Destabilization of antibonding MO vs stabilization of bonding MO My textbook writes that

  The net energy stabilization due to the occupation of a bonding MO is equal to the net energy destabilization due to the occupation of the corresponding antibonding MO.

Essentially, it claims that these two quantities are equal.

(Image source: SparkNotes)

Shouldn't the antibonding orbital be destabilized more than the bonding orbital is stabilized, though?

TL;DR

  "The destabilisation of an antibonding MO is equal to the stabilisation of a bonding MO, relative to the constituent AOs."

Several key assumptions are inherent in this statement.

Born–Oppenheimer approximation (not discussed further)
"Orbital approximation": wavefunction of $\ce{H2, He2, ...}$ can be expressed as an antisymmetrised product of one-electron wavefunctions (called MOs)
Said MOs take the form of the stationary states of $\ce{H2+}$, without modification
LCAO approximation: stationary states of $\ce{H2+}$ can be expressed as a linear combination of 1s orbitals on both hydrogens
The overlap integral $\langle \mathrm{1s_A} | \mathrm{1s_B} \rangle$ is neglected

If we make all five assumptions, then the statement above is correct.

If we only make assumptions 1 through 4, then MOs will still have the same form as above (i.e. everything else about the MO diagram is correct), but the statement above will be incorrect: the antibonding orbital will be more destabilised than the bonding orbital is stabilised.

The question linked in the comments does give a qualitative explanation, so depending on what you are looking for, it may be worth checking out.

Assumptions

The MOs that you see are one-electron wavefunctions that are obtained by looking at the single-electron species $\ce{H2+}$. Finding the exact stationary states of this system (by analytical solution of the Schrödinger equation) is a bit complicated, so the strategy is to find a set of approximate stationary states for the electron in $\ce{H2+}$, which we can label as $\{|i\rangle\} (i = 1, 2, 3 \cdots)$. These states are called molecular orbitals.

For multi-electron molecules, like $\ce{H2}$, a further approximation is usually invoked, in expressing the electronic wavefunction of $\ce{H2}$ as a properly antisymmetrised product of these states $\{|i\rangle\}$. Therefore, the MOs, which were only "rigorously" derived for $\ce{H2+}$, may be extended to molecules such as $\ce{H2}$. Very loosely speaking, we may assign one stationary state $|i\rangle$ to one electron, which corresponds to placing an electron in an orbital. Because of the Pauli exclusion principle, each stationary state may only be assigned to a maximum of two electrons.

These are obviously crude approximations and it is a surprise that it even works decently well. In particular the issue of electron-electron repulsions in multi-electron systems is glossed over; in this very simple model, the MOs, being derived from a one-electron system, are entirely unaffected by them, which doesn't really make sense. As the comments on the question have alluded to, there are much more sophisticated ways of obtaining MOs nowadays, but since the claim in the question is derived using this model, we have to use it.

MOs of hydrogen molecule ion

Once we have assumed that the stationary states (i.e. MOs) are of the form

$$\psi = c_1\phi_1 + c_2\phi_2$$

where $\psi$ is an MO and $\phi_1, \phi_2$ 1s atomic orbitals on the two hydrogens, then one can use the variational principle to optimise the two coefficients $c_1, c_2$. The idea is that one wants to minimise the energy of the wavefunction $\psi$:

$$\mathcal{E} = \frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$$

with respect to both $c_1$ and $c_2$. To this end, both the following partial derivatives must be equal to zero:

$$\left(\frac{\partial \mathcal{E}}{\partial c_1}\right)_{\!c_2} = \left(\frac{\partial \mathcal{E}}{\partial c_2}\right)_{\!c_1} = 0$$

The maths will be omitted and if one is interested, one may find it in a variety of textbooks. For example it is covered in Atkins & Friedman's Molecular Quantum Mechanics 5th ed. (search for Rayleigh–Ritz method), or in Coulson's Valence 3rd ed. It is also described thoroughly in this handout from QMUL. The key result is the secular equations, which are a set of generalised eigenvector equations:

$$\mathbf{Hc} - \mathcal{E}\mathbf{Sc} = \mathbf{0}$$

where the $2 \times 1$ vector $\mathbf{c}$ simply contains the coefficients $c_1$ and $c_2$, the elements of the Hamiltonian matrix are $\mathbf{H}_{ab} = \langle a | H | b \rangle$ and the elements of the overlap matrix are $\mathbf{S}_{ab} = \langle a | b \rangle$. The basis set used is $\{|\phi_1\rangle, |\phi_2\rangle\}$. For a non-trivial solution (i.e. $\mathbf{c} \neq \mathbf{0}$) we require the secular determinant to vanish:

$$|\mathbf{H} - \mathcal{E}\mathbf{S}| = 0$$

Now we can use symbols to fill in these matrix elements:

$H_{11} = H_{22} = \alpha$ (this must be so, as both hydrogen atoms are equivalent by symmetry)
$H_{12} = H_{21} = \beta$ (assume that $\beta$ is real, so by the hermiticity of $H$, we must have $H_{12} = H_{21}$)
$S_{11} = S_{22} = 1$ (the atomic orbitals are normalised)
$S_{12} = S_{21}$ (again, assume that this is real. Assumption 5 is that this goes to zero, but for now we will not neglect it; later we can set it to zero to see what happens.)

The exact form of the Hamiltonian, as well as more explicit expressions for these quantities, are given in Atkins MQM. For now they need not concern us. Solving the determinant we obtain two values of $\mathcal{E}$ and two associated vectors $\mathbf{c}$, which can simply be written as the MOs $\psi = c_1\phi_1 + c_2\phi_2$. The MOs (i.e. approximate stationary states) are now quoted without proof, with their eigenvalues (i.e. their approximate energies):

$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2 + 2S}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \frac{\alpha + \beta}{1 + S_{12}} \tag{1} \\
\psi_2 &= \frac{1}{\sqrt{2 - 2S}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \frac{\alpha - \beta}{1 - S_{12}} \tag{2}
\end{align}$$

$\mathcal{E}_1$ is the ground state energy (i.e. $\mathcal{E}_1 < \mathcal{E}_2$) since $\alpha$ and $\beta$ are both negative quantities (again MQM has more info).

Stabilisation and destabilisation energies

Now we have actually reached the part where the original question can be answered. What is the stabilisation of the bonding MO? The energy of the constituent AOs is $H_{11} = H_{22} = \alpha$. So, the stabilisation must be:

$$\begin{align}
E_{\mathrm{stab}} &= \alpha - \mathcal{E}_1 \\
&= \alpha - \frac{\alpha + \beta}{1 + S_{12}} \\
&= \frac{\alpha S_{12} - \beta}{1 + S_{12}}
\end{align}$$

Similarly, the destabilisation of the antibonding MO is:

$$\begin{align}
E_{\mathrm{destab}} &= \mathcal{E}_2 - \alpha \\
&= \frac{\alpha S_{12} - \beta}{1 - S_{12}}
\end{align}$$

The quantity $S_{12}$ is positive (again it should be in MQM), so $E_{\mathrm{destab}} > E_{\mathrm{stab}}$.

The "incorrect" piece of information comes from a further simplifying assumption - namely, that $S_{12} \ll 1$. That is not the greatest assumption, since the values of overlap integrals can sometimes be quite non-negligible (for a $\ce{H-H}$ bond length of $74~\mathrm{pm}$, $S_{12} = 0.753$!), but it is a useful way to predict approximate energies and the general forms of the MOs.

If $S_{12} \ll 1$, then $1 \pm S_{12} \approx 1$ and equations $(1)$ and $(2)$ become:

$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \alpha + \beta \tag{3} \\
\psi_2 &= \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \alpha - \beta \tag{4}
\end{align}$$

Now it turns out that, with the additional neglect of overlap approximation, $E_{\mathrm{stab}} = E_{\mathrm{destab}} = -\beta$. That's where the "destabilisation of antibonding MO = stabilisation of bonding MO" idea comes from.

The arguments above can be extended to the case where you have constituent AOs which are different. The main difference is that $H_{11} \neq H_{22}$, so you have to write $H_{11} = \alpha_1$ and $H_{22} = \alpha_2$. The result you obtain will be rather similar: the destabilisation of the antibonding MO is greater than the stabilisation of the bonding MO, but in the limit of $S_{12} \to 0$, they become equal.


Reactions in water radiolysis - producing hydroxyl radicals There’s an excellent question on Stack Exchange regarding the radiolysis of water here but I have one simple outstanding question that perhaps someone might now; in radiobiology, a common DNA damaging free radial is caused by the radiolysis of water. This reaction is something like

$$\ce{H2O + $h\nu$ -> H2O+ + e- }$$

$$\ce{H2O+ + H2O ->  H3O+ + OH. }$$

which produces a hydroxyl radical which can prove fatal to cells. I follow the logic here, but given the massive energy of the ionizing radiation, how come a cleaving reaction like the following isn’t common?

$$\ce{H2O + $h\nu$ -> H. + OH. }$$

I’ve seen the energy quoted to break water this way at around $\pu{493.4 kJ/mol}$, which is way below the energy ($h\nu$) of an x-ray photon, even after compton scattering. Is there a reason why this reaction isn’t valid or common?

H (atomic hydrogen radical) is in fact formed.  It is 9% of the radial species formed by radiolysis.

See Reactivity of the Hydroxyl Radical in Aqueous Solution and references cited therein for further information.


Why does the Leidenfrost effect (seemingly) not apply to the case of molten NaCl when it is poured into water? On pouring molten sodium chloride into water you can see that when molten $\ce{NaCl}$ (table salt, i.e. sodium chloride) is poured into water ($\ce{H2O}$), the Leidenfrost effect appears to be nonexistent, whereas it clearly manifests itself in instances where other molten salts such as sodium tetraborate (1) (borax) and sodium carbonate (2) are poured into water.

Why does the Leidenfrost effect in molten NaCl last much shorter than that of the other salts, appearing to be nonexistent?

Input from paracetamol:

  "The Leidenfrost effect (physical phenomenon) does show up. See the "cushion" of water vapour around the blob of molten salt in the video? Though short-lived, it's pretty apparent. Now why molten table salt would react with water faster than molten borax or soda (as implied in your post) is an interesting question."

In the video, TheBackyardScientist offered an explanation for the huge explosive effect when 50 grams of molten NaCl is poured into water. He said that some water penetrated the blob of molten NaCl and blew it apart from the inside. A Leidenfrost effect on the outside of the blob is visible, but not the major effect.

The borax blob did not explode, so it must not have been penetrated by a significant amount of water. Borax is a tetrahydrate; when it melts and then solidifies, it goes thru a glassy stage which might be expected to be somewhat cohesive, perhaps resembling Silly Putty. In addition, it loses a lot of water on melting: an initial amount of 50 grams melts to only 26 grams. (Boric acid solidifies similarly.)

NaCl does not exhibit a glassy stage on cooling (note the crystalline surface on the patty from solidified NaCl). Perhaps 26 grams of NaCl would also not explode but merely show the expected Leidenfrost effect. As a 50-gram blob of NaCl cools, solid (strong) regions form over weak molten interior; the weak part of the blob splits open, allowing water into an interior hot enough to boil the water quickly and blast the blob skins apart. The hardened skins of the blob serve as a partial container - like a pipe bomb.

50 grams of washing soda (sodium carbonate decahydrate) will melt down to 18 grams; again, perhaps this is too little to give an explosion.

The Leidenfrost effect should be expected for small blobs of anything. The explosive effect might be expected only for larger quantities, but including even materials cooling thru a glassy stage, if the amount were large enough.


Calculating Ka values from std. state thermodynamic data My professor showed us a sample calculation of the Ka value for a strong acid. He used standard state thermodynamic data.

I have an issue with this:

1) Wouldn't this give us an understated Ka value?

2) Standard state conditions start with 1 molar solutions of solute. So in writing the reaction equation of an acid with water and by using that equation to find the delta G std. of reaction and then using the relationship between delta G and K to find K, wouldn't we be running into a sort of "common ion" effect issue? There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions. Finding the K of an acid dissolving in water that has a pH of 0.0 doesn't give us a Ka value (or does it?)

EDIT to address Martin:

This is what he did:

$\ce{HNO_3 + H2O ->H_3O^+ +NO_3^-}$

Find ${\Delta G^o_r}$ for above reaction.

Use ${\Delta G^o_r=-RTlnK}$ to solve for K.

My question: Is this K that you find a $\ce{K_a}$ value?

The definition of $K_a$ for $\ce{HNO3}$ is $$\frac{[\ce{H3O+}][\ce{NO3-}]}{[\ce{HNO3}]}$$ (concentrations in mol/L), which is precisely the $K$ that you get from the $\Delta G_r^0 = -RT\ln K$ formula, assuming you use the right free energies (which is not a trivial matter since they must take into account the interaction of each species with surrounding water molecules, an effect known as solvation).

You may be wondering why the concentration of water does not appear in the equation for $K_a$ (or more generally the equation for $K$ in a reaction in aqueous solution).  Because that concentration remains essentially unchanged throughout, it is kept out of the defining expression for $K$, for simplicity. From the point of view of $G^0$ this is correct since the standard state to which the $G^0$ of water refers is pure liquid water, while the standard state for the other three species is 1M concentration.

  There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions.

You're correct that for specifying the $G^0$ of $\ce{H3O+}$ the standard state is 1M. But the standard state of $\ce{H3O+}$ is not where the reaction takes place. The same $K$ will be found whether you add the $\ce{HNO3}$ to pure water or to a solution of $\ce{HCl}$ of pH zero -- the final concentrations of $\ce{H+}$, $\ce{NO3-}$ and $\ce{HNO3}$ will be different in the two cases, but $K$ will be the same.

You can find a good discussion of the "standard state" for ionic species in solution here:
http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set10/lecture10.doc

I'll quote a couple of paragraphs.  You can see from this discussion that the calculation of these standard free energies is based exactly on the equilibrium constants, so the formula your professor used works almost by definition!

  As always, we can talk about the Gibbs free energy, enthalpy and entropy of such reactions. However, if we try to break these reactions down into the Gibbs energy, enthalpy and entropy of formation of each species in the reaction, we end up defining things like the Gibbs energy of a sodium cation in solution. Yet there is no way to generate just a sodium cation in solution (that would violate charge balance). Thus, we can only determine the Gibbs energy of formation of pairs of ions. This does not cause any real physical chemistry problems, but it causes a book keeping problem: how do we write down the Gibbs energy of formation of a Sodium cation in a book so that we can use it in any reaction where sodium cations are formed?

  The answer is that we come up with a sort of standard ion, $\ce{H+}$ (really something more like $\ce{ H3O+}$) and we define the free energy of formation of all other ions relative to this one. For example, I can measure the Gibbs energy change when I dissolve HCl in water to form $\ce{H+}$ and $\ce{Cl-}$.
  $$\ce{HCl}\text{(g)}\to\ce{H+}\text{(aq)}+\ce{Cl-}\text{(aq)}
 \quad\Delta G^0{}_1
$$
  I can now just define the Gibbs energy of formation of $\ce{H+}$ as zero and then determine the relative Gibbs energy of formation of $\ce{Cl-}$.
  $$\begin{align}&
\Delta G^0{}_1=\Delta G^0_{\mathrm{F}}(\ce{H+})+\Delta G^0_{\mathrm{F}}(\ce{Cl-})-\Delta G^0_{\mathrm{F}}(\ce{HCl})
\\&
\Delta G^0_{\mathrm{F}}(\ce{H+})=0\quad\text{(definition)}\\&
\Delta G^0_{\mathrm{F}}(\ce{Cl-})= \Delta G^0{}_1+\Delta G^0_{\mathrm{F}}(\ce{HCl})
\end{align}
$$


Separation techniques for different organic compounds Select the best (most appropriate, easiest, fastest) separation technique for each pair of compounds shown below:

  Separation techniques:

  evaporation under reduced pressure
  column chromatography
  water-diethyl ether extraction
  acid-base extraction

This requires knowledge of what compounds are separated most efficiently/appropriately by what separation technique. Since I did the labs separately I can only attempt to try to meld the concepts together.

Pair 1 is a carboxylc acid and chlorodiene. Both are polar.
Pair 2 is an ether and hydrocarbon. One is polar the other nonpolar.
Pair 3 looks to be an ester and cyclodiene. One is polar and the other slightly polar.

My attempt:

Based on the acid I'd use acid-base on pair 1.
Pair 2, since one is polar and the other nonpolar, seems like it needs liquid-liquid extraction (water-diethyl ether extraction)
Pair 3 are polar. Unsure of this one. They'd be slow on a column. Evaporation is based on a low boiling point.

You have correctly identified the carboxylic acid in pair 1. The second conpound is 9-chloroanthracene. Calling it a chlorodiene isn't correct. It's an aromatic compound. Anyway, your conclusion to use acid-base extraction is reasonable.

In pair 2, you seem to overestimate the polarity of the first compound. But there is another, more striking difference between the dipentyl ether and n-hexane.

   Dipentylether, $\ce{C5H11-O-C5H11}$, has a much higher molar mass than n-hexane, $\ce{C6H12}$. As a rule of thumb, you can assume that the compound with the higer molar mass will have a much higher boiling point. (There will be exceptions when strong interactions between molecules come into play, such as in the case of hydrogen bonds. But this isn't the case here!) If you look up the boiling points (bp) for both compounds, you'll see that our rule of thumb is supported here. The bp of n-hexane is 68 °C, that of the much heavier ether is > 180 °C. Consequently, distillation (under reduced pressure) is the way to go here.

In pair 3, you seem to misinterpret the nature of the second compound. It's not a diene, but an aromatic hydrocarbon named naphthaline. You might want to read about the properties of these compound and/or the comcept of aromaticity some other time. But for now, let's concentrate on another detail in the structure of the first compound in this pair.

   There is an $\ce{OH}$ group in the the first molecule. Here, the term polar is justified! Think about the typical setup of column chromatography. The stationary phase is usually silica (or aluminium oxide), while the mobile phase typically consists of mixtures of not-so-polar organic solvents. Compound 1 can strongly interact with the silica through hydrogen bonding, while compound 2 is an unsubstituted aromatic hydrocarbon. Separation by colunm chromatography is a very promising method for this pair.


What is the difference between ignition temperature and flash point? I am trying to clarify this for Class 8 Science. Their textbook says "It is the lowest temperature at which a substance catches fire". However, it does not clarify whether it is on its own or due to an external source. I read through various websites and there are both kinds of definitions put up! Some also mentioned a flash point. So wanted to know what the proper definition of ignition temperature is and how it differs from flash point. Thanks!

From Wikipedia:

Autoignition Temperature: the lowest temperature at which a substance spontaneously ignites without a source of ignition (such as a flame or spark).

Flash Point: the lowest temperature at which the vapours of a (volatile) material will ignite, with an ignition source.

Ignition temperature means the same thing as autoignition temperature according to Merriam Webster.


How do I view a DSN6 file as text? I am trying to isolate certain pieces of information within a dsn6 file. Specifically, I want to find the electron density around two atoms within a protein. I want to then run a python code to determine if there is positive electron density between the two of them. In order to do so, I believe I need to convert the dsn6 file, which is stored in a very compressed format, into a format I can understand.

I have searched around found ways to view dsn6 files (such as through PyMOL or COOT in CCP4) as a model. However, I do not know a way to view this decompressed file as text. I have tried, but have been completely unsuccessful in trying to figure out how these programs decompress/read the files.

This is what a piece of a dsn6 file currently looks like:

˘>˝‰‡>™†≠>I14>)B=*À¢<™ÅK=i§=‰≤Ñ=ê8ºO∑ææÑ¸ΩﬁZ
Ω2…=wÇ=œjÖ=ÿm=◊6=â(:ÿ◊ôΩl≥æ=sËΩ,vzºËm
>dg>ÄàG>>hº&>è,>SªÂ=

I hope that the decompressed dsn6 file will look similar to what a pdb file looks like. For example,

ATOM    228  OG1 THR A  58       7.843  45.672  59.760  1.00 15.56   O
ATOM    229  CG2 THR A  58       6.672  44.132  61.223  1.00 14.90   C
ATOM    230  N   PRO A  59       9.292  42.373  62.754  1.00 13.16   N
ATOM    231  CA  PRO A  59       9.409  40.982  63.160  1.00 13.07   C

I asked this question in Stack Overflow, but thought that it might be better suited here. Thank you for the help!

There is a program called mapman that can convert between different formats. The .xplor format is a text-based one, the others are all binary, so that's the one you need. The program runs on linux and mac platforms.
Other crystallographic software suites (ccp4, etc) might have similar capabilities, but this one might be the easiest to install (standalone program).

I want to find the electron density around two atoms within a protein.

This is much easier with a molecular browser. You could just set the contour levels to the ones your are interested in, and take a look. If you need to do this programmatically for thousands of structures though, that will not help you.
It will be kind of tricky to find the right spot in the density, though, because they use fractional coordinates (makes generation of symmetry much easier) and your coordinates are in orthogonal coordinates. If this is a problem, just ask another question.
Here is the description of how data is organized in an .xplor map:

Following the map header, the density matrix is then written section-by-section with c moving slowest (in z-sections). Each section of the density map is preceded by a section number.
Thus, for the formatted map file each section of the density map is written using FORTRAN statements of the type
WRITE(OUNIT,'(I8)') KSECT
WRITE(OUNIT,'(6E12.5)') ((SECTON(I,J),I=1,ISECT),J=1,JSECT)
and the resulting map is written with six pixels on each line. The binary format is identical except the format statements are missing, so that each line that is written contains the entire length of map along the `fast' (a-axis) direction.


Why is Cl2 only formed in acidic media? If aqueous $\ce{NaCl}$ and $\ce{NaClO}$ are added together no reaction occurs, but if you add dilute acid to the solution $\ce{Cl2}$ gas is formed. Why does this reaction only happen in acidic media? I know that there's something to do with the reduction potentials but I couldn't formulate a full explanation for this effect.

Basically this is due to two reasons.

The disproportionation of chlorine in water looks like this:

$$
\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}
$$

So you need the $\ce{H+}$ to shift the equilibrium to the halogen site and the halogen produced is a gas and will leave the reaction.

The other explanation is that, as $\ce{HOCl}$ is an oxidizer and a $\ce{Cl+}$ transfer-agent, both of these actions happen in an $\ce{S_{N}2}$-type fashion. For the $\ce{Cl+}$ transfer it is called associative activated substitution of a nucleophile, here $\ce{Cl-}$. So it's pretty much like an $\ce{S_{N}2}$ reaction, where
$\ce{H-Cl}$ and $\ce{Cl-OH}$ react.


Can the plane of symmetry for a compound be the plane of the paper? 1,4-Dichloro-2,5-dimethylbenzene is symmetric about the plane of paper.

I have never seen a plane of symmetry positioned like this. Would it be correct to call it a plane of symmetry?

In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.

If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring.  This $\sigma_h$ plane then could be described as in the following example by the yellow disk:

(credit to symmetry@otterbein, example symmetry $C_{6h}$, hexakis(Me2N)benzene).

Your example however is about a methyl-substituted benzene. While the carbon atoms of the methyl groups are in the plane of the symmetry, you no longer can apply the same operation (reflection on this mirror plane) on the hydrogens of these methyl groups to yield corresponding, symmetry equivalent hydrogens on the other side of the mirror.  This is only possible if both the methyl groups share a special conformation: for each exactly one hydrogen must reside in this $\sigma_h$ mirror plane; then, the other two hydrogens of the methyl group in question are symmetry equivalent.

In all other conformations, the three hydrogens of one methyl group were no longer symmetry equivalent in respect to this $\sigma_h$ plane.  And it would require only one methyl group to be «off» to erase the $\sigma_h$ plane for the whole molecule.

I would suggest to venture out the interactive tutorials around symmetry@otterbein.  All you need is an internet browser with javascript enabled.  Later, if interested in applying the concepts of symmetry into crystallography (including the transition of Schoenflies notation used in IR-spectroscopy by the  Hermann-Maguin about point groups), you may find its sibling crystals@otterbein a valuable complement.


Why is it possible to use lipids to recover diamonds in mining? Diamond mining

This link says that diamonds will stick to "anything oleaginous" and that other materials slide down the fat that is applied to an inclined plane used in the mining apparatus.  I infer that the surface of the diamond interacts via dispersion forces and that other minerals are more polar.  Is this the case?  If so, would diamonds stick to wax or plastic?

According to the Wikipedia piece on diamonds:

  Diamonds are naturally lipophilic and hydrophobic, which means the diamonds' surface cannot be wet by water but can be easily wet and stuck by oil. This property can be utilized to extract diamonds using oil when making synthetic diamonds.

It looks like the surface of diamond has dangling bonds terminated with hydrogen, so your non-polar hypothesis is correct, as applied to terminal moieties on the surface. See this article about the material properties of diamond where it is stated:

  Diamond surfaces are hydrophobic when the surface carbon atoms terminate with a hydrogen atom and hydrophilic when the surface atoms terminate with an oxygen atom or hydroxyl radical.

and

  Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen and the behavior is moderately hydrophobic. This allows for separation from other minerals at the mine using the so-called "grease-belt."

You ask:

  If so, would diamonds stick to wax or plastic?

I believe the answer is yes, but certainly factors such as the transport of diamond with grit in a slurry over/through a hydrophobic material such as the two you name might come into play: I think mixing the mined slurry with goop is more effective than sheeting it across a 2-D material, and a higher extract yield results.

Note that at higher temperatures and with other reactants, the surface properties can change (as stated in the linked article immediately above):

  Treatment with gases or plasmas containing the appropriate gas, at temperatures of 450 °C or higher, can change the surface property completely.


Explaining the colour change in the potassium permanganate titration of iron(II) ions In the redox titration:

$$\ce{MnO4- + 8 H+ + 5 Fe^2+ -> Mn^2+ + 4 H2O + 5 Fe^3+}$$

the colour change which occurs is purple to colourless, because of the decreased concentration of permanganate ions. But doesn't this fail to account for the visible iron ions? $\ce{Fe^2+(aq)}$ is green and $\ce{Fe^3+(aq)}$ is brown.

Anyone have an explanation for why the end result is colourless and not brown solution?

I think this is because $\ce{MnO4-}$ ions have a very distinct colour. At the end of the reaction, the solution is colourless. I think that the $\ce{Fe^3+}$ ions give the solution a brown colour, but it is in a very low concentration. The brown $\ce{Fe^3+}$ ions are too few to give the whole solution a brown colour.


Why isn't ionization energy called cationization energy? I'm learning chemistry and ionization energy is "[...] the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation." It seems more intuitive to me to call this cationization energy. Other than historical reasons and convention, are there any reasons why it's not?

Atoms can have electrons added to them or removed from them, which in turn ionises the atoms. However, electron addition and removal aren't the same thing in terms of enthalpy changes. The key here is the difference between the
process and the enthalpy changes during the process - as much as an atom is ionised during the production of a cation or an anion, the enthalpy changes are different. Because of this, there are two different terms used for the enthalpy changes during these processes:

'Ionisation energy' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous positive ions (cations), under standard conditions,
'Electron affinity' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous negative ions (anions), under standard conditions.

As seen from the definition for 'ionisation energy', it's already described that there's a production of cations, so it would be redundant to call it 'cationisaion energy'.  The word 'ionisation' in itself is typically used to describe the removal of electrons from an atom, which could also contribute to the term's name.

The reason why there are two different terms is due to how the electrons are dealt with during ionisation. The removal of electrons is always endothermic, whereas the addition of electrons is only sometimes endothermic.

When removing electrons, more energy is required than what's released to remove the electrons from the atom, hence why it's always endothermic. This is due to the Coulombic attraction between the negatively charged electron(s) in the atom's orbit and the positively charged proton(s) in the atom's nucleus. The required energy will increase with each successive removal, since the electrons will get closer to the nucleus and the attraction will increase because of it. This enthalpy change is explicitly called 'ionisation energy'.

When adding electrons, only the first addition is exothermic whilst the rest are endothermic. The first addition is 'easy'; you're just adding an electron to a neutrally charged atom, and the energy released during this process is greater than the energy required. However, with each successive addition, the energy required will increase. This is due to the Coulombic repulsion between the anion and the electrons, so more energy is required than what's released to 'force' the electrons into the atom's orbit, hence why successive electron additions are endothermic. This enthalpy change is explicitly called 'electron affinity'.


Redox couple notation How is the redox couple as it is written for the oxidation of $\ce{MnO4-}$, as $\ce{MnO4-,H+/Mn2+,H2O}$ correct?

I agree that $\ce{MnO4}$ is being oxidized as it loses electrons (going from -1 charge to +2) but I don't see why $\ce{H+}$ is also being oxidized. Could someone explain?

The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written :  MnO4-/Mn2+, without mentioning Hydrogen

And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.


Generating molecular graph Laplacian based on Hückel theory While studying graphs and graph Laplacians from "Laplacian eigenvectors of graphs: Perron-Frobenius and Faber-Krahn type theorems.", I encountered a type of graph Laplacians that rise from quantum chemistry. This matrix is consists of off-diagonal elements called resonance integrals and diagonal terms called Coulomb integrals. This is a sparse matrix, meaning that the matrix elements corresponding to non-connected atoms are zero. It is further mentioned that

  ... the entries of this matrix, H, are tabulated for different atoms and
  bonds.

I was looking online to find some table that gives this information and write a package to compute this matrix given a SMILES entry. As an alternative, I also looked for an already existing package that provides this matrix. The only thing I found was the implementation of the extended Hückel method in RDKit, which requires molecule conformation as input (apparently, eHM needs atom coordinates).

I was wondering if anyone can point me to such a table or python package or let me know if I'm missing something.

You're describing simple Hückel theory. This is usually applied to $\pi$-conjugated systems to understand the stabilization in conjugated and aromatic molecules.

There are a variety of Python packages, for example:

Felix Plasser wrote one for his blog and the current location is at his homepage
Randle Taylor has a SHMO package on GitHub, including a GUI called Orbis

I believe the version from Plasser will read in files using Open Babel's python interface, so certainly you could handle SMILES.

People would perturb the $\alpha$ (atom site energies) and $\beta$ (bond interaction) parameters in various ways. It's hard to find good tabulations of these, but the best appears to be:
"A brief review and table of semiempirical parameters used in the Hueckel molecular orbital method" J. Chem. Eng. Data 1967 122 pp. 235-246

You mention the somewhat related Extended Hückel theory (EHT). This considers both $\sigma$ and $\pi$ bonds and is thus a lot more useful, although as you mention you need coordinates to calculate overlap values, etc.


Insoluble salts and Gibbs free energy Is it true that according to gibbs free energy equation, insoluble salts such as AgCl don't dissolve because the process would be too endothermic? The right side of the equation, TΔS, would have to be positive because entropy increases when salts dissolve, which means that ΔH would have to be high enough to make the equation positive. Is this correct?

Yes, this is correct. Assuming that you are talking about solubility in water, it is discussed by JF Liebman, Structural Chemistry, 15, 165 (2004) which needs a subscription to read in full, but much of the key information is given on the first couple of pages which are available as a preview. The relevant thermodynamic quantities (in kJ mol$^{-1}$) for dissolution of one mole in water are given as a table, with AgCl compared with NaCl and KCl:

  $$ \begin{array}{cccc}  & \Delta H & \Delta G & T\Delta S \\ \hline \text{NaCl} & +4 & -9 & +13 \\ \text{KCl} & +17 & -5 & +22 \\ \text{AgCl} & +45 & +37 & +8 \end{array} $$

The author got the original data from NBS Tables of Chemical Thermodynamic Properties.
All the entropy terms are favourable.
All the enthalpy terms are unfavourable, but the AgCl has a much more unfavourable enthalpy of solution than the alkali metal chlorides, and the
entropy cannot compensate for this. The Ag$^+$ ion is poorly hydrated in solution
compared with Na$^+$ and K$^+$.


Structure of NO2 compound? What is the correct structure for the $\ce{NO2}$ compound (not ion)? I always thought it was like:

since the negative charge would be residing the on the highly electronegative oxygen, but was reading this blogpost by the guys at WolframAlpha, who state that contrary to popular belief, the correct structure is actually:

They explain that it has to do with molecular orbital theory, but don't explain in depth.

Can someone please explain why this occurs?

According to Electronic Structure of NO2 Studied by Photoelectron and Vacuum-uv Spectroscopy and
Gaussian Orbital Calculations J. Chem. Phys. 53, 705 (1970) :

The highest molecular orbital $4a_1$ is occupied by the 1 unpaired electron.

The electron population of this orbital is (see table VI):

0.53 on the N atom (0.16 2s, 0.37 2pz)

0.24 on each O atom  (0.24 pz)

Experimentally, Oxides and Oxyions of the Non-metals. Part II.
C02- and NO2 of the Chemical Society (1962): 2873-2880 collects values for the partition of unpaired electron density among N and O atomic orbitals.

The most purely experimental values are:

N 2s  0.094
N 2pz 0.364
N 2px 0.054
O 2pz 0.33

and

N 2s 0.103
N 2pz 0.471
O 2pz 0.33

(where the values for O are both O atoms added together)

There is extensive discussion of the various Lewis structures of NO2 from a dimerization point of view (that the opposite spin electrons should be on the atoms that form the dimer bond) starting on page 90 if the 2015 book Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increase-valence Structures , and also elsewhere in the book from a monomer point of view.

Overall, it is clear theoretically and experimentally that the unpaired electron is delocalized, having significant density on both N and the two O atoms.  It is wrong to say that a particular one of the Lewis structures is "correct".


What is the reason behind electron transition in the flame test I am a beginner in this topic but I am aware that when flame test (for alkali and alkaline earth metals) is done there is a s-s transition of electrons from the ground to their excited state.

But I don't understand why the electron needs to transit back to their ground state if I consider that the temperature of the oxidising or non luminous flame (that is the energy supplied to the electron) remains constant.

So my question is

Why does the electron release the absorbed energy and transit back to the ground state even though a constant external energy is supplied to it?

In other words, what is the need for the electron to absorb the energy , then reach the excited state and release the energy to transit back to the ground state. Instead why does it not remain in the excited state after absorbing the required amount.

The electron releases its energy because it can. It happens because it has a non-zero probability to do so. If it would never happen it would have zero probability.

Whether the electron just got excited prior to emitting some light does not matter. Particles do not have a memory. Instead such processes simply happen based on whether the requirements are met (e.q. absorbing energy is only possible if energy is available) and on probabilistic rules (quantum mechanics).

Consider some particles in thermal equilibrium: Even the though the temperature does not change on a macroscopic level, the particles still constantly exchange heat with each other. This is called dynamic equilibrium.

The difference to the flame test is, that the particles exchange energy not only by heat, but by light as well.

Exchange of heat is basically a collision of two atoms, so this is a 2-particle process: its rate dependents on the likelihood of those 2 particles "hitting" each other.

Emission of light is a 1-particle process: it can just happen spontaneously. However, absorbing light is a 2-particle process: a photon needs to intercept an atom in order for the atom to absorb its energy.

Heat is the source of energy, and it allows for redistribution between all particles. But for the light, the emission happens much more often than absorption, because 1-particle processes are much more likely to happen than 2-particle processes. Therefore we see the overall emission of light until the heat source is gone.


Clarification on usage of dU =nCvdT In many places I have come across the following derivation
Work done in isothermal expansion

By first law of thermodynamics

$\mathrm dU=Q+W$

As $\mathrm dU=nC_V\,\mathrm dT$, and in isothermal process $\mathrm dT = 0$

$\therefore\mathrm dU=0$
$W=-Q$
i.e heat absorbed is completely converted into work

The second step doesn't make any sense at all. As far as my understanding $\mathrm dU=nC_V\,\mathrm dT$ is true only when volume is constant, but it's clearly mentioned expansion in the statement.
So how to justify the use of $\mathrm dU=nC_V\,\mathrm dT$  expression even when volume is not constant?
Please consider the fact that I'm a school student so please answer in a simple manner.

The reason why statement #2 doesn't make sense to you is because it is not generally true!
For a closed system of constant composition, with P-V work only,
$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT +  \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$
The difference between that statement and your statement is the extra term at the end, which is the amount by which U changes with V at constant T, times the change in V.
[Minor note: I've dropped the "n" you use, as I'm using total heat capacities, not molar heat capacities.]
However, if we have an ideal gas, U is independent of V, which means:
$$\left(\frac{\partial U}{\partial V}\right)_T=0,$$
Hence, with an ideal gas, the term on the right goes away, giving us your statement #2:
$$dU = C_v dT$$
I.e., your statement #2 doesn't follow directly from #1.  Instead, it's specific to ideal gases.
So:  If a process invoving an ideal gas is carried out at constant pressure instead of constant volume, it is true that $dV$ may not be zero.  But since that non-zero $dV$ doesn't matter when it comes to calculating $dU$ for ideal gases, we can use $C_V dT$ even when $dV \neq 0$.


Measuring lead content in pewter I've bought pewter at the thrift store and I want to try casting metal. This pewter may contain toxic metals. A ready-made testing kit I've found cost €25 for each sample.

In comparison pewter of know origin (98% tin/2% antimony) costs €40 per kg,
pewter from the thrift store is €4 per kg. If possible I want to test the pewter before buying.

Is there an easy way to measure the lead content of a pewter sample?

I'm not a chemist but do have a basic understanding of chemistry.
I've calculated the density of the sample to be: $8.55~g~cm^{3} ± 0.5~g~cm^{3}$.

You are likely looking for trace levels, so I do not suggest any at home kit. At best, is it colorimetric and sensitive to interferences.
If you really want to know, find an environmental lab and ask them to run ICP-MS to determine the lead content. Atomic absorption spectroscopy would be the second best method.


Determining phase from intermolecular interactions and kinetic energy If the energy of intermolecular attractions are much less than the average kinetic energy of the molecules, what is the physical state of the substance?

Most likely gas.

In fact, the definition of an ideal gas is a gas in which its molecules have no intermolecular interactions (except elastic collisions).

As the strength of intermolecular attraction increases, the molecules will be brought closer together. Molecules in liquids are much more densely packed than molecules in gases.


What determines a substance's energy of activation with air? Ice will melt when heat is applied; paper will catch fire. In trying to figure out why — what the difference is between things that melt and things that catch — I found "Burn, Char, Melt" by Roberto Gregorius, who says it depends on

  whether the substance easily forms compounds with some oxidant [usually oxygen]. This term "easily" can be taken to mean that the energy required to start the reaction (called the energy of activation) is low enough so that the reaction starts (as in striking a match), or the compounds produced in the burning are low enough in energy that there is a tendency to form it. Of these two, energy of activation is usually the controlling factor in deciding if a reaction will take place.

All right, so what determines a substance's energy of activation with air? (Wikipedia isn't helping.) In case it isn't obvious, I'm uneducated in chemistry, and an answer tailored accordingly, to the extent possible, would be ideal.

Energy of activation is pretty much an experimental thing. You can try to explain obtained results, but predicting exact values without numerical analysis is not really possible.

Basically, given a compound to react with air, you first devise a reaction mechanism -- how the reaction proceeds. Now, using some rules, you determine the highest energy transition state. Finally, you try to approximate the energy difference between this and the reactants by comparing with known quantities.

About the rules that are used to compare energies and energy differences: There are many of them, and out of the scope of this answer (indeed, one would be hard pressed to fit them all on a single bookshelf). Basically, one uses general rules that compare stability. For example, a positively-charged carbon is more stable when it has more alkyl groups attached to it (the reason usually given for this is hyperconjugation). A carbon conjugated with a double bond is considered even more stable, due to resonance.

Regarding paper, paper is made of molecular fibres which are effectively infinitely long. To melt paper, one would have to disentangle these (as molecules need to be much farther apart and with much less attraction in liquid form), and even then one would be left with a very, very viscous liquid (which is not too different from a solid). So paper would be extremely hard to melt, if not impossible.


Charge and spin multiplicity change in IRC search I have successfully isolated the TS for a radical cation catalysed C-C bond formation using Gaussian with input

1 2

for charge and spin multiplicity.

But when I do an IRC search with the correct input for charge and spin multiplicity (1 2 as above), the IRC log files terminate normally but when seen on GaussView, it says the molecule is neutral with singlet spin.

Is this a problem with GaussView? Will it affect the interpretation of the IRC in any way? In other words, does the IRC search still occur on the doublet PES or has it (somehow) gone to the singlet PES?

In other words, does the IRC search still occur on the doublet PES or has it (somehow) gone to the singlet PES?

You could check that by looking at the output. Right at the end of each and every SCF optimization Gaussian prints a block that starts with SCF Done and contains information about energy of the state to which SCF converged and its total spin. For instance, below is one such block in the results of test170 calculations from Gaussian 09 test set:

 SCF Done:  E(UHF) =  -40.1568456632     A.U. after   10 cycles
             Convg  =    0.6432D-08             -V/T =  1.9978
 <Sx>= 0.0000 <Sy>= 0.0000 <Sz>= 0.5000 <S**2>= 0.7832 S= 0.5165

Now, regarding GaussView, interestingly, it also wrongly shows that the state is singlet and not a doublet for that test job. Besides, GaussView reports the following when opening the output:

  Unresolvable inconsistency between charge and multiplicity and orbital occupancies. Orbital data will be ignored.

Honestly speaking, I suspect that is a bug, because while reading the output I could not see any inconsistencies. I've tried few other IRC jobs on radicals form the Gaussian test set (test171, test452, test453) and GaussView always reports that a state is singlet when in fact it is a doublet. Moreover, in some cases GaussView even failed to read all the steps of a normally terminated IRC calculation. So, the only inconsistency I've noticed so far is that GaussView could not always correctly read Gaussian outputs of IRC calculations on states that are not singlets.


Why perborate is greener bleaching agent? All I know is, perborate degrades into supposedly less harmful compounds.  But what exactly happens that it is considered greener? Consider say perborate vs. hypochlorite.  Why perborate is "greener"?

According to Wikipedia, "Chlorination... can oxidize organic contaminants, producing chloroform and other trihalomethanes, which are carcinogenic, and many hundreds of [other]... by-products.

[W]hen household bleach and wastewater were mixed, 1–2% of the available chlorine was observed to form [halogenated] organic compounds."

Perborates, percarbonates and similar bleaching agents, while strong oxidizers, will not create these chlorocarbon chemicals (unless the chlorine is also supplied).

However, boron and perborates are not entirely "green". Boron compounds are somewhat toxic. If the prices were the same, I'd prefer percarbonate bleach to perborate.


Why is HNO3 a stronger oxidising agent than H3PO4? Why is $\ce{HNO3}$ a stronger oxidising agent than $\ce{H3PO4}$? $\ce{N}$ and $\ce{P}$ have the same oxidation number. Is it the electronegativity difference between $\ce{N}$ and $\ce{P}$?

Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)

$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.

That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction

So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced.

But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:

Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as  in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.
The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.

$\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$


Ideal gas law derivation from kinetic theory Consider the derivation of the ideal gas law from kinetic theory presented here: http://en.wikipedia.org/wiki/Kinetic_theory

I have some questions

This derivation assumes the container is a cube, but the law holds for containers of any shape, right?
This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?
The derivation assumes that the particle intercepts a give side with frequency $2L/v_x$. Why can we make this assumption?

This derivation assumes the container is a cube, but the law holds for containers of any shape, right?

The resulting law holds for any shape of container. However, the details of the derivation depend on the geometry. A cube is easy, so we use it.

  This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?

Yes, it holds for a mixture of any ideal gases - the key is that only the kinetic energy of the gas molecules matters. Heavier molecules would be moving slower at the same temperature, but they would exert the same force (and therefore pressure) on the walls.

  The derivation assumes that the particle intercepts a give side with frequency 2L/vx. Why can we make this assumption?

This assumption works for only one particular case. If you notice the third sentence at the beginning of the pressure derivation (emphasis is mine):

  When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed

They are looking at the special case of a collision at a 90 degree angle, which bounces off the opposite wall. In that case, it will take the particle t = L/V time to reach the opposite wall, and t = L/V time to bounce back - so the total is 2L/V. This illustration may help explain it better:

Now this might sound like an unreasonably strict condition, but the reason that it works is that the component velocity distribution for each particle is symmetrical (it's a normal distribution), which means that for every particle that impacts the wall at 95 degrees (for example) there is another that impacts 85 degrees. As a result, the average angle of impact is 90 degrees for each wall, and therefore the average collision time is given by looking at 90 degree impacts for each component of the velocity vector.


Given the volumes: determine the pH and the final temperature of a mixture knowing only the initial pH and the temperature of the un-mixed components I have a pretty basic question but the last time I took Chemistry was in 2007.  I am studying the Navy's Nuclear study guide for their interviews and one of the question I am faced with is below.

Determine the final pH and temperature when the following two solutions are mixed together in a $\pu{3L}$ container.
Sol A: $\pu{2L}$, pH = 3, and $\pu{80 ^\circ F}$
Sol B: $\pu{1L}$, pH = 5, and $\pu{40 ^\circ F}$

Would this be as simple as $(2/3) \cdot 3 + (1/3)\cdot5$ for the pH and $(2/3)\cdot80 + (1/3)\cdot40$ for the temperature?

For the temperature, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this enthalpy of mixing yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.

For the pH, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is logarithmic, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (I will make the assumption here that you have simple liquids, which are not buffered.) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the amount of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.

Short discussion on the pH: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.


Dissociation of zinc anode during electrolysis Example electrolysis setting:
Cathode: copper
Anode: zinc
Electrolyte: dilute aqueous sodium chloride [$\ce{NaCl(aq)}$]

During electrolysis, zinc is more reactive than copper, hence electron flows from zinc anode to copper electrode.

At the zinc anode:
$\ce{Zn(s)->Zn^2+ (aq) + 2e-}$
Q: What is this process known as?
("Dissociation" is usually applied for salts that dissolve to form mobile ions, and "discharge" is usually used for ions in the electrolyte itself that gain/lose electrons. Hence, what would a proper term be?)

When the zinc anode "dissociates" in the  electrolyte, $\ce{Zn^2+}$ ions are formed.
Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?
(Because $\ce{H+}$ions are selectively discharged at the copper cathode, this leaves the $\ce{Zn^2+}$ions  $\ce{Na^+}$ions are bonded to  $\ce{Cl-}$ions, so does the  $\ce{Zn^2+}$ions form just move into the electrolyte just like that, with an unstable electronic configuration?)

Q: What is this process known as?

This is a particular case of dissolution. Call it electrolytic dissolution, if you want to be more specific. The word "dissociation" certainly does not apply here. It is used when talking about salts and other ionic compounds.

  Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?

Well, $\ce{Zn^2+}$ just goes into the solution and gets solvated (hydrated), that is, surrounded by interacting water molecules, pretty much like any other ion. Some people in some contexts would probably call these ions "free", though they are surely not as free as in vacuum. The same applies to the $\ce{Na+}$ and $\ce{Cl-}$ ions out there, and no, they are not linked to each other in any way except for electrostatic interaction. Have you heard of dissociation? That's what it is about. A $\ce{Cl-}$ ion does not "know" that it "belongs" to $\ce{Na+}$ and not $\ce{Zn^2+}$. They all just coexist there together, probably with some other ions like $\ce{OH-}$. As the process goes on, I expect some $\ce{Zn(OH)2}$ precipitate to appear.


Is the Hartree-Fock determinant always the minimum energy single Slater determinant solution to a molecular Hamiltonian? Is it always true that the Hartree-Fock slater determinant is the minimum energy single slater determinant solution to a given molecular Hamiltonian $H$?
By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals). The Hartree-Fock slater determinant $|\Phi_0\rangle =|\chi_1\ldots\chi_N\rangle$ is constructed by choosing the lowest $N$ energy solutions of the Hartree-Fock equation.
Now, $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ is an eigenstate of the Hartree-Fock Hamiltonian $H^{HF}$ with energy $\sum_{i=1}^N \epsilon_{i_j}$, so constructing $|\Phi_0\rangle$ from the $N$ lowest energy solutions of the HF equation clearly minimizes the energy of $H^{HF}$.  But the molecular Hamiltonian $H$ is given by
$$H = H^{HF} + \sum_{i=1}^N\sum_{j>i}^N \frac{1}{r_{ij}} - \sum_{i=1}^N v^{HF}(i)$$
where $v^{HF}$ is the HF potential.  This satisfies
$$\langle{\Phi_0}|H|\Phi_0\rangle = \sum_{i=1}^N \epsilon_i - \frac{1}{2}\sum_{i=1}^N \sum_{j=1}^N \langle ij|\,|ij\rangle $$
where
$$\langle ij|\,|ij\rangle:=\int dx_1 dx_2 \chi_i^*(x_1)\chi_j^*(x_2)\frac{1}{r_{12}}(\chi_i(x_1)\chi_j(x_2) - \chi_j(x_1)\chi_i(x_2))$$
Due to the $\langle ij|\,|ij\rangle$ term, it is not clear to me that constructing $\Phi_0$ from the $N$ lowest energy solutions of the HF equations still results in the lowest energy single slater determinant for $H$.  An excited determinant would have a higher energy contribution to the $\sum_i \epsilon_i$ term, but could this is offset by a greater deduction from the $\langle ij|\,|ij\rangle$ term?

By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals).

This is not true: only the converged Hartree-Fock determinant $|\Phi_0\rangle = |\chi_{i_1}\ldots \chi_{i_N}\rangle$ has stationary energy with respect to orbital perturbations.
Stationarity of energy requires all orbitals in the determinant to fulfill the Hartree-Fock equations
$$
\hat f \chi_i = \varepsilon_i \chi_i
$$
where the Fock operator
$$\hat{f} = \hat h + \sum_j^N(\hat{J_j} - \hat{K_j})$$ depends on the $N$ occupied orbitals $\{\chi_j\}$ included in the determinant.
The Hartree-Fock determinant $|\Phi_0\rangle$ fulfills this condition and is therefore stationary with respect to orbital perturbation, because the constituting orbitals are converged (in SCF iterations) solutions to these equations, where the Fock operator is calculated from the $N$ orbitals with lowest energies.
Any other (excited) determinant $|\Psi\rangle$ is generally not stationary with respect to perturbing the orbitals. Since other orbitals are occupied in this determinant, it corresponds to a different Fock operator $\hat {f'}$ where the sum is over the orbitals in the determinant $|\Psi\rangle$ which differs from the set of orbitals included in $|\Phi_0\rangle$.
Using this Fock operator $\hat{f'}$ in subsequent SCF cycles, the total energy of the excited determinant is again minimised by tweaking the orbitals. In this course, I suppose, it is quite likely that the excited determinant relaxes back to the ground state Hartree-Fock determinant  $|\Phi_0\rangle$. The iterations might also converge to a state with higher energy, but one-determinant representations of excited states are rather inaccurate anyway.
In principle, in should be possible for an excited determinant to converge to a determinant with lower energy if and only if the initial converged solution is merely a local minimum. However, this is very unlikely if using a reasonable initial guess.


How can a catalyst not be included in a rate equation if, by definition, it speeds up a chemical reaction? I thought that anything not in a rate equation was automatically zeroth order and therefore did not affect the reaction. However, I have heard that catalysts can be involved in a reaction while not being in a rate equation. How can this be possible?

Enzyme kinetics is one area of catalysis that has been studied extensively. The Michaelis-Menton equation being one of the central pillars and explicitly states the roles of concentration for both the substrate and enzymes.

Also, to answer your question: the rate equation is at times more interested in establishing equilibrium conditions than obtaining the time it takes to get there. The time to reach equilibrium is slightly neglected as the question "are we there yet" seems trivial when distance = rate times time.


Why is there an exponent 4 after the brackets in sp3? My professor wrote an electron configuration for carbon as: 1s2 (sp3)4

I thought it was just 1s2 sp3 where did the 4 come from?

The electron configuration of carbon is canonically written as 1s22s22p2. What you now can do is invoke orbital hybridization and combine the electrons in the 2s and 2p orbitals. Then you get sp3 hybrid orbitals with a total of 4 electrons in them. The 1s electrons stay where they are. To mark the new hybrid orbitals as separate, you put brackets around them.

You then end up with an electronic configuration of 1s2(2sp3)4.


Sum of spins of electrons in a wavefunction I wanted to ask a question about spin in an electron.

We were presented earlier today with the following information:

Using my synoptic knowledge, I recall that spin up $\alpha$ had a value of $+\frac{1}{2}$ and spin down $\beta$ had a value of $-\frac{1}{2}$.

Also, using the information above, $\alpha(1)\alpha(2)$ would have a total Spin quantum number $S$ of $\frac{1}{2} + \frac{1}{2} = 1$, $\beta(1)\beta(2)$ however I thought would result to being $S$ of $-\frac{1}{2} - \frac{1}{2} = -1$ and even so, one spin up and one spin down $\alpha(1)\beta(2) +  \alpha(2)\beta(1)$ would total $\frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2}= 0$. These calculations contradict the data given above for the triplets and singlets states.

I'm confused. How are the values of $S$ for $\beta(1)\beta(2)$ and $\alpha(1)\beta(2) +  \alpha(2)\beta(1)$ exactly derived?

I think the reason you're getting confused is because when we talk about "spin", we actually are referring to two separate observable values, the z-component of the spin (properly designated $\hat{S}_z$) and the square of the magnitude of the three-component spin vector (properly designated as $\hat{S}^2$).

The commonly used values of $+\frac12$ and $-\frac12$ are shorthand for $+\frac12 \hbar$ and $-\frac12\hbar$, the only values possible to measure for $\hat{S}_z$.

Because of quantum uncertainty requirements, we can know the values of both the z-component and the square of the magnitude of spin at the same time, but we cannot know either the x- or y-component while we know the z-component. As a result, we can only say that the spin vector lies somewhere on a the surface of a cone centered on the  z-axis that expands upwards from the origin ($+\frac12$) or downwards from the origin ($-\frac12$).

For two uncorrelated spins (like in a triplet state), the z-components either add (2 positive) or (2 negatives) or cancel (1 positive and 1 negative), but that only means that the net z-component is 1, 0, or -1. Even if $\hat{S}_z=0$, there is still a net spin from the x- and y-components. The value of the magnitude (or technically its square) comes from $\hat{S}^2$, which is conserved as 1 for all three possibilities. That's the S value given in your notes. For the $\hat{S}_z=0$ case, that means the spin vector lies in the x-y plane as one of the possible radial vectors of a circle with radius 1 centered on the origin in that plane.

For the correlated spins (singlet state), the magnitude squared is 0, which occurs because the two spin vectors exactly oppose each other, and all three components cancel out, not just $\hat{S}_z$, though we have no measure of the x- and y- components that are cancelling. We only know the resulting magnitude is 0.


Directional nature of the acetamido group in electrophilic substitution reaction Here, in $\ce{CH3CONH2 -}$ group, the nitrogen has a lone pair and can thus activate the benzene ring along with having an ortho-para directive effect in the following manner:

However at the same time, both nitrogen and oxygen are electronegative. Hence, they should attract (roughly speaking) the electron cloud of the ring to deactivate the ring and may result in a meta directive effect in the following manner:

So which effect is applicable here and why?

First of all, your second diagram concerning canonical structures are INCORRECT. There $\ce{N}$ has formed five bonds. $\ce{N}$ can't form five bonds because of unavailability of valance $\ce{d}$ subshells. It can only have the maximum covalency upto $4$ and how do you even think of keeping $\ce N$ atom neutral after forming five bonds which is anyway wrong. So, please make that correction.

There will be only one resonance structure:

(source)

Due to that delocalisation with carbonyl carbon, $\ce N$ will acquire a $\delta$+ charge which will increase its electron withdrawing effect (similar to $\ce{-I}$ effect) to some extent. Still, due to large no. of canonical structures of delocalisation in the benzene ring from the $\ce N$ atom, its $\ce{+R}$ effect is much stronger (strong $\ce{2p$\pi$-2p$\pi$} $ overlap) than its $\ce{-I}$ effect and thus the $\ce{-NHCOCH3}$ group still acts as an ortho-para directing group. But, its activating ability is slightly decreased from $\ce{-NH_2}$ group due to that $\ce{-I}$ effect.


Identify what alloy of metal tool The title is more general than my particular question, in particular I bought a "allen key" and it broke after doing a lot of force with it trying to turn a hard screw.
In order to get a harder tool I need to know the material of this one. Tell me how to test if this allen key is chrome vanadium. If i should weight it, measure volume thus density, and this picture:

broken allen
The other broke like this, it deformed:

deformed allen key, because forcing

I want to be also a little educated on this topics: probably the tool is chrome vanadium but of low quality, or maybe it is extremely expensive or hard to test for me if the new tools I bought are of another material and if it's harder than the broken ones, the new ones are chrome vanadium and "stanley" (a corporation).

You can tell by the dark gray appearance that it is not vanadium chrome, but rather an alloy of hardened steel.  If it were made of vanadium chrome, it would not have broken, and it would also have a silver appearance to it.  Density may help tell if you are close, but with alloys the density does not necessarily conform to the rule of mixtures.


In the wet medium of an ocean, how does a hard shell form? In general, if you were to try make something hard in a liquid medium (especially water) it is quite difficult to make the material solid.

Things such as mollusks though, have no problem generating hard shells in a wet medium.

What exactly are the physics behind this formation?

Carbon dioxide from the atmosphere (or from decaying matter in the ocean) reacts with water to form carbonic acid:

$$ \ce{ CO2 + H2O -> H_2CO_3 }$$

and this reacts with calcium ions to form calcium carbonate:

$$ \ce{H2CO3 + Ca^{2+} -> CaCO3 + 2H+ }$$

The solubility of calcium carbonate is about $13~\mathrm{mg \over L}$, so if the concentration of calcium carbonate is greater than this the excess will precipitate out as solid calcium carbonate.

Shell-forming organisms actively absorb calcium from the water around them, so they are able to increase the concentration of calcium carbonate to above $13~\mathrm{mg\over L}$ and precipitate the excess to form their shells. They can get the carbon dioxide from the water around them or from their own metabolism. The actual details of shell formation is far more complex than this as it's controlled by processes with the cells of the organism rather than just being uncontrolled precipitation. I'm not sure how well the details are understood even today.

Incidentally, this is why shelly fauna are not keen on the acidification of the oceans that results from increased atmospheric carbon dioxide. The solubility of calcium carbonate is strongly $\mathrm{pH}$-dependent and rises sharply as the water gets more acid. Given that the shell is usually intended to stop other animals eating you, having your shell dissolve is generally not good for life expectancy.


Why do atoms get smaller moving left to right on a periodic table? I get that it's because of the increase number of protons, but I still don't understand the concept. When you move left to right on a periodic table, the protons increase at the same rate as the electrons. In fact,  the number of electrons end up being the same as the number of protons, no matter how far you are across a period (unless it's an ion).

So shouldn't the attractive force of the nucleus and the forces of electron-electron repulsion be cancelling out due to the equal number of protons and electrons in an atom? Or is it because protons have a greater force of attraction compared to the repulsion force between electrons?

Atomic size across a period, in fact, has everything to do with electron shielding and the effective nuclear charge (charge felt by the electrons).

Across the period, electrons are added to the same outer level, meaning that shielding by inner electrons does not change, however, the numbers of proton increases. This results in the effective nuclear charge rising significantly (as no. protons increases while added electrons are shielding inefficiently), which of course, means that the atom has a smaller size.


Unexplained peak cluster in proton NMR trace? I am working on a straight-forward NMR graph as part of my test prep. I have a pretty limited understanding of NMR, but I thoroughly read the chapter leading up to this practice question and am having difficulty explaining all of the peaks that I am seeing. Let me show the question and my understanding:

Here is the chemical and the trace that results.

My understanding of this problem is that there are four groups of equivalent hydrogens, and I numbered them 1-4. The 1st group has 3 equivalent hydrogens, with two adjacent non-equivalent hydrogens (group #2). Using the n+1 rule, the H group #1 should create 2+1=3 peaks in a cluster. Same goes for H group #4, since there are two adjacent hydrogens, so another 3 peaks. When considering H-group #2, the three hydrogens from H-group #1 should be 3+1=4 peaks. Same goes for H-group #3, which has the three hydrogens from H-group #4 so 3+1=4.

So, I am expecting to see 2 groups of 3 peaks and 2 groups of 4 peaks. I see the two groups of 4 peaks, but not the 2 groups of three peaks. Instead, I see 1 group of 6 peaks.

Am I supposed to believe that the 2 groups of 3 are just located so near it appears to be 6? Or am I missing something?

Thanks for any help!

You are correct. The chemical shift of methyl protons 1 and 4 are close to one another.  You still see the two expected triplets, they just have similar chemical shifts.  That's not too surprising since they are both methyl groups attached to a methylene group.  We could tell which triplet belongs to which methyl carbon by measuring and comparing the 2 sets of CH2-CH3 coupling constants.


Why is the electronegativity of potassium and rubidium same? The electronegativity of potassium and rubidium is reckoned at 0.82 for both. Why is it same for both of them? Shouldn't it be less for rubidium as compared to potassium owing to the addition of one extra shell in rubidium?

There are a few ways that electronegativity can be measured/calculated, the original being that of Pauling, followed by others scales such as the Allred-Rochow and Mulliken electronegativity. When looking at the values obtained by Pauling, one must keep in mind that these values aren't grounded in quantum mechanics (i.e. electronegativity does not have an observable that can be measured). However, these values seen on the Pauling electronegativity charts are relative values estimated from bond energies.
Anyways, defining Pauling's electronegativity as the "ability of an atom to attract electrons" (Matthew Salem, Chem Libre Texts), one can look at two factors to estimate these values.
1. Nuclear Charge/Atomic radius - As we know from Coulombs law,
$\large F=k\frac{q_{1}q_{2}}{r^2}$
the attraction between to oppositely charged points is a function of the magnitude of each charge (q) and the distance between them ($r^2)$. So the number of protons in the nucleus along with the distance of the electron from the nucleus determine the attractive force it feels from the nucleus.
2. Electron Shielding –  The electrons that are closer to the nucleus can shield bonding electrons from the  attractive forces of the nucleus. The ability to shield outer electrons is greatest in the s orbitals, a little weaker in the p orbitals, and poor in the d/f orbitals.
With these factors in mind, one can approximate the effective nuclear charge $(Z_{eff})$ of an atom, which is an estimation of the attraction certain electrons will feel. Effective nuclear charge is estimated with the equation below,
$(Z_{eff}) = Z - S$
where Z is the atomic number and S is the shielding constant (number of non-valence electrons).
If you calculate the $(Z_{eff})$ for rubidium and potassium, one would find that they are equal, suggesting that the electronegativity of each atom is are approximately the same.
This link discusses $(Z_{eff})$ in more depth if you are interested.


How much pressure is needed to make dry ice? By compressing $\ce{CO2 (g)}$ the gas gains a lot of heat. When the hot compressed carbon dioxide is left to cool, it attains the temperature of the surrounding material. When it is released into the pressure outside (about 100kPa), it will get so cold (actual temperature depending on how much the carbon dioxide was compressed), that it will freeze (below 78.5 degrees celcius).

How much pressure is needed to make dry ice at 293K (20 degrees celcius)?

Isn't that a bit different?

At 293 K, you need about 56 bar to liquefy carbon dioxide. If you then release the pressure, a part of the liquid evaporates again while cooling down the rest until it solidifies at around -80 °C.


Why does K have a greater electron affinity than Ca? Why will $\ce{K}$ have a greater electron affinity than $\ce{Ca}$? I know that $\ce{Ca}$ has a greater first ionization energy, and I also know that $\ce{K}$ has a greater second ionization energy, but I am confused on how to explain this.

Neither K nor Ca "wants" to ionize.  Both ionization energies are endothermic.  This question refers to eelctron affinity, which is gaining an electron.

K has a more exothermic electron affinity because the electron gained fills the 4s orbital.  In Ca the electron goes to 3d.  The energy gap between 4s and 3d is enough to make the process barely exothermic.

Note that the numbers given in tables are positive but that the process is exothermic.


Priority in naming organic compound I'm confused if ethyl and methyl can get any of the number 4 and 5 then what will be the order?

The most important simplified criteria for the numbering of simple compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Therefore, if no decision can be made about the locants for an ethyl and a methyl group using the previous rules, the lowest locant is given to the ethyl group according to Rule (g) because ‘ethyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).

For example, a correct systematic name is 4-ethyl-5-methyloctane rather than 5-ethyl-4-methyloctane since ‘ethyl’ is cited first in the name.


Tools to draw molecular structure (wireframe) along with atom numbering I am looking for open-source tools which can generate 2d molecular structure from the MOL or SMILES file. Further, it SHOULD number the atoms in the order of occurrence of atoms in the connection table of MOL file or SMILE string. I used MarvinSketch 6.1.2 but the position of numbering is not very good, as it get cluttered with the bonds.
Please refer some tools with which I can draw molecule similar to the one shown in figure but numbering is also clear. In the figure, all numbering except 1-7 are ok.

To answer the question, I assumed you would like to stick with Marvin if possible. Likely, the options described and found functional in version 5.2.4 met at this site are functional at later versions, too.

So I drew from scratch a similar representation of your molecule.  And indeed, there is overlap by labels and the drawing.

Yet the software allows you to print the drawings.  So, thanks to the installation of HP-1102P printer and printing into a file I got a postscript file.  This vector format is of course beneficial, you may import it in inkscape, for example.  Note two potential pitfalls: select "import" (instead of "open") and "import text as text".  Twice you have to ungroup the drawing before being able to move the labels around as you prefer.  Or to change their colour (vide infra), or substitute them.  Eventually, you may either save it in numerous file formats or like in this case, export the resulting bitmap.


How to apply Henry's law down the sea? Background: Henry's law states that:

  At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

And mathematically:

$$S=k\times p_{\text{gas}}$$
where $S$ is solubility, $k$ is Henry's constant and $p$ is the partial pressure of the gas.

Question: I want to calculate the oxygen solubility down the sea. Is partial pressure the atmospheric one? If it is so, we have same solubility on all sea deepness, which is contradictory. What am I missing?

Edit

I know this law can be thought as depending on this equilibrium:

$$\ce{A {(g)} <=> A {(s)}}$$

What does it even mean to calculate or measure solubility of something down the sea?

Why, there is a precise and well-defined meaning, and it is hinged on equilibrium between the gas and the solution, which supposedly should be established at the said conditions. Our imaginary gas bubble, like everything else, must withstand the immense pressure of all those layers of water up above.

So no, the partial pressure is not just the atmospheric one.

Deep abyss of the seas is undersaturated in all gases, including oxygen, and that by a wide margin.


One mole of solute (NaCl) is dissolved in 1 litre water. The molarity of solution is One mole of solute ($\ce{NaCl}$) is dissolved in $\pu{1 L}$ of water. The molarity of the solution is:
A) $\pu{<1M}$
B) $\pu{>1M}$
C) $\pu{=1M}$
D) $\pu{=2M}$

My answer: It’s a fairly simple question so, I did what anybody would do.
$\mathrm{Molarity}=\frac{\text{No. of moles}}{\text{volume of solution}}$
$\mathrm{Molarity}=1$
That is the obvious answer, and it’s probably right, but the answer given is $\pu{<1M}$. I just wanted to confirm that the answer given is right or wrong. If it is right, why?

The answer $<\pu{1M}$ is correct.

Suppose you have $\pu{1L}$ ($\pu{1dm^3}$) of water and then you add your salt. When you add your salt the volume of the solution increases so now you have $\pu{1 mol}$ of $\ce{NaCl}$ in more than $\pu{1L}$ of solution. That means that you can't have a $\pu{1M}$ solution anymore.

If you want a $\pu{1M}$ solution then:

Add $\pu{1 mol}$ of salt in a flask.
Add water until you reach $\pu{1L}$ of volume.

EDIT: when you add salt, the volume of the solution increases slightly and less than what you expect(volume of water + volume of dry salt).


Interpretation of miscibility curves I stumbled across miscibility curves in the section Solubility of Liquids in Liquids (Organic Chemistry, by Wallwork and Grant) today. I'm not sure if my problem is with the way the content was laid out in the book or whether I'm not fully understanding the material, but I feel the book has dealt with the topic in a rather, obscure way.

I've taken snapshots of it and uploaded a portion of my text below.

What I want to know is if I interpreted the text (the bit about miscibility curves) correctly, which I seriously doubt.

Here's my interpretation:

This is the miscibility curve for phenol in water blah-blah-blah. All the points present in the area bound by the curve with the x-axis (concentration axis) are points of immiscibility. The curve itself represents the threshold of miscibility.

For example, at 20°C, phenol is soluble in water up till the concentration corresponding to point A on the curve. However, for phenol concentrations corresponding to points between A and B (lying on the line AB), the phenol-water mixture is immiscible and the phenol and water form separate layers. Although, at concentrations corresponding to point B (and beyond), the phenol is completely soluble. When for a constant temperature, the miscibility curve maps to two concentrations on the x-axis (eg; points A and B in the graph), which represent thresholds of miscibility at that temperature, the two mixtures (i.e- at A and B) are called conjugate pairs.

I labored for over two hours to come to the above conclusion, but when I saw the portion of the text which mentions that the liquids A and B exist together (at equilibrium) for the given temperature, I had to throw in the towel.

I have a very strong feeling that a large part of my 'interpretation' is horribly incorrect. If anyone's got the time, could you explain what the text actually means?

This is no different from what is observed in many binary systems. Here, the phenol and water exhibit a miscibility gap. As you noted, above point C, the liquid is one liquid with whatever composition it has. If one were to have a liquid at the composition corresponding to C, at a temperature above 66C. you would have one liquid. As you dropped the temperature to 66C, nothing would change.

Drop below 66C, and the liquid will phase separate in to two liquids of different compositions, given by the points on the miscibility curve. To figure out how much of each liquid you have, you do the appropriate weighted average as given by the book. That weighted average can be seen to be correct, since, say at 20C, sitting on point A you would be 100% liquid with the A composition, and sitting on point B you would be 100% liquid with the B composition. In between, you are balancing between the two liquids, which are in thermodynamic equilibrium.

If you have a book of phase diagrams handy, look at almost any phase diagram involving Pb, and you will likely find a liquid-liquid miscibility gap. For fun, look at Ag-Ni, two metals with fcc crystal structures, and you will see that the solids are immiscible, and there is even a liquid-liquid miscibility gap.


How can I avoid crystallisation of sugars in vinegar at low temperatures? I produce high density balsamic vinegar, when it's winter and the temperatures are very low, high density products tends to crystallize. think this is a problem of over saturation of the glucose. The main sugars are glucose and fructose the ratio is 1:0.95.

I try to make products that are lower than 70 brix, or 1,35 density and 6% acidity, composed by $\ce{CH3COOH}$ and tartaric acid.

How can I exactly calculate the solubility threshold of this mixture? How can I measure it?

How can I prevent the crystallization to happen?

Thank you very much
I can't sleep at night because of this problem !!!

It is difficult to correctly measure the solubility, especially if it goes to multicomponent solutions, as your vinegar. It is possible to find tables with some components of your mixture, but you won't get the full picture, and it looks as the moment, when the hands-on experience is most important.

Nonetheless, I found some facts for you, namely:
the solubility of a-D-glucose in all the solvents increases with
increasing temperature and decreases with increasing volume fraction
of acetic acid, from Solubilities of {α-d-glucose in water + (acetic acid or propionic acid)} mixtures at atmospheric pressure and different temperatures

Another point is that the supersaturated solution can be stabilized by various means, and I think it already happens, as your vinegar contains lot of "impurities" from the grapes.

As for the second point, how to prevent it, the simplest solution will be to keep the temperature high enough, so that the crystallization does not occur. If this is not possible, the supersaturated solution will crystallize, sooner or later. If you than ask, how to slow down the process, until spring comes, filtration could help to remove the crystallization nuclei.


Why is chemistry unpredictable? Disclaimer: I am not a chemist by any means, and I only have knowledge limited to what I learned in my university's Chemistry III course. Basic understanding of everything up to valence electron orbitals.
Why is there no set of rules to follow which can predict the product of chemical reactions? To me, it seems that every other STEM field has models to predict results (physics, thermodynamics, fluid mechanics, probability, etc) but chemistry is the outlier.
Refer to this previous question: How can I predict if a reaction will occur between any two (or more) substances? The answers given state that empirical tests are the best way we've gotten to predict reactions, because we can discern patterns or "families" of reactions to predict outcomes. Are we only limited to guessing at "family" reactions?
In other words, why am I limited to knowing my reactants and products, then figuring out the process? Can I know the reactants, hypothesize the process, and predict the product?
If the answer is "It's complicated", I would enjoy a push in the right direciton - like if valence orbitals actually do help us predict, or any laws of energy conservations etc, please give me something which I can go research.

First of all, I'd ask: what do you admit as "chemistry"? You mentioned thermodynamics as being a field where you have "models to predict results". But thermodynamics is extremely important in chemistry; it wouldn't be right if we classified it as being solely physics. There is a large amount of chemistry that can be predicted very well from first principles, especially using quantum mechanics. As of the time of writing, I work in spectroscopy, which is a field that is pretty well described by QM. Although there is a certain degree of overlap with physics, we again can't dismiss these as not being chemistry.
But, I guess, you are probably asking about chemical reactivity.
There are several different answers to this depending on what angle you want to approach it from. All of these rely on the fact that the fundamental theory that underlies the behaviour of atoms and molecules is quantum mechanics, i.e. the Schrödinger equation.*
Addendum: please also look at the other answers, as each of them bring up different excellent points and perspectives.
(1) It's too difficult to do QM predictions on a large scale
Now, the Schrödinger equation cannot be solved on real-life scales.† Recall that Avogadro's number, which relates molecular scales to real-life scales, is ~$10^{23}$. If you have a beaker full of molecules, it's quite literally impossible to quantum mechanically simulate all of them, as well as all the possible things that they could do. "Large"-ish systems (still nowhere near real-life scales, mind you — let's say ~$10^3$ to $10^5$) can be simulated using approximate laws, such as classical mechanics. But then you lose out on the quantum mechanical behaviour.
So, fundamentally, it is not possible to predict chemistry from first principles simply because of the scale that would be needed.
(2) Small-scale QM predictions are not accurate enough to be trusted on their own
That is not entirely true: we are getting better and better at simulating things, and so often there's a reasonable chance that if you simulate a tiny bunch of molecules, their behaviour accurately matches real-life molecules.
However, we are not at the stage where people would take this for granted. Therefore, the ultimate test of whether a prediction is correct or wrong is to do the experiment in the lab. If the computation matches experiment, great: if not, then the computation is wrong. (Obviously, in this hypothetical and idealised discussion, we exclude unimportant considerations such as "the experimentalist messed up the reaction").
In a way, that means that you "can't predict chemistry": even if you could, it "doesn't count", because you'd have to then verify it by doing it in the lab.
(3) Whatever predictions we can make are too specific
There's another problem that is a bit more philosophical, but perhaps the most important. Let's say that we design a superquantum computer which allowed you to QM-simulate a gigantic bunch of molecules to predict how they would react. This simulation would give you an equally gigantic bunch of numbers: positions, velocities, orbital energies, etc. How would you distil all of this into a "principle" that is intuitive to a human reader, but at the same time doesn't compromise on any of the theoretical purity?
In fact, this is already pretty tough or even impossible for the things that we can simulate. There are plenty of papers out there that do QM calculations on very specific reactions, and they can tell you that so-and-so reacts with so-and-so because of this transition state and that orbital. But these are highly specialised analyses: they don't necessarily work for any of the billions of different molecules that may exist.
Now, the best you can do is to find a bunch of trends that work for a bunch of related molecules. For example, you could study a bunch of ketones and a bunch of Grignards, and you might realise a pattern in that they are pretty likely to form alcohols. You could even come up with an explanation in terms of the frontier orbitals: the C=O π* and the Grignard C–Mg σ.
But what we gain in simplicity, we lose in generality. That means that your heuristic cannot cover all of chemistry. What are we left with? A bunch of assorted rules for different use cases. And that's exactly what chemistry is. It just so happens that many of these things were discovered empirically before we could simulate them. As we find new theoretical tools, and as we expand our use of the tools we have, we continually find better and more solid explanations for these empirical observations.
Conclusion
Let me be clear: it is not true that chemistry is solely based on empirical data. There are plenty of well-founded theories (usually rooted in QM) that are capable of explaining a wide range of chemical reactivity: the Woodward–Hoffmann rules, for example. In fact, pretty much everything that you would learn in a chemistry degree can already be explained by some sort of theory, and indeed you would be taught these in a degree.
But, there is no (human-understandable) master principle in the same way that Newton's laws exist for classical mechanics, or Maxwell's equations for electromagnetism. The master principle is the Schrödinger equation, and in theory, all chemical reactivity stems from it. But due to the various issues discussed above, it cannot be used in any realistic sense to "predict" all of chemistry.

* Technically, this should be its relativistic cousins, such as the Dirac equation. But, let's keep it simple for now.
† In theory it cannot be solved for anything harder than a hydrogen atom, but in the last few decades or so we have made a lot of progress in finding approximate solutions to it, and that is what "solving" it refers to in this text.


Ionic Mechanism for Wurtz Reaction Instead of going for the normal free radical mechanism for Wurtz reaction, What reasons prevent us from believing in the following ionic mechanism?

\begin{align}
\ce{R-X + 2Na &-> R^--Na+ + NaX}\\
\ce{R-X + R^--Na+ &-> R-R + NaX}\\
\ce{2R-X + 2Na &-> R-R  + 2NaX}
\end{align}

Related but doesn't answer my question here

The actual mechanism is believed to be strongly dependent on the nature of halide, metal (not only $\ce{Na}$ is used in this reaction), and solvent. The 'ionic' mechanism you have written here is the part of the most accepted mechanism which involves both, an 'ionic' and a radical reaction.
First step. Through the single-electron transfer from $\ce{Na}$ to the halogen atom, an alkyl radical $\ce{R^.}$ and sodium halide is formed:
$$\ce{Na + R-X -> Na+X^- + R^.}$$
Alkyl radical further acts as an electron acceptor from another $\ce{Na}$ atom to form highly nucleophilic intermediate (can be isolated in some cases):
$$\ce{R^. + Na -> R^-Na+}$$
This alkyl sodium intermediate reacts with another alkyl halide in the aliphatic nucleophilic substitution:
$$\ce{R^-Na+ + RX -> R-R + NaX}$$


Container to minimise radiation exposure from U-238 Consider a sample of uranium-238 that is undergoing decay to thorium-234. During the transmutation, a gamma ray is emitted from the nucleus. However, the thorium-234 daughter particle does not decay quickly enough to be detected.

What kind of storage container should the uranium sample be stored in to minimise radiation exposure?

Due to its very long half-life, U-238 is not particularly radioactive; the specific activity is only about 12 kBq/g. The emission probability of gamma radiation is less than 0.1 % per disintegration of U-238. Furthermore, the gamma energy is relatively low. Therefore, the direct gamma radiation from pure U-238 (i.e. disregarding Th-234 and further decay products) is usually insignificant.

For example, the gamma dose rate in 1 m distance from 1 kg of pure U-238 (about 12 MBq) is only approximately 0.1 nSv/h – probably less due to self-absorption. By way of comparison, the dose rate from natural background radiation amounts to about 100 nSv/h.

Nearly all of the alpha radiation from U-238 is absorbed within the sample. Only a small fraction originating from the surface of the sample can escape but is stopped by air within a distance of about 3 cm.

Therefore, you do not need any shielding. Just use a suitable closed container to prevent contamination or incorporation.


Pauli exclusion principle Pauli exclusion principle states that 'No two electron can exist in same state' or 'No two electron can have same set of quantum numbers'.

But in reality there is no state of electron. The state is of whole system which is composed of many electrons. It is our approximation that each electron is described by a single wave-function which we called spinorbital. So what does that statement even mean in reality?

Quantum numbers is only exact for hydrogen atom. What does quantum number even mean for a electron in a many electron system.

Wikipedia has the correct definition:

  A more rigorous statement is that with respect to exchange of two identical particles the total wave function is antisymmetric for fermions, and symmetric for bosons. This means that if the space and spin co-ordinates of two identical particles are interchanged, then the wave function changes its sign for fermions and does not change for bosons.

The textbook shorthand you quote, 'No two electron can have same set of quantum numbers', only makes sense when you are using the one-electron approximation. As you move beyond that approximation, you also have to move beyond this shorthand definition.