HOMO/LUMO and charge distribution I am a little bit confused between charge distribution and HOMO/LUMO site in determining chemical reactivity. In charge distribution, the sites which are deficient and abundant in electrons are localized (e.g. in carbonyl molecules, we would expect that the carbonyl carbon to be partially positive and we would predict that a nucleophile would attack into that deficient carbon) but in HOMO/LUMO orbitals the site for reactivity differs on the orientation and sign of the orbitals. It may be in HOMO this carbon has an orbital for an overlap but in LUMO a node may appear in that site. My question is where would the nucleophile react in a species, by charge distribution or HOMO/LUMO orbital?

You need both an empty LUMO for the electrophile with an in phase HOMO for the nucleophile at the site of attack AND the proper charge distributions.

The LUMOs are critical; you can't shove electron density into more electron density, and LUMOs represent a lack of electron density.  However, the electron density for a nucleophile will be attracted to the partial positive charges.

Organic chemistry normally requires you make judgments about the relative impacts of various phenomena, and this is another example.


What is SPDF configuration? Recently in my chemistry classes, the teacher spoke about SPDF configuration and then said that we'll be taught about it in higher classes.

But I'm sorta curious to know that what is SPDF configuration and is there something like - obtaining electronic configuration in SPDF format?

s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms.  These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p).

(image source)

So for example,

a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital
a lithium atom with 3 electrons would be $\ce{1s^2~2s^1}$
fluorine has 9 electrons which would be $\ce{1s^2~2s^2~2p^5}$

Again, as we read from left to right the energy of the orbital increases and the superscript shows the number of electrons in the orbital. You can read more here.


Why are methanol flames less visible than other flames? The combustion of methanol tends to generate flames that are less visible than the flames generated by the combustion of other substances?

My guess is that methanol burns quickly as it has a relatively simple structure that provides some of the oxygen needed for combustion.  I am assuming that the flame comes from a mixture of fuel and air that is undergoing reaction.  If the reaction is fast enough, then there will be less flame.  However, methane and hydrogen both have visible flames, much more visible than methanol, leading me to wonder how the one oxygen in methanol makes so much of a difference.

Here's a link to a YouTube video showing a dish of methanol burning next to a dish of ethanol.  The methanol flame is more difficult to see.  Methanol was also widely used as a fuel in the auto racing circuit, until fires that resulted from crashes or fuel filling accidents resulted in "invisible fires" that were difficult to see during daylight, resulting in severe burns to crews and drivers ("you can't fight the fire, if you can't see it", this is one reason why it was replaced by ethanol as a fuel for Indy racing).

Flame color (visibility) is dependent upon the elemental composition of the material being heated and the temperature to which it is heated.  This is because heat will excite electrons in the element(s) to an excited state; when these electrons return to the ground state visible light (flame test) will often be emitted

If we look at the carbon content of methanol and ethanol we find the following weight percents:

methanol - 38% carbon by weight

ethanol - 52% carbon by weight

If we compare the heats of combustion for methanol and ethanol we find:

methanol $\ce{~~}$ -726 kJ/mol (exothermic)

ethanol $\ce{~~}$ -1368 kJ/mol (much more exothermic)

There's just not much carbon there to burn in methanol!  Consequently burning methanol doesn't get that hot and its flame will not be as colored as materials that burn more exothermically and result in higher temperatures.


Heats of formation of neutral molecules and homolytic vs heterolytic bond dissociation in mass spectrometry I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

Great efforts have been made to generate accurate and reliable ion thermochemistry data. Once such data is available, it can be employed to elucidate fragmentation mechanisms and in addition, it is useful for obtaining some background on the energetic dimensions in mass spectrometry.
Heats of formation of neutral molecules, $\Delta H_{f(RH)}$, can be obtained from combustion data with high accuracy. Bond dissociation energies can either be derived for homolytic bond dissociation
$$\ce{R - H \rightarrow R^. + H^., \Delta H_{Dhom}} \tag{2.12}$$
or heterolytic bond dissociation
$$\ce{R - H \rightarrow R^+ + H^-, \Delta H_{Dhet}} \tag{2.13}$$
The homolytic bond dissociation enthalpies give the energy needed to cleave a bond of the neutral molecule which, in the gas phase is in its vibrational and electronic ground states, to obtain a pair of radicals which also are not in excited modes. Homolytic bond dissociation enthalpies are in the range of $3-5 \ eV$ (table 2.2). The heterolytic bond dissociation energies apply for the case of creating a cation and an anion in their ground states from the neutral precursor which means that these values include the amount of energy needed for charge separation; they are in the order of $10-13 \ eV$ (Table 2.3). Due to the significantly altered bonding situation, breaking of a bond is clearly less demanding in molecular ions than in neutral molecules.

I think this is good explanation. However, since this is my first exposure to these concepts, I'm interested in a higher-level explanation. Specifically,

What does "heats of formation of neutral molecules, $\Delta H_{f(RH)}$", mean in the context of mass spectrometry, and why is it relevant?

What is the relevance of homolytic vs heterolytic bond dissociation in mass spectrometry?

The Wikipedia articles for these two concepts, 1. Standard Enthalpy of Formation and 2. Bond Dissociation Energy: Homolytic Versus Heterolytic_Dissociation, offer more general explanations, but I'm interested in understanding in a general explanation in the context of mass spectrometry.
I would greatly appreciate it if some of the more knowledgeable contributors would please take the time to clarify these concepts.

The heat of formation of a neutral molecule is important in mass spectrometry because in order to measure the characteristics of individual molecules, a mass spectrometer has to convert them to relevant ions. Once ions form, they can be moved about and manipulated by external electric and magnetic fields as depicted in following sketch of mass spectrometer (MS):

The Ion Source, the Mass Analyzer, and the Detector are the three main components of the mass spectrometer. The Ion Source, which can be considered as the heart of the spectrometer, is the one needed to ionize individual molecules. Electron Ionization (EI) is the most common ionization technique used for mass spectrometry (Note: Older literature usually refer to EI as Electron Impact, but Electron Ionization is the currently accepted term; EI works well for many gas phase molecules, but it does have some limitations as well, such as causing extensive fragmentation so that the molecular ion is not observed for many compounds). In this ionization technique, molecules of the sample are bombarded by electrons generated from a heated filament (see the insert red box in the diagram). The electrons are produced by passing a current through the filament. The amount of current controls the number of electrons emitted by the filament. An electric field accelerates these electrons across the source region to produce a beam of high energy electrons. When an analyte molecule (usually organic molecule) passes through this electron beam, a valence shell electron can be removed from the molecule to produce an ion, mostly a positive ion (Note: Ionization does not occur by electron capture, which is highly dependent upon molecular structure).

Since the ionization is produced by a single electron that is accelerated to $\pu{70 V}$, the technique is commonly referred to as $\pu{70 eV}$ EI. This high energy electron collides with a molecule (analyte), it transfer more than enough energy to over come ionization energy of the particular site (usually a electron from a lone pair of oxygen of carbonyl group or alcohol group) of the molecule, often ionizing the analyte by knocking away one of the molecular electrons (either bonding or non-bonding). The resultant ion is a radical cation. This web page describes the possibility of the fragmentation happens in mass spectrometer:

  The amount of energy transferred during this process depends upon how fast the electron (from ion source) is traveling and how close it passes to the molecule (analyte). In most $\pu{70 eV}$ EI experiments, approximately $\pu{1400 kJ/mol}$ ($\pu{15 eV}$) of energy is transferred during the ionization process. There is, however, a distribution of energy and as much as $\pu{2800 kJ/mol}$ ($\pu{30 eV}$) is transferred to some molecules. Since approximately $\pu{960 kJ/mol}$ ($\pu{10 eV}$) is required to ionize most organic compounds and a typical chemical bond energy is $\pu{290 kJ/mol}$ ($\pu{3 eV}$), extensive fragmentation is often observed in $\pu{70 eV}$ EI mass spectra. The distribution of energy transferred during ionization and the large number of fragmentation pathways results in a variety of products for a given molecule. Other electron voltages may be used to vary the amount of fragmentation produced during ionization. For most organic compounds the threshold energy for EI is about $\pu{10 eV}$. After ionization of the molecule, the residual energy from the collision may cause the molecular ion to fragment into neutral pieces and smaller fragment ions. This fragmentation can be through either homolytic cleavage or heterolytic cleavage. The molecular ion is a radical cation, but the fragment ions may either be radical cations or carbocations, depending on the nature of the neutral fragment. For example, Fragmentation of Toluene is depicted in following diagram:

Important: In Mass spectroscopy, the electron volt ($\pu{eV}$) is frequently used as an energy unit (for single molecules or atoms). Keep in mind that $\pu{1 eV} = \pu{1.6022 x 10^{-19} J}$ (per molecule or atom). Thus, $\pu{1 eV}$ (per molecule or atom) $= \pu{96.4152 kJ/mol}$.

A recent publication described why most fragmentations in mass spectroscopy preferred homolysis over hererolysis (Ref.1):

  Bond strength is an essential parameter for discussion of bonding and reactivity. While bond fission can occur by homolysis or heterolysis, for neutral compounds such as alkanes the term “bond strength” generally denotes rupture by the lowest energy homolytic pathway. For example, the $\ce{C–C}$ bond strength in ethane is listed as $\pu{359 kJ mol–1}$,$^{[Ref.2]}$ representing homolysis, which requires one-third of the energy for heterolysis ($\pu{1297 kJ mol–1}$).$^{[Ref.3]}$

In addition, Ref.4 is a great review on mass spectrometry‐based analysis of protein structure.

References:

Karlee L. Bamford, Saurabh S. Chitnis, Rhonda L. Stoddard, J. Scott McIndoe, Neil Burford, "Bond fission in monocationic frameworks: diverse fragmentation pathways for phosphinophosphonium cations," Chem Sci. 2016, 7(4), 2544–2552 (doi: 10.1039/c5sc03804a).
Donald F. McMillen, David M. Golden, "Hydrocarbon bond dissociation energies," Annual Review of Physical Chemistry 1982, 33(Oct), 493–532 (https://doi.org/10.1146/annurev.pc.33.100182.002425).
Pratim K. Chattaraj, Andres Cedillo, Robert G. Parr, Edward M. Arnett, "Appraisal of Chemical Bond Making, Bond Breaking, and Electron Transfer in Solution in the Light of the Principle of Maximum Hardness," J. Org. Chem. 1995, 60(15), 4707-4714 (https://doi.org/10.1021/jo00120a010).
Frederik Lermyte, Dirk Valkenborg, Joseph A. Loo, Frank Sobott, “Radical solutions: Principles and application of electron‐based dissociation in mass spectrometry‐based analysis of protein structure,” Mass Spectrometry Reviews 2018, 37(6), 750-771 (https://doi.org/10.1002/mas.21560).


Can the following compound having a centre of symmetry be called both optically inactive and meso? The Centre Of Symmetry (COS) has been shown by a blue dot.

My questions are:

Can this molecule be called optically inactive? I have this doubt because single bonds permit free rotation and the molecule hardly
remains in the given form for long.
Can this molecule be called a meso-compound as it optically inactive?
Slightly unrelated: I'm actually a bit confused with the definition of meso. I learnt that meso compounds must have a plane of symmetry. Am
I correct?

According to Wikipedia a meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
Update: I realized my mistake in thought procedure after asking the question.(I fell into the "Meso Trap" :-P )
I would like to add this link as it might help future viewers.
http://www.masterorganicchemistry.com/2011/01/12/the-meso-trap/

You are correct that the molecule is not chiral. It is indeed a meso compound due to the internal symmetry plane. Also, don't worry about free-rotation of single bonds.


Entropy changes during transfer of heat between two bodies Consider two metal blocks, one at 500K and another at 300K, they are bought to contact with each other until they reach a common temperature.
Find the entropy change of each object and the total change considering there is no change in the volume of the metal blocks in the process and no the other interactions excluding the heat transfer between the two blocks.Also is this process irreversible, why?

also
$C_v= 10 J/K$ for both the metal blocks

My attempt:
This process is irreversible, as once it reaches the equilibrium temperature it cannot go back to the initial state without external intervention that is to say heat wont flow from cold body to hot body and also when two bodies are at the same temperature heat doesn't flow from one to the other to increase the temperature of one and to decrease the other.

now to compute the $\Delta S$ we can perform the same process very very slowly such that the two systems are always at equilibrium.
thus we can write $dS=\frac{\delta q_{rev}}{T}=\frac{dq}{T}=\frac{C_vdT}{T}$
thus for system 1: $\Delta S_1=C_v ln\frac{T_f}{T_i}$
and thus $\Delta S_{TOTAL}=C_v ln\frac{T_{f}}{T_{i1}}+C_v ln\frac{T_{f}}{T_{i2}}$

My questions :
I have a feeling my reasoning is not rigorous enough and my definitions are going in circles .
and my analytic solution is wrong as the i'm basically using $\delta q$ of the irreversible process and using the same information in the assumed reversible process.

This answer is correct.  After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature reservoirs running from its initial temperature to its final temperature.  At least, that is what your equation implies.  This is definitely the correct way to determine the entropy change.


Which of the these SN1 solvolysis of two alkenyl bromides reactions is faster? The SN1 solvolysis of one of the following alkenyl bromides is in MeOH approximately 100000 times faster than the other one. Which is the more reactive one? Draw the mechanism including the transition state, determine the rate determining step and draw the reaction product of each compound.

Now what is apparently important here is the rotation of the substituent in each relative to the aromatic ring. In the second compound (with two methyl groups), the rotational profile shows destabilized molecule except when 90 degrees. That means the rotation in the second molecule helps with the leaving of Br and this (rate determining) step is much faster than when there are no methyl groups (first compound).
Is that a complete explanation or I am missing something important? Also is there anything special about the mechanism of the two reactions and the products in each case?

[...] SN1 solvolysis [...]

Let's recall what we know about these. We have a two-step reaction, consisting of

heterolysis of the starting material
association of a nucleophile $\ce{Nu}$

We remember that the heterolysis is not assisted by $\ce{Nu}$, there is no backside attack of $\ce{Nu}$, pushing out the leaving group. We also remember that the association step (addition of $\ce{Nu}$ to the carbocation intermediate) is usually pretty fast. Consequently, heterolysis, i.e. the formation of a carbocation is the rate-determining step.

Thank's to good old Arrhenius, we know that the rate is determined by the activation energy of this step.

The activation energy is lower when the resulting cation is more stable.

The cation is more stable when the delocalization of the charge is better.

Distribution of charge on the vinyl cation is better when it's actually a
styryl cation.

What is the geometrical requirement for this interaction and what might prevent it?


Reversible & irreversible adiabatic processes Work done in reversible is more than that of irreversible

Suppose we are going from state $(P_1, V_1, T_1)$ to some other state say $(P_2, V_2, T_2)$ for reversible and $(P_2, V_2, T_2')$ for irreversible. Which should be greater, $T_2$ or $T_2'$?

My attempt

Let us say heat supplied is same i.e q for both reversible and irreversible. We already know that $U = q - W$

Reversible as work done is more so $U$ becomes less and in irreversible $W$ is less so $U'$ is more than $U$ i.e $U' > U$

As $U'$ is greater $T_2'$ should be greater. Is this correct explanation?

For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$.  Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$.  It doesn't matter what process is used to get from state 1 to state 2.  Only 2 intensive variables are required to specify the state of the system (assuming no phase change).  So $T_2=T_2"$.

Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.


Significant figure rules when plugging a value into an equation or formula I have not been able to find an answer to what may be a silly question so I am asking it on here as a last resort.

If I am to plug in say a number like 14.0 Celsius into the formula that solves for temperature in Fahrenheit.  So I would do 9/5 times 14.0 to get 25.2.  When I then add the 25.2 to 32 does the rule for addition matter here?

Would the answer be 57.2 or 57?

Is there a general rule for when I am plugging in a measured value into a formula that is given to me?  I believe that only the measured values are significant and all numbers in the formula are exact with infinite significant digits.

If the conversion is exact, all uncertainties carry over. Arguing the same for significant digits is very tricky and can lead you on erroneous paths. It is thus easier for you to convert the uncertainty that is present in the originally measured value into a new uncertainty for the converted value and then infer from that the number of significant digits.

Let's assume that your measurement is $\vartheta = 14.0 \pm 0.1 ~ \mathrm{^\circ C}$.

The conversion to Fahrenheit is
$$ \phi(\vartheta) = \left(\frac{9}{5} \frac{\vartheta}{\mathrm{^\circ C}} + 32 \right) ~ \mathrm{^\circ F} \,,$$
which we can straightforwardly apply to the mean of the measurement which gives us $57.2~\mathrm{^\circ F}$.

To do the error propagation under the assumption that the errors are independent, we do the following:
$$\sigma_\phi = \sqrt{\left( \frac{\partial\phi}{\partial\vartheta} \right)^2 \sigma_\vartheta^2} = \frac{\partial\phi}{\partial\vartheta} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \cdot 0.1~\mathrm{^\circ C} = 0.18~\mathrm{^\circ F}$$
The term under the root would in principle be a sum, but here we only have one variable with an uncertainty associated with it. For more details, look up this Wikipedia article on error propagation.

Anyway, our newly converted value is correctly expressed as
$$ \phi = 57.20 \pm 0.18 ~ \mathrm{^\circ F} \,.$$


Why are reversible reactions spontaneous? If a reversible chemical reaction is spontaneous in one direction, shouldn't it be non spontaneous $(ΔG > 0)$ in the other? In other words for the reaction:

$$\ce{A + B ⇌ C + D}$$

if

$$\ce{A + B -> C + D}$$

has $ΔG < 0$ and thus proceeds spontaneously, doesn't this imply that

$$\ce{C + D -> A + B}$$

would have $ΔG > 0$ and thus proceed non spontaneously?

A negative $\Delta G^0$ implying that a reaction will proceed spontaneously or a positive $\Delta G^0$ implying that a reaction will not proceed spontaneously is just a "rule of thumb" to give you a feel for what to expect.  But all a negative $\Delta G^0$ really means is that the equilibrium constant for the reaction is large (and formation of the products is highly favored), and a positive $\Delta G^0$ means that the equilibrium constant for the reaction is small (and formation of the products is not highly favored).  However, if you start out with only reactants present initially, irrespective of the sign of $\Delta G^0$, the reaction will proceed spontaneously.  However, if $\Delta G^0$ is negative, at final equilibrium, the extent of the reaction will be greater.


About combining a cation and an anion I’m trying to learn about ions. There is a slide with examples about combining cations with anions. It goes like this:

  $$\text{For} \ \ \ce{Al^3+} \ \ \text{ and } \ \ \ce{O^2-} \ \ \text{you get} \ \  \ce{Al2O3}$$

Wait what? So you have an atom of aluminum (in this case a cation because it loses three electrons) and an atom of oxygen (an anion since it receives two electrons). You combine them. Why do you now have TWO atoms of aluminum and THREE atoms of oxygen? Why does combining a cation and an anion suddenly spawn new atoms? And what happened to their charges?

The next example is this:

  $$\text{For} \ \ \ce{Ca^2+} \ \ \text{ and } \ \ \ce{Br-} \ \ \text{you get} \ \  \ce{CaBr2}$$

Alright. I see a pattern. For some reason, if the charge of an atom is positive or negative, such number will define the number of atoms for the other element. So since calcium has $2{+}$, there would be two bromine. And since bromine has $-1$ there will be one calcium. I don’t get why, though.

And a third example,

  $$\text{For} \ \ \ce{Na+} \ \ \text{ and } \ \ \ce{CO3^2-} \ \ \text{you get} \ \  \ce{Na2CO3}$$

Ok so this is a bit weird since there are now three elements involved.

But according to the pattern, since sodium has charge $+1$, there should be just one $\ce{CO3}$ molecule. And since $\ce{CO3^2-}$ has a charge of $-2$, there should be two of sodium.

Great, I suppose it’s an useful pattern, but my questions are:

Why? Why does losing/gaining an electron cause the other element to gain/lose atoms?
Does this pattern apply only when you combine a cation with an anion? What if it is the other way around? (An anion with a cation, or does it make no sense at all?)
After combining a cation with an anion, do both atoms lose their charge? Observe how the results no longer have $+$ or $-$ superscripts.

I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.

The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.

(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)

The total charges need to balance out per "unit cell" (an intermediate-level chemistry concept that means the "smallest repeating subunit of a crystal lattice.")
So this means that you need to balance the total charges by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.

In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.

Ionic compounds that have an overall neutral charge are written without the charge superscripts,  thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$

If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.

Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -> Al_2O_3$

Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -> CaBr_2$.

For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) molecule instead of an atom.

Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -> Na_2CO_3$.


Solvent for reaction mixture I'm looking for some suitable solvent for ether cleavage of Guaiacol (o-Methoxyphenol) using aluminium triiodide ($\ce{AlI3}$) with $\ce{DMSO}$. The original paper calls for using hot acetonitrile as a solvent, but I wonder what might be the alternatives. The ideal solvent would be some solvent with only mild toxicity.

What hydrocarbon solvents are a suitable replacement for this preparation?

Aluminium iodide ($\ce{AlI3}$) is first introduced in 1984 as an easily accessible, a highly regioselective versatile ether-cleaving reagent with novel cleavage pattern (Ref.1). Application of $\ce{AlI3}$ in organic synthesis has been reviewed (Ref.2) and it seems like acetonitrile is the best solvent to be used in catechol ether demethylation (e.g., 2-methoxyphenol or 1,2-dimethoxybenzene).

For example, recent article discussed an one‐Pot cleavage of aryl alkyl ethers by aluminum and iodine in Acetonitrile (Ref.3). Their proposed mechanism for the cleavage is depicted in following diagram:

Accordingly, since it needs acid work-up for final product, I'd suggest you may use acetonitrile as the solvent of reaction, and use your solvent of choice to extract the final product from aqueous work-out.

If you are not happy with that, you may use either carbon disulfide ($\ce{CS2}$), which is also toxic, or toluene (their are references have been in literature using benzene as a solvent with mixed results) (Ref.2). Use of $\ce{CS2}$ is applied for specific reactions such as substrates having allyl functionality. Yet, it should work with simple compound as yours.

I'm little curious of your intend to use non-toxic solvent in the reaction since DMSO is also toxic in certain concentrations. In that case, you may use solvent-free conditions (you need to use enough DMSO for convenience, which is not critical according to the mechanism)(Ref.4).

Note: Cyclohexane is not a suitable reagent for this reaction, although it was involved in some publications (Ref.3). For example, when demethylation of eugenol reaction has bee done in acetonitrile at $\pu{80 ^\circ C}$ for $\pu{18 h}$, hydroxychavicol was achieved in 78%. However, when the reaction was run in cyclohexane under identical condition, no hydroxychavicol was achieved.

References:

M. Vivekananda Bhatt, J. Ramesh Babu, "New reagents 3: aluminium iodide - a highly regioselective ether-cleaving reagent with novel cleavage pattern," Tetrahedron Letters 1984, 25(32), 3497-3500 (https://doi.org/10.1016/S0040-4039(01)91058-5).
Juan Tian, Dayong Sang, "Application of aluminum triiodide in organic synthesis," ARKIVOC 2015, (vi), 446-493 (http://dx.doi.org/10.3998/ark.5550190.p009.309).
Juan Tian, Huaxin Yue, Pengtao Yang, Dayong Sang, "One‐Pot Cleavage of Aryl Alkyl Ethers by Aluminum and Iodine in Acetonitrile," Chemistry SELECT 2019, 4(1), 38-41 (https://doi.org/10.1002/slct.201803469).
Mehran Ghiaci, Jila Asghari, "Dealkylation of Alkyl and Aryl Ethers with $\ce{AlCl3 - NaI}$ in the Absence of Solvent," Synthetic Communications 1999, 29(6), 973-979 (https://doi.org/10.1080/00397919908086060).


without using electronegativity, which one of these molecules is polar? If we are given 5 molecules without their electronegativity

BrF5

PBr5

CCl4

XeF2

XeF4

How to determine that one of them is polar or not ?

I know by using the difference in electronegativity we can determine the polarity
But again, electronegativity is not given  ?? /:

When we say something is polar, it has a very specific meaning. It means that there is a spatial separation of charges, or a dipole moment. In the simplest case, this is given by:

$$
\mu = Qr
$$

Where $\mu$ is the dipole moment, $Q$ is the charge on each pole (equal and opposite) and $r$ is the distance between the charges.

Anything with charges can have a dipole moment, and therefore be polar. In chemistry, we typically worry about whether or not two things are polar:

Bonds
Molecules

Bond Polarity

Chemists are concerned with the polarity of bonds because polarity affects the character of the bond - the more polar the bond is, the more it behaves like an ionic bond. The less polar, the more it behaves like a covalent bond. This has implications for everything from naming rules to reactivity, but all you need to know for this problem is:

  All bonds between two atoms are polar, unless the atoms are identical.

(This is including whatever else the atoms may be bonded to). The only question after that is "how polar is the bond," and that is where tables of electronegativity come in. Using an arbitrarily defined scale, we can measure the relative polarity of bonds by comparing the difference in electronegativity between the two elements.

Molecule Polarity

For molecules, polarity isn't quite so simple. We know that all bonds between dissimilar atoms are polar, but in a molecule, sometimes the dipole moments add up to form a net dipole moment of zero. This is a little bit difficult to explain if you don't have a mathematics and physics background that includes vectors and summations of forces/moments.

If you do, then when I say something like:

  The net dipole moment for a molecule is equal to the sum of dipole moments over each bond.

It will make sense to you. If you don't, then you have to get a little more creative. I tell my students to imagine that there are ropes connecting each outer molecule to the central molecule, along each bond. Then imagine pulling each rope towards the side that is more electronegative (has a higher electron density). The more polar the bond, the harder you pull on the rope. Then imagine whether the molecule moves. If the forces balance, it stays put, so the net dipole moment is zero and it is not polar. If it does move, there was a net dipole moment, so it is polar.

This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).

Let's look at each one, using wikipedia's geometric pictures.

$\ce{BrF5}$

As you can see, there is a lone pair at the bottom (this is a square pyramidal geometry). The net dipole moment will be pointing "up", which makes this a polar molecule.

$\ce{PBr5}$

This one has a trigonal bipyramidal geometry, with each bond symmetrically opposed to each of the others. If you add up the vectors, they result in a net dipole moment of zero. Therefore, it is non-polar.

EDIT: As ron points out in the comments below, this is not really a binary molecular compound, and in reality forms an ionic crystal structure with $\ce{PBr4+}$ and $\ce{Br-}$. I don't think the author of the question intended you to worry about that, but if they did, the answer is still "non polar," although we wouldn't really call it a "molecule" any more.

$\ce{CCl4}$

Carbon tetrachloride has a tetrahedral geometry, and all the dipole moment vectors cancel. Therefore, it is non-polar.

$\ce{XeF2}$

This one is linear - no net dipole moment, so it is non-polar.

$\ce{XeF4}$

I could only find the ball-and-stick model for this one. It is square planar - there is a lone pair of electrons (not shown) on each "face" of the square on Xe. There is no net dipole moment, so it is also non-polar.

Out of all of these, the only polar molecule is $\ce{BrF5}$ (the first one).

To review the steps:

Draw the Lewis structure
Figure out the geometry (using VSEPR theory)
Visualize or draw the geometry
Find the net dipole moment (you don't have to actually do calculations if you can visualize it)
If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.


How can you tell if a sugar is D/L in its chair form? I'm having trouble determining if sugars are D/L in their chair forms. My textbook has generously provided this image of D-Glucose.

So I know that from the fischer, when I put everything into a chair view, the things on the left point up, and the things on the right point down.

However, what about the bottom $\ce{CH2OH}$ group. Which way does it point, and does it determine D/L in a ring? I'm imagining "rotating" all the groups 1 position over clockwise (does it matter if its clockwise/counterclockwise? How do you know which way to spin it?) such that the hydroxyl is now where the $\ce{CH2OH}$ was, H is on the right and $\ce{CH2OH}$ is on the left.

Does that sound right? I can vaguely remember this from high school, but it's become very clouded.

D-glucose stays D-glucose regardless of whether it is in chair (hemiacetal) form, open chain form, or any other minor form. These different forms (if you could isolate them) would likely have different specific rotation values (interaction with plane polarized light). There are two chair forms-one with an axial OH and the other equatorial (probably favored)--but both are forms of D-glucose.


Why Use Lewis Structures to Determine Radicals As far as I know, Lewis Structures are convenient for representing molecules, but give no insight on the actual electron orbitals.

For example: The Lewis Structure of $\ce{O_2}$ does not reflect paramagnetism. Therefore, we use MO theory to explain it.

Then, why are Lewis Structures used to show if a molecule is a radical?

For example: hydroxyl is drawn as $\ce{.OH}$ to reflect the unpaired electron. That is a Lewis Structure.

Why do Lewis Structures work in this case?

Lewis structures are excellent for showing radical character for an odd number of electrons.

The reason they fail for molecular oxygen is that they Lewis representation carries no intrinsic information about the molecular spin, and as such it cannot distinguish the unpaired electrons in molecular oxygen's antibonding orbitals as spin matched (triplet) or spin opposed (singlet).

However, when there are an odd number of electrons, it is clear there will be one unpaired spin, and it is not relevant whether what the spin orientation is for these cases.

That being said, and somewhat outside the scope of this question, Lewis structures cannot distinguish between doublet and quartet states.


Is palladium really that poisonous? In Iron man 2, the protagonist Tony Stark was anxious about his blood toxicity due to palladium core that was keeping him alive was slowly poisoning his blood. It is written here that:

  The palladium core in the arc reactor that keeps Stark alive and
  powers the armor is slowly poisoning him, and he cannot find a
  substitute.

  The palladium core

  Palladium poisoning Tony Stark

But in the wikipedia article of Palladium, it is clearly written that:

  Palladium is a metal with low toxicity. It is poorly absorbed by human
  body when digested.[...] High doses of palladium could be poisonous;
  tests on rodents suggest it may be carcinogenic, though no clear
  evidence indicates the element harms humans.

So, Palladium has no clear evidence that it is harmful to humans, why it was chosen for the film?

To quote user paracresol's question from a very, very related question:

  Is there anything realistic[...], or is it just another movie gaffe?

Intoxication can be acute (short time of exposure, effects appear fast), or chronic (longer exposures, the onset toxic effects will take days, months or years). For Iron Man it seems clear that this would be a chronic intoxication.

I found this safety guideline from the EMA, which states:

  Mice given $\ce{PdCl2}$ in drinking water (5 mg palladium/l) for a lifetime
  showed suppression of body weight and an increase in amyloidosis of
  several inner organs. In another study of questionable value
  (chloropalladosamine was given enterally to rats for about 6 months) a
  dose of 0.08 mg/kg body weight (corresponding to 0.041 mg $\ce{Pd^2+}$/kg) was
  determined as the no-observed-effect-level (NOEL), above which changes
  in body weight, hemoglobin content, blood serum parameters as well as
  functional and morphological changes have been reported. There is
  insufficient histological information available from this study.

Extrapolating human toxicity from mice data is a long shot, but since this is all we have it will have to do. Also, bear with me for pretty inaccurate math from here on (it's fiction anyway).

It's been a while since I watched the movie, but if I recall correctly each palladium piece was roughly 7cm x 4cm x 1cm. That's a volume of 28cm$^3$. The density of palladium is 12.023g/cm$^3$, therefore this piece weighs a bit under 350g.

I'm going to assume that he burns through a palladium piece every two days. Good times.

The NOEL (see above) is 0.08mg/kg/day. For an 85kg human, that's 6.8mg/day,  roughly 0.02% of the mass in the reactor. I recall that the palladium was heavily corroded when he extracted it, so a loss of 0.5-1% of palladium seems pretty reasonable. Being conservative, the reactor would release about 0.9g a day. If more than one tenth of a thousandth of this palladium is absorbed by Tony's bloodstream (which doesn't sound like much), then he will be in trouble.

To sum up, yes, I do believe it possible to suffer from palladium poisoning if you have a chunk of that thing dissolving inside your chest day and night.

Answering the other half of your question, Tony Stark used palladium for his first arc reactor in the first Iron Man. I don't think it was planned, they just went with it.


Same monomers, different water content? We're currently studying "the chemistry of contact lenses", which is mostly polymers and gels. I just looked at the package of two pure hydrogel lenses made out of the same monomers, but they have different water content? Why?

There are a few things that influence the properties of polymers:

Molecular weight of the chains, i.e. how many monomers are joined in each chain, can drastically alter the polymer's properties. For example, for polyethylene, UHMWPE has extremely long chains and produces a very strong material suitable for bullet-resistant vests, whereas LLDPE has much shorter chains that produce a weaker material that is much more ductile (think about how easy it is to stretch a plastic shopping bag or the like).
Another factor that affects a polymer's properties is branching. Branching is when a polymer is not made up of nice straight chains, but chains that occasionally fork off into several different chains. The difference between LLDPE and LDPE is that LDPE has a lot more branching. This keeps the chains from packing as nicely so LDPE isn't as strong.
The other big factor (other than adding another monomer to make a copolymer) is the introduction of crosslinking. Crosslinking is where different chains are connected to one another, often by addition of a molecule that can bond to a chain at both ends, e.g. rubber vulcanization crosslinks the different chains with linear sulfur chains. Natural rubber is very different than what most people think; it's a soft sticky material that doesn't hold its shape. The polyisoprene chains it's composed of aren't crosslinked and can easily slide past one another, but once crosslinked, the chains are connected and aren't free to move as much. Depending on the degree of crosslinking, you can go from something somewhat soft, like a bouncy ball, to something very rigid, like a car tire.

Hydrogels are not simple linear polymers. While they're mostly made from a single monomer, small additions of related molecules are used to control how they polymerize. N.B. These apply generally to silicones, other polymers may have other branching/termination processes. If the monomer only has two sites at which it can connect to other monomers, naturally the only structure that can be formed is a linear chain. One can control the length of the chains by adding a small amount of a terminator—a monomer that can only make one connection. If a terminator reacts with a growing chain, it stops the growth as it can't connect to anything else. The more terminator added, the shorter chains are, on average. Similarly, if you add a monomer that has more than two reactive groups, branching and crosslinking will occur. Often, a non-crosslinked polymer is supplied as a liquid, then a crosslinking agent is added to harden the polymer. Dow Corning has a great primer on the actual chemistry involved.

What makes a hydrogel special is that it's a polymer network that encapsulates water, i.e. it's an extensively crosslinked polymer that is hydrophilic. If the polymer was just a series of individual chains, it would just dissolve in water since nothing is holding the different chains together. When crosslinked, water can get in, but the mesh of polymer is linked together enough to hold its shape, though it may swell to accommodate the water. Depending on the degree of crosslinking, more or less water can fit within the structure. A less linked polymer can fit more water since there are fewer crosslinks resisting, but is less rigid when hydrated for the same reason (think Jell-O). A more crosslinked polymer won't absorb as much water, but will be much more rigid (like a contact lens). Contact lenses have to balance holding their shape, oxygen permeability, surface hydrophilicity, etc. so there are many reasons why, despite being based on the same primary monomer, two different brands may have a different blend of the other stuff that lets one control the hydrogel's properties.


Why is the bond order in the SO₃ molecule 1.33 and not 2? $\ce{SO3}$ molecule has three double bonded oxygen to the central sulfur atom.

Sulfur has $\ce{sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen.
So shouldn't the bond order be 2?

The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: Hypervalency and the octet rule.

I stated in Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

  The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.

There are a couple of molecules that belong roughly to the same class. I call them Y aromatic systems: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see Why is there a need for resonance?)

I have explained the delocalisation in the nitrate ion, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare here). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.

However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.

Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme here.

Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions.


What makes many water soluble fertilizers blue? I've noticed that water soluble fertilizers tend to be blue. What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes? I haven't seen this in organic fertilizers. I figured chemists might be able to examine the ingredients of Miracle Gro or something and tell me. I'm guessing it has something to do with copper.

What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes?

I have some Miracle Gro on hand, the water soluble kind: The blue stuff is copper sulfate, or ${\rm CuSO}_{4}$. Here is the MSDS for Miracle Gro Water Soluble All Purpose Plant Food, which lists the chemicals in the mixture.

  Is this necessary or for aesthetic purposes?

It's necessary in these formulations for purposes detailed (in part) here and here. In short, it is included in these formulations because plants need copper for biosynthesis; and because it acts as a fungicide, presumably leading to a nicer ecosystem in one's lawn/garden.


Diastereoisomer of quinic acid and its proton NMR? Question:

  Quinic acid, A, has the gross structure shown below:

  (i) Draw all the possible diastereoisomers of A and indicate which are chiral.
  (ii) Quinic acid is optically active and the marked hydroxyl groups are cis to one another; which of your structures is consistent with this data?

  Using the $\ce{^{1}H}$-NMR data below, draw a likely conformation for quinic acid, making sure that you fully assign all these data.

  $\ce{^{1}H}$-NMR for quinic acid in $\ce{D2O}$: δ 4.09 (1H, q, 3 Hz), 3.99 (1H, ddd, 11 Hz, 9 Hz, 5 Hz), 3.39 (1H, dd, 3 Hz, 9 Hz), 2.12 (1H, dd, 13 Hz, 5 Hz), 2.05 (2H, d, 3 Hz), 1.86 (1H, dd, 13 Hz, 11 Hz).

My working so far:
$\hspace{29ex}$

Firstly, have I missed any diastereoisomers? And what does it mean by 'conformation' of quinic acid? Does it mean a chair conformer?

But I'm not sure about the shift at 4.09 ppm since I made the assumption that the quartet is actually a doublet of doublets. Also surely $\ce{H_{f}}$ would vicinal couple with $\ce{OH}$ on the same carbon?

I see one major issue in your working immediately: The NMR has been recorded in $\ce{D2O}$, hence there will be no more $\ce{OH}$ groups present in your molecule, only $\ce{OD}$ groups. Since deuterium is not hydrogen, you cannot observe any coupling to the $\ce{OD}$ groups as if they contained hydrogen.

Also, you said:

  I made the assumption that the quartet is actually a doublet of doublets.

This is an incorrect assumption. They have multiple dd’s that they reported as dd which gives us less reason to suspect an incorrect report. Note that typical dd signals have two different coupling constants, e.g. the proton at 2.12. Now you could imagine that the two doublets fall together. But in that case, they would fall together to become a triplet, not a quartet. You can try it with splitting trees: Either a dd stays dd, or it collapses into a t. And finally, the quartet cannot be a set of two doublet signals from distinct hydrogens since it only has an integral of 1.

That said, the attribution of different hydrogens to their position in the ring is remarkably simple. Let’s label the quaternary carbon $\mathrm{C1}$ and then go around the ring in any direction. Our first stop is $\mathrm{C2}$.

   Since this is a $\ce{CH2}$ group, its chemical shift must be lower than $3~\mathrm{ppm}$. We have two choices. Either we take the two hydrogens that exhibit a $^2J$ coupling or the other two. Since we can easily identify the $^2J$ coupling, let’s take those.

With the hydrogens of $\mathrm{C2}$ in hand, we can check, which hydrogen must be on $\mathrm{C3}$.

   Again it’s pretty simple. The two coupling constants that are not $^2J$ constants must all be present in the hydrogen’s signal. There is only one in question.

Once we have gotten that sorted, we can go back and determine the stereochemistry that must be present here.

   We have one very large $^3J$ coupling and one mediocre one. The very large coupling is, of course, due to an axial-axial relationship between the two hydrogens.

We now have three hydrogens firmly in place: Those on $\mathrm{C2}$ and the one on $\mathrm{C3}$. Let’s move on to $\mathrm{C4}$.

   There is one coupling constant on the hydrogen at $\mathrm{C3}$ which is not yet assigned and which thus must belong to the coupling to this hydrogen. It’s magnitude again tells us pretty clearly which dihedral angle we expect and thus what the relationship between the two must be.

With $\mathrm{C4}$’s stereochemistry sorted out, $\mathrm{C5}$ is next.

   The coupling constant is rather different this time, indicating the opposite relative stereochemistry. Note that the hydrogen at $\mathrm{C5}$ is reported as a quartet of the same coupling. We’ll get back to that in a second. For now, we know what must be substantially different here.

And now we can move on to $\mathrm{C6}$, the one carbon we are still missing.

   As I just mentioned, we have an apparant quartet for the hydrogen on $\mathrm{C5}$. But that’s okay. It just means that there are three rather similar coupling constants to neighbouring hydrogens. One is taken from the coupling to $\mathrm{C4}$’s hydrogen, meaning the two of $\mathrm{C6}$ must be the other two. Thankfully, there is one signal with an integral of two left, which also has the same coupling constant. And also thankfully, if you did everything correctly, there should be no questions as to why this stereochemistry along the $\ce{C{5}-C{6}}$ bond be correct.

Finally, we have a hint in the question as to the overall structure:

   Namely that the two hydrogens on $\mathrm{C1}$ and $\mathrm{C4}$ are syn. Thus, the carboxylic acid must be in equatorial position and the hydroxy group axial.

A note on finding the diastereomers: Enantiomers are only enantiomers if all stereocentres are reversed. Thus, if you take one centre and keep it the same all the time and only exchange the direction of the others, you are bound to find diastereomers only. I suggest immobilising the $\mathrm{C1}$ carbon since it is rather prominent up there all alone; it gives the entire thing less hassle. Also, any achiral isomers will always have a plane of symmetry through it which helps, too.


Carboxyllic acid Test I'm currently in high school doing a project in chemistry. So I had to make a project where concepts from physical chemistry will be used alongside organic chemistry.
After going through my book I decided to create a test for carboxylic acids using electrochemistry. The basis of the test is:

Aldehydes, ketones and alcohols are non ionic in aqueous solution
while carboxylic acids are ionic. Therefore  carboxylic acid can be
used as an electrolyte.

So I finalized the following:
Lead, copper (my school doesn't have much chemicals and reagents and mostly available one was lead and copper so I stuck with it). The thing is, boiling the given organic compound with elemental lead, hydrogen peroxide and taking the precipitate obtained and measure the weight and dissolve it in test tube with water this acts as an electrolyte for lead half cell and make a copper half cell and connect the copper electrode and lead electrode with wires and put a suitable salt bridge. If there is reading in ammeter connected to the wires then the given organic compound contains carboxylic acid.
Is my method okay?
Edit:

1)since someone pointed out a lot of flaws ,  can I do like this:
Taking the aqueos solution of the organic compound as electrolyte and putting Zn,Cu electrodes in it and connecting the electrodes using a wire and connecting the wires to an ammeter ,(will this work like a lemon battery?)

Let's be positive. The edited proposal has merit. You can pass a current thru a solution of a carboxylic acid and measure it with an ammeter (a more sensitive milliammeter would be better).
How about this: check for conductivity. Commercial conductivity meters use AC current and an AC ammeter. Use two inert, identical electrodes for this, and use DC if you can't arrange an AC apparatus. Do a conductivity measurement (i.e., measure the ohms resistance of the solution with the two electrodes taped to a certain distance separation (e.g., two copper wires (or stainless steel) 1 cm apart, 1-2 cm immersion, taped to a plastic ruler) so you can do the identical measurement on another solution.
Then, for a second test, examine the conductivity of that same solution diluted by a factor of 10. Now here's where you are on your own: the conductivity will be less, but not exactly 10 times less, because as the carboxylic acid is diluted, it will ionize to a greater percentage, and therefore carry more current at the same voltage than its concentration reduction predicts.
I say you are on your own, because the experimental results will not be as clear as you might wish (lots of opportunity for scatter), but this will enable you to explore the physical chemistry idea of the ionization of carboxylic acids and dilution and the question of voltage vs current, DC vs AC, electrode composition (reactivity); also bring up pH.
Above all, keep it simple! Consider two variables and plot one against another. You would be surprised how difficult that will be and how there are many more variables just waiting to hop into your experiment!
But you will have fun learning about all the variables and writing up the results and impressing your teacher with your comprehension. Good luck!


Is there any difference between thermal energy at room temperature in fluids(water) vs air? Context of the question: I was reading on hydrogen bonds, specifically how close they are to the thermal energy of the fluids. So I googled the value of thermal energy at room temperature, which was $\pu{2.5 kJ/mol}$. Which is a difference of $10$ order of magnitude (H bonds are $\pu{23 kJ/mol}$), which makes sense to me given the time taken to break ($1$ to $\pu{20 ps}$) and reform a bond ($\pu{0.1 ps}$). That is how the question I posed in the title arised.

Since sea water can retain heat for longer periods of time, and they release it more slowly as well – what is the difference between thermal energy at ambient room temperature for gas (atmosphere) vs fluids (water), and how does this change with volume?

First of all, just a minor thing, the thermal energy itself is always the same. You can't grab it, you can just measure it. What is different is the amount of energy "stored" in air or water, but its no different "energy".

The retainment of heat in the sea is due to the heat capacity.
https://en.wikipedia.org/wiki/Heat_capacity
This is a specific property of each molecule.
To increase the temperature of 1 g water by 1 K you need more Energy than for 1 g of air. That is the reason why the temperature near the sea is more moderate (absorbing heat in the summer and releasing heat in winter).

This is also a reason, why you can go into a sauna with 90 °C without burning yourself, but you can't grab into 90 °C hot water. This is also due to thermal conductivity but also due to the capacity. When your skin touches the 90 °C hot air, it is absorbing the heat providing you with cooler air around your skin. The same happens in water, but first, your skin has to absorb much more energy to cool down the water, second of all the surrounding water heats up the cooler water layer around your skin much quicker, than hot air does. That's why air is considered an isolation.


Finding the nearest neighbour, from an atom within a crystal Is there any software can give me the information of nearest neighbour, nest nearest neighbour,....(and their distance) from an atom within a crystal (described by a .cif file)? Perhaps some visualization software like VESTA?

I don't use VESTA and there are probably a few tools that can do this.

My suggestion would be to use the open source Python package ASE or "Atomic Simulation Environment." developed in part by the group of Karsten W. Jacobsen.

There are several algorithms available for generating neighbor lists and iterating through them:
https://wiki.fysik.dtu.dk/ase/ase/neighborlist.html

Here's an example from the documentation:

# get neighbors of atom 42
indices, offsets = nl.get_neighbors(42)
for i, offset in zip(indices, offsets):
    print(atoms.positions[i] + dot(offset, atoms.get_cell()))


Conversion of formamide to isocyanide Isocyanides are the only class of organic compounds which possess a divalent carbon. They react like a geminal nucleophilic/electrophilic synthon with oxidation of the C(II) atom to C(IV), which parallels the chemistry of carbenes and carbon monoxide.

Some of the most common methods for synthesizing isocyanides involve the dehydration of formamides. Various dehydrating agents (phosgene, phosphorus oxychloride) are used in the presence of a base (pyridine/TEA). In the follow reaction, however:

$\hspace{2.4cm}$

I'm unable to understand the role of p-TsCl, and can't work out the reaction mechanism.

Just like phosgene or phosphorus oxychloride, p-TsCl is acting as a dehydrating agent. Its affinity for oxygen atoms is fairly apparent when you consider its common use as an alcohol protecting agent. The mechanism is as follows:

$\hspace{2.6cm}$


About the meaning of the names of f orbitals Our chemistry teacher encouraged us to study the history and naming of the orbitals on the web. (actually, our textbook did, but that's irrelevant to the problem) I easily could find the reason behind their naming, and what their names stood for. However, some sources stated that f as in f orbitals stands for fine, while others claimed it to be fundamental. Which of these two is correct? And why did scientists gave f orbitals that name? (as you're aware, they did it due to the f block elements' emission spectrum; my question is, what was the specification of their emission spectrum?) I'm sorry if the question looks stupid, but I can't figure it out myself.

I would have to say "f" overwhelmingly stands for "fundamental" (not "fine").

An early (1915) reference is "Some Recent Discoveries in Spectrum Series" Astrophysical Journal, vol. 41, p.323

  Within each of these three systems there are, in general, four different types of series, known as "principle", "sharp" (second subordinate), "diffuse" (first subordinate), and "fundamental" (Bergmann).  These names are not always very appropriate; for instance the "sharp" series is often diffuse, the "fundamental" may be composed of the faintest lines in the spectrum, but this is not vital.

The fundamental series was discovered by Arno Bergmann in 1907, and is sometimes called the Bergmann series. It was named "fundamental" by William M. Hicks, I beleive in A Critical Study of Spectral Series. Part I: The Alkalies H and He (1911): "F is the symbol for the new series (fundamental) as P.S.D. stand for those already known" (page 57) and "It was my first impression that this would be the same for all the alkalies and the .987 of the earlier discussion. It would be the basis of a fundamental series for each
element, and the letter F which has been attached to it had its origin in this idea." (page 94).

The names existed before orbital theory.  There is no need to investigate f-block elements to observe excitations to f-orbitals.  Electrons of any elment can be excited to an f-orbital. In fact, Bergmann discovered the series in the sodium spectrum.  What was orginially known as the sharp series, were transition to s-orbitals, the primary series transitions to p-orbitals, etc.

See The Origin of the s, p, d, f Orbital Labels for additional information.


Can a charge smaller than that of an electron exist? If so how? In school I've always been taught that the smallest charge possible is that of an electron( or proton) however I recently solved a question regarding a dipole and the charge came smaller than that of a proton, after  the solution given in the book however, there was a little statement written saying this can also exist in real life as well which seems off to me.. like how can you get less charge then the smallest unit of charge? If the answer is quarks well don't they exist in a combined state and wouldnt they combine to form protons and stuff (don't know too much about this topic tho.. I know a little thru YouTube videos but that's it) and hence change the charge to a multiple of that of a proton? Any insight is welcome
*Please note that this is not a homework question, it's just out of my curiosity.. thx for understanding!
Also not rly sure if I should ask this on chem stack exchange or phy..

Dipole moments have a distance times an implicit charge, but you could give the moment as unit charge separation times a distance which is not the bond length. This would eliminate fractional charges and suggest some distance, which suggests that an electron and the nucleus are point charges separated by something less than the bond length. This is difficult to visualize. The electron is not really a point charge. Dipole moment is not separable into fractional charges or tiny distances, it measures a distortion of charge.
And when you think about it, an undistorted charge distribution (like electron density in benzene) is still "distorted" in that electron density is between atoms and above/below the atomic plane.
Other fractional charges might be imagined in compounds like magnetite: $\ce{Fe3O4}$. Each iron has a charge of $\pu{2.666}$. But if you describe magnetite as $\ce{Fe2O3.FeO}$, then two thirds of the iron atoms have a charge of +3, and one third have a charge of +2. More satisfying.
And then there is cyclopentadienyl anion:

where each carbon has an extra 0.2 extra electronic charge, and there are other ions like cycloheptatrienyl cation, cyclopropenium cation, etc. that have a surplus or deficiency of electrons.
But there is no fractional electric charge, just fractional molecules, or combined valences. Although, I still wonder how quarks can have a partial charge!


Reverse-Engineer a material Let's say I am seeking to find an alloy with certain properties, is there any method or process that enables one to backtrack from desired properties to actual structure and composition of a material?

In general no.

What you're describing is known as inverse design, that is, designing something with particular properties.

I agree completely with the comment above, that you can find many things from databases. But when you mix together multiple components, into new inter-metallic compounds or alloys, the results do not necessarily match the individual components.

Consider, for example, "Galinstan", the various alloys of gallium, indium, and tin. While each of these have low melting points, the eutectic point for the ternary alloy is $\:\mathrm{-19 ^{\circ} C}$.

There are many other examples, but the point remains that without careful experiments or theory for materials design, it's tricky to do inverse design.


What are examples for state functions? List some examples of state functions. Check all that apply:

  internal energy
  work
  volume
  pressure

I know that a state function is independent of the path taken, so I think work and volume are state functions. Am I correct?

Let's say you go to Stonehenge.
You see a big rock there.
You can observe the state of the rock.
But how much work was done to bring the rock there?
How far away did the rock come from? Was it a downhill or uphill journey?  Was the stone dragged, moved with wheels, or pulled on a sled over snow?

So you can know the exact state of the rock, but not how much work was done to bring it to that state.

Therefore work is not a state function. The others given in the exercise, i.e. internal energy, pressure, volume, temperature are all state functions.


Naming ester as a substituent to carboxylic acid Can carboxylic acid functional group -COOH and ester's functional group R-COO-R' be in one structure? If yes, carboxylic acid will be the parent chain and ester will be the substituent as c.a. is higher than ester in the priority table. Then, how do we name the structure where ester becomes a substituent? Thank you.

As you have correctly stated in the question, the $\ce{-COOH}$ group has higher priority than the $\ce{-COOR}$ group. In fact, the $\ce{-COOH}$ group has the highest priority (excepting cations) in IUPAC nomenclature.
The $\ce{-COOH}$ group should be included as the suffix of the parent chain, while the $\ce{-COOR}$ is indicated using a prefix.
In fact, Section P-65.6.3.2.2 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names, 2013 (aka the Blue Book) specifies the rules for this.

When, in an ester with the general structure R-CO-O-R' or R-S(O)XO-R', another group is present that has priority for citation as the principal group or when all ester groups cannot be described by the methods prescribed for naming esters, an ester group is indicated by prefixes as 'acyloxy' for the group R-CO-O—, and 'alkyloxy(alkany1)...oxo' or 'alkyl(allcanyl)oxycarbonyl' for the group —CO-OR'.

I quote the example provided in the text:

And another one (not from the Blue Book though)

3-Ethoxy-3-oxopropanoic acid

References: Favre, H. A., Powell, W. H. Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 R. Soc. Chem 2013 DOI: 10.1039/9781849733069


What is the reaction between a copper anode and a zinc cathode? I prepared a battery with $\ce{Zn}$ screws and a $\ce{Cu}$ wire, with water as an electrolyte. The $\ce{Zn}$ screw acted as the cathode, and the $\ce{Cu}$ wire acted as the anode. I got nearly $0.6\:\mathrm{V}$ as the emf. I placed numerous amounts of such cells in an ice cube tray. Each ice cube acts as a cell and i got nearly $6\:\mathrm{V}$ of emf. What is the reaction happening here? What is the reaction involved with the electrodes and the electrolytes?

Well the standard galvanic cell with $\ce{Zn}$ and $\ce{Cu}$ would usually look something like this, assuming there's a salt bridge.

and the reaction could be written as

\begin{align}
\ce{Cu^{2+}_(aq) +2e^{-} &<=> Cu^0} & E^\circ &=0.34\\
\ce{Zn^0 (s) &<=> Zn^{2+} (aq) +2e^-}& E^\circ &=+0.76\\
\ce{Cu^{2+}(aq) + Zn^0 (s) &<=> Cu^0 (s) + Zn^{2+}(aq)} & E^\circ &= +1.10
\end{align}

As for your measured emf, that can be calculated using the Nernst Equation.


Why are isotopes of hydrogen the only ones with special names? Why does no other element but hydrogen get special names for its isotopes?

Harold Urey and George Murphy used spectroscopy to identify deuterium late in 1931, announcing it at the 1931 Christmas meeting of the American Physical Society. Picking up out of 'From Nuclear Transmutation to Nuclear Fission, 1932-1939" by Per F. Dahl:

  If anything, the naming of the new isotope proved more problematic than its isolation.  At a special session on heavy hydrogen at the general June meeting in 1933 of the APS in Chicago, organized in conjunction with the Century of Progress Exposition, the ensuing discussion on its naming 'threatened to become acrimonious,' according to Francis Aston of the Cavendish Laboratory - the great authority on atomic weight measurements and a guest speaker at the discussion.  The argumentation had to do with whether to retain the name 'hydrogen' for the isotope, as Niels Bohr preferred; after all, it was not a new element and had the atomic number 1.  Both Gilbert Lewis and Ernest Lawrence opted for 'dygen' for the H$^{2}$ isotope and 'dyon' for its nucleus, wheras Rutherford preferred 'diplogen' and 'diplon' instead.  In the end, Urey had the last word, as he was entitled to, settling on 'deuterium' for the isotope and 'deuteron' for its nucleus.

Given the heavyweights in the field wanting a separate name for the isotope, even Neils Bohr could not hold back the tide.  Again, though, you have to remember that this was in the very early days of nuclear physics. While the existence of the nucleus dated back to Rutherford's experiments pre-WWI, they were still very unclear just what the nucleus was constructed out of.

This was all occurring just as Cockroft and Walton were using their new ion accelerator to perform the first human-induced nuclear fission ($^{7}$Li + p -> 2 $^{4}$He), published in the Proceedings of the Royal Society on July 1st, 1932. Alpha, beta, and gamma particles had been identified since the late 1800s, and natural sources were used for nuclear physics experiments. It was readily recognized that a source of energetic protons would be highly desired.

Also, remember that 1932 was the year that Chadwick discovered the neutron as a separate entity. Thus, also from Dahl:

  The discovery of the neutron would soon revise Urey's view of his deuteron as consisting of two protons and an electron.  However, as late as October 1932, it was still not obvious among physicists that the deuteron consists of 'one proton and one Chadwick neutron' instead of two protons and one electron.

Against this rapid-fire series of experiments on nuclear physics, it really isn't unusual that deuterium got a separate name - it was still not universally appreciated that it wasn't something weird. Than, once it was accepted, it was too useful to readily distinguish experiments with protons vs deuterons vs tritons to revert their names back to normal isotope nomenclature.


Cheap silocone casting resin (1 or 2 component resin)? I searched in many places and on Youtube and I found ways
to make silicon mold putty from cheap silicone using ordinary
household stuff and kitchen stuff.

But nowhere is a recipe on making silicon casting resin out of the cheapest everyday silicone.

Clearly, casting resin has a much better usage spectrum than molding putty and it is the way molds should be created. Ethylene-Acetate dissolves silicone and results in very long
hardening times (weeks or months) and I suspect that it also
could make the molds shrink a lot while hardening. Any ideas?

I am not dependent on kitchen materials and I can buy some
chemicals like Ethylene-Acetate in the local painter's store.

The problem with attempting this is the nature of the cheap silicone.  Most store bought silicone is already polymerized before use.  For example, the silicone caulk which can be purchased in home improvement areas is a finished product which  has been infiltrated with solvent.  These caulks contain no catalysts to facilitate reactions.  Any curing is the result of solvent drying from the caulk and solidifying the resin.

If you were to attempt to further dissolve one of these materials with something like ethyl acetate (see here for a simplified list of solvents), it probably would not go well.  All you would be doing is further impregnating the silicone with solvent.  While this would decrease the viscosity, it would increase your drying time.  Shrinking would occur during your casting, and I expect that you would also experience a very messy result, no matter the quantity of release agent you applied.

Consider a very simple silicone (aka polysiloxane).  Two silanol molecules react by hydrolysis to form a disiloxane and a water molecule.  In order to reverse the process and obtain your starting material, you would need to put the water back into the disiloxane to cleave it.  This is energetically expensive.

Even the full polymerization process is not a spontaneous one.  When you purchase a silicone casting resin from a supplier, you are actually buying a Part A composed of n-siloxane (where n is a small enough number to be a liquid) and a Part B composed of crosslinker and catalyst (this is often silicic acid and tin because they are inexpensive).  So what you are actually doing is taking a short chain polymer and crosslinking it into a solid mass.

Adding silicic acid and tin to a store-bought silicone product such as caulk would probably cause a reaction.  The result would probably be a very hard rubber with incompatible properties for casting.  However, I have not attempted this.  Furthermore, I am not sure silicic acid would be easily accessible to a home chemist.

Despite how fun it would be to synthesize your own silicone casting resin at home from everyday chemicals, it probably isn't feasible.  There are some inexpensive options out there for resin casting, and I encourage you to go with those.


Why do most explosives have symmetrical chemical structures? I was just turning the pages of my chemistry dictionary, when I found the structure of TNT (Trinitrotoluene):

...there after I got interested in its symmetry as an explosive, and started looking for structures of other explosive compounds and found that almost all of them have very symmetrical structures.... Here are some that I found:

RDX

HMX: High Molecular Weight RDX (cyclotetramethylene-tetranitramine)

PETN [Pentaerythritol tetranitrate (PETN)]

Heptanitrocubane

HHTDD (hexanitrohexaazatricyclododecanedione)

TATB, triaminotrinitrobenzene or 2,4,6-triamino-1,3,5- trinitrobenzene

There are many others which can be found here: wikipedia/chemical-explosives

I wonder why they generally tend to be so symmetrical (as compared to any other random non-explosive molecule)?

The main reason to synthesize symmetric molecules are related to their synthesis:

Some symmetry is due to the nature of the molecular structure itself, like RDX and HMX. This is related to tricyclic structure of hexamethylenetetramine. Similarly, for Tri/tetramers of peroxyacetone. For these structures, they are created rather spontaneously (HMX with some help) than by directed multistep synthesis.
Some symmetry is due to reaction thermodynamics and kinetics, that does not prefer asymmetrical structures. E.g. TNT is produced easily by direct toluene nitration, as methyl and nitro groups direct coming groups to 2,4,6 positions, forming the symmetric molecule. An asymmetric 2,3,4-TNT is not created this way, its synthesis would be much more complicated, expensive and product may not be safe to produce.

Many asymmetric molecules are not stable enough to be synthesized, or are not chemically stable enough to be kept.

For the keeping stability, Picric acid and its salts are famous for their disintegration products making the explosive oversensitive.

Some symmetry is due low control over the reaction quotient, combined with the advantage of reaction undergoing all steps. See nitroglycerine being trinitrate ester, PETN being tetranitrate ester.

If they are symmetrical, their instability is evenly distributed over the molecule, so they are relatively stable, compared to eventual asymmetric variant, considering their explosive potential.
That allows their synthesis, safe and cheap production, safe manipulation, safe aging.

If they were asymmetrical, their instability would be localized and concentrated to some molecule locations and the molecule would be much less stable.

Additionally, at big asymmetry of bigger molecules, there is question, why to keep the "non-explosive molecule part"?

Consider 2,3,4,5,6-pentanitrodiphenyl versus 2,3,4,5,6-pentanitrotoluene.

Another example of an asymmetrical unstable explosive is the renowned $\ce{NI3 . NH3}$

Summary

I would summarize the answer as kinetic and thermodynamic instability, together with synthesis complexity, discriminates existence and production of asymmetric explosives, comparing to much broader group of non-explosive compounds.

Additionally, the symmetry origin is often coined in symmetry of molecules chosen to be converted to explosive, or to be the structural basis for explosive.

As their symmetry makes promise to form an efficient explosive. E.g. pentaerythritol is perfectly symmetrical alkohol with 4 OH groups to be nitrated.


Reaction of carbonyl cyclic diene with Grignard reagent Predict the major product of this reaction:

The two compounds on the right are the possibilities that I thought of (weren't given in the question). According to me:

attack on carbonyl carbon by $\ce{MeMgBr}$ and then hydrolysis with $\ce{aq. HCl}$
first tautomerise to form phenol. $\ce{MeMgBr}$ will remove acidic proton of phenol. The $\ce{aq. HCl}$ workup will then add the hydrogen back to phenol.

I an confused as to which product would be the major product. I thought phenol would be the major product. But to my surprise, the answer given is 1-methylcyclohex-2-enol. I am not able to understand how it is formed, and need help here.

Source: Joint Entrance Exam (JEE) 1997 India

The given answer makes no sense with the substrate actually presented (but see "alternative possibility" below). You will get phenol.

If you start with phenol then, of course, the Grignard reagent just yanks off the hydroxyl hydrogen, only to have that hydrogen restored by the acid workup.  Now suppose you were to have the keto form cyclohexa-2,4-dien-1-one.  This tautomerizes to phenol precisely because the methylene hydrogen is more strongly acidic than it would be as a hydroxyl hydrogen in phenol.  So the Grignard reagent would if anything, go even faster for the methylene hydrogen in cyclohexa-2,4-dien-1-one than it would go the hydroxyl hydrogen in phenol - unless the ketone enolizes first.

That latter issue is academic.  Either way you get the same phenolate ion product, which upon acid workup gives phenol.

An alternative possibility

If the substrate structure in the question was misdrawn, then the given answer could be correct. Cyclohexa-2-en-1-one, with only one double bond in the ring conjugated to the carbonyl group, does not have any (relatively) strongly acidic hydrogens and so the Grignard reagent will add to the carbonyl group.  After acid workup the ketone is thereby converted to the tertiary alcohol in the given answer.

Grignard reagents do not react with carbon-carbon double bonds, so after adding to the carbonyl group or extracting an acidic proton that's all there is.


How to find the possible valence states of elements in the lower end of the d block and the f block? For silver, why don't the electrons in the 4d orbital move to the 5p orbital?

Is it possible for any elements to move their electrons between $n$d and $(n+1)$p orbitals?

When the f orbital is completely filled with 14 electrons, should we take it in consideration to find the valence states?

When Z=58, can you show me the electronic configuration and how do we form the valence state(s)?

[For silver], why don't the electrons in the 4d orbital move to the 5p orbital?

In the case of silver, this is likely because the energy separation between the 4d and 5s orbitals are quite large (absorbing in ultraviolet region, which is why silver appears white/silver), so the energy separation between 4d and 5p would be even larger.

In general, when you write down the electron configuration you are writing down the ground state, i.e. the most stable configuration. While it is possible in principle to move electrons from 4d to 5p, it will not give you the most stable configuration as moving that electron requires a large amount of energy.

For inferring the likely valence states, you can only use orbitals with unpaired electrons, and/or move the electrons that are not separated by large energy differences between different orbitals to make hybrid orbitals with unpaired electrons. In this case, as you cannot move 4d electrons to 5p (which would make some kind of spd hybrid orbital), you can only have a valence state of 1 from the unpaired electron in 5s.

  Is it possible for any elements [to move their electrons between $n$d and $(n+1)$p orbitals]?

Yes, but only for elements nearer the top of the periodic table (and not at the very top, as you will then not have any accessible d orbitals). Period 3 and period 4 (I think) would be most likely to do this. For example, phosphorus can have a valence state of 5, and sulfur can have a valence state of 6, both by involving their accessible 3d orbital, to make hybrid $\ce{sp^3d}$ and $\ce{sp^3 d^2}$ orbitals respectively.

  When the f orbital is [completely] filled with 14 electrons, should we take it in consideration to find the valence states?

In general, no. When the f block is completely filled, the electrons in the f orbital are generally assumed to be core-like, i.e. stable, or low in energy, therefore will have a large energy gap to the other electrons in 6s, 5d and 6p (or 7s, 6d and 7p), and do not tend to participate in bonding/valency. There are possibly one or two exceptions, but I cannot recall them right now.

  When Z=58, can you show me the electronic [configuration] and how do we form the valence state(s)?

Z = 58 is Cerium, which has the following electronic configuration: $\ce{[Xe] 4f^1 5d^1 6s^2}$.

On paper, it will appear that due to the diffuseness of the orbitals 4f, 5d and 6s, it is probably unusual to form covalent compounds to cerium, and you would expect a valence state of 3 (if you only use the 5d and 6s electrons), or a valence state of 4 (if you also use in addition the 4f electron).

It was found the 4f is actually near the energy of 5d and 6s, giving rise to a variable electron structure and a valence state change of 3 to 4 when cerium is cooled or compressed. Moreover, it is possible to form organocerium compounds with a valency of 3. By the way, this is not something that you can deduce simply by looking at the electronic configuration. You have to find this out empirically, using experiments!


Isomers of Octahedral Coordination Compound with Two Bidentate Ligands Draw all possible isomers of $\ce{[CoNO2(en)2NH3]+}$

My answer:

I reckon that there are 4 isomers, as drawn above. Am I correct?

Note: I am answering this as a homework question, i.e. not actually giving a definite answer.

As a first idea, reduce the degrees of freedom in your complex. You could start by assuming what the number of isomers is for $\ce{[CoCl2(en)2]}$, in which the two non-ethylenediamine ligands are identical. And you can further reduce that complex’ complexity by going to something like $\ce{[CoCl2(NH3)4]}$, in which the ethylene bridges have also been removed. (It may have made more sense to stick with $\ce{NH3}$ ligands in the first step, but then I cannot simply substitute for ammine in the second step, so I chose something entirely different there. What you choose is up to you, as long as you don’t create ambiguity.) For the tetraammine complex, it should be clear that there are two different isomers: cis and trans. Upon readding the ethylene bridge, these two isomers still exist. Do your four structures account for this isomerism?

Next up, ask yourself whether any of the isomers of $\ce{[CoCl2(en)2]}$ can be optically active, that could give you another set of isomers. This is important because of the next step.

This next step involves readding the difference between the two monovalent ligands, i.e. going back to $\ce{[Co(NO2)(en)2(NH3)]+}$. Remember to consider the topicity of the two chlorides you had before in case you run into enantiotopic or diastereotopic chlorides that would give you a new set of isomers.

Finally, remember that nitrite has two or three possible means of attaching to a metal centre: κN, κO and κO,O'. Remember to consider whether all theoretical possibilities are actually possible.

The final answer I came up with is given in a spoiler tag.

   The trans-isomer has a high symmetry of $D_\mathrm{2h}$ if I did the transforming correctly, meaning that it is does not have enantiomers. Also, it renders the two chlorides homotopic, meaning that we get one trans isomer.

   There are two cis-isomers that behave as enantiomers of each other: a Δ and a Λ isomer. These two each have homotopic chlorides again, meaning we get a set of two cis-isomers. To be honest, I only saw that when I tried it out with a modelling kit.

   This means that substituting one of the two chlorides with something else will result in identical compounds, giving us three different isomers. (Note that you have drawn two pairs in which one transforms into the other by simple rotation. You only have two different complexes drawn in the question.)

   The chelating nitrite ion is not possible in your complex since you have a rather small metal centre that only allows octahedral coordination. Therefore, each of the three isomers we established earlier can exist in a κN and a κO form. That gives us a grand total of six possible isomers whereof two pairs are enantiomers of each other and three are constitutional isomers of the other three.


Tridecane isomer name What is the name of this $\ce{C13H28}$ isomer:

$\hspace{5.5cm}$

I think it must be "4-ethyl-6,7-methylnonane," but I saw somewhere that its name is "6-ethyl-3,4-methylnonane". Why? The ethyl number must be smaller, isn't it? (because E is before M). Am I wrong?

Alphabetic order dictates which order the substituents are written in, but not the numbering of their positions.  Numbering is determined by what minimizes the numbers.  E.g. 3-ethyl-2-methylhexane is correct, but 4-ethyl-5-methylhexane is not.


Naming ionic compounds with multiple cations and anions I have seen complex ionic compounds that have mixed anions and/or mixed cations. For Example I have seen this:

$$\ce{NaKCl2}$$

Also known as Sodium Potassium Chloride. The only information I can find on this is an ion cotransporter for sodium potassium and chloride ions.

Anyway the structure of a single monomer of this would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{K+}$ $\ce{Cl-}$

I have also thought of this as a possibility:
$$\ce{Na3ClO}$$

This looks very similar to sodium hypochlorite in terms of its formula but has a different structure. The structure of it would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{Na+}$ $\ce{O^2-}$ $\ce{Na+}$

I know that naming mixed ionic compounds is not easy unlike naming simple ionic compounds(which itself can be hard if you don't know the name of a particular polyatomic ion). But how would I go about naming these mixed ionic compounds?

I propose you consult the recommendations of the International Union of Applied Chemistry (IUPAC), in particular the Red Book as reference.  Obviously, you deviated for a reason from the Hill formula of $\ce{NaKCl2}$ (that were $\ce{Cl2KNa}$) some data bases adhere to.

As already presented in rules IR 4.4.2.1, and IR 4.4.2.2, the order of the elements within the formula may be arranged either by their relative electronegativity (starting by the element of least electronegativity), or by alphanumeric order, respectively.  So rules in section IR 5.4 subsequently discriminate between electropositive constituents to be named prior to electronegative ones in the "overall name".  In addition (quote):

"The order of citation is alphabetical within each class of constituents (multiplicative prefixes being ignored), except that hydrogen is always cited last among electropositive constituents if actually classified as
an electropositive constituent."

The same reference provides numerous examples of application useful to look at, too.  Page 76 (of the book, not the pdf) is a good place to start.  As in nomenclature of organic compounds, there are instances you want to point the attention not only to the stoichiometric composition of the compound, rather than to different constitutional isomers.


Change in boiling and freezing point I recently encountered a question while solving previous year question papers for JEE.

This is a JEE 2005 question.

The question demands to complete this statement:

  Equimolar solutions in the same solvent have _______

There are four options and the correct answers are

  same boiling point

and

  same freezing point

I don't think this is correct as the change in boiling and freezing point depends on molality and not molarity. Hence, according to the options, the one saying different boiling and freezing points seems to be correct to me, however, such exams very rarely have mistakes which are not corrected later.

Please clarify.

At low concentrations for which the ideal bp elevation/fp depression expressions usually apply, molarity is linearly proportional to molality, therefore the statements are equivalent.

This webpage explains nicely why low concentrations are important:

  Raoult's law only works for low concentration solutions. Why? Well, in order for our approximation to work, the interactions between the solute and solvent molecules must be nearly identical. If the interactions are stronger, then the heat of vaporization of the solvent will change and thus our whole approximation falls apart. Since we know that intermolecular forces vary greatly between molecules, the affect of those forces must be kept at a minimum by keeping the solute concentration real low.

In any case (as explained by additional statements in the abovementioned website, which discusses dissociation of ionic solutes), the statement made in the answer to your exam would not be true generally even if molality instead of molarity was mentioned.

The point of the question seems to be that a student should associate a universal response of the solvent that is independent of solute with colligative properties as described using Raoult's law, which is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, btw), and that in this realm of application molarity is approx linearly proportional to molality.


Half-equations for the combustion of sulfur dioxide Given the following redox equation:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$

$\ce{SO2}$ is being oxidized to $\ce{SO3}$ and $\ce{O2}$ is being reduced to $\ce{SO3}$.

I figured a way to write the half equations:

Oxidation half reaction: $$\ce{SO2 + H2O -> SO3 + 2H+ + 2e-}$$

Reduction half reaction: $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$

After balancing and summing, I get: $$\ce{2SO2 + 2H2O + O2 + 4H+ -> 2SO3 + 2H2O + 4H+}$$

By cancelling $\ce{H2O}$ and $\ce{H+}$ we obtain the desired redox equation.
The used half equations necessitate the presence of water and acidic medium at first, but in reality we need neither water nor an acidic medium.
Is there a better way of constructing half-equations that do not require any extra materials at the beginning? I mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it.
In the "preparation" section here they didn't mention water or acid.
In addition, I guess oxygen must have sulfur trioxide as the corresponding reduced form instead of water.
Am I mistaken?

Orthocresol already sufficiently pointed out, that half-reactions are a book-keeping tool for electrons. When talking about electrolysis there might be coincidental overlap between half-reactions and the overall mechanism, but as the wording is supposed to indicate, this is not necessarily true.
Many reactions look simple at first glance, but on the second look they appear as highly complex systems of different equilibria. Actually the most simple looking reactions are often the ones that are painstakingly complex.

  I actually mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it. Am I mistaking?

Since you have been asking about "the real half-reactions" I thought I give a little more insight into the mechanism.

I found a fairly old publication on the mechanisms of the formation of sulfur oxides during combustion.[1]
For the purpose of illustrating the complexity this shall be enough, but I am fairly certain there are more modern approaches and kinetic and mechanistic studies available today.

The reaction itself proceeds only at moderate temperatures (400-600 °C) and is slightly exothermic.
$$\ce{SO2 + 1/2O2 <=> SO3 (g) + $99.0$~kJ}$$
Because the reaction proceeds quite slowly, catalysts are necessary. Here comes into play what you find in Orthocresol's answer: you need a species that provides you with oxygen(-2) ions. In praxis you use $\ce{V2O5}$ catalysts, e.g. in the synthesis of sulfuric acid.[2]
Sulfurtrioxide is very hydroscopic, which indicates, that the reactions that you proposed cannot happen, since it would hydrolyse immediately in the presence of water.
$$\ce{SO3 + H2O -> H2SO4(l)}$$

Sulfurtrioxide has many modifications, most commonly it exists as $\ce{(SO3)3}$ units in solid phase. Condensation is also one of the driving forces of this reaction, in the gas phase the equilibrium between $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ would not be entirely on the product side.
There is also a variety of different other sulfuroxides that are possibly involved in this complex equilibrium. This is due to the fact, that sulfurtrioxide is a powerful oxidiser.

Let's have a glance at a few reactions that are involved in the formation. The reaction itself most likely proceeds via a radical pathway in the gas phase, with $M$ being a catalyst of some sort (it can be the wall or some inert molecule):
$$\begin{align}
\ce{SO2 + O2 &<=> SO3 + O}\\
\ce{SO2 + O + $M$ &<=> SO3 + $M$}
\end{align}$$

One important reaction that disturbs the equilibrium is
$$\ce{SO_{$n$} + O <=> SO_{$n-1$} + O2}.$$

I also found a publication about the mechanism in the troposphere and things get even more complicated there.[3] The reason for that is dilution, a reasonable amount of molecules in excited states, and various other species involved. This topic became interesting because of air pollution and acid rain.
I am only giving a few examples and leaving out a lot of details, as stated previously, this is for illustrative purposes.
$$\begin{align}
\ce{SO2 + O2 &<=> SO4}\\
\ce{SO4 + O2 &<=> SO3 + O3}\\
\ce{SO2 + SO2 &<=> SO3 + SO}\\
\ce{SO2 + O3 &<=> SO3 + O2}\\
&\text{etc.}
\end{align}$$

As soon as you consider species that are usually available in the troposphere, e.g. $\ce{N2}$, $\ce{NO}$, $\ce{H2}$, $\ce{CO}$, $\ce{CO2}$, etc, things get really complicated.

Conclusion

You cannot write half reaction equations for this equilibrium. Half reactions in general do not represent the reaction mechanism; they are a just a tool to generate overall reaction equations.
The mechanism of the reaction is significantly more complex than only two half-reactions.
You are not mistaken, you don't need water to perform the combustion, however, you need some sort of catalyst, that is able to absorb the excess bonding energy.

References

Arthur Levy, Earl L. Merryman, William Thomas Reid, Environ. Sci. Technol., 1970, 4 (8), 653–662.
Arnold F. Holleman, Nils Wiberg, Egon Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. De Gruyter: 2008. (in German)
English version: Nils Wiberg, A. F. Holleman, Egon Wiberg  (Eds.): Holleman-Wiberg's Inorganic Chemistry. 1st Edition. Academic press: 2001. (Amazon, Google books)
Jack G. Calvert, Fu Su, Jan W. Bottenheim, Otto P. Strausz, Atmospheric Environ., 1978, 12 (1-3), 197-226.


What is the magnetic moment of tris(oxalato)nickelate(IV)? I'm really struggling with finding the magnetic moment of this complex:

In the solutions, it says that it is 2.83 BM (which corresponds to 2 unpaired electrons). However, in drawing out the d orbital diagram (using crystal field theory) I found that there are actually 4 unpaired electrons (since there are 6 d electrons and the complex will be octahedral and high spin). Am I missing something or is the solution wrong?

If the compound is really nickel(IV), then there are only two choices: a high-spin $\mathrm d^6$ or a low-spin $\mathrm d^6$ configuration, corresponding either to 4 or 0 unpaired electrons. As higher oxidation states make low-spin configurations more likely (due to the lower energy of metal orbitals which means the energy is more similar to the corresponding ligand orbitals leading to greater stabilisation and destabilisation of bonding/antibonding orbitals; $\mathrm{t_{2g}}$ is nonbonding while $\mathrm{e_g^*}$ is antibonding if π effects are ignored, so the energy difference increases), I would expect a nickel(IV) compound to be diamagnetic rather than paramagnetic.
However, oxalate is easily oxidised to carbon dioxide and likewise transition metals in unusually high oxidation states (which is true for nickel(IV)) are typically very good oxidation agents as their reduction is preferred. Therefore, I highly doubt this complex would be stable and suggest it immediately decomposes to carbon dioxide and a nickel(II) or nickel(III) complex. To quote Wikipedia:

Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

Thus, I am inclined to suspect some kind of typo. Two unpaired electrons would correspond to the well-known nickel(II) $\mathrm d^8$ system and tris(oxalato)nickelate(II) would seem much more likely.