Same Activation Energy for Forward and Backward Reactions While reading about equilibrium, I read that the equilibrium constant, $K_\mathrm{c}$, changes with temperature due to the forward and backward reactions having different activation energies. Is it possible for a reaction to have the same activation energy for the forward and backward reactions, such that $K_\mathrm{c}$ does not change when temperature changes? If so, please provide an example.

The van't Hoff equation describes how the equilibrium constant changes with temperature:

$$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$

So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at least if entropy is not considered (the typical diagram explaining activation energy is for a single molecule with potential energy as the y-axis, enthalpy as the difference in energy between reactant and product, and activation energies as the difference between reactant or product and the activated complex).

  If so, please provide an example.

Take the transfer of protons across a membrane. Equilibrium is reached when the concentrations on either side are equal. A change in temperature will not change the equilibrium constant (which is 1, with a reaction enthalpy of zero).

Another example would be a reaction that turns one enantiomer into the other, e.g. D-alanine to L-alanine. In the absence of chiral co-solvents or solvents or other unusual circumstances, again K = 1 and enthalpy = zero over a range of temperatures.


Why is the oxidation state of Co -1 in [Co(CO)4]-? In $\ce{[Co2(CO)8]}$, most people say the charge of $\ce{Co}$ is 0. How does it become $\ce{[Co(CO)4]-}$?

Most answers would say that $\ce{Co}$'s charge becomes -1. How does this work? -1 is not listed as the oxidation state of $\ce{Co}$ or any transitional metals.

Ignore list of "allowed oxidation states" for metal complexes. "allowed oxidation states" might be the most common, but definitely not the only ones the atom is allowed to be in. For example, for $\ce{Fe}$ all oxidation state from -2 to +6 are known, for $\ce{Mn}$ -3 and from -1 to +7 are known, for $\ce{Ir}$ it is -3 and from -1 to +8  with tentative +9 reported for gas-phase monocation.

  In $\ce{[Co2(CO)8]}$, most people say the charge of $\ce{Co}$ is 0. How does it become $\ce{[Co(CO)4]-}$?

It is traditionally assumed that ligands do not change oxidation state of the central atom and all atoms in them have same oxidation state as in a free molecule. At the same time, the total sum of all oxidation states of the particle must be equal its charge. Assuming total charge of the particle $-1$ and neutral ligands, the charge by necessity means that oxidation state of  $\ce{Co}$ in $\ce{[Co(CO)4]-}$ is -1.

This is not the only case, there are numerous carbonyl anions with negative charges, like $\ce{[Mn(CO)5]-}$, $\ce{[Fe(CO)4]^{2-}}$. Compounds with even higher formal negative oxidation state of the central metal atom were reported, with anions like $\ce{[Mn(CO)4]^{3-}}$ and $\ce{[Cr(CO)4]^{4-}}$.

It would be a gross misunderstanding to assume that extra charge is located on the metal atom. It is siphoned onto $\ce{CO}$ ligands via $\pi$-backbonding. However, again, metal-ligand interactions are traditionally ingnored in calculation of oxidation state.

Truly negatively charged metal atoms are much rarer, but still happen. Example would be salt $\ce{[Na(15-crown-5)]Na}$ with sodium anion. Such salts are extremely powerful reducers.


Why is the value of specific rotation different in different cases? An unknown compound with a mass of $4.5~\mathrm{g}$ is dissolved in enough carbon tetrachloride to make a total volume of $250~\mathrm{cm^3}$. The observed  rotation of this solution is $+357.75^\circ$ in a $25~\mathrm{cm}$ cell using the sodium D line. But if $4.5~\mathrm{g}$ is dissolved in $125~\mathrm{cm^3}$ we observed  rotation is $+355.50^\circ$. Calculate specific rotation for this compound. (assuming length of polarimeter tube is $1~\mathrm{dm}$)

My approach is starting with the formula stated on Wikipedia.

$$\text{Specific rotation (in first case)} =
\frac{+357.75^\circ}{\left(\frac{4.5}{250}\right)(1)}=19875^\circ=75^\circ$$

$$\text{Specific rotation (in second case)} =
\frac{+355.50^\circ}{\left(\frac{4.5}{125}\right)(1)}=9875^\circ=155^\circ$$

Any idea why the two answers are different? Did I miss out something?

A polarimeter is typically a long cylindrical tube with flat ends. I think the 25 cm was referring to the diameter of the tube. The other wrinkle here is that the angle measurement may be + or -. So 357.75 may in fact be -2.25.

Specific Rotation (in first case)=$$\frac{-2.25^{0}}{(\frac{4.5}{250})(1)}= - 125^{0}$$

Specific Rotation (in second case)=$$\frac{-4.50^{0}}{(\frac{4.5}{125})(1)}=-125^{0}$$


Difference between white sugar, brown sugar and red sugar I always get confusing why the appearance of normal sugar (the white one), brown sugar and red sugar are different although they are sugar. Is it because they each have different molecular formulae or different number of bond or maybe the brown and red sugar have some transition metal inside?

I really don't. So, I wish someone will help on this question.

The white sugar you refer to as "normal" sugar is mostly pure sucrose, which forms translucent whitish crystals.

White sugar doesn't start out white --- it's actually refined from a plant extract (primarily from sugarcane) that is a mixture of sucrose, glucose, fructose, and some minerals including calcium, iron, and magnesium. The unrefined extract is "raw" sugar, and these minerals give it a brown color.

Brown sugars are partially-refined sugars which still have some of the other components besides sucrose. When it's fully refined, the appearance is white.

When we refine the raw sugar, the remainder is referred to as molasses -- a dense, dark brown syrupy mixture of sucrose, glucose, fructose, and some minerals which give it the signature dark brown color.

Red sugarcane doesn't produce red sugar as @LDC3 stated. "Red sugar" is actually refined sugar with a red pigment added to it... unless you're talking about what the Chinese refer to as "red sugar" -- which is just their name for brown sugar!


How to calculate the rate of formation of a product when given the rate law? The following data are given for the reaction of $\ce{NO} \text{ and } \ce{Cl2}$:

$$\ce{2NO + Cl2 -> 2NOCl}$$
The reaction is second order in $[\ce{NO}]$ and first order in $[\ce{Cl2}]$,
and the  initial rate equals $\pu{1.43E-6 mol L^-1 s^-1}$ at the instant when $[\ce{NO}]_0 = [\ce{Cl2}]_0 = \pu{0.25 mol L^-1}$.

The problem then says:

Calculate the rate of formation of $\ce{NOCl}$ when $[\ce{NO}] = [\ce{Cl2}] = \pu{0.11 mol L^-1}$.

I thought it's just $(2) × (\pu{9.152E-5})× [0.11]^2[0.11]$, which is $\pu{2.44E-7}$,  but the answer provided is half that $(\pu{1.22E-7})$, so I must have gotten the rate equation incorrect.
Is the rate of formation of $\ce{NOCl}$ equal to twice the rate of the reaction, as when I consider this reaction equation: $\ce{2NO + Cl2 -> 2NOCl}$?
Or is the rate of formation of  $\ce{NOCl}$ directly equal to the rate of the reaction, as when I consider this reaction equation: $\ce{NO + 1/2Cl2 -> NOCl}$?
I'm really confused how to tell which of these two approaches to use.


Best way to clean a gold-plated yoyo bearing? I have a gold-plated yoyo bearing which needs cleaning. Most people clean regular non-gold-plated bearings with mineral spirits (naphtha), lighter fluid (hexane/heptane), and acetone, but these gold bearings are somewhat new and we don't have a consensus on a cleaning method. There's debate as to whether the gold plating influences spin time or if it's the precision manufacturing, but they do spin forever and are quite silent. Anyway, these bearings are $22, so constant replacement is an enormous barrier to using these outside of competitions.

I've researched gold cleaning techniques, and all the results say to rinse with water. That's not a very good option since all yoyo bearings have tight spaces where dust gets trapped. To understand this particular bearing's sensitivity: I did try rinsing with tap water last night, and now it barely spins. It seems even the contaminants in the tap water were enough to block up this bearing. Also, while most bearings can be used with a very thin lubricant to prolong life and decrease noise, this one cannot. I could try distilled water, but we really need a precision solution.

In most of the case distilled water is enough, but you should be sure that there are no salts inside your object if you can perform a Conductometry test on the washing water follow this procedure: Wash the item inside a buck for 3 minutes with distilled water measure Conductometry, take another bucket with the same amount of water wash the item for 3 minutes, measure Conductometry repeat these steps until conducibliy drop down.

This is a Conductometry test I've made while restoring an ancient metal object. You can see that I needed 5 washing to get good result.

If you have high conductivity you should identify the salts with some spots test to understand where they came from.

If you have grease you can use a surfactant like Plysorbate 20 2% in distilled water not more!
Eventually if there to much salts and grease you can use: Thiourea solution 8-10% Sodium thiosulphate 30% EDTA 10 % (reference here).

Remember to dry very carefully the object with a hair dryer, NEVER leave the object wet for more than a couple of minutes!

Eventually you can use a protective like Soter or Zapon these normally cause in terms of colour change a $\Delta E$ of 2-4 so ask to your self if you really need it for a gold object.

In my opinion avoid ultrasonic cleaning for composite materials! If you really care about the object ask for a metal restorer!


Does Cl- Have more -I effect or +M effect ( Resonance effect)? In this question my answer was (A) because i though $\ce{NH2}$ group would give more electrons than $\ce{Cl-}$ . But the correct answer is (D) and my teacher's reason is that chlorine has more negative Inductive effect than positive Resonance effect. Is that true?

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, while $\sigma_\mathrm{para}$ is a combination of the strength of both inductive and mesomeric effects (more resonance effect than inductive because the substitution is one carbon away compared to meta-position). For comparison, when the substitution is $\ce{H}$, both effects are zero while when the substitution is $\ce{NO2}$ group, $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ are $+0.710$ and $+0.778$, respectively (resonance effect predominates). Based on Hammett's original definition, when $\sigma$ of any group has positive sign,it is electron withdrawing group (EWG) and the numerical value is an indication of hoe big electron withdrawing ability compared to that of $\ce{H}$ (value of which is always zero). Note that Hammett has assigned these sign and value according to his research on the acidity of substituted aromatic carboxylic acids (EWG substitution makes acid stronger).
Contrary to carboxylic acids, EWG substitution at aromatic nucleus makes amino bases weaker because EWG dilutes the electron density on active $\ce{N}$ atom by inductive and/or resonance effects. Therefore, largere the positive number of $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of EWG, weaker the base. Also keep in mind that $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of $\ce{NH2}$ group are $-0.161$ and $-0.66$, respectively. Hence, basic strength of given benzamidine are decreased in order of their 4-substitution of $\ce{NH2} \gt \ce{H} \gt \ce{Cl} \gt \ce{NO2}$ meaning your correct answer is $\bf{(D)}$. It is noteworthy that this order remains the same even if these substitutions are at 3-position of the aromatic nucleus (based on $\sigma_\mathrm{meta}$ values).
I can't really find the $\mathrm{p}K_\mathrm{a}$ of relevant compounds. However, I found $\mathrm{p}K_\mathrm{a}$ values of some guanidines, $\ce{X-C6H4-N=N(N(CH3)2)N(N(CH3)2}$ (parent compound: 2-phenyl-1,1,3,3-tetramethylguanidine, $\ce{C6H5-N=N(N(CH3)2)N(N(CH3)2}$) which would prove my point (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\ce{X} & \sigma_{(m,p)}^a & \mathrm{p}K_\mathrm{a} \ \text{of 2-Ar-guanidine} & \mathrm{p}K_\mathrm{a} \ \text{of aniline}^b  \\
\hline
\ce{H} & 0 & 12.18 & 4.62 \\
\text{3-}\ce{CH3} & -0.06 & 12.25 & 4.70 \\
\text{4-}\ce{CH3} & -0.14 & 12.37 & 5.11 \\
\text{3-}\ce{OCH3} & 0.11 & 11.96 & 4.23 \\
\text{4-}\ce{OCH3} & -0.28 & 12.57 & 5.34 \\
\text{3-}\ce{Cl} & 0.37 & 11.47 & 3.52 \\
\text{4-}\ce{Cl} & 0.24 & 11.70 & 3.98 \\
\text{4-}\ce{NO2} & 1.26 & 9.78 & 0.99 \\
\hline
\end{array}\\
^a: \text{Ref.2 and } ^b: \text{Ref.3.}
$$

References:

Przemyslaw Pruszynski, "Synthesis and properties of phenyl substituted derivatives of 2-phenyl-1,1,3,3-tetramethylguanidine," Canadian Journal of Chemistry 1987, 65(3), 626-629 (https://doi.org/10.1139/v87-107).
D. D. Perrin, Boyd Dempsey, E. P. Serjeant, In $\mathrm{p}K_\mathrm{a}$ Prediction for organic acids and bases; Chapman and Hall: London, England, 1981 (ISBN 978-94-009-5885-2).
D. D. Perrin, In Dissociation constants of organic bases in aqueous solution, Volume 1; International Union of Pure and Applied Chemistry. Commission on Electroanalytical Chemistry, Butterworths: London, England, 1965 (Suppliment 1972)(ISBN-13: 978-0080208275).


Entropy change of isothermal irreversible expansion of ideal gas Two mole of an ideal gas is subjected to isothermal expansion from $\pu{2 atm}$ to $\pu{1 atm}$ at $\pu{300 K}$. Calculate entropy change of the system, surrounding and total, if the process is irreversible.

Since

$$ΔS_\mathrm{system} = nR\ln\left(\frac{P_1}{P_2}\right),$$

I got the $ΔS$ for the system as $1.4R$. I know that for reversible process the entropy should not change, hence the entropy of surroundings decreases by $1.4R$. But how do we calculate the same for an irreversible one?

Here is a cookbook recipe for determining the change in entropy for a system that has suffered an irreversible process:

THE RECIPE

Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system
Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states.  This is the most important step.
Devise a reversible (alternative) path between the same two thermodynamic equilibrium states (end points).  This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path.   For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic.  You can even separate various parts of the system from one another, and subject each of them to a different reversible path, as long as they all end up in their correct final states. Plus, there are an infinite number of reversible process paths that can take you from the initial state to the final state, and they will all give exactly the same value for the change in entropy.  So try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4).
For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path.  This will be your change of entropy S.  That is, $\Delta S=\int{\frac{dq_{rev}}{T}}$ , where the subscript rev refers to the reversible path.

For further details and worked examples, see:
Reference https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


Degree Of freedom Of A Critical point What's the degree of freedom of a critical point of a single component system?

In the critical point the difference of density between the liquid phase and the vapour phase is so small the are indistinguishable, hence considered as single phase. So according the gibbs phase rule (. F=C - P +2) the degree of freedom should be 2.

But critical point is a point. And the degree of freedom of a point in the phase diagram should be zero( for example, the triple point).

So what's the correct degree of freedom   of the critical point?

There are 3 phases at the critical point. Sub-critical vapour, Sub-critical liquid and critical fluid. 3 phases -> DoF = 0

That is one way to see it - and for convenience that is how I see it. It might be a bit wobbly to regard that as rigorous analyse though.

You can also say, more rigorously, that there is 1 phase and 2 constraints:
$$\left(\frac{\partial\rho}{\partial V}\right)_T = 0 $$
$$\left(\frac{\partial^2\rho}{\partial V^2}\right)_T = 0 $$
Subtract the number of constraints from the DoF and you get 0 again.

This defines a single point on the P/T curve.


Why is oxygenated water not H2O2? Oxygenated water has become popular for a variety of uses recently - for health purposes internally as well as enhanced wound healing.

Anyhow, my dad asked me recently if I knew how it is that oxygenated water does not become hydrogen peroxide. I don't know the answer to it; but it got me thinking... How does this work? Does the oxygen not bond to the water molecule? If so, how does the bottle of water not just out gas when ever it is opened until the amount of oxygen contained in the bottle is equal to that of the room air.

The oxygen is dissolved in the water, just as salt can be dissolved. It does not (to any appreciable degree) combine with water molecules to form hydrogen peroxide.

The reason that oxygenated water is not fizzy like soda water (CO2) is the solubility of oxygen in water is about 2% that of CO2. So the amount of bubbles formed when the cap is removed is only about 1/50 as much as for soft drinks. Or champagne.


Obtaining activity coefficients of conjugate acids of some common carboxylic acid molecules Recently, I have been wondering about justifying the notion on how some common acids (e.g. carboxylic acids) are themselves weaker bases than water to the point that we don't usually consider them to act as base. To begin, we use the formula for $\mathrm{p}K_\mathrm{b}$ and consider for example acetic acid (written in organic chemistry convention, with $\ce{OAc^-}$ being the acetate group $\ce{CH_3COO^-}$ for unfamiliar readers)

$$\mathrm{p}K_\mathrm{b} = -\log \left(\frac{a_{\ce{H2OAc+}}a_{\ce{OH-}}}{a_{\ce{HOAc}}}\right)$$

where everything is measured at room temperature, 1 atm.

As usual, if the solution is dilute enough, we can approximate the activity with concentration.

However, the species $\ce{H2OAc+}$ seemed to be not very popular as NIST have few data on its thermodynamic parameters. While it is in principle possible to perform an experiment to measure $a_{\ce{H2OAc+}}$, if it is indeed very weak as a base, then the measurement result will only be reflecting the acetate ionisation equilibrium and masked any tiny result of $a_{\ce{H2OAc+}}$ if any.

  How does one obtain the necessary thermodynamic parameters for a potentially highly unstable protonated species in aqueous medium in order to work out the $\mathrm{p}K_\mathrm{b}$ of the common carboxylic acids?

If no specific activity coefficient data are available, I would use the general formula for ionic compounds of small enough ion cincentations:

$$
\begin{align}
I &= \frac{1}{2}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2} \\
\log{\gamma} &= -0.509 . ( z_+^m . |z_-|^n )^{\frac{1}{m+n}}. \sqrt{I} \quad (*)\\
\end{align}
$$

$c_i$ is the ion concentration.
$z_i$ is the ion charge.
$I$ is the Ionic strength.
$\gamma$ is the activity ciefficient by the Debye-Hückel equation.
(*) - See the link for more info and more complex formulas for the medium $I$.


How to dissolve the build-up from a paraffin based oil? I have a CNC lathes that use "oil" for cutting fluid during the machining of mostly metal materials.  Typically a heavy paraffinic based cutting oil (petroleum distillate) is used during machining.  My company switched suppliers of this cutting fluid and went to another cutting oil, supposedly with compatible chemistry.  This happened months ago.  Within weeks of making the switch the oil (the two oils mixed together) became slightly "sticky" or tacky.  This negatively affected the machine tools and processes and the decision was quickly made to go back to the original product from the original supplier.  Over a period of several months the reaction between the two products has left a byproduct of build-up of some type inside the machine tools.  I'm thinking it's the paraffin separating from the oil (somewhat) and along with other components of the chemical make-up that's left this brown, sort of sticky, almost "glue-like" material all over the internal surfaces of the machines.  I'm looking for some solvent(s) that will quickly dissolve this material.  Acetone seems to work but by itself it's not a great option.  The fumes are a problem, it evaporates too quickly.  Any other ideas?

Naphtha, paint thinner or mineral spirits are common names for the type of solvent suggested by @NotNicolau, and should be effective. Caution: all are flammable and could be ignited by a spark from machining, from a motor or from a switch. These solvents are considered only mildly toxic, though.

Chlorinated solvents such as trichloroethylene are less flammable and even more effective at removing this "varnish" (probably not paraffin, which would be greasy but not tacky) but they are more toxic and the fumes are more annoying.

Ventilate the area well both during cleaning and afterwards until all solvent has evaporated and blown away.


Confusion in half filled or full filled electronic configuration At the end of electronic configuration, we were taught that, electron orbitals are most stable when they are either fully filled or half filled.
E.g., the final valence configuration of chromium is $\ce{(4s)^1 (3d)^5}$ and not $\ce{(4s)^2 (3d)^4}$.
But the final electronic configuration of chlorine is $\ce{(3s)^2 (3p)^5}$ and not $\ce{(3s)^1 (3p)^6}$.

Why doesn't chlorine follow the half filled or full filled theory?

Having a half-filled or filled subshell is stabilising, as you say. But for Cl, the difference in energy between 3s and 3p is greater than the additional stabilisation from filling the 3p subshell. 3d and 4s are closer in energy, so for Cr it is favourable to promote a 4s electron and have a 4s1 configuration.


Measuring the % of pectin in a fibre after enzymatic treatment with pectinase I will remove the pectin from hemp fibres with Bioprep 3000L pectinase. How can I measure the % of pectin before and after for loose fibres? I only found a way to measure it for fabric, but I will perform this test on just loose hemp fibres.

Note that I do not have professional experience in the textile industry!

Supposed that the stems have been (mechanically) treated, and the "bark" is removed, the ruthenium red method should work.

In Crystal-Structure of Ruthenium Red and Stereochemistry of its Pectic Stain by Clarence Sterling, Am. J. Bot. 1970, 57, 172-175, the specific binding of Ruthenium Red (RR, CAS-RN: 11103-72-3) along a pectin chain, that is, between a $\ce{COOH}$ of one galacturonic acid unit and the $\ce{OH}$ of the next, has been described.

In a more recent publication, Qualitative distinction of carboxyl group distributions in pectins with ruthenium red, published in Bot. Bull. Acad. Sin., 1999, 40, 115-119 (PDF), the method was applied to orange peel pectines of different origin. Here, the dye-pectin adducts were precipitated by addition of $\ce{CaCl2}$ and centrifugated. Then, the absorbtion of the supernatant solution and a stock solution of the dye at $\lambda = 534\,\mathrm{nm}$ were compared.

Consequently, it should be possible to take fibre samples, suspend them and detemine whether they decrease the absorption of RR (= still contain pectin).

Are you planning to bleach these fibres in a way recently discussed in another question? It might be interesting to figure out whether these conditions (alkaline hydroperoxide) are sufficent to (partly) degrade the pectin.


Why does heating an atom make it emit certain frequencies? We're going over quantum basics in chemistry right now and I'm very confused.

Electrons can only accept in discreet quanta to move up an energy level, right? And they reflect other forms of light that don't supply energy in their specific quanta, right? And flame is just infrared electromagnetic radiation, right?

Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?

Thanks for any help.

Absorbing photons is just one of ways atoms and their electrons, or molecules, can gain energy. Other ways are e.g mechanical collisions and chemical reactions.

Note that atoms do not reflect light they do not absorb, but let it pass through like if they were not there. If they had reflected it,  air would have been an opaque fog. A very small number of photons are dispersed by the Rayleigh scattering, what gives the sky it's blue colour.

A flame  contains mixture of

atoms
atomic ions
free electrons
molecules
radicals(molecules with unpaired electrons}
molecular ions ( and ion radicals )

Atoms and atomic ions do have spectral lines in their emission or absorption spectra, related to particular energies, frequencies and wavelengths ( somewhat broaden, mostly  by the Doppler effect ).

But molecules, radicals and molecular ions have rather broad, diffuse bands.

Photons in UV, VIS and near IR have energies typical for electron excitations/transitions.

Far IR and MW have energies needed to change molecular bond vibrations and bond/molecule rotations, as these are quantized as well.

High energetic collision and mainly chemics reactions frequently lead to a  molecule with excited electron. This energy gets dissipated either as mechanical energy passed to other molecules, or, it can be emitted.

Special cases are free electrons and solid matter in flames, that are not limited to particular frequencies at all.

Electrons can emit photons of any frequency (limited by electron energy ).

Solids ( smoke, large scale atomic structures ) have continuos emission spectrum.


How much pressure does Gallium exert when it freezes? Assuming the liquid Gallium is in a sealed container, how much pressure will it exert on its container when it freezes?

Enough to break a glass container.

Gallium is one of the few substances that expands on freezing (like water). The solid is about 3% less dense than the liquid. This is enough to shatter a glass bottle full of gallium on freezing (see this video).

Solids that expand on freezing can exert a lot of pressure (clearly more than enough to break glass) but the exact amount of pressure probably depends on the vessel and the circumstances of freezing.


What anions does the loss of a proton from HOCN and HNCO produce? Are they same or different? What anions does the loss of a proton from $\ce{HOCN}$ and $\ce{HNCO}$ produce? Are they same or different?

Once, I had asked my teacher if $\ce{HCN}$ and $\ce{ HNC }$ are different or same compunds, he said they were different and yes, they had different structures. Applying the same logic here, $\ce{OCN-}$ and $\ce{NCO-}$ are different anions because they'll have different structures.

When they are bonded to hydrogen that is $\ce{HOCN}$ $\ce{HNCO}$. The two compounds will be different:

From Wikipedia

  Organic cyanates are called isocyanates when there is a $\ce{C−NCO}$ bond and cyanates when there is a $\ce{C−OCN}$ bond.

Due to resonance after deprotonation the two anions will be the same.


Can sodium bicarbonate be considered an acid? I was reading about acids and bases today and finally decided to question the statement that "baking soda is a base."

Let's start with the dissolving of baking soda, $\ce{NaHCO3}$.  The equation to represent its dissolution in water is $\ce{NaHCO3 + H2O <=> Na+ + H3O+ + CO3^{-2}}$.

In this "reaction," baking soda is a proton donor, creating hydronium ions and a conjugate base, $\ce{CO3^{-2}}$.  Doesn't this make baking soda an acid?

If what I have written here is correct, then baking soda will create a base when dissolved, but baking soda itself is not the base.  Is this accurate?

The statement you read that "baking soda is a base" comes from the fact that a solution of sodium bicarbonate (baking soda) and water has a pH of around $8.3$.

However, sodium bicarbonate is amphoteric with respect to Brønsted–Lowry acid/base theory, which means that it can act as either an acid or a base. More specifically, the bicarbonate ion is amphiprotic, meaning that it can either donate a proton, acting as an acid, or accept a proton, making it act as a base. The reason that a sodium bicarbonate solution is slightly basic is that it's ability to take a proton from a water molecule is greater than its ability to donate a proton to a water molecule. In other words, the chemical reaction causing the solution to be basic is:

$$\ce{HCO3- + H2O <--> H2CO3 + OH-}$$

The other equilibrium reaction at play, the reaction "trying" to make the system more acidic, is:

$$\ce{HCO3- + H2O <--> CO3^2- + H3O+}$$

Again, in an otherwise pure aqueous system, it is the first reaction that dominates, and thus giving the observed slightly basic solution.

Note that the $\ce{H2CO3}$ produced in the first reaction is also part of yet another equilibrium process:

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

Where the $\ce{CO2}$ is liberated as a gas. This is the cause of the well-known "fizzing" when baking soda is added to vinegar (acetic acid).


Conversion of isonitrile (isocyanide) group into cyanide group? In an isonitrile (isocyanide) group, the carbon atom has a negative charge (I know that this is due to the lone pair of electrons) instead of the nitrogen atom, but in my point of view, this seems a little unstable. So, why doesn't the isocyanide group convert into a cyanide group in which carbon doesn't have a negative charge (carbon valency fully satisfied)?

$$\ce{R-N+\bond{3}C- -> R-C\bond{3}N}$$

Isonitriles will, given the correct conditions, spontaneously transform into nitriles. The reaction mechanism is very interesting; it is basically a cationic [1,2]-migration of the $\ce{R}$ group under retention with a three-membered cyclic transition state. To reach this transition state, the $\ce{C-N}$ bond must be broken and the alkyl residue must form a cation-π interaction with the $\ce{C#N}$ triple bond. This is a non-classical carbonium cation, as shown in scheme 1.

Scheme 1: Mechanism of the [1,2]-migration and sp³-π interaction in the intermediate carbonium ion.

The reaction rates were extensively studied by Meier et al.[1] They note that the typical migration preference of $\text{tertiary} > \text{secondary}>\text{primary} > \text{methyl}$, indicative of cationic carbenium intermediates and observed e.g. in Criegee, Beckmann and Lossen rearrangements, does not apply here. Instead, a migratory aptitude similar to the Curtius or pinacol rearrangements is observed. Another interesting observation is ‘the almost complete lack of a bridgehead effect on the [reaction] rates’, further indicating that a planar carbenium transition state cannot be part of the mechanism and that bond angles and lengths will probably remain unchanged en route to the transition state.[1] When phenylic isonitriles react, next to no substituent effect is observed, indicating that the cation is not delocalised into the phenyl ring.

All of this evidence indicates towards the reaction being ‘properly prototypical for a sigmatropic 1,2-rearrangement via a nonpolarized, tight, hypervalent three-centered transition state.’[1] The authors go on to state that this is in perfect agreement with ab initio calculations, but do not provide an appropriate source.

Isonitriles were known in the 19th century; I am not able to find a definite first description but A. Gautier noted their reactions with acids in 1867.[2] In 1868, he goes on to be the apparant first to describe an isonitrile–nitrile rearrangement.[3] Weith, however, was probably the first to actually observe the reaction as such; he notes:[4]

  Zu dem Ende wurde nach der  H o f m a n n ’schen Methode dargestelltes und durch Destillation möglichst von Anilin befreites Cyanphenyl 2—3 Stunden lang im geschlossenen Rohr auf 200–220° erhitzt. Nach dem Erkalten fand sich eine beträchtliche Menge dunkel gefärbter nadlicher Krystalle vor, der Röhreninhalt besass nicht mehr den Geruch des Cyanphenyls, dagegen den des Benzonitrils. [sic! to all apparant errors]

  Thus, phenylisonitrile prepared by the Hofmann method and purified from aniline by distillation was heated in a sealed tube to 200–220 °C for 2–3 h. After cooling, a significant amount of dark-coloured, needle-like crystals were found; the tube’s content no longer smelt like phenylisonitril but like benzonitrile.

Therefore, for the rearrangement to take place temperatures above $150~\mathrm{^\circ C}$ appear to be required — which also explains why isonitriles are kinetically inert enough to be isolated under standard conditions and even with 19th century methods. (Gautier used $\ce{AgCN}$ to synthesise them from iodoalkanes.)[3] An activation energy as high as this makes sense, since, as noted above, the $\ce{C-N}$ bond must be broked and a significant rearrangement must take place for the carbonium ion to form.

Some isonitriles, namely those which upon dissociation form very stabilised cations such as benzylisonitrile, can also react according to a different path that is basically an $\mathrm{S_N1}$ rearrangement or can be considered in a chain-like structure. This, however, is a rather special case as Meier et al. note.[1]

References:

[1]: M. Meier, B. Müller, C. Rüchardt, J. Org. Chem. 1987, 52, 648. DOI: 10.1021/jo00380a028.

[2]: A. Gautier, Liebigs Ann. Chem. 1867, 142, 289. DOI: 10.1002/jlac.18671420304.

[3]: A. Gautier, Liebigs Ann. Chem. 1868, 146, 124. DOI: 10.1002/jlac.18681460108.

[4]: W. Weith Ber. Dtsch. Chem. Ges. 1873, 6, 210. DOI: 10.1002/cber.18730060180.


Why are fluorides of transition metals unstable in low oxidation states? Why are fluorides of transition metals unstable in low oxidation states?

I would think that since fluorine and oxygen are highly electronegative, it's obvious that they be stable at high oxidation states. However, $\ce{Cl}$  is also highly electronegative, but it does exist in low oxidation states with transition metals.

For example $\ce{CuCl}$ exists  but not $\ce{CuF}$ and  $\ce{TiCl2}$ exists but not $\ce{TiF_{2}}$. Why is it so?

  Fluorides of transition metals unstable in low oxidation states

This is the statement given in my high school chemistry text book but no explanation provided. I am interested to know the reason for it.

You can take the example of $\ce{CuF}$ which is not known.

Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:

$$\ce{Cu + F2 -> CuF2}$$

$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$.  $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:

$$\ce{2CuF2  -> 2CuF (unstable) + F2}$$

$$\ce{2CuF  -> CuF2 + Cu}$$

Similarly, you can illustrate this for $\ce{VF2}$.


Electrolysing vinegar Yesterday I made a mixture of vinegar and salt and then electrolyzed it. I was just wondering if anyone could tell me what the process created. For instance, would it create ethane?

As @Amritansh Singhal said, $\ce{H2}$ would be produced at the cathode from the electrolysis of both the water and some from the acetic acid (main ingredient of vinegar after water) and $\ce{O2}$ would accordingly be produced at the anode.

The question in your comment "Would it create ethane?" is where some more interesting chemistry might take place.  The short answer is "very little at the most" since in the presence of water, the cell potential will be limited by the electrolysis of water as described above.

However, if you could obtain pure acetic acid and produce the proper cell potential, then you could see the Kolbe electrolysis reaction, which would produce ethane.  According to excerpts from this Wikipedia article":

  "The Kolbe reaction is formally a decarboxylative dimerisation of two
  carboxylic acids (or carboxylate ions)."

With respect to acetic acid specifically the above article states:

  "The reaction mechanism involves a two-stage radical process:
  electrochemical decarboxylation gives a radical intermediate, then two
  such intermediates combine to form a covalent bond. As an example,
  electrolysis of acetic acid yields ethane and carbon dioxide:"
  $$\ce{CH3COOH → CH3COO− → CH3COO· → CH3· + CO2}$$
  followed by:
  $$\ce{2CH3· → CH3CH3}$$

Where the latter product is ethane.


Why is the HOMO-LUMO gap in N2 and CO of different sizes? Why is the HOMO-LUMO gap in $\ce{N2}$ and $\ce{CO}$ different even though they have the same number of valence electrons?

As correctly noted in the question, $\ce{CO}$ and $\ce{N2}$ are iso-electronic, so they should be comparable. But a difference remains: $\ce{CO}$ is hetero-nuclear, $\ce{N2}$ is homo-nuclear. The latter case typically means that there is better overlap between the orbitals (because they are of similar extent/energy). This translates into a greater gap between the bonding and the anti-bonding combination of the orbitals. This will definitely change the HOMO-LUMO gap. However, without performing a detailed quantum chemical calculation, it is not clear to me whether the HOMO-LUMO gap should be bigger or smaller for the homo-nuclear case, because one has to take the entire molecular orbital diagram into account.


How to differentiate between iron and sodium flames? According to my textbook, when a flame test of an iron salt is performed, it produces an orange, mostly yellow flame. Sodium salts also produce a yellow flame.
As the colours of these two flames are too similar, how do I differentiate an iron flame from a sodium flame?

Sodium flame is yellow, but all the light is due to two lines in the yellow region. If you look a sodium flame through a purple glass (cobalt glass), the yellow color is absorbed, and you do not see the sodium flame any more. The flame looks dark and colorless through a cobalt glass.
On the contrary iron flame looks yellow, but it is made of a huge number of lines belonging to all regions of the spectrum. The sum of these colors look yellow, but it's a visual effect. The flame contains all colors. If you look at such an iron flame through a cobalt glass, the flame is visible. It looks purple or violet, but it is visible.


Ketone to alkene regiochemistry What are the reagents for the following reactions:

Typically, when thinking about ketone to alkene reactions, one thinks of the Wittig reaction. However, I don't see an appropriate form of that reaction here. This was posed as a question on my problem sheet and I suspect - after having given it some thought - that it's just a Hoffman/Saytzev question.

For reaction A, I think reaction with $\ce{MeMgBr}$ to make the tertiary alcohol. At this point, one has to think of a way to get the least thermodynamically stable product (the kinetic product). Perhaps sulfuric acid? It would protonate the hydroxyl so the transition state wouldn't have significant alkene character and the "base" is bulky. Correct?

For reaction B, I think it should be reacted with $\ce{NaBH4}$ to make the secondary alcohol then elimination can be performed using whatever. Maybe sulfuric acid again or just tosylate it then use $^t\ce{BuOK}$. Does this sound right?

Part (a)

I agree with the previous comments that a Wittig reaction is the best way forwards especially in terms of regioselectivity of the double bond. PPh3, MeI and NaH would be an excellent combination. Perhaps draw the mechanism and you'll see how/why this work so well?

If you are interested, you could use (Cp)2Ti=CH2 also known as Tebbe's Reagent. Klaus Warzecha mentioned this already (above).

Part (b)

I like your idea of reducing, tosylating and using the non-nucleophilic base tBuOK


About Fe(III) and Fe(II) coordination complexes Why does Fe(III) form octahedral coordination complexes if it has 5 electrons in its d-orbitals? I understand that Fe(II) has 6 electrons in its d-orbitals and 6 lone pairs from 6 ligands as the 12 electrons fill up the 3d, 4s and 4p orbitals (18-electron rule, $6+12=18$). However, Fe(III) has 5 electrons and 6 lone pairs is not enough. I know basic crystal field theory and thought that some of the metal-ligand interactions were covalent bonds, but it doesn't add up for both low and high spin states.

Also, do the ligands need to line up with the orbitals form coordinate bonds? I know it sounds counterintuitive since they repel and that's the basis of the crystal field splitting, but how do coordinate bonds form in coordination complexes then?

Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ complex).

So how are they still stable? Well, coordination compounds quickly exceed the simplistic bonding pictures that students are taught at the beginning of their chemistry classes. And the ligand field (or crystal field) model—while it does a good job of explaining some observations at a low level—also doesn’t exactly capture what we consider the current knowledge of coordination compounds. A much better description of how they are bonded lies within molecular orbital theory and scheme 1 below.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left of the scheme, the metal’s 3d, 4s and 4p orbitals are placed, while the right-hand half contains the ligand orbitals; simplified as one σ-type orbital per ligand in an octahedral complex. After taking a look at the scheme and finding it perplexing, we can attempt to find the stabilising and destabilising factors in forming a complex.

The most obvious stabilising effect is that all ligand orbitals form bonding interactions that lower their energy. For this, they make use of the typically unoccupied metal’s 4s and 4p orbitals alongside the often occupied 3d orbitals. As we are stabilising occupied orbitals at the expense of unoccupied orbitals, this already results in a net stabilisation.

The second bit of stabilisation results from the interaction of the ligands with parts of the 3d orbitals: precisely the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals, often collectively termed $\mathrm{e_g^*}$ orbitals according to their symmetry group. These may be occupied, but these also have better overlap with the ligand orbitals so the bond formed is stronger than in the case of 4s and 4p.

Finally, the electrons in the other three metal d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$; collectively termed $\mathrm{t_{2g}}$ according to their symmetry group) are, in fact, non-bonding in this simplified picture as their symmetry does not match any ligand orbital. Thus, any electrons in here do not incur any stabilisation but also—arguably more importantly—are not destabilised.

The 18-electron rule derives from this simplified picture and assumes that all bonding and non-bonding orbitals are occupied (that means: all ligand orbitals and the $\mathrm{t_{2g}}$) while all antibonding orbitals (most notably: $\mathrm{e_g^*}$) remain unoccupied. If you count electrons and add up, this gives 18.

However, as I have established the $\mathrm{t_{2g}}$ electrons don’t actually destabilise the complex (under the simplified scheme). Thus, it is relatively irrelevant if there are 0, 2 or 4 of them; it doesn’t change the general scheme. Likewise, even though any $\mathrm{e_g^*}$ orbitals are antibonding, their overall effect is not extreme, so there can be electrons occupying them while still allowing for a stable complex.

The picture on the metal is further complexified (pun intended) by considering high spin versus low spin. In an uncomplexed metal ion, all five d orbitals are energetically equivalent, so they should be filled according to Hund’s rule. In a complex, you have two options:

Hund’s rule is observed across all five d orbitals regardless of the energy difference
the energy difference is large enough that spin pairing in the $\mathrm{t_{2g}}$ orbitals occurs before the $\mathrm{e_g^*}$ orbials are filled according to Hund’s rule.

The first case is termed high spin and should be treated as the standard case for 3d metals, while the second case is termed low spin and is the standard case in 4d and 5d metals. Note however, that clever choice of ligands may cause a 3d complex to adopt a low-spin configuration and some complexes are even capable of flipping between high and low spin.

While $\ce{Fe(H2O)6]^2+}$ may look like a prime example of the 18 electron rule, it is actually a high-spin complex so the outline I drew above is no longer valid. On the other hand, high-spin $\ce{[Fe(H2O)6]^3+}$ has 5 d-electrons which allows for every d orbital to be populated by exactly one electron—a ‘slightly more favourable’ state. Therefore, high-spin iron(II) complexes are easily oxidised in aquaeous solution but high-spin iron(III) complexes are not as easily reduced. (Obviously, the picture changes when you consider low-spin complexes such as $\ce{[Fe(CN)6]^3-}$, which is a good oxidising agent.)

tl;dr:

The 18-electron rule is more of a guideline than an actual rule; it states that more stable adducts can be formed, if, but that’s all;
All ligand–metal interactions can and should be considered covalent bonds;
Technically, the orbitals and the ligands do need to line up but because electrons are volatile we can assume they have lined up automatically;
Essentially, the crystal field model isn’t entirely wrong when it predicts the $\mathrm{e_g^*}$ orbitals to be raised to a higher energy level. It just provides a bad explanation. The orbitals end up being antibonding; that should equally explain.


Why isn't there a third buffering zone in the pH curve of this titration? I'm in chem 30 right now and we were learning about pH curves and buffering zones. I was looking at a question on the titration of $\ce{H3PO4}$ with $\ce{NaOH}$. The curve showed two qualitative reactions occurring as seen in the two sudden changes in pH. It also showed two buffer zones. Shouldn't there be 3 buffer zones not 2? My understanding is that once the reaction of $\ce{HPO4^{-2}}$ and $\ce{PO4^{-3}}$ is occurring, a third buffer zone would occur.

According to this, we have three pKas, as you observed: 2.12, 7.21 and 12.67 (at 25°C).
Your values may be different due to temperature.

The third value (12.67) can't show up in your graph because it doesn't go beyond pH 12.


Why are optically active compounds abundant in nature? Enantiomers have most physical properties exactly same, which makes them hard to seperate from one another. Due to large amounts of physical similarities and chemical similarities (when dealing with optical non active compounds) anyone will assume that if I eat any one of enantiomers of some compound I will get same amount of energy and they will perform exact same task in my body.

Given that many optically active compounds are extracted from nature, why do organisms go through pain of separating enantiomers for their consumption? Why do optical isomers are found in nature that are hard to separate in a lab?

The title is actually a different question than the post, which contains several questions, but I'll answer all of them since they are closely related.

Why are optically active compounds abundant in nature?

Nature needs some kind of complexity to work. To build a living system you need hundreds, thousands or even millions of different molecules, just a few won't do it. If you just draw random molecules you'll end up with a lot of chiral ones. Unless you are sticking to linear chains without functional groups or aromatic (flat) systems you will almost always end up with chiral molecules. It actually would be surprising if there would be no chirality.

Given that many optically active compounds are extracted from nature, why do organisms go through pain of seperating enantiomers for their consumption ?

As Zhe already said in the comment: they don't need to separate those. All the animals and plants basically use only one enantiomer.

Why do optical isomers are found in nature that are hard to seperate in a lab?

Well, you are right. Enantiomers have the same physical properties unless they are in a chiral environment. Organisms on earth are able to produce only one enantiomer because the whole environment is chiral which means both enantiomers have different physical properties.

If we want to separate enantiomers in the lab we are using the same system. We take a chiral phase for chromatography (those are quite expensive) which means both enantiomers behave differently.

We are also able to synthesize enantiomerically pure compounds if we use a chiral environment during synthesis. But this is often expensive and we aren't nearly as good as biological systems doing this.


Calculating Fluoride in toothpaste Question
  A linear regression line was obtained; $$\log(\ce{[NaF]})~(\mathrm{mol~L^{-1}})~ \mathrm{vs~EMF}:
y = 59.8x +600.8$$
  $0.7540~\mathrm{g}$ of toothpaste was diluted in a$50~\mathrm{mL}$ and tested to obtain a EMF value of $390$. what is the ppm of $\ce{F-}$ in $198~\mathrm{g}$ of toothpaste?

I plugged in $390$ into the eqn to get $-3.525$, taking the antilog to obtain $0.0002985$, and times it by the volume $0.05~\mathrm{L}$ to get the moles of $\ce{NaF}$.

Due to stoichiometry, this also equals $1~\mathrm{mol}$ of $\ce{F-}$. I multiplied by $18.998~\mathrm{g~mol^{-1}}$ to obtain $0.000283527~\mathrm{mol}$ of $\ce{F-}$. I then multiplied by the factor: $\frac{198}{0.750}$ to obtain $0.074454015~\mathrm{g}$ of $\ce{F-}$ in the $198~\mathrm{g}$ of toothpaste.

I'm stuck here, how can I calculate the ppm?

The expected value of fluoride in toothpaste is $1000~\mathrm{ppm}$

Working with your numbers gives me the same result through moles of $\ce{NaF}$.  However, at this point you can multiply by the molar mass of fluoride, 18.998 g/mol, to get grams of fluoride, not moles, specifically, 0.0002835 g $\ce{NaF}$.

Parts per million is a mass-based concentration measure.  It's similar in concept to percent mass, but instead of parts per hundred, it uses parts per million.  Since we know the original mass of this sample, we can calculate the concentration in ppm based on that value.  It will be the same concentration as in the larger sample.

I get the following:
$\frac{0.0002835 \textrm{ g } \ce{F-}}{0.7540 \textrm{ g toothpaste}}\times 1000000 = 376.0\textrm{ ppm }\ce{F-}$


Is the pz orbital wave function three- or four-dimensional? The wave function of the $2\mathrm{p}_z$ orbital is
$$Ψ = \frac{1}{4\sqrt{2π}}\left(\frac Z a\right)^{5/2} r \mathrm e^{-Zr/a}\cos θ.$$
I'm confused if this function will be a three-dimensional function or a four-dimensional function?
My professor told me that
$$z/r = \cos θ,$$
where $θ$ is the angle made by $r$ with the $z$ axis. So, I guess it will be four-dimensional since we would have to then plug in
$$r = \sqrt{x^2 + y^2 + z^2}$$
in the equation.
But when he plotted the function, he plotted a three-dimensional one! What is the correct answer and why?

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$.
There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, spherical coordinates $(r,\theta,\phi)$, cylindrical coordinates $(z, r, \theta)$, to name the most common. We have these different ways because functions might look "nice" in one coordinate system and not nice in another.
As it turns out, the wave function for the $p_{z}$ orbital looks "nice" in spherical coordinates, that is, $\Psi(r, \theta, \phi)$. It looks less nice in cartesian coordinates, but you can do the replacement for $\theta$ and $\phi$ as noted in the question. However, people do not usually mix coordinates systems together because they are redundant. For example, no one writes $r\sqrt{x^{2}+y^{2}}$. They write either $r^{2}$ or $x^{2}+y^{2}$.
To conclude, the $p_{z}$ orbital you have written is a function of 3 variables: $r, \theta, \phi$. It just happens not to depend on $\phi$, so it looks like just a function of $r$ and $\theta$. Forget this concept of dimensionality because it's not doing you any favors.


What is Redox reaction? Is it really possible? I was a reading a book that a senior gave me, it said that redox is a type of reaction in which both reduction and oxidation take place. I know about oxidation but not reduction. After googling reduction reaction I found it's simply just the opposite. But then how can a reduction and oxidation take place together? It would be better if you put forth some examples.

I think you mean Redox, which is a portmanteu of Reduction-Oxidation. An easy example of redox is the Haber process (nitrogen and hydrogen form ammonia)

$$\ce{N2 + 3 H2 <=> 2 NH3}.$$

In the Haber process the nitrogen is reduced (the oxidation state changes from $0$ to $-3$), and the hydrogen is oxidized (the oxidation state changes from $0$ to $+1$). In that redox reaction, both oxidation and reduction take place, as you said.

There's no oxidation without reduction and vice-versa. The electrons lost in oxidation have to go somewhere! If they wouldn't, then there's no reason for the reaction to occur in the first place.


How is the pH scale constructed? How is the pH scale constructed , I mean the strength of an acid is determined by its pKa value and this one can be determined in different solvents if water doesn't suits it but how do you determine the pKa value of a substance in water after you have found the value of its pKa in another solvent ? Also I don't think I quite understand why the pH interval in water is running from 0 to 14 (I know that it has to do with the leveling effect but when you take this scale for other solvents the starting value can be different from 0 , if I am not mistaken for liquid NH3 the starting value is 5 ) can someone explain to me from where does this values come from ?

The pH is determined experimentally with a glass electrode. A glass electrode is a small glass tube limited by a thin membrane and filled with an aqueous KCl solution with a silver electrode inside. There is also a similar silver electrode out of the tube, in a compartment in contact with the outer solution. When you dip a glass electrode into an aqueous solution, the experiment shows that the tension E read between the two electrodes is proportional to the logarithm of the H+ concentration of the external solution. It is :  E  = Constant + 0.059 pH. The Constant depends of the particular glass membrane.
And the experiment shows that pH is -log[H+] at low acidic concentrations, and is measured as

7 in pure water,
3 in 0.001 M aqueous solutions
2 in 0.01 M aqueous HCl solutions,
11 in 0.001 M aqueous NaOH solutions
12 in 0.01 M aqueous NaOH solutions.

These values are in agreement with the usual expression : pH = -log[H+], or : pH = 14 - log[OH-]. At higher concentration, the measured pH values are not exactly equal to the logarithm of [H+]. The measured pH is said to be the logarithm of the "activity" of H+.

The pH does not go only from zero to 14. It may be smaller than zero in concentrated solutions of HCl and of H2SO4.

In non-conducting solutions, the glass electrode does not give any stable values. The measured tension is fluctuating without any reason and can be ± 3, 5, 10 or 0, and this changes constantly.


Does 99.8% acetic acid cause severe skin burns like formic acid? I've so far handled only concentrated fuming nitric, sulfuric & hydrochloric acids from the inorganic side. Even during an accidental spill on the hand I did say only the $\ce{HNO3}$ affects the skin by turning it yellow and the top layer sometimes peels off.
But in the organic acid side I've had once experience in the past long long ago with formic acid and it was pretty nasty. A really pungent smell and if it gets on the hand then immediately the top layer of skin will peel off. Worse than there inorganic ones. Is 99.8% acetic acid this strong?
I'm about to prepare some homemade stop bath solution for Black and White film development. I'm going to prepare a 10% acetic acid solution. Just wanted to know though I will wear nitrile gloves, in case a few drops come on the skin how bad will it be? I can't seem to find any video or much info of it's effect on YouTube or anywhere either? Just want to take all precautions.

Yes it is corrosive to skin.  The Safety Data Sheet identifies it as Category 1A for skin corrosion, which means:

A substance that, according to animal data acquired from a scientifically validated method, produces irreversible damage to the skin after an exposure of three minutes or less, and within one hour of observation, in at least one of three animals.

Don't let the "organic" part fool you.  In terms of handling hazards glacial acetic acid rivals concentrated hydrochloric acid, and actually beats it in skin corrosion classification.


Calculate the number of atoms of oxygen present in 1.3 mol of sulfuric acid Calculate the number of atoms of oxygen present in $\pu{1.3 mol}$ of $\ce{H_2SO_4}$

1 mol has $6.02\times 10^{23}$ atoms

So $\pu{1.3 mol}$ must have $1.3 \times 6.02 \times 10^{23} = 7.826 \times 10^{23}$ atoms

Since there are 4 oxygen atoms  out of 7 atoms in total in $\ce{H_2SO_4}$,

${4 \over 7} \times 7.826 \times10^{23}$

So there are $4.472 \times 10^{23}$ oxygen atoms present.

However the answer key says that $3.1 \times 10^{24}$ atoms is the answer.

What went wrong?

The key about the unit mole is that it does not reference which type of entity is specified. Using the mole only makes sense if you define what you are using the mole for. This could be:

protons
atoms
ions
molecules
formula units of an ionic or metallic solid
aggregates
apples
stars
…

The mole is basically just a large number that you can use instead of e.g. the number 1,000,000 or the number 0.5.

So in your question, $\pu{1.3 mol}$ of $\ce{H2SO4}$ means that you have $1.3 \times 6.022 \times 10^{23}$ molecules of $\ce{H2SO4}$, each of which contains one sulphur atom, two hydrogen atoms and four oxygen atoms. Thus, in one mole of $\ce{H2SO4}$ you can find:

$\pu{2mol}$ of hydrogen atoms
$\pu{1mol}$ of sulphur atoms
$\pu{50 mol}$ of protons
$\pu{50mol}$ of electrons
and $\pu{4mol}$ of oxygen atoms.

The correct solution is thus:

$$n(\ce{O}) = 4\times n(\ce{H2SO4}) = 4 \times 1.3 \times 6.022\times 10^{23} = 3.13 \times 10^{24}$$


What is the name and formula of this structure? Could anyone please let me know the formula and name of this structure? Any answer is helpful. Could you also summarize how to deduce the formula?

Here is an image:

That's Atorvastatin, a cholesterol lowering drug (works by inhibiting the enzyme HMG-CoA reductase).
https://en.wikipedia.org/wiki/Atorvastatin
For more data, see e.g. https://www.trc-canada.com/product-detail/?CatNum=A791725


Why are smoke odors persistent in fabrics? Obviously smoke is very complex, but apparently phenol compounds are the main components of wood smoke that produce the characteristic odor.  (One of more significant of these compounds seems to be syringol.)  The human nose seems to be pretty sensitive to these compounds.

Smoke seems to be very easily absorbed by various fabric-like materials, which then release the compounds back into the air over long periods of time, making the odor quite persistent.  I'd like to know what is going on with these compounds and the fabric-like materials to cause this behavior.

Why is the smell of smoke so persistent, while many other strong smells are not?
In fact this is a tricky question, because you can't make such a comparison. According to IUPAC Gold book Smoke is:

An aerosol originating from combustion, thermal decomposition or
thermal evaporation. Its particles may be solid (magnesium oxide
smoke) or liquid (tobacco smoke).

So it is a fluid colloidal system in which the "the dispersed phase is a solid, a liquid or a mixture of both and the continuous phase is a gas" others strong smells that you smell during your every day life, are mainly already gasses. The interaction areosol-material an a gas-material is completely different.

Interaction between fabric and gasses
We can decompose the interaction in two different phenomena:

gas trapped inside the little voids inside your fabric
gas adsorbed in the fibers surface

The first is governed by Fick law of diffusion this process reach the equilibrium very quickly e.g. if you enter in a room with a gas after a couple of minutes the concentration of gas inside the voids of your fabric will be the same as the outside concentration.
The second is more hard to determine, however is more slower because the adsoprtion occur if there are free sites where the molecule can be placed, and this is a competitive process between all the others gasses present in larger quantities in the atmosphere, that already occupy these sites.
However if you stay a day in a room with a gas is possible the some gas molecule will be ad-sorbed in your fabric. This interaction is more persistent, however I think that the amount of substance is quite negligible.
So interaction between gasses and fabric is not very persistent.

Interaction between aerosol and fabric
This is completely different because colloidal suspensions are unstable. Aerosol are high energy system that tend to lower their energy forming bigger particles or bonds with other materials. So the deposition of these particles over the surface of others materials will be most of the times a spontaneous process. The high surface of the fabrics indeed make it easier.
So what happen when a particulates/aerosols settles on the fabric? This is not  gas trapped inside the little voids of your fabric or gas adsorbed in the fibers surface is a mixture of liquids and solids: many big (relatively!) "clusters" of molecules spread all over the fabric, from the tiny pores to the fiber's ravines. What matter here is the vapor pressure of the substance  you got this equilibrium (for a liquid):
$$\ce{Molecule_{(l)} <=> Molecule_{(g)} }$$ and the analogous with $_{s}$ instead of "l" for the sublimation. For both of these equilibria only the molecules in the liquid-gas or solid-gas interface can evaporate\sublimate, so the cluster surface is very important!

The case of Syringol

In the case of Syringol you have a solid that melt at 50–57 °C. So the sublimation is the only path that allows you to smell it. For most solids the vapor pressure is very low so the equilibrium will keep going to the right for a long long time, releasing syringol, until the source end.


Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't? Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem.

TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot.

First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting:

The natural ionicity, $i_\ce{SF}$, of each $\ce{S-F}$ bond [in $\ce{SF6}$] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$, resulting in a sulfur center of charge $+2.69$. [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$. The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization.

If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$. So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram.
You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way:

The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation, although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak.
(An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910; see also refs. 12 and 13 in that article.)
Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$, then those "correct" resonance forms that we drew would involve $\ce{H-}$. I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable.
Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$, the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised.
In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$. Is there anything better? Well, there's the species $\ce{CH6^2+}$, which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$, and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258. While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$.
Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much. Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details.


Why do the two methylene protons in ethylbenzene have a chemical shift at 2-2.9? The data sheet does not have any chemical shifts in this range for compounds containing only carbon and hydrogen

Protons attached to a carbon that is connected to a benzene ring typically resonate in the range 2.2-2.9 ppm (see the Ar-CH range in this table).

Here is the proton nmr spectrum for ethylbenzene.  The methylene protons absorb around 2.7 ppm, right in the upper end of the expected range.

image source


What is the difference in the reducing power of various ionic hydrides? By definition "Ionic or saline hydrides are composed of hydride bound to an electropositive metal, generally an alkali metal or alkaline earth metal."

What is the relative order of reactivities of all the ionic hydrides of alkali and alkaline earth metals?Is there any general trend?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{LiAlH4}$ ?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{NaBH4}$ ?

I know reducing nature of $\ce{LiAlH4}$>$\ce{NaBH4}$.

This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)

The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.

Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.

In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).

In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.

Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).

At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....

Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.

As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant.

So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...

As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.

Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.


Physical properties of phenols Order of increasing boiling point is p-nitrophenol > m-nitrophenol > o-nitrophenol. The difference between bp of p- and o-nitrophenol is that one form intermolecular hydrogen bonding while other form intramolecular hydrogen bonding. Then what is the reason for high bp of p- than m-nitrophenol as both form intermolecular hydrogen bonding?

In case of p-nitrophenol the lone pair on oxygen of OH takes part in an extended resonance upto NO2 , more delocalisation of lone pair results into more delta minus on H atom ,so stronger H bonding, whereas  in case of m-nitrophenol it cant have resonace structures upto NO2. Since less delocalisation results in comparitively less delta minus, H bonding is comparitively weak.


How can chemical energy be stored in chemical things like candle wax, gasoline,...? I have learned about a type of storing energy is chemical energy. But energy isn't sth we can see, can touch. It just simply gives the ability to work. So how can energy be contain in such chemical things like above?

Yes, energy isn't a "substance" it is the ability to do work. But chemicals can still store it.

The key to understanding how is the relative amount of energy in different chemicals. Some chemical reactions release heat because the products of the reaction are lower energy than the chemicals that react (reactions are a bit more complicated than this in reality and some are driven by changes in entropy but that isn't important for understanding simple reactions). A candle, for example, consists of a mixture of long chain hydrocarbons with a formula approximating CH2. A candle will burn in air (reacting with molecules of oxygen) to give a mixture of CO2 and H2O. This reactions releases heat (simplifying slightly) because the bonds making up the products have lower energies than the bonds making up the candle wax.

This is why we can say that chemicals "store" energy. But we should always be careful to specify the reaction involved as chemical energy is stored relative to the possible products from a reaction not in some absolute sense.


Lead's affinity for oxygen in the cupellation refining process My inorganic chemistry textbook1 says:

Cupellation is a process, where crude gold is taken in a small bowl called cupell and melted in the presence of air. Due to high oxygen affinity of $\ce{Pb}$, it is converted to $\ce{PbO}$ (volatile) and escapes from the system.

Now, the affinity for sulphur and lead is explained by HSAB (soft acid-soft base). But how to account for the affinity of oxygen (hard base) and lead (soft acid)?
References

Guha, Sudarsan. Concise Inorganic Chemistry for JEE (Main and Advanced). 4th ed., Wiley India Pvt. Ltd., 2017.

Indeed if you do the cupelation, the lead will form litharge and some parts of the litharge will volatilize.

But usually the cupel is made out of bone matter - this is very porous.

While the molten lead and the gold will stay in the cupel, the litharge will be absorbed by the cupel like a sponge.

If all lead is converted to litharge and all litharge is absorbed by the cupel only the gold shall be left - so far to the theory.


Reactions in the presence of sunlight Some reactions in my textbook specify the need for the presence of sunlight,
eg-photolytic decomposition of silver bromide and substitution reaction of methane with chlorine.

What is going on in these reactions?
is there something special in light or can these reactions take place if provided with another source of energy like heat?

Also, can thermal reactions(reactions requiring heat) be done in the presence of sunlight?

There is nothing magical about light; it just brings in some energy. What's important is that it brings a huge lump of energy to one point, at once. If you'll heat the system in the hope that the heat will provide the necessary energy for the reaction, then you might or might not succeed. In the worst case, before initiating your reaction, heat will initiate some other reactions you don't want.

I expect the radical bromination of some huge branched alkane to be a particularly bad case. Put it under sunlight, and it will go smoothly; heat it, and you'll have all sorts of side reactions, fragmentations, isomerizations going on at once. Of your examples, decomposition of silver bromide is relatively "well-behaved" (after all, $\ce{AgBr}$ must decompose since it doesn't have many other things to do), and $\ce{CH4 + Cl2}$ is somewhere in between.

If you try it the other way around, it is the same: some thermal reactions can be started by using sunlight instead of heat, some can't.


Molecular connectivity index I would like to know what is valence molecular connectivity index of 3rd order cluster. Please provide an example?

Thanks for your help.

It sounds like you're needing help with chemical graph theory.

I'm no expert in this topic, but it is one that has interested me in the past. Hence I can recommend 'Topological Methods in Chemistry' by R. E. Merrifield and H. E. Simmons—a book which explicitly deals with chemical graph theory in chapters 3 and 4; various topological indices feature throughout.

As a complement to this book, I would recommend the book on Topology by Munkres. The above book isn't rigorous enough to approach this topic perhaps, and so a mathematical treatise is also needed. I'm currently working through these topics myself for my own research.

It doesn't quite answer your question, but I hope it helps in your lofty quest!


Antacid tablet is acid? Doing a grade 9 experiment comparing the alkalinity of antacid tablets.
Using bromothymol blue indicator.
Dissolved tablets in both tap water and distilled water and in all situations antacid came out  yellow in with bromothymol blue indicator ... so the antacid is an acid?
I was intending to have students compare brands by titrating with HCl and comparing the volume of HCl used, but have nowhere to go as the antacid is already an acid?
Can anyone explain what is wrong?

You can do your experiment. The active ingredient in most antacids is bicarbonate, and bicarbonates are amphoteric: they can be both acids and bases. In most cases, bicarbonate is a weak base, so despite your bromothymol blue outcome, bicarbonate is a much weaker acid than $\ce{HCl}$, and you should be able to do your titration.


Oxidation of methane In the reaction $\ce{CH_4 + 2O_2->CO_2 + 2H_2O}$ the $\ce{C}$ atom is oxidized. I can clearly see the addition of $\ce{O}$ atoms, however I don't see a change in Oxidation state. In the products, $\ce{C}$ oxidation state (O.S) was +4, and in the reactants, $\ce{C}$ O.S is +4. I believed an oxidation meant an increase in oxidation state.
What am I missing?

Actually in $\ce{CH4}$ with slightly more electronegtaive carbon it is in $-4$ state. With $\ce{CO2}$ the oxygen is more electronegative oxygen carbon is $+4$. Therefore oxidation is also defined as addition/replacement of a more electronegative element.

Electronegativities $$\begin{array}{}\bf H&2.2\\\bf C&2.55\\\bf O&3.44\end{array}$$


Chair conformation. Wedged lines vs no wedged lines? As the title suggests, I have been wondering why some chair conformations utilize wedged lines for some of the bonds while others have regular thin lines for all bonds, you can take cyclohexane as an example.

Cyclohexane is often drawn as a regular hexagon even though its prevailing chair conformation is not planar.

Perspective diagrams are preferred when discussing conformation (since there is no other reasonable way to depict boat and chair forms) and acceptable when it is just desired to emphasize configuration. (For the depiction of carbohydrates and structurally related molecules, Haworth projections can be also preferred.)

The use of wedged and bold bonds in such perspective diagrams is discussed in Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006):

  ST-3.3 Use of bold bonds to emphasize perspective

  Since perspective diagrams imply a single absolute configuration, it is not necessary to use solid wedged or bold bonds to emphasize that configuration further. In some cases, such bonds may help an inexperienced viewer to understand the perspective used. (…)

However, note that

  Only solid wedged and bold bonds should be used for the purpose of emphasizing perspective. Hashed wedged bonds and hashed unwedged bonds should not be used, either in isolation or in combination with bold bonds.


Does Br2/H2O oxidize all aldehydes including carbohydrates? In many books, it is mentioned that aldehydes are easily oxidized even by mild oxidizing agents. However, I have never seen bromine water explicitly mentioned. But in the carbohydrates chapter, it is mentioned that glucose etc. are oxidized by bromine water.

So is it that only carbohydrates are oxidized by bromine water or all aldehydes?

Also, do they react with bromine water under the conditions of the bromine water test for unsaturation? Or do they react only under vigorous conditions?

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic brown color in the presence of aldoses, but will not in the presence of ketoses.

The general mechanism for the oxidation of an aldehyde under these conditions is:

Though ketones will be able to form a hydrate like aldehydes, no deprotonation can occur to eliminate the bromide ion as we see in aldehydes. Hence no further oxidation can occur.


Rearrangement of an oxyallyl cation I am supposed to predict the product of rearrangement of this cation, having been told that it's an oxyallyl cation:

I did have a look at the Wikipedia entry (since it mentions "oxyallyl cation") and not even once does it mention such a compound as I have shown above. This question is from my high school text book chapter on "Aldehydes and Ketones", and I'm not sure I'm even supposed to know what to do here in the first place.

Anyways, I did guess that the negatively charged oxygen atom will donate its lone pair to the positively charged carbon atom1, forming a three member hetero-ring. But, my book gives the correct answer as:

I'm not sure where they dumped the second oxygen atom (and from where they brought an extra carbon atom) to form this all-carbon ring. Also, is that $\ce{-OMe}$ at the ortho position only for ornamental purposes, or does it affect the mechanism in some way?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; page 218 in solution manual, Q91 chapter on Aldehydes and Ketones

1: Yeah I know that all these atoms are in conjugation and the charges are only partially negative/positive.

Count carbon and oxygen atoms in the book's answer versus the original molecule, and infer that the book got it wrong.  One of the carbon atoms in the book's answer should be oxygen.  Question for you:  Which one?


Dependence of the rate of a catalysed reaction on substrate concentration? In my answer to this question: Master equation for catalyst and substrate binding?. I made the assumption along the following lines:

If we have a substrate concentration $[S]$ then the rate of binding of any substrate molecule with a given enzyme is proportional to $[S]$. Such that $$\text{rate}=\alpha [S]$$
I have however, read that it is rather a hyperbolic relation. So are there any (and if so what) assumptions required so that we can assume that the rate is proportional to $[S]$ rather then it been a hyperbolic relation?

There is always a maximum convertion rate $V_{max}$ for every given enzyme-substrate system. In some cases the substrate itself can inhibit the enzyme's activity at high concentrations, because of the formation of some ESS-complexes, which can not dissociate into product and enzyme.

The frequent correlation is described by the Michaelis-Menten kinetics, which introduces the Michaelis constant $K_M$ as the substrate concentration at which half of all enzyme is bound to substrate $-$ and thus the reaction is running at a speed of $\frac{V_{max}}{2}$.

As you can see, the relation is described by a saturation function. But for loose approximations, where $[S] \leq K_M$ you could say its roughly proportional: $$rate = \dfrac{V_{max}}{2 \times K_M} \times [S]$$


How would I neutralize a PEP (Phosphoenolpyruvic) acid? PEP (Phosphoenolpyruvic acid) is acidic but I need to make it neutral. I need a solution (pH 6-8) of PEP in a final 1 mL volume with a concentration of 100 mM.

How would I make this?

For our work, we also need to make 100 mM PEP and we neutralize it using KOH. Traditionally, using Na+ based salts isn't ideal for biological assays so K+ salts are better for our applications.

Alternatively, if you're adding the PEP to a different solution, you can buffer the reaction solution. While the Phosphate does have a good amount of buffering capacity, effective concentrations of the molecule will be low so you ultimately won't be adding much.


Determine most electronegative element based on successive ionization energy data X, Y and Z are three unknown elements whose first 5 ionization energies are given below. Which of the 3 is the most electronegative?$$
\begin{array}{|c|c|c|c|c|c|}\hline
&\text{IE}_1&\text{IE}_2&\text{IE}_3&\text{IE}_4&\text{IE}_5\\\hline
\ce{X}&577&1815&2740&11600&15000\\\hline
\ce{Y}&800&2427&3660&25026&32827\\\hline
\ce{Z}&738&1451&7733&10543&13630\\\hline
\end{array}
$$

Because of the IE ratios, I concluded that X and Y are in the third column and Z is in the second column. I know that EN increases from left to right for elements in the same row. Is there anything else I should know in order to solve this?

Let's look at what amounts of energy we get from IEs.

$$\ce{X}: 577 \rightarrow 1815 \rightarrow 2740 \rightarrow 11600 \rightarrow 15000$$

After taking out three electrons, we see a first jump in IEs. Thus, there are three electrons in the valence shell.

$$\ce{Y}: 800 \rightarrow 2427 \rightarrow 3660 \rightarrow 25026 \rightarrow 32827$$

Again, after three electrons out, we have the first jump. The result is the same as X: Three electrons in the valence shell.

$$\ce{Z}: 738 \rightarrow 1451 \rightarrow 7733 \rightarrow 10543 \rightarrow 13630$$

You see the first bump in IEs in between IE2 and IE3. So, the third electron comes out of a lower shell and thus, two electrons in the valence shell.

The more the electrons, the more shells they will fill and the farther they'll be from the nucleus (among other lesser important factors). That means X has more electrons than Y because of relatively lower IEs, and therefore Y is more electronegative as it's closer to the fluorine.

You're correct that from left to right we get more electronegativity for the atoms, but note that
though with exceptions, such as Al and Ga, from top to bottom you get lesser electronegativities.

Even in the case of Al and Ga, the one that's more electronegative will get relatively more IE to lose electrons. So that's not what you should worry about. So let's conclude:

  If you see two elements that are in the same group, the one with a generally higher IE is more electronegative.


Is there a theoretical way to predict the coordination number a metal can have in a complex? We know that metals can form complexes with different coordination numbers, but I was told that certain metals have a greater probability to have a certain coordination number.

But I was not able to find any such list that correlates a metal with its most favorable coordination number.

As an example, I guess $\ce{Fe^3+}$ forms complexes with coordination number 6, but can I actually predict the coordination number in a reaction or not?

Summary of molecular orbital theory for coordination compounds-Ligand Field Theory

Molecular Orbital Diagram can explain what you are looking for. First, you choose the geometries you expected to have with corresponding metal cation and ligands. You need to do this since different complex structures will have different more and less energetic d-orbitals.

Second, do MO diagrams for all geometries, and compare the energies, or more properly, stabilities of compounds. Looking at antibonding interactions with ligands you can even predict which bonds of the ligands are becoming more vulnerable to attack.

Additionally, there are such lists in inorganic chemistry textbooks in general. Ferric ion; for instance, can make stable 7-ligand-complex. However, ligands are also very crucial on this aspect and there is ongoing research, like this one:

8 coordinate ferric ion


What are good representatives for the concepts of endergonic and exergonic? We must demonstrate two reactions: endergonic and exergonic. I'm wondering if thefollowing reactions are accurate.

Exergonic: I have sodium bicarbonate and vinegar. Would it be catabolic, spontaneous, and $-\Delta G$ as well?
Sodium bicarbonate and vinegar is your basic science-fair volcano reaction. You have the baking soda in a container of some sort and pour the vinegar. The formula would be something like
$$\ce{Bicarbonate of soda + vinegar -> Carbonic acid + Water + Carbon dioxide gas}.$$

Endergonic: I have ammonium chloride and water. Would it be anabolic, nonspontaneous, and $+\Delta G$?
Ammonium chloride would be dissolved into the water. The formula would be something like
$$\ce{NH4Cl + water -> NH4+ + Cl-}$$

I've just barely touched on the $G$ value thing. I don't exactly know what's going on there. What I mean I guess is that if it's $+\Delta G$, the enthalpy increases, and if it's $-\Delta G$, the enthalpy decreases. To be more precise, we're on the Gibbs free energy lesson at the moment.
Endergonic means energy is absorbed during the reaction. Exergonic means that energy is released during the reaction. Is energy released or absorbed during a vinegar and baking soda reaction? Is energy released or absorbed when dissolving ammonium chloride and water?

As you say you are starting with Gibbs free energy, just pay attention to the $\Delta G$ values. Remember that $\Delta G= \Delta H -T\Delta S$ so it doesn't depends only on enthalpy but also on entropy. A process where $\Delta G<0$ is exergonic and in the opposite case it is endergonic.

About the examples we can say that:

Salts that disolve in water are exergonic processes, $\Delta G_\mathrm{disol}<0$.
Sodium bicarbonate will disolve in vinegar to produce $\ce{CO2}$ (because of the high concentration of acetic acid), it is also exergonic.


Reaction of (RhCp*Cl2)2 with butadiene This is the reaction that took place:

This is the information from the $\ce{^1H}$ NMR spectrum:

$$\begin{array}{|c|c|c|}
\hline
\delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline
1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline
1.73 & 15 & \text{s} \\ \hline
2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline
3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline
3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline
3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline
\end{array}$$

What is the product formed?

I know the product only contains carbon, hydrogen, rhodium, and chlorine, and I think its molecular formula is $\ce{C14H21ClRh}$ (perhaps $\ce{H22}$ instead of $\ce{H21}$). My initial thought was that the diene could displace one chloride ligand to form $\ce{[RhCp^*Cl(C4H6)]}$.

The precatalyst system [RhCp*Cl2]2 was developed by Peter Maitlis and his group at the University of Sheffield in the 1970's and 80's. It has been used as a hydrogenation catalyst under a number of conditions, and is readily commercially available. Under the right conditions, ethanol can act as a suitable transfer hydrogenation source (not quite as good as 2-propanol, but better than methanol), but in this case, in the presence of a suitable base (sodium carbonate), the rhodium dimer will react to give a stable product.

$$\begin{array}{|c|c|c|}
\hline
\textbf{label} & \delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline
\textbf{A} & 1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline
\textbf{B} & 1.73 & 15 & \text{s} \\ \hline
\textbf{C} & 2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline
\textbf{D} & 3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline
\textbf{E} & 3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline
\textbf{F} & 3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline
\end{array}$$

From the NMR, it is possible to deduce what structure is likely, if we construct a valid spin system from the available data.

signal B: δ1.73 integral of 15 : we still have a Cp*
signal A: δ1.65 integral of 3 : looks a lot like we have a methyl group. It has a doublet coupling of 6.2Hz, which corresponds to signal E at δ3.58.
signal E: δ3.58 is single 1H which appears as a dq. We can account for the q coupling; it is next to a methyl group, which leaves the doublet coupling of 7.0Hz, which corresponds to signal F at δ3.96
signal F: δ3.96 is a single 1H with additional coupling to 2 other single 1H nuclei; one of 11.0Hz (signal C), which is pseudo-trans, and the other at 6.6Hz (pseudo-cis, signal D). This two signals have a mutual coupling of 1.5Hz, typical of a pseudo-gem couping on a double bond.

So that gives a spin system skeleton that looks like:

which, is able to form a stable π complex.
Homework spoiler alert:

and a useful reference from many years ago....

   K. Moseley,   J. W. Kang and    P. M. Maitlis J. Chem. Soc. A, 1970, 2875-2883. DOI: 10.1039/J19700002875


VSEPR geometry of Bromate Ion After doing the Lewis structure of Bromate Ion and in order to find the molecular geometry using VESPR method, we have:

central atom: Br

Electrons of the central atom: 7

Electrons that contribute the 3 Oxygens : 3

Electrons that contribute the central atom for the π bonds: -2

Charge of ion (negative) : 1

Total electrons : 9

But if we divide 9 with 2 in order to find the σ-bonding pairs, we get 4,5. what is wrong? Because after searching the molecular geometry of BrO3- it says that it is trigonal pyramid.

There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

How many valence electrons do we have? (add up)

For bromate, the calculation is:

$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?

This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.

$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$

$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$

$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{Br-O}$ bonds are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:

(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image. Originally taken from an answer about iodate hence the central iodine. Replace it with bromine to get bromate.)


why are amines more water soluble than alcohol for similiar molecular weight? kindly someone please explain the reason behind this behaviour despite the fact that oxygen is more electronegative and forms stronger hydrogen bond.

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong.


Will this reaction form magnesium carbonate? I have magnesium oxide pills I take. I know if I dissolve them they'll form magnesium hydroxide, and I'm trying to make pure magnesium from my pills. I've come up with a double replacement reaction, will it do what I think it will?

$$\ce{Na2CO3(aq) + Mg(OH)2(aq) -> MgCO3(s) + 2NaOH(aq)}$$

Will this precipitate out magnesium carbonate?

Another thing: I was planning on using electrolysis to separate the magnesium, but if it's insoluble then I'm not sure what to do. How can I separate the magnesium from the carbonate?

If such a reaction takes place at all it is not very efficient because magnesium hydroxide is not very soluble.

And if you want to get the metallic magnesium neither the carbonate nor the hydroxide will really work.  Both decompose at a relatively low temperature to the oxide, which is then solid up to $\pu{2800\!^\circ C}$.  Convert the magnesium to the chloride and melt that, preferably as a eutectic with alkali metal chlorides.  Then electrolyze the molten salt.  An aqueous solution of magnesium chloride  (or anything else containing magnesium) does not work.