and 4 electrons respectively and thus would require 2 s-functions. I can not, however, understand why.

By your definition, any minimal basis set would only include as many orbitals (including spin state) as there are electrons. This implies that each atom would be described as a single determinant and therefore the entire system would be described by a single determinant. Without unoccupied orbitals, there is no binding. Hence, a minimum basis set has to fulfill a few more constraints. A common requirement is the inclusion of the full valence shell and invariance under spin-flip. For second period elements, this includes p-orbitals.


What is the proof for saying all the compounds-of-an-element (except few), to have the same oxidation number? Following statements are extracted from Puri, Sharma, and Kalia's Principles Of Inorganic Chemistry book:

In all hydrogen compounds (except hydrides of active metals), the oxidation number of hydrogen per atom is +1.
In all oxygen compounds (except hydrogen peroxide and other peroxides and superoxides), the oxidation number of oxygen per atom is -2.

Have Chemists tested the oxidation state of every compound (of hydrogen, Oxygen or any element considered to have fixed oxidation sate) in the universe, to say all the compounds (with few mentioned exceptions) of the considered element to have the fixed oxidation state?
Is there any proof or law for supporting above statements? Or is the author sloppy in not saying "In all the known hydrogen/oxygen compounds"?

This statement is a generalisation. As every generalisation it makes things simpler to remember but even here they give exceptions. Of course, chemists couldn't test all oxygen or hydrogen compounds. The assumption is based on electronegativity of atoms in the "usual" compounds. In OF2 oxygen is not in -2 oxydation state. Also in BH3 hydrogen is not +1. I would not even go as far as "In all the known hydrogen/oxygen compounds". "In most of known" would be enough.


How are chemicals prilled? What does the process of prilling something like, say, calcium nitrate, involve? Do you need other chemicals to prill the substance?

I don't know the answer for sure, but will take a guess based on what I read at the link provided by Mithoron.

Non-chemists outside the laboratory, and persons who have to handle and work with those chemicals would certainly benefit from prilling to reduce dusting which could be detrimental to their health in breathing and on the mucous membranes.

Gardeners use Ammonium, potassium and calcium nitrates in fertilizers and stump removers, and do not always wear eye protection or dust masks. Thus prilling helps to mitigate exposure to these chemicals.


Computing force fields using Open Babel I am trying to use Open Babel (OB) to compute universal force field (UFF) given an xyz file. I read each xyz file and turn it into a pybel molecule. Here is the code I have so far

def xyz_to_smiles(fname: str) -> str:
    mol = next(pybel.readfile("xyz", fname))
    smi = mol.write(format="smi")

    return smi.split()[0].strip(), mol

def get_qm9_OB(target_index=np.arange(0, 133884)):

    smiles = []
    molecule = []
    for i in target_index:
        mol_idx = f"{i+1:06d}"

        tmp_smiles, tmp_molecule = xyz_to_smiles('../dsgdb9nsd.xyz/dsgdb9nsd_' + mol_idx + '.xyz')

        smiles.append(tmp_smiles)
        molecule.append(tmp_molecule)

    return molecule, smiles

molecules, smiles = get_qm9_OB(target_index=np.arange(100,200))

for i in range(len(smiles)):
    mol = molecules[i].OBMol

    ff = ob.OBForceField.FindForceField('UFF')
    ff.Setup(mol)

    ff.GetAtomTypes(mol)
    ff.GetCoordinates(mol)
    for at in ob.OBMolAtomIter(mol):
        type = at.GetData("FFAtomType") # an OBPairData object
        print(type.GetValue())

Now, I am trying to compute bond stretching interactions. My questions are:

How can I get the UFF constants for bond stretching? (IOW, is there an OB equivalent for RDKit function GetUFFBondStretchParams?)
Is there a function to calculate bond stretching interactions for each bond? (IOW, is there an OB implementation of either harmonic oscillator or the Morse function?)
How can I get coordinates from OBAtom object? I found GetCoordinate() but the output is

<Swig Object of type 'double *' at 0x1a1eb4a9f0>

Alternatively, GetVector() outputs

<openbabel.vector3; proxy of <Swig Object of type 'OpenBabel::vector3 *' at 0x1a1fbddbd0> >

How can I get the actual values?

P.S. I posted the problem on OB mailing list, with no answers.

Coordinates can be accessed using

[OBAtom.GetX(),OBAtom.GetY(),OBAtom.GetZ()]

look at this Github discussion.

If you have a better answer please don't hesitate to answer the question.


Why does density increases when adding nanoparticles in a mixture of water and some clay? I want to know that why density of drilling fluid which is a mixture of water, barrite, bentonite and carboxymethyl cellulose increases when i am adding silica nanoparticles. My ain of adding these nanoparticles is to improve the viscosity.

If your question is why density increases with the addition of nanoparticle then i have to recommend this quora post.

  If the attractive forces between the constituent particles of the
  solid with the water molecules are high enough, the total amount of
  intermolecular space is reduced and the total volume of the solution
  can be smaller that the combined volume of its components, leading to
  an extra increase in density....

Surface of silica nanoparticles have siloxane($Si-O-Si$) and silanol ($Si-OH$) groups. Because of the presence of these functional groups silica nanoparticles tend to form hydrogen bonds between themselves and/or with other polar molecules such as water. Its these interactions which reduces the intermoleculr space which further reduces the volume thus increase in density.

If the silica nanoparticles interect with themselves and form a linked three dimensional networks then it will also increase the viscosity of the solution.


How do I determine stereochemistry of the product When pure (L) lactic acid is esterified by racemic-2 butanol then what is the nature of the stereoisomers so obtained i.e. whether they are diastereomers or enantiomers.

I studied the mechanism of esterification but I didn't find anything regarding the stereo isomerism that can affect the formation of the product.Any ideas.Thanks.

The reaction doesn't change the stereo configuration for either reactant, so we can assume that the product is enantiomerically pure with respect to the carbon attached to the hydroxyl group (in orange below), and a racemic mixture with respect to the tertiary carbon (in blue).

Since there are multiple stereocenters, you can either have both the same (same molecule), both different (enantiomers), or have some different and some the same (diastereomer).  The products would be diastereomers.  In particular they would be epimers, which differ in only one stereocenter.


How to obtain sodium oxide from sodium chloride? Under what conditions can $\ce{Na2O}$ be made from $\ce{NaCl}?$

I know $\ce{NaCl}$ doesn't oxidise under normal ambient conditions, but in the presence of what temperature and pressure ranges is this reaction possible?

There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I).
In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride.

Method 1

Electrolysis of molten sodium chloride:
$$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$
Oxidation of sodium metal to oxide by burning:
$$\ce{4 Na + O2 ->[>\pu{250 °C}] 2 Na2O}$$
Drawback: pure sodium oxide cannot be obtained by direct oxidation of sodium.
Instead, a mixture of sodium peroxide and sodium oxide is formed.
In order to suppress the formation of peroxide, sodium metal or sodium nitrate is added in excess to the mixture in inert atmosphere:
$$\ce{Na2O2(s) + 2 Na(l) ->[\pu{150 °C}] 2 Na2O(s)}$$

Method 2

Convert sodium chloride to sodium bicarbonate using Solvay process:
$$\ce{NaCl(aq, conc) + H2O(l) + NH3(g) + CO2(g) -> NaHCO3(s) + NH4Cl(aq)}$$
Thermal decomposition of bicarbonate first yields in sodium carbonate:
$$\ce{2 NaHCO3(s) ->[\pu{250 - 300 °C}] Na2CO3(s) + CO2(g) + H2O(g)}$$
which is subsequently calcined to form an oxide:
$$\ce{Na2CO3(l) ->[>\pu{1000 °C}] Na2O(s) + CO2(g)}$$
This appears to be a preferred method as it is less energy- and resources-consuming and allows to obtain $\ce{Na2O}$ selectively.


More effective fire extinguisher liquid than water I join robot fire fighting contest.  One of the mission is to extinguish candle by spraying it with liquid (water is default option). For spray mechanism I use pump with Venturi effect principle.  Is there any liquid substance that is more effective extinguishing fire than water by Venturi spraying mechanism? Homemade is preferred.

You could consider an aqueous film forming foam (AFFF) system, this is water with some additives to form a foam. Originally foam was something intended for class B fires, but there are now foams which are intended for both class A and class B fires. They are even considered to be acceptable for use on gas fueled fires where there is no risk of explosion (class C). Thus there are ABC foams.

A word of warning some ABC foams such as those based on PFOS are horrible and harmful. It might be best if you stick to a water spray system. The PFOS foams are considered to be problem when they enter the environment, if I recall correctly they are very harmful to fish and the perfluoro compound PFOS only breaks down very slowly in the environment.


Oxygen generator, heat driven, from air What chemical can be used to absorb $\ce{O2}$ from the air at normal temperature and pressure, then release the $\ce{O2}$ at normal pressure and elevated temperature, below $\pu{900 ^\circ C}$ ($\pu{1650 ^\circ F}$), that can be cycled many times?
Closest solutions I've found are silver, but its quite expensive, and absorbs oxygen slowly at normal temperature. And possibly phosphorous, but it is toxic and needs a solution.
Same question about $\ce{N2}$. I'm interested in separating $\ce{N2}$ and $\ce{O2}$ without the use of single-use chemicals or pressure pumps. Using cycled heat as a source to drive the reaction.
I expect there is some oxide that can switch oxidation levels back and forth with temperature, like hematite and magnetite.
Idea solution will be something like $\ce{CaCO3}$ is for $\ce{CO2}$. A cheap material, that quickly reacts with $\ce{CO2}$ at normal pressure and temperature. Quickly releases the $\ce{CO2}$ at elevated temperature and normal pressure.
I did check the search, it seems people usually offer to use electrolysis for this task. But water is way too costly to electrolyze in terms of power. Is there something that can be electrolyzed with $10$ times less energy, to get oxygen, that can be safely oxidized again later? Hydrogen is rather dangerous too.

The only substance that absorbs $\ce{O2}$ to form an oxide at low temperature and release it at higher temperature is mercury. This experience was done by Lavoisier, Scheele and Priestley in the years around $1780$. They were able to produce pure oxygen $\ce{O2}$ by first heating pure mercury in air at about $300$°C to slowly produce mercury oxide $\ce{HgO}$, and then overheating this oxide $\ce{HgO}$ to about $400$°C. At this temperature, $\ce{HgO}$ is decomposed into mercury and pure oxygen $\ce{O2}$. This was the way oxygen was first discovered and produced.
Unfortunately, mercury ¡s unique in this respect, and cannot be replaced by another element. No metals have been found later on to have a similar property. It is a pity : presently this experiment cannot be repeated, because of the toxicity of mercury and its compounds.
This why your dream will remain at the state of dream.


Is CO2 produced when KHCO3 reacts with H2O? We have been doing experiments on the rate of photosynthesis to find out the importance of $\ce{CO2}$ and we used $\ce{KHCO3}$ in one beaker and found that the plant kept in that beaker produced greater amount of oxygen that those without it. Sources say that $\ce{KHCO3}$ acts as a source of $\ce{CO2}$ but I'm not sure of it. What is the reaction taking place to produce $\ce{CO2}$?

Compiling the comments by Max and orthocresol:

  What is the reaction taking place to produce CO2?

$$\ce{HCO3− (aq) + H2O (l)
<=> H2CO3 (aq) + OH− (aq)
<=> H2O (l) + CO2 (aq) + OH−(aq)}$$

or
\begin{align}
\ce{HCO3− (aq) + H2O (l) &<=> H2CO3 (aq) + OH−}\\
\ce{H2CO3 (aq) &<=> H2O (l) + CO2 (aq)}
\end{align}

  $\ce{KHCO3}$ does not readily decompose into $\ce{CO2}$ when dissolved in water.
  It is just that the plant is using the $\ce{CO2}$, so that pulls the equilibrium over, Le Chatelier's principle.


Why Cu looses 2 electrons to become stable instead of gaining one electron to fill its 4s2 3d10 configuration? Cu has 4s1 3d10 configuration so why it does not gain 1 electron and complete its suborbital. Why it looses its 2 electrons to become Cu++ which is more stable. Why?

Copper can gain 1 electron to become Cu-.

Copper has an electron affinity of 1.236 eV, so Cu- is stable to electron loss by 1.236 eV.

See also "The Dissocationi Energy and Heat of Formation
of the Molecule NaCu" Z. Naturforsch. 28a, 316-317 which says:

  But unlike as for NaAg a noticable ionic contribution
  to the bonding is observed for NaCu

Also there is Resonant two-photon ionization spectroscopy of LiCu :

  The unusual shape of this curve is thought to derive from avoided
  crossings between the $\ce{Li^+Cu^-}$ ion pair state and covalent states, with the $\ce{Li^+Cu^-}$ ion pair state
  ultimately correlating to the ground electronic state of the molecule.

So if Cu is to make a compound with an electronegative element, Cu should lose electrons, but with an electropositive element it can gain an electron.


Can molarity ever exceed molality Work so far:
Molality = mol solute/kg solvent
Molarity = mol solute/L solution

So for molarity to exceed molality in the same solution, the numerical value of
kg of solvent > L of solution

From my experience this never happens. However by applying this to a very dense solvent, could the mass of solvent actually exceed the combined volume of solvent and solute. I am aware that volumes are not additive, and so if the density of a solvent is 10 g/mL the density 10 times of water and 0.5 M 2 L of HCl is added to 4 L of solvent, mass of solvent would be 40 kg solvent could the volume actually be greater then 40 L? Is there an example of a physical system where molality is lower then molarity?

There is one possibility with glycerol With density 1.261 g/cm 3.Here the exact numerical value of mass is greater than that of volume of glycerol in kg and litres respectively.Moreover glycerol is soluble in water and vice versa.


Cobalt electroplating over nickel-plated steel From my research, I understand that cobalt can be used on top of a nickel layer to improve durability of the nickel plating on a steel object.
I am using a nickel anode and nickel acetate as electrolyte.

Would I be able to use a cobalt anode and cobalt acetate as electrolyte?
If I buy $\ce{CoSO4 * 7H2O}$ but don't have a cobalt anode, can I use a stainless steel or titanium anode?

in the past cobalt was used as a brighterner for Nickel plating electrolyites, now since Cobalt is about 5 times more expensive than Nickel, is hardly used but you can add about 30 grams per litter of CoSO4 to your regular nickel plating solution and you will end up with a Nickel/Cobalt alloy that is stronger and whiter than plain Nickel, also keep in mind that in these type of baths Cobalt plates out faster than Nickel, that means that a 10% Cobalt Sulfate(90% Nickel Sulfate) plating bath will produce about a 70% Nickel/ 30% Cobalt alloy plating.


In the gold foil experiment why didn't thallium form? When alpha particles hit the gold nucleus shouldn't they form thallium.
More over shouldn't it also ionize the gold or any atom it hits how does it remain unchanged unionized?

The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of course, folks rapidly wanted both other energies (or controlled energies) or other charged particles (such as the protons used by Cockroft and Walton for their first accelerator measurements).
These days, much higher energy particles are available, and are commonly used in nuclear physics. A nice recent paper on the topic is in Physics Review C, "Cross section of $\alpha$-induced reactions on $^{197}$Au at sub-Coulomb energies". Here, the authors measure the reaction cross-sections for $^{197}$Au($\alpha$,$\gamma$)$^{201}$Tl, $^{197}$Au($\alpha$,n)$^{200}$Tl, and $^{197}$Au($\alpha$,2n)$^{199}$Tl. Although mainly of interest to astrophysical heavy-element nucleosynthesis, the results bear on this question.
The bottom line is that high (> 13MeV) $\alpha$ energies are needed to drive these reactions. At the lowest energy used (~ 13.7 MeV), the cross-section for the $^{197}$Au($\alpha$,n)$^{200}$Tl reaction is 0.67 $\mu$b (micro barns). At 20 MeV it has increased to 37 mb (milli barns).
Anyway, it is quite clear that these reactions do not occur to any measurable extent in the few MeV regime.


Most useful reagent for gravimetric analysis of silver My question reads as follows:

"What's the most useful reagent for the gravimetric analysis of silver from the options below:
A. Sodium nitrate
B. Barium carbonate
C. Potassium sulphate
D. Ammonium chloride"

I can rule out A and C as they do not form a silver precipitate with a solution containing silver. However, both B and D can form a silver precipitate so I am not sure which of them would be the most useful.

You are missing something: before you even put the silver in, NH4Cl is soluble, while BaCO3 has already precipitated (insoluble)!

This means that Cl- will precipitate more completely with NH4Cl.  Therefore, D.


Detailed kinetic explanation for vapor pressure reduction by dissolved solute? The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

An open beaker of water in equilibrium with water vapor is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:
(A) leaves the solution increases
(B) leaves the solution decreases
(C) leaves the vapor increases
(D) leaves the vapor decreases

According to the NCERT for Class XII, Part I, pg. 46, para 3,

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If anon-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.

So, after reading the last lines of above quoted text, I think that the vapor pressure is reduced, because the rate at which solvent molecules leaves the solution decreases, due to a decrease in exposed surface area. So, acc. to me option (B) should be correct. But, it was incorrect, acc. to the key.
Is there anything else, I'm missing?

The answer given is,

 (C) leaves the vapor increases

I'd like to specifically commment on this:
According to the NCERT for Class XII, Part I, [pg. 46, para 3][1],

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.
[emphasis mine]

NCERT's explanation for why vapor pressure is lowered due to the presence of a dissolved solute is incorrect, because vapor pressure is independent of the surface area accesible to the solvent (assuming it's non-zero).  Consider these two examples:

You have two sealed containers, both of which contain an open beaker of identical solvent.  In container A, the beaker is low and wide.  In container B, the beaker is tall and narrow.  The surface area of solvent in container B is thus smaller.  Yet, assuming the conditions in the respective containers are identical, the equilibrium vapor pressures will be the same.

You have two sealed containers, both of which contain an open beaker of identical solvent.  The beakers are identical, so the surface area of liquid is the same. The conditions in the containers are also identical.  Glucose is dissolved into the solvent in container A, while sucrose is dissolved into the solvent in container B.  The final concentrations are the same.  Sucrose is larger than glucose.  Consequently, the fraction of the surface area covered by solvent in container A is larger than in container B.  Yet, ignoring non-ideality, the vapor pressures are the same; they depend only on the concentration of solute, not its nature.

The surface area only matters for kinetics, i.e., for how quickly the solvent can escape the container to reach equilibrium.  Clearly, with a larger surface area, the equilibrium vapor pressure will be reached more rapidly.
Feel free to send an email to NCERT with a link to this comment.
Here's the thermodynamic explanation: The addition of solute lowers the chemical potential of the solvent, due to the entropy of mixing.  [There can also be energetic effects, in either direction, but with a soluble solute the entropy term is dominant.]
At equilibrium, the chemical potential of the solvent in the liquid and vapor phases must be equal.  Thus, since the chemical potential of the solvent in the liquid phase has lowered, that of the the solvent in the vapor phase must lower as well.
And since the chemical potential of a gas increases with its partial pressure, the vapor pressure of the solvent in the gas phase will decrease until its chemical potential reaches the new (lowered) chemical potential of the solvent in the liquid phase.
As far as the question itself goes, I'm uncertain, because the question is asking for a kinetic, rather than a thermodynamic, explanation.  But I can offer two alternative pictures:
Let's represent the movement between the liquid and vapor phases of the solvent ("X"), as follows:
$$\ce{X_{(l)}<=>X_{(g)}}$$
Picture (I): Both "B" and "C" are correct.
Let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase.  This shifts the reaction to the left, which means that the rate of the forward reaction initially decreases (which is answer "B") and the rate of the backward reaction initially increases (which is answer "C").  Eventually the system reaches its new equilibrium point, at which point the rates of the forward and reverse reactions are again equal.
Picture (II): "B" is correct.
Again, let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase. The rate at which gas enters the liquid phase depends only on the concentration of gas, so the backwards reaction will be unaffected by the presence of the solute in the liquid.  However, the rate of the forward reaction will be reduced, since it is now more favorable to for the solvent to remain in the liquid state.  Hence the answer is "B".
As I mention in the Comments:
In Picture (I), the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute.
But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then Picture (I) would not apply.
I'm afraid I don't know enough about the microscopic kinetics to have certainty on this.  Perhaps someone that has done simulation work on this could provide an answer.  [But to get this, you would want to vote to reopen the question.]


Molecular geometry of BaH2 according to VSEPR theory I am asked to predict the correct molecular geometry of $\ce{BaH2}$ according to VSEPR theory.
I assumed that it is linear due to barium being an alkaline earth metal and therefore not having any lone pairs when bonded to two hydrogen atoms, but apparently the bond angle between the two hydrogen atoms is about 119°.
Why is that the case and how can one explain this observation with VSEPR theory?
$\hspace{3cm}$

There is a d-Orbital-Contributions to HOMO in d0 MX2-Complexes.
All higher‐level calculations show SrH2 and BaH2 to be bent with angles of ∼140° and 120°, respectively, while CaH2 is linear with a flat potential‐energy surface for the bending motion.
The use of a core‐polarization potential together with the 2‐valence‐electron pseudopotential approach allows an investigation of the relative importance of core‐polarization vs direct d‐orbital bonding participation as reasons for the bent structures. The calculations strongly suggest that both contribute to the bending in SrH2 and BaH2. Thus, the bent structure of [BaH2] may be considered to arise from a stabilization of the b2 HOMO upon bending by mixing in metal dyz character in second-order perturbation theory.
Reference:
https://core.ac.uk/download/pdf/232777714.pdf
http://www.chimdocet-inorganica.it/SITO_ESERCIZI/Complementi/COMP1/VSEPREccezioni.pdf
This observation has been rationalized successfully by an increased involvement of (n - 1)d-orbitals in the bonding as M gets larger,19-21
and by the influence of core-polarization interactions between the M and X sites.2 Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs.


Connection between irreducible representations and electronic states in diatomic molecules I am trying to understand the connection between irreducible representations and molecular electronic states.
To explain the problem: I found some potential energy curves (PECs) for $\ce{N2+}$ in the form of irreducible representations such as $\mathrm{A_g}$, $\mathrm{A_u}$, $\mathrm{B_{2g}}$, ...

However, in the literature, I normally find the PECs in the form of $\Sigma$, $\Pi$, $\Delta$... states. How do we connect these two pictures? I am not an expert in quantum chemistry calculations. A simple explanation will be highly helpful.

In the linear $C_{\infty \mathrm{v}}$ and $D_{\infty \mathrm{h}}$ point groups there are two notations for the irreducible representations which are equivalent, in that

$$\begin{align}
\mathrm{A_1} &\equiv \Sigma^+ \\
\mathrm{A_2} &\equiv \Sigma^- \\
\mathrm{E_1} &\equiv \Pi \\
\mathrm{E_2} &\equiv \Delta \\
\mathrm{E_3} &\equiv \Phi \\
&\,\,\vdots
\end{align}$$

For $C_{\infty \mathrm{v}}$, these are the actual irreps; for $D_{\infty \mathrm{h}}$ you simply need to tack on a g or u label to denote the symmetry with respect to inversion. See e.g. the $D_{\infty \mathrm{h}}$ character table found at http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=1001&option=4 which contains both notations.


Why do we need the identity operator, E In group theory, in order to assign a point group, one must first identify the symmetry elements present in the molecule.

To give an example, water has:

a rotational axis, labelled C2 since it is a two fold axis, with a π/2 rotation giving back (what appears to be)the same thing
two mirror planes, labelled σv and σv' (with additional labels xy/xz to define what plane the mirror plane lies in)

Symmetry elements present in water

(Note that for the C2v point group, the axis are defined 'arbitrarily', different tables/books might swap x and y)

From this, it would look like water has three unique symmetry elements, however this is not the case, in fact, water has four due to the presence of an additional symmetry element, E, known as the identity operator (sometimes seen as I in certain textbooks).

This E operator corresponds to 'do nothing' (i.e. leave atoms where they are), which is easily seen in the character table for the C2v point group (to which water belongs) in which the characters under E are all '1':

$$\begin{array}{c|cccc|cc} \hline
C_\mathrm{2v} & \color{red}{E} & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline
\mathrm{A_1} & \color{red}{1} & 1 & 1 & 1 & z & x^2, y^2, z^2 \\
\mathrm{A_2} & \color{red}{1} & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_1} & \color{red}{1} & -1 & 1 & -1 & x, R_y & xz \\
\mathrm{B_2} & \color{red}{1} & -1 & -1 & 1 & y, R_x & yz \\ \hline
\end{array}$$

$\,$

C2v character table, from orthocresol's Group Theory Tables

  The identity, E, consists of doing nothing; the corresponding symmetry element is the entire object. Because every molecule is indistinguishable from itself if nothing is done to it, every object possesses at least the identity element. (Taken from Atkins' Physical Chemistry).

Since transforming a molecule by the identity operator just gives back itself, the use of 'E' in a qualitative sense appears redundant, however it appears in every character table, and as such much have a purpose.

Why do we need the identity operator, E, and what use does it have in chemical group-theory?

It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory.

I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group, reprove all theorems, etc.) without explicitly using $\rm E$. That would be a titanic amount of work, though, and quite pointless at that. Also, many things like "order of a group is divisible by..." would start to sound less natural.

Come to think of it, why do we need the number $0$ in math? It just sits there and does nothing, after all. Well, doing nothing may be quite important: it makes a lot of other things simpler.


Optimization and DFT-D3 energy calculation in GamessFireFly version 8.1.1 My input was

$CONTRL SCFTYP=RHF DFTTYP=B2PLYP DFTD=.T. d5=.T. RUNTYP=OPTIMIZE $END
$DFTD VERSN=3 $END
$SYSTEM TIMLIM=3000 MEMORY=30000000 $END
$BASIS EXTFIL=.T.  GBASIS=acc-pvdz $END
$STATPT HSSEND=.T. NPRT=-2 $END
$FORCE NVIB=2 VIBSIZ=0.005 $END
$DATA
 H2O
C1
O  8.0                   -3.46278   2.54045   0.
H  1.0                  -2.50278   2.54045   0.
H  1.0                  -3.78324   3.44539   0.
 $END

I am getting the following error:
FATAL ERROR: DFTTYP = B2PLYP   DOES NOT YET HAVE GRADIENTS PROGRAMMED.

So is it not possible to perform optimization with this functional?

In original GAMESS-US code there are no analytic gradients for "double hybrids" and it looks like it is true for Firefly as well. You can use numerical gradients by adding NUMDER=.T. in $CONTRL section. See this for details.


How can a chemical structure be stored in a matrix? How do I represent a chemical structure in a matrix format in Matlab? Can I input a chemical structure into Matlab using the S.M.I.L.E.S notation?

I have to specify the connectivity of different atoms with others and type of bonds between them. I have to write the structure in matrix form, so that it can be utilized in some linear programming. For example, if one atom is connected to another atom I have to store 1 in matrix otherwise 0 (binary variables).
I need to develop back end code for my purpose to design some molecules using this approach. As of now I am dealing with 2D structures.

Are the use of MATLAB (or OCTAVE) and a self-designed molecular representation mandatory for you project?

Established 2D- and 3D representations for molecules beyond a connectivity matrix, i.e. with coordinates, (partial) charges, annotation of isotopes, etc. exist, such as in the form of the Chemical Markup Language (CML).

In addition, well-documented toolkits, frameworks and modules are available to operate on these data are available for C++, Java, Perl, Python, and Ruby.


How polarization effects from cations affect H-bond strength The average hydrogen-bond strength in pure water is higher than the average H-bond strength between the first and second hydration shells around a $\ce{Al^3+}$ ion (or any cation, for that matter). This leads to a shift in the OH stretch toward the lower frequencies.

This has something to do with the induced polarization resulting from the positive charge, no doubt, but I fail get why the H-bond becomes weaker. The H-bond would get stronger if the electron density around the H-atom participating in the H-bond gets smaller (larger coulombic attraction between H and O). When $\ce{Al^3+}$ approaches the O-atom of a water molecule, electron density will be forced toward the positive charge, ultimately affecting the H-bonded H-atom, which would give stronger H-bond, at least in my head.

So what's wrong with this picture?

Okay, I realized what was wrong with my picture above.

It is not the hydrogen bond that is weakened, but the OH bond within a water molecule that is weakened, and then stretches at a lower frequency.
When the cation is introduced, and is near the oxygen atom, the electron cloud in the water molecule is moved toward the cation (polarised), leading to electron density in the OH bond being reduced => weaker bond => lower frequency vibration.

The hydrogen bond is not really of interest here, although the same effect takes place when two water molecules are H-bonded. Hence, the OH stretch of an isolated water molecule will vibrate a bit faster than the OH stretch in a H-bonded water molecule.


Quantity of water in a redox problem $\pu{76 g}$ of $\ce{FeSO4}$ are oxidized by $\pu{0.04 M}$ solution of $\ce{KMnO4}$ in presence of $80\%$ $\ce{H2SO4}$. What quantity of water must be added to the final solution of $\ce{Fe2(SO4)3}$ for the concentration to be $\pu{0.25 M}$?

I wrote the reaction:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

But I have no idea how to find the amount of water. I found the quantity of $\ce{Fe2(SO4)3}$ to be $\pu{100 g}$, but I have no idea to find all the water.

To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:

$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$

where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.

From the balanced redox reaction, which it seems you have established correctly:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:

$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$

$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$

$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$

From (1):

$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$

Combining (2–4) into (5), and taking into account that

$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$

one can obtain the following expression for the volume of water to be added:

$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$

The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:

$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$


Mathematics in inorganic and quantum chemistry Is mathematics important to learn concepts in inorganic chemistry, such as point groups, point symmetry, because doing these topics requires one to determine the symmetry of the elements and to visualize 3D (such as determining the symmetry of a water molecule?)

As these concepts are derived from mathematics, I wonder whether taking maths courses would be helpful for me to understand these concepts better and if so which maths course would be most useful? Such as linear algebra, abstract algebra?

For better understanding of molecular symmetry you indeed need some mathematics, namely, some basics of group theory, but aside from that I doubt that you need some mathematics apart form arithmetic and calculus for general, inorganic, organic, and even physical chemistry.

For quantum mechanics you absolutely need to familiarize yourself with the concept of complex numbers and you have to learn some basic approaches for solving differential equations. For an advanced insight into quantum chemistry you need some linear algebra, by which I mean the theory of vector spaces both finite- and infinite-dimensional, not just finite-dimensional part of it, which, in a sense, reduces to matrix algebra.

P.S. Abstract algebra is nothing but a common name for different areas of mathematics which studies different so-called algebraic structures. Say, group theory studies groups, linear algebra studies vector spaces, etc.


Why do strong acid-weak base reactions "essentially go to completion"? This is what my textbook says. I understand that the strong acid would readily donate a hydrogen ion, but I can't see how all the weak base will readily accept it (at least not to "essential completion").

This is a matter of equilibrium. Let's consider the  acid-base reaction of $\ce{HCl}$ and $\ce{NH3}$:

$$\ce{NH3 + HCl <=> NH4+ + Cl-}$$

Which can be broken apart into:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

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

For these reactions at equilibrium:

$$K_\mathrm{eq}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \hspace{1cm} K_\mathrm{b}(\ce{NH3})=\ce{\frac{[NH4+][OH-]}{[NH3]}} \hspace{1cm} K_\mathrm{a}(\ce{HCl})=\ce{\frac{[H3O+][Cl-]}{[HCl]}} $$

$$K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) = \ce{\frac{[H3O+][Cl-]}{[HCl]}} \cdot \ce{\frac{[NH4+][OH-]}{[NH3]}}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \cdot \ce{[H3O+][OH-]} = K_\mathrm{w}K_\mathrm{eq}$$

$$K_\mathrm{w}K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) \Rightarrow$$

$$K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot \frac{K_\mathrm{b}(\ce{NH3})}{K_\mathrm{w}}=\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \Rightarrow$$

$$\mathrm{p}K_\mathrm{eq} = -\log \left(\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \right) = \log \left(\frac{K_\mathrm{a}(\ce{NH4+})}{K_\mathrm{a}(\ce{HCl})} \right) = \mathrm{p}K_\mathrm{a}(\ce{NH4+}) - \mathrm{p}K_\mathrm{a}(\ce{HCl})=9.24-(-8)=17.24$$

Just from the $\mathrm{p}K_\mathrm{a}$ difference of the acid and conjugate acid of our reaction, we figured out that at equilibrium the products are present in a $10^{17.24}:1$ ratio to the reactants, meaning the reaction has essentially gone to completion.


Rotovibrational selection rules for symmetric tops The selection rules for symmetric top molecules are

$\Delta J = 0, \pm1$ and $\Delta K = 0$ for $K \neq 0$ (parallel transition)

$\Delta J = \pm1$ and $\Delta K = 0$ for $K = 0$ (parallel transition)

$\Delta J = 0, \pm1$ and $\Delta K = \pm1$ for $K = 0$ (perpendical transition)

My question concerns the possibility of a $\Delta J=0$ transition. For example, the only way how $\Delta J =0$ is possible for linear molecules is with degenerate normal modes (e.g. $\ce{CO2}$ bend), as these can "evolve" into one another and give rise to vibrational angular momentum, which ensures that conservation of angular momentum can still hold regardless of $\Delta J$.

The fact that $\Delta J =0$ is allowed for all transitions of symmetric tops means that there is vibrational angular momentum associated with all of them. My questions are:

What is the origin of vibrational angular momentum if the normal modes involved are not degenerate? Can you help me visualise it, for example for the $\ce{C-Cl}$ stretch of $\ce{CH3Cl}$ (a parallel non-degenerate transition)?
Why is the selection rule different for $K=0$ and $K\neq0$?

Some of the statements you make above are not quite correct, rather that correct individual points I think that it is simpler to start from scratch, as it were, and try to explain what happens.

Two quantum numbers are needed to define the rotational energy of a symmetric top molecule, such as CH$_3$I or benzene. The total angular momentum is characterised by $J$ and this generally does not lie along the 'figure' axis of the molecule which is the C-I bond axis in CH$_3$I for example. This quantum number has integer values starting at zero, i.e. $0,\, 1,\,2\cdots$  The total angular momentum is $ \hbar\sqrt{J(J+1)}$. The second quantum number $K$ is the projection of $J$ along the figure axis and so has $2J+1$ values ranging from $-J$ to $+J$ and therefore has values $0,\,\pm 1,\, \pm 2, \cdots \pm J$. The size of the projection quantum number is $K\hbar$ and clearly as it is the projection along the figure axis $K$ cannot exceed $J$.

The energy levels of any vibrational level form stacks of $J$ levels rather like those of a rigid rotor but where each $K$ value has its own stack beginning with $J=K$. Each type of vibration has sets of $K$ and associated $J$ levels for each vibrational quantum number.

We are considering rotation-vibrational transitions. The selection rule for parallel transitions, i.e. those with a dipole moment along the figure axis always have $\Delta K=0$ and then $\Delta J = 0, \pm 1$ unless $K=0$ then transitions with $\Delta J=0 $ are forbidden. The reason that $\Delta K = 0$ is that radiation cannot induce transitions between different $K$ values because rotation about the figure axis induces no rotating dipole moment. (The dipole lies along the axis in this type of transition). When $K=0$ there is no angular momentum component along the figure axis and so if the total angular momentum is also zero there is no rotating dipole to couple to the radiation.

In the case of perpendicular transitions the dipole has a component perpendicular to the figure axis. Bending vibrations can induce perpendicular transitions. The selection rules are $\Delta K=\pm 1$ and $\Delta J = 0,\pm1$. In this case changing $K$ does induce a change in the dipole and so transitions can occur.


Energy of eclipsed and staggered conformations of ethane and propane Which points on the given energy diagram represents eclipsed confirmation of ethane and staggered confirmation of propane respectively?

Attempt: Eclipsed configuration is unstable because of repulsions between filled $\sigma$ or bonding orbital and the anti or staggered configuration is more stable due to interactions between filled $\sigma$ orbital and empty $\sigma^*$ orbital.
From the given diagram, either R or S could represent eclipsed confirmation of ethane and P or Q could represent staggered confirmation of propane.
Since they are different compounds, they should lie on different energy diagrams. So the answer could be (R,Q) or (S,P). Which one is correct?

That diagram is horribly drawn.  The way it is drawn, the 13.6 kJ/mol number is meaningless - it measures from the maximum of one curve to the minimum of the other curve.

I think that you are supposed to assume that for one of the molecules, the one for the lower curve, the torsional strain is 12.6 kJ/mol and for the other molecule the torsional strain is 13.6 kJ/mol.

Do you have some reason to think that the barrier to free rotation for propane would be higher or lower than for ethane?  Perhaps your book gives the value for the torsional strain of ethane/propane? (through a quick google search I was able to find the torsional strain for both of these)


Numerical simulation and calculations of pH a. Reaction of $\ce{CO2}$ addition in water is as follows:

$\ce{CO2}$ + $\ce{H2O}$ $\ce{->}$ $\ce{H2CO3}$

b. Then, pH of $\ce{H2CO3}$ solution is increased to 12 by addition of 1M $\ce{NaOH}$.

So, the reaction would be:

$\ce{2NaOH(aq)}$ + $\ce{H2CO3(aq)}$ $\ce{->}$ $\ce{Na2CO3(aq)}$ + $\ce{2H2O}$

c. This new solution at pH 12 is added to $\ce{CaCl2}$ solution; the final reaction of the system would be:

(  $\ce{Na2CO3(aq)}$ + $\ce{2H+}$ + $\ce{2OH-}$  ) + $\ce{CaCl2}$ $\ce{->}$ $\ce{CaCO3(s)}$ + $\ce{2 NaCl}$ + $\ce{2H2O}$

My questions are:

1. Is the final chemical reaction correct?

2. By setting the pH of step c equal to 12, which is ( $\ce{Na2CO3(aq)}$ + $\ce{2H+}$ + $\ce{2OH-}$ ), and addition of 50 ml of that to $\ce{CaCl2}$ solution, is it possible to make some sort of pH calculations?

$\ce{[H+]} = 10^{-12} M$

$\ce{[OH-]} = 10^{-2} M$

$\ce{[CO3]}$ = assuming 0.05 M in 50 ml

3. Overall, is it possible to do a numerical simulation (e.g., MATLAB) of change in the overall solution pH? OR, predict the final pH by calculations?

Thanks and Best Regards,

As an answer to your overall question: it is indeed possible to do numerical simulations of pH in different solutions or mixtures, and there is a number of programs or spreadsheets available if ou search the net. My personal favorite is Curtipot made by Prof. Gutz: http://www.iq.usp.br/gutz/Curtipot_.html


Decrease in molar conductivity with increase in concentration Why does molar conductivity $Λ$ for strong electrolytes decrease with increase in concentration? I mean, more concentration would mean higher number of solute particles present in the solution, and as it is a strong electrolyte all of them would  disassociate and can help in conduction of electricity, so the $Λ$ should ultimately increase, but $Λ$ is given by

$$Λ_\mathrm{m} = Λ^°_\mathrm{m} - Ac^{1/2}$$

What am I missing here?

The equation you quote is the empirical relationship discovered by Kohlrausch (1907) for the conductivity of strong electrolytes, and is valid for concentrations up to $\sqrt{c} \approx 0.3$. $\Lambda_m$ is the molar conductivity usually given with units $\mathrm{\Omega^{-1}\,cm^2\,mol^{-1}}$ ($\Omega$ = ohm).  The $\Lambda_m^0$ is the molar conductivity at infinite dilution which effectively means that of pure water.

The reason that the conductivity falls with concentration is due to inter-ionic forces on the ion mobilities. There are two related effects, the Electrophoretic and Relaxation Field effects.

Each ion is surrounded by a 'cloud' of oppositely charged ions and when a field is applied there is simultaneous movement of an ion in one direction and of the opposite movement of the ionic atmosphere; the electrophoretic effect. Both the central ion and its atmosphere drag solvent molecules along with them  and this results in a retardation of the movement of the central ion. The effect can be evaluated using Stoke's Law $v=f/6\pi\eta L$, where velocity is $v$ and force $f=zeE$ for an ion of charge $z$ in field $E$. The solution viscosity is $\eta$ and $L$ the Debye length which is the effective radius of the ionic atmosphere, typically a few nm in millimolar solution. The velocity is identified with the deceleration of the ion caused by the ionic atmosphere moving in the opposite direction.

The second effect is a relaxation effect. Suppose the central ion is randomly moved, the overall spherical symmetry that existed before any movement occurred will be broken, i.e. the central ion is asymmetrically located with respect to its atmosphere, and because of this a 'relaxation field' is set up. The symmetry is restored by the effect of the relaxation field which takes only ten to hundreds of nanoseconds.

When the ion is under the influence of the applied field the time average of the forces on the ion are therefore no longer zero. The external field driving the ions through solution is reduced by the continual effect of the relaxation field. This seems strange since this field relaxes so very rapidly compared to motion of the central ion through its atmosphere, $\approx 10^{-3}$ sec, however, its magnitude is huge $10^5$ V/cm compared to the applied field and even small changes in symmetry do have a measurable effect in acting against the external field.


Why add acid to rid of soluble carbonate or hydroxide impurities? Will it not also react with halide I'm testing for When identifying metal halides with silver ions, the positive ion $\ce{Ag+}$ will form $\ce{AgX}$ with halide in a solution. However, $\ce{H+}$ is added (typically in the form of an acid like $\ce{HNO3}$) to solution to get rid of $\ce{OH-}$ and $\ce{CO3^{2-}}$ ions first. Will the $\ce{H+}$ not also react with the halide $\ce{X-}$?

Hydrogen halides are strong acids, meaning that their conjugate bases $\ce{X-}$ are extremely weak. Thus, adding $\ce{H+}$ will neutralize stronger bases such as $\ce{OH-}$ first before $\ce{X-}$.


Avogadro's Number: 6.023 x 10^23 or 6.022 x 10^23? I was taught that Avogadro's number was $6.023 \cdot 10^{23}$. Now, the accepted value is allegedly $6.022 \cdot 10^{23}$. Has there been a change? If so, when and why? There are still some sites that use $6.023 \cdot 10^{23}$, so if this is wrong, why have those sites not been corrected?

Currently, the definition of Avogadro constant depends on the definition of the kilogram, and thus has an uncertainty. The recommended value[1][2] is

$$N_\pu{A} = \pu{6.022140857(74) \cdot 10^{23} mol-1}.$$

It will probably be fixed, however, coming May 20th, 2019 on World Metrology Day to exactly[3][4]

$$N_\pu{A}^{2019} = \pu{6.02214076 \cdot 10^{23} mol-1}.$$

See the references for a thorough review of the history, reasons for redefinition, how some relations become exact and others inexact, and to which degree, etc.

References

[1] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'A Critical Review of the Proposed Definitions of Fundamental Chemical Quantities and Their Impact on Chemical Communities (IUPAC Technical Report)'. Pure and Applied Chemistry, 2017, 89 (7), 951–981. DOI: 10.1515/pac-2016-0808.

[2] Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Rev. Mod. Phys., 2016, 88 (3), 035009. DOI: 10.1103/RevModPhys.88.035009.

[3] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'Definišion of the Mole (IUPAC Recommendation 2017)'. Pure and Applied Chemistry, 2018, 90 (1), 175–180. DOI: 10.1515/pac-2017-0106.

[4] 'A new definition of the mole has arrived'. IUPAC press release. archived link: goo.gl/5haPwJ


Allotropes of carbon conductivity Give reason how can i made diamond a conductor by doping  charcoal and graphite which are allotrope of Carbon act as conductor. Whereas diamond which has a different crystal structure act as an insulator

In charcoal and graphite, conjugated pi systems (the molecule orbital configuration found in carbon-carbon double bonds) act as the conductive band, where electrons can move freely.

In diamond, all bonds are SP3 hybrid bonds (the molecule orbitals found in single bonds). Therefore, no pi systems and thus no conductive band exists.

Edit (after  Jon Custer's comment, thanks): Wikipedia says: Except for most natural blue diamonds, which are semiconductors due to substitutional boron impurities replacing carbon atoms, diamond is a good electrical insulator, having a resistivity of 100 GΩ·m to 1 EΩ·m[28] (1011 to 1018 Ω·m). Thus it is possible to dote diamonds with boron to make them more conductive.


Are the 2s and 2p subshells degenerate in the He+ ion? I know that the 2s and 2p (as well as 3s, 3p, and 3d) electron subshells are degenerate in the hydrogen atom. I was wondering if they are also degenerate in other one-electron species, such as the $\ce{He+}$ atom.

Yes, because this system is identical to the original if we simply rescale the charge (or, if you'd like, modify the units of our new system). On a more fundamental level, degeneracies are caused by symmetries of the system, and these are affected not by the magnitudes of various constants or parameters, but rather by the form and structure of the relevant interactions.

The degeneracy of the subshells in the hydrogen atom corresponds to the SO(4) symmetry of the hydrogen atom, which leads to the conservation of the quantum-mechanical form of the Laplace-Runge-Lenz vector. This symmetry depends on the inverse-square law nature of the Coulomb force and appears also in celestial mechanics, where the corresponding inverse-square law force is now gravity.

Changing the charge of your hydrogenic atom will not break this symmetry. On the other hand, fine structure corrections to the hydrogen atom will: if you start looking at relativistic and spin-orbit corrections, you change the form of the Hamiltonian, break the symmetry, and hence resolve the degeneracy.


Saturated and unsaturated elements? I was reading about Group 15 and came upon the following:

  The penultimate shell in N contains 2 electrons (saturated), in P contains 8 (saturated), in Arsenic 18 (saturated) while antimony and bismuth contain 18 electrons (unsaturated) each.

What does the author mean by saturated and unsaturated?

Although I've never heard this terminology before, I'm pretty sure I can see what it means.

As you can see from this diagram, each increase in the principal quantum number (n), adds a new type of orbital to that shell.

The electron configurations of the group 15 elements are:

\begin{array}{|c|c|} \hline
Element & Electron~configuration \\ \hline
\ce{N} & 1s^2~2s^2~2p^3 \\ \hline
\ce{P} & 1s^2~2s^2~2p^6~3s^2~3p^3 \\ \hline
\ce{As} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^3 \\ \hline
\ce{Sb} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~5s^2~5p^3 \\ \hline
\ce{Bi} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~4f^{14}~5s^2~5p^6~5d^{10}~6s^2~6p^3 \\ \hline
\end{array}

$\ce{N, P}$ and $\ce{As}$ all have full penultimate shells, whereas $\ce{Sb}$ and $\ce{Bi}$ do not -  the $4 f$ and $5 f$ orbitals are unnocupied respectively. Therefore 'saturated' seems to refer to elements where the penultimate shell is fully occupied, and 'unsaturated' to elements where it is not.


Experiment using hydroxides of alkaline earth metals In three different test tubes we add a very small amount of:

1)In the first test tube, we add oxide of magnesium

2)In the second test tube, we add hydroxide of calcium

3)In the third test tube, we add hydroxide of barium

Then we add deionized water in the three test tubes. We shake them and after we filtrate them, we measure the pH of each filtrate.
I'm curious why we use hydroxides of calcium and barium but we use oxide of magnesium instead of hydroxide of magnesium?

I would attribute this two 2 reasons:

Reason 1: Insolubility of Magnesium hydroxide.

Magnesium hydroxide is almost insoluble in water and will not easily dissolve in water.

Reason 2: Safety is one of the reason of the choice of compounds.

The formation of hydroxides of alkaline earth metals are generally exothermic and this intensity increases down this group.

Reaction of magnesium oxide with water will produce the
corresponding hydroxide:

$$\ce{MgO + H2O → Mg(OH)2  (very slight reaction)}$$

(The reaction quickly stops because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction).
The reaction of calcium oxide to produce its hydroxide is
exothermic:

$$\ce{CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔH_r = −63.7 kJ/mol of CaO)}$$

Barium oxide is  harmful to human skin and may cause irritation on exposure for prolonged periods or in large quantity.

Its reaction with water is also know to be violent.

So in my opinion it is much better to user hydroxides of Ba and Ca.


What is a topological isomer? I'm working on a project where I have to understand the concept of topological isomer. I found a review or two containing the idea but I couldn't find a book or a detailed description of what a topological isomer is. I would like a suggestion for a book, article or other resource that gives me a clear definition (Wikipedia doesn't do a good job either). Or I would like someone to give me a clear description of what a topological isomer is.

"Isomers" are compounds that are equivalent but for a certain change in structural property. There's different types of isomers depending on which sort of structural property you need to change in order to inter-convert the two molecules.

Structural isomers are equivalent in their atomic makeup but differ based on their chemical connectivity. For example, 1-propanol and 2-propanol are structural isomers - they share the same formula ($\ce{C3H8O}$) but differ in how those atoms are connected. You can change one to the other by rearranging how the atoms are connected to each other.
Stereoisomers are identical in chemical makeup and connectivity, but differ in the spatial orientation of their atoms. There's several different subtypes, which vary based on what sort of change to the molecules can interconvert them.

E-Z isomers can be interconverted by a theoretical rotation around a double bond - something that normally doesn't happen.
Optical isomers can be interconverted by mirroring the molecule (again, something that doesn't happen)
Diastereomers can be interconverted by mirroring the arrangement around a certain subset of atoms.

Conformational isomers are like stereoisomers, but with the distinction that they differ based on changes which do happen under normal conditions. For example, by a rotation around a single bond, or by the umbrella-like inversion of a nitrogen. Different conformational isomers are normally not isolatable from each other, as they rapidly interconvert.

There's a subset of these isomers called atropisomers which would otherwise be interconvertable, but local conditions (e.g. steric hinderance) means that the interconversion is slow under standard conditions, such that the different isomeric forms can be isolated from each other.

Okay, that brings us to your question on topological isomers. These are conceptually very closely related to conformational isomers and atropisomers. Like conformational isomers, topological isomers are equivalent in chemical formula, connectivity, and even the normal stereochemical properties like E-Z and R-S designations. However, like atopisomers - and unlike most standard conformational isomers - two different topological isomers cannot interconvert on an appreciable timescale, and thus each isomer can be isolated from the other. But unlike atropisomers, the difference between two topoisomers is not due to rotation around a single bond.

Instead, the difference between two different topoisomer is due to the different topology of the molecules. Topology is a somewhat complex mathematical topic, but for our purposes here, you can think of topology as a description about the invariant way something structured on a global scale, that can't be changed by bending, twisting, folding, etc.

For example, take a rubber band. No matter how you bend, twist, fold or tie the rubber band, it's still topologically the same. (That's part of the definition of "topology".) However, if you cut the rubber band, pass it through itself and reattach it, you can get something that's called a trefoil knot. If you look locally at the rubber band, all the connectivity is the same. There isn't any twists in the trefoil that couldn't also be present in the original circular band form. But on a global scale the trefoil and the band are very different. There's no way you can bend, fold, twist, tie, etc. the two to interconvert them without cutting. Because of this, we say the trefoil and the circular band are topologically distinct, in a way the circular band and a twisted form of a circular band aren't.

This is the difference that's being expressed in the case of topological chemical isomers. All of the local structure between the two is the same, but the large-scale global structure of the molecule is such that you cannot interconvert the two isomers without breaking bonds. You can't make one into the other simply by rotating a bond -- not even by rotating a bond that is energetically disfavorable to rotate.


How can cooling gases do not change their physical states? It is stated in my book "All gases condense to form liquids or solids before -273.15 degree Celsius  is reached?

could you explain this statement generally and my question in the title specifically?

Well since the lowest temperature that can exist is -273.15 C.

All gases condense to liquid or solids almost all to solids because at this temperature molecular activity ceases.

And the title of your question seems contradicting with what you have written. Probably you might have mistaken interpreting "before".

also only ideal gas can remain gas at this temperature as they have no force of attraction.
http://en.wikipedia.org/wiki/Absolute_zero


Quantifying reactivity in substrates and electrophiles What is "quantitative treatment of reactivity in substrates and electrophiles"?
I read about it in Jerry March's Advanced Organic Chemistry textbook, but I'm finding it difficult to comprehend.

One large body of work aiming to quantify electrophilicity was accumulated by the Mayr group. Numerous reagents were tested against typical substrates, recording the kinetic rate constants allowed to attribute electrophilicity / nucleophilicity along the equation of

$$ \log k (\pu{20 ^\circ{}C}) = s_N (N + E) $$

with
$E$ = electrophilicity parameter
$N$ = nucleophilicity parameter
$s_N$ = nucleophile-specific sensitivity parameter

Beside the selection of publication mentioned on the group page, work was invested to create a map, and a public database with currently "1075 nucleophiles and 276 electrophiles".


Why does vibrational frequency influence quantum tunneling of a proton? Furthermore, consideration of the tunneling in the framework of the Wigner theory assumes that the tunneling factor is small. However, the $\ce{O-H}$ frequency is very high ($\gg k_\mathrm{B}T/h$), and the H atom tunneling under a potential energy barrier when proceeding either from a ground or from other excited vibration levels may be significant.

Source: German, E. D.; Sheintuch, M. Kinetics of Catalytic OH Dissociation on Metal Surfaces. J. Phys. Chem. C 2012, 116 (9), 5700–5709. DOI:10.1021/jp2106499.

Why does a higher frequency influence the quantum tunneling effect?

It is not the frequency per se but the mass that is important,; the probability of tunnelling an energy E is given by

$$p(E)=\exp \left(-\frac{4\pi\sqrt{m}}{h^2}\int_{x_a}^{x_b}\sqrt{V(x)-E}\;dx \right)$$

where $x_a$ and $x_b$ are two points either side of the potential barrier through which tunnelling occurs. The integral is in effect a measure of the area of the potential $V$ above position $x_a$ to $x_b$ through which tunnelling can occur.

As m increases, for a given potential barrier and energy, the tunnelling probability falls.


Chirality of buta-1,2-diene My teacher says buta-1,2-diene is symmetric and hence optically inactive but I think it's the opposite, because allenes with an even number of double bonds are optically active because they have no plane of symmetry. Which answer is correct?

Let's see if a picture can help.  Newman projections can be handy in analyzing the stereochemistry of allenes.  On the left is a Newman projection of methyl allene (buta-1,2-diene).  There is a sigma plane of symmetry that is perpendicular to the screen and contains the 3 allenic carbons plus the methyl group.  Any molecule with a plane of symmetry is achiral.  On the other hand, look at the Newman projection to the right, 1,3-dimethyallene (penta-2,3-diene).  There is no plane symmetry in this case, only a $\ce{C_2}$ axis.  Draw the mirror image of the pictured allene and you will find that it is not superimposable on its mirror image. 1,3-dimethylallene is a chiral molecule.


Empirical and Molecular formula: Homework I'm working on a high school chemistry book and want to double check one of the homework assignments. The question is for students to write the empirical and molecular formula for 1,2-dichloroethane.
I can't find this online. The molecular formula is given as C2H4Cl2, but could this also be the empirical formula? I don't find any sites as listing the empirical formula as CH2Cl.

The molecular formula and empirical formula are right. As we known, empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

If the compound molecular formula is C2H4Cl2, so we divide it with 2, 4, and 2 GCD / greatest common divisor. The GCD is 2, so the empirical formula is CH2Cl.


Nucleophilic addition reaction of ketone to form alcohol - no acid treatment? I'm preparing for a chem prac on reduction of a ketone group (in particular, the C=O in benzoin) using a nucleophilic addition reaction.

In the experimental method, all we do is react benzoin with $\ce{NaBH4}$, the nucleophile. The solvent we're using is a mix of ethanol/water. From what I've learned, the reaction forms a $\ce{C-O^-}$ alkoxide group which then needs to grab a proton from something to form the alcohol.

In the procedure, we mix ethanol/water, and that's it - no addition of dilute acid to then protonate the alkoxide. We're told to add hot water. Am I right in saying that addition of hot water provides a source of protons to protonate the alkoxide?

Thanks :)

The nucleophile is the H-, a hydride. In your SP3 ketone, the tiny Hydride will have no issue producing a high yield of the alkoxide.

The heat will simply make the reaction faster and drive the equilibrium further.

Ethanol + water is a mild acid, so no issue with protons to complete your alcohol, and the heat makes that mix slightly more active as a proton donor.

You end up with a new equilibrium of your 2 alcohols competing for the proton, however with plenty of methanol and some careful washing with hexane or toluene you should be able to get your alcohol out of that mixture...


Redox reaction with hydrogen peroxide with potassium iodide I think I've managed to write down the correct reaction:

$$\ce{H2O2 + KI + H2SO4 -> K2SO4 + I2 + H2O}$$

Now I need to balance this using the ion-electron method in an acidic environment.

Normally this is no problem, but I'm a bit confused on how to work with the reduction of $\ce{O}$ from $\ce{-I}$ to $\ce{-II}$ and I'm not sure how to write down the total ionic equation. I've tried the following:

$\ce{2H^{+1} + O_2^{-2} + K^{+1} + I^{-1} + 2H^{+1} + SO_4^{-2} -> 2K^{+1}  +SO_4^{-2} + 2I^{} + 2H^{+1} + O^{-2}}$

But I'm not sure if that is correct, and if it is correct, I'm not sure what to do from here.

After this reaction and after diluting (with water?), I also need to know what will happen if I add a few drops of a starch-solution (I will actually have to perform this experiment so if it's merely something visible and not necessarily related to the reaction, then this part is less important.)

The final question is "what property of $\ce{H2O2}$ is demonstrated here?" and I suppose the answer to that is it's function as an oxidizer?

With all redox reactions, we should first look at all oxidation states of our reactants and products.

$$\begin{array}{ccc}
\text{Element} & \text{Oxidation state in reactants} & \text{Oxidation state in products} \\
\hline
\ce{H} & +1 & +1 \\
\ce{O} & -1\text{ in }\ce{H2O2}; -2\text{ in }\ce{H2SO4} & -2 \\
\ce{K} & +1 & +1 \\
\ce{S} & +6 & +6 \\
\ce{I} & -1 & 0
\end{array}$$

Now, we have to look at what is being oxidized and reduced:

$$\ce{H_2O_2 + KI + H_2SO_4 -> K_2SO_4 + I_2 + H_2O}$$

Hydrogen has the same oxidation state of 1+.

Potassium has the same oxidation state of 1+.

Iodine is oxidized from -1 to 0.

Oxygen is reduced from -1 to -2.

For redox reactions, we have to write out our half-reactions.

$$\ce{H2O2 -> H_O}$$

$$\ce{KI + H2SO4 -> K2SO4 + I2}$$

For our two half-reactions, we first have to make sure everything is balanced other than hydrogen and oxygen.

For our second half-reaction, we need two moles of potassium iodide in order to balance out the potassium and iodines.

$$\ce{H2O2 -> H2O}$$

$$\ce{2KI + H2SO_4 -> K2SO_4 + I2}$$

In order to balance oxygens, we add molecules of water.

$$\ce{H2O2 -> 2H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2}$$

Since we are in acidic solution, we add protons to balance hydrogen.

$$\ce{H2O2 +2H+ -> H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2 + 2H+}$$

We add electrons to balance charge:

$$\ce{H2O2 +2H+ -> 2H2O + 2e-}$$

$$\ce{2KI + H2SO4 + 2e- -> K2SO4 + I2 + 2H+}$$

Now we add 'em(the half-reactions) both up and cancel out necessary terms:

$$\ce{2KI + H2SO4 + H2O2 -> K2SO4 + I2 + 2H2O}$$

This differs from your original equation since I think you forgot to balance it.

Hydrogen peroxide here is reduced to water in acidic solution.


How can I safely handle a concentrated (fuming) nitric acid spill? I will shortly be using $90~\%$ or higher concentration nitric acid in a home environment for the purpose of decapping ICs. I understand the corrosive dangers of nitric acid, and most of the important dangers of its reactivity (e.g. exothermic with water, don't make nitrobenzene, explosive nitrations are bad) and will be performing the decapping in a well-ventilated clean area. I'll also be wearing the best thick acid-resistant gloves I could find, since it seems nothing is really rated for RFNA handling.

My primary concern is handling a spillage. I know a common disposal technique for liquid waste is to feed the $\ce{HNO3}$ very slowly into water to dilute it, then add a base such as sodium bicarbonate to level out the pH, using an indicator mixture to monitor it. However, that's not so easy if the spillage is a few splashes on the workbench — I need a reasonably safe way of wiping it up and disposing of it. Someone told me that acetone wipes are an option for cleanup, but they couldn't remember where they'd read it, so I'm wary of accepting that advice without having it verified.

So, what're my options? How can I clean up a splash of RFNA off a work surface without doing myself (or the environment) any real harm?

Of course, you’ll want to prevent a spillage as well as possible (making sure the bottle doesn’t fall over etc.) but sometimes you just run into bad luck. Let’s hope that your work bench is at least capable of withstanding the acid for a minute or two — if you have a tiled working surface that would be preferred since they will almost certainly withstand.

If the spill is small — a few drops or so — I would take care of it by quickly adding large amounts of water (more than ten times the volume of spilt acid). I would suggest a wash bottle or something similar. Once you have added water to dilute the acid, you can wipe it up with standard kitchen wipes.

Yes, people are always told to slowly add acid to water and not the other way around. However, if it’s only a small amount of acid and using a large amount of water (and if the acid is not sulfuric acid) you can ‘get away’ with doing it the wrong way; the water will absorb the heat quicker than it can boil.

In case you spilt larger amounts, I would also prepare some baking soda to have around. Carefully (really carefully: it will bubble vigorously) sprinkle some of the baking soda onto the spill until the gas evolution is markedly reduced. Then continue as above, i.e. add water and wipe up with kitchen wipes. This intermediate baking soda step would mainly be to reduce the acid concentration by evolving $\ce{CO2}$ — it would be an overkill to use it on small spills, too.

Have all the required material ready, before you unscrew the cap of the acid bottle.

Finally, I would not use acetone wipes. Acetone can react with nitric acid in may unpredictable ways that you really don’t want to run into. Use water and simple, dry kitchen wipes.

By the way: Having eye protection (safety goggles) is greatly recommended.


Is there a soluble salt, or compound that when dissolved in water results in a solution with density less than pure water? Generally water soluble salts tend to 'fit between' the molecules of water such that the volume of the resulting solution does not increase much relative to the volume and added mass of the salt. So the result is that the solution generally has a higher density than pure water.

But are there any special cases; any salts or compounds that result in an expansion of the resulting solution volume such that its density actually decreases?

Sodium polyacrylate comes to mind, but if I'm not mistaken, the expansion doesn't take place until a significant amount of the compound is added, and then only after the liquid is changed to rather a gel or solid. My question relates to the resulting solution actually being a liquid with density < 1.0 g/cm³.

So then if not for water, are there solutions of other solvents that might behave this way?

No, you can't dissolve a salt to make a lower density solution.

Instead of thinking about salt ions just fitting into water molecules, think about it as if the salt ions are letting everything fit together better. The water molecules are achieving their current density because of the presence of hydrogen bonding. When you add the ions you introduce better bonds between the charged ions and the polar water molecules. These bonds allow everything to fit together tighter leading to a rise in density.

This decrease in overall bond length is a key to the spontaneity of the solvation reaction. If you introduce an ion that increases the bond length you would see no solubility because they system would be more energetically stable as a bulk solid and water with hydrogen bonding.

The polyelectrolyte you bring up raises a good point, but it is not an exception to this rule. Instead, the polyelectrolyte is so large that it "solvates" into water to form a gel. This new material is not really a solution so much as just a new material with water in it. Hydrated salts would be another example.

This should be the case for all solute-solvent systems. However, this is not to say what you want can't be done. You may have to looks at some reversible reaction or cis/trans type switching material. A simple solute-solvent system won't get it done.


Effect of doubling kinetic energy on phase transitions The kinetic energy of the molecules in a sample of $\ce{H2O}$ in its stable
  state at $ \pu{–10 ˚C}$ and $\pu{1 atm}$ is doubled. What are the initial and final phases? [US National Chemistry Olympiad, 2010]

Answer: $\ce{solid -> gas}$

My attempt: Since water freezes at $\pu{0^\circ C}$ at $\pu{1 atm}$, then the water clearly begins in a solid state. In the solid state, most of the atoms are locked within a crystal of hydrogen bonds and therefore have extremely small kinetic energies. Thus, doubling the kinetic energy should not have any impact on the phase, since gases have significantly more kinetic energy than the solid/liquid phase.

Why does the solid have enough kinetic energy such that doubling it results in sublimation?

The atoms are not locked in place.   They vibrate back and forth.  They do so with plenty of velocity and thus kinetic energy, even though the vibrations switch back and forth so rapidly that the atoms do not travel a large distance in any one direction.


Does >2e contribution of an atomic orbital (AO) to a hybrid AO of a molecule violate Pauli exclusion? This question is inspired by On the Role of d Orbital Hybridization in the Chemistry Curriculum J. Chem. Educ., 2007, 84 (5), p 783.

The article calculates 6 bonding sulfur atomic orbitals in SF6 as each having a 0.768s term (s meaning 3s). This implies $6\times0.768^2 = 3.5$ of the 6 valence electrons corresponding to 3s character.

Is it valid to say there is that much 3s character?

The answer to your question is yes, such a contribution violates the Pauli principle, and is hence not a meaningful wave function.

The demonstration of this calculation (paper) aimed at a different statement and is unfortunately very thin on the calculation details. Furthermore it is a bit difficult to distinguish between the different models they use.
However, I believe the key is actually given in the first sentence of the "Results and Discussion" section (see emphasis below). As far as I understand it, the formula presented in equations (2) are not normalised. Furthermore it neglects any ionic contribution and/or occupation numbers. This further manifests itself that the structure is about 600 kcal/mol higher in energy than the resonance hybrid.

  When the $\ce{3s}$, $\ce{3_{$px,y,z$}}$, and $\ce{3d_{$x^2−y^2$}}$ sulfur AOs are free to mix in any proportion to form six bonding orbitals on the central sulfur, the resulting energy is 1305.8 kcal mol1 (entry 1,
  Table 1) with sulfur orbital mixing as described by [..., equation (2)].
  John Morrison Galbraith, J. Chem. Educ. 2007, 84 (5), 783-787. (emphasis mine)

While the wave function constructed in equation (3) seems to obey the criteria of a physically meaningful wave function,  but it produces an energy even higher.

After that they suddenly jump into the MO picture of the argument, but not really explaining that. In this description one can clearly see in figure 3, that the s orbital only contributes to one molecular orbital.

While I completely agree with the basic premise of the paper, from my point of view it completely fails to present a compelling argument. For example: how did they come up with an energy of >300 kcal/mol of $\ce{SF6}$ relative to the $\ce{S + 6F}$ fragmentation? (And who uses Gaussian98 in 2007?)

I personally would rather recommend our own discussion:

Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

Last but not least I would like to recommend some educational papers that I find more enlightening on the matter (no order):

Ronald J Gillespie, Bernard Silvi, Coord. Chem. Rev. 2002, 233-234, 53-62. The octet rule and hypervalence: two misunderstood concepts;
Ronald J. Gillespie, Edward A. Robinson, Inorg. Chem. 1995, 34 (4), 978–979. Hypervalence and the Octet Rule;
Thomas A. Halgren, Leo D. Brown, Daniel A. Kleier, William N. Lipscomb, J. Am. Chem. Soc. 1977, 99 (21), 6793–6806. Apparent octet rule violations, fractionality, and the interpretation of localized molecular orbital structures. Polarization and hybridization functions in chemical bonding.


Confidence in conformation search results from molecular mechanics Conformational searches using molecular mechanics can be either systematic or 'random' using Monte Carlo methods (or other non-systematic methods) to sample conformational space.

In general, my understanding is that Monte Carlo methods perform just as well as systematic searches # assuming that the right parameters are used. This is often described in (organic) computational papers along the lines of :

  A sufficient number of conformations were explored such that all low-energy conformers within  a X kj/mol window of the global minima were found at least X times.

There is a lot of ambiguity over how many times a low energy conformer needs to be found in order to 'guarantee' that the conformational space has been fully explored. Certainly, in my own experience I can generate an ensemble that fulfils this criteria but is clearly non-complete (if I run the same search from a slightly different starting geometry I generate another ensemble with some different low-energy conformers).

My question is two-fold:

Is there any mathematical basis for the statement about finding a low energy conformation x times, or is it largely a case of convenience that works well empirically in the majority of situations ?
Is there really any way to know that a search is comprehensive without systematically exploring the entire PES ?

#: There was a report that Ken Houk lost a bet due to his belief that systematic sampling was the only true way to explore a potential energy surface, though I cant find the paper where this was discussed.

We have published² a probabilistic treatment of this question, but for the case of MD simulations (and not MC). The principal idea is to use Good-Turing statistics to estimate the probability of unobserved species (ie conformations that have not -as yet- been observed). I have no idea whether this treatment can be transferred to MC, but there is an open source program³ -based on R- that you could try to use.

² https://pubs.acs.org/doi/10.1021/ci4005817

³ https://github.com/pkoukos/GoodTuringMD


How can scientists be certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium? Scientists are certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium. How? and why?

You are talking about the first 103 elements.  An element is by definition identified with the number of protons in its nucleus, the so-called atomic number.  Any atom with 6 protons is carbon, it cannot be anything else.  So there cannot be any missing elements out to 103, as all of these elements have been confirm discovered by accepted scientific methods.

For example, here is a link to a CHEM Study film showing how transuranium elements (i.e. atomic number > 92 and man-made) can be separated from one another and verified as distinct.

http://archive.org/details/transuranium_elements


Free energy and kinetic energy for reacting gases I am confused about the energies we are talking about when speaking in regards to the Maxwell Boltzmann Distribution and activation energies.

Is there an relationship between kinetic energy and free energy for ideal gases? I am wondering if it’s possible to plot the x-axis of a Maxwell Boltzmann Distribution in terms of free energy as opposed to kinetic energy or velocity.
Is the activation energy the difference in the free energy of the activated complex and the average free energy of the gas (or any reactant)?
From the answer here, the activation energy mark on Maxwell Boltzmann Distributions like this one refers to a minimum kinetic energy level. Since the activated complex/transition state is a state, I do not understand why having a minimum kinetic energy (and right collision orientation) means reactants will reach the transition state. How come there is no free energy requirement?

My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,

In nature there are three main distributions that "things" follow:

1) Boltzmann

2) Bose-Einstein

3) Fermi-Dirac

The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the "way" energy is distributed among the energy levels of a molecule/atom/generic particle.

Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular "state". The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale.

At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.

Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme:

"The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to
\begin{equation}
e^{-\frac{E}{KT}}
\end{equation}
Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state."

You have already come across this factor in kinetic theory. !!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!.

You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$.
\begin{equation}
f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}}
\end{equation}

The Maxwell velocity distribution is just a special case of a Boltzmann distribution.

With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can.

This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question.

If you have any questions I will try to answer them when I can! :)


Is there a simple assay to determine the elemental carbon content of biochar? A little bit of background. I make charcoal for use as biochar: putting carbon into the ground where it retains water and soluble nutrients. It also provides a good habitat for the friendly soil bacteria and fungi, along with low-tech carbon capture and storage for the masses.

I would like to know how much carbon is going back into the ground, and be able to say with confidence what difference I'm making to my carbon footprint, but the two – at least – unknown factors are the ash content of the char, and the water content from quenching and dust abatement sprays. I can say with confidence that a kilo of dry char will retain 3 kilos of water, which makes it difficult to judge by the look and feel.

I'm looking for a simple assay which will tell me with reasonable confidence how much elemental carbon a sample contains.

FWIW, my understanding of chemistry is rusty: I've got an A-level (the UK university entrance qualification from 1976) but my degree is in Physics, and my background is in engineering. Thanks for any help.

Assuming your charcoal is mostly:

water (evaporates)
carbon (burns)
ashes (does neither)

I would :

weigh a sample of your charcoal to get the mass $\ce m_0$
dry it in an oven at low-ish temperature (50°-80° or more, whatever doesn't burn your charcoal) for a few hours.
weigh it again to get $\ce m_1$
burn the sample in air as thoroughly as possible
weigh what's left to get $\ce m_2$

Now you can calculate :

mass of water (+ other volatiles) = $\ce m_0$ - $\ce m_1$
mass of carbon (+ other non-volatile burnables) = $\ce m_1$ - $\ce m_2$
mass of ashes = $\ce m_2$


Available orbitals for promotion I am under the impression that elements from the third period onwards can promote electrons into d orbitals to form hybridised orbitals. Why would an element in the third period promote an electron to the 3d orbital when we learn that 4s orbitals are filled first because they are lower in energy (is it to do with orbital penetration?)
Why also can a 2nd period p block element not promote to 3s?

There is a considerable amount of debate as to whether d-orbitals of main group elements (I assume we’re talking about them here; the discussion is rather pointless for d-block elements) are used in hybridisation or not.

Originally, the octet expansion, a.k.a. using them, was postulated by Linus Pauling because he was unhappy to use charge separation to describe molecules such as sulphur trioxide. However, it is entirely possible to draw Lewis structures that do not violate the octet rule and hence do not require d-orbital hybridisation for practically every case. There may be more mesomery required in some cases and less in others.

I’m not too good at repeating the arguments of supporters of main group elements’ d-orbital hybridisation theory, because I belong to the other group. The main argument against d-orbital participation is, as you noticed, that they are far removed on an energy scale; even behind the next s-orbital (3s, 3p, 4s, 3d) — hence the order of elements in the periodic table. I also have troubles understanding why phosphorous would only allow one d-orbital to hybridise, sulphur two and chlorine and the heavier halogens three. This makes no sense to me. (It seems to make sense for others, though.)

In my opinion you are absolutely correct in asking that if a 3p-element can use 3d orbitals, why should it not use 4s for hybridisation, and consequently why 2p cannot use 3s. Assuming 3d orbitals to be used, I could not answer your question satisfactorily.