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Classifying acid-base reaction between aluminium hydride and lithium hydride
So I was dealing with a reaction the other day, which is also briefly written in the title:

$$\ce{AlH3(s) + LiH(aq) -> Li+(aq) + AlH4-(aq)}$$

The strange thing is:

Arrhenius definition cannot explain the reaction, since there is no $\ce{OH-}$ in the reaction formula.
Also Brønsted–Lowry definition cannot explain the reaction, because the stuff which $\ce{LiH}$ gives to $\ce{AlH3}$ is not a proton (hydrogenium, $\ce{H+}$) but a protide (hydride, $\ce{H-}$).
And as far as I know, this is the exact definition of Lewis acid and base written in my chemistry textbook:

Lewis acid: A substance which can receive unshared electron pair
Lewis base: A substance which can give unshared electron pair

leading to a conclusion that the above reaction also cannot be explained through the Lewis definition. Below are my questions:

Is there anything wrong in my thought?
Is that really an acid-base reaction? If it is, what kind of acid-base reaction is it?

"(aq)" really?

@Mithoron Yes, that is how it was written.

Then better get a new book, 'cause you'd sooner get a nuke in your basement then aqueous LiH.

This is indeed a Lewis acid-base reaction.
The hydride ion has a lone pair which is, in general, not bound to anything. The aluminium atom has at least one unpopulated orbital (typically taken to be a $\mathrm p_z$ orbital in a planar $\ce{AlX3}$ environment). This makes hydride a Lewis base and aluminium a Lewis acid. The association of hydride to aluminium is thus a Lewis acid-base reaction.

@PenPoint Hydride is the Lewis base. Lithium is technically also a Lewis acid, but it is not receiving here.


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Can we run PBEh-3c in Gaussian?
Can someone help with with setting up a calculation to run the new composite scheme PBEh-3c in Gaussian?
The code has been implemented in Turbomole and ORCA and it's pretty easy to run it there. Unfortunately I need to run a supramolecular stack in Gaussian because it's what we have in the cluster.
Thanks

I'm voting to close this question as off-topic because this could be solved by reading the manual: http://gaussian.com/dft/

I'm voting to close this question as off-topic because it is seeking direct input help to a quantum chemistry program. We are not the support hotline.

I would not have closed this, especially not as homework.

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What is "chemical pressure"?
While reading some publications on solid superconductors, I encountered a term "chemical pressure" a few times, which is usually attributed to the changes of superconducting transition temperature $T_\mathrm{c}$ alongside with the hydrostatic, or physical, pressure. For the reference, consider this paper:
Chemical Pressure and Physical Pressure in $\ce{BaFe2(As_{1-x}P_{x})2}$ [1].
Whereas I can understand what the physical pressure is and how it's defined and estimated, I found no sources which would explicitly define either what chemical pressure is, or how it's measured (and what units are used) and/or analyzed.
All explanations are somewhat expansive and usually point to the effects of isomorphic substitution in the crystal lattice (ionic structure), or intermolecular repulsion between ligands (molecular structure). So, what chemical pressure is in the nutshell, how it's measured and expressed?
Reference

Klintberg, L. E.; Goh, S. K.; Kasahara, S.; Nakai, Y.; Ishida, K.; Sutherland, M.; Shibauchi, T.; Matsuda, Y.; Terashima, T. Journal of the Physical Society of Japan 2010, 79 (12), 123706 DOI: 10.1143/JPSJ.79.123706. (Open Access)

I have heard similar terminology before used in describing hydrogen-rich compounds under pressure as a surrogate for metallic hydrogen (which may be a high temperature superconductor). See here: "[...] demonstrating the potential to observe metallization and superconductivity in hydrogen within a potential silicon-hydrogen alloy at higher pressures, but still much lower than would be necessary for solid hydrogen, due to chemical precompression by the silicon." Not sure if that can help steer you anywhere.

@NicolauSakerNeto Oh, this is something new to me, I've seen "positive" and "negative chemical pressures", now there is "chemical precompression". Good find, but still, what on Earth does this term mean? :)

A crystal's structure is dependent on physical pressure. When applied pressure is changed, lattice constants and solid parameters are altered as well. Instead of messing with physical pressure (e.g., hydrostatic), consider tampering with something else entirely. What happens if one changes a phosphorous atom for silicon? What if you add impurities to an interstitial lattice site? Lattice constants transform, cell's volume expands or contracts. Chemical pressure is that thing which brings about the new state.

Effects of chemical and physical pressure may vary, hence the distinction. See D. M. Friedman. New Research on YBCO Superconductors, Nova Publishers: $2008$, link, Chapter 6;  Gaudet et al. Effect of Chemical Pressure on the Crystal Electric Field States of Erbium Pyrochlore Magnets ($2017$) link. (This is from a 15 minute read, assume no expertise on my part.)

(Apologies: not chapter 6, but chapter 1, subsection 6. Starting page 58. This is freely available, hence the former link.)

@LinearChristmas I flipped through the references, but I still have no idea what chemical pressure is. "The relation among site disorder, chemical pressure and physical pressure effects has been poorly underdtood" doesn't sound very promising. Do you mean that chemical pressure is the electrostatic repulsion (+ve) or attraction (-ve)? It would be great if the source explicitly says what this thing is in 1-2 sentences, what the units are, and how it's measured.

Yes, I do think it's electrostatic repulsion and attraction in the fitting sense. Chemical pressure = internal pressure of solid from or rather by the substitution or addition. '...by chemical substitution of the  smaller $\ce{Ca^2+}$ ions into the bigger $\ce{Sr^2+}$ sites (applying the chemical pressure)...' link. (continues)

Can be measured by effects on electronic transitions, e.g., rate constants. '...As a result, the driving force for the high-spin->low-spin relaxation increases...[hence] the corresponding rate constant also increases. Thus the effect of the different host lattices on the guest can be viewed as a variation of an internal or chemical pressure...[C]hemical pressure can be compared to the effect of external pressure' (Andreas
Hauser et al, Chemical Pressure, CHIMIA
2002, 56, No. 12, p 685 link.

Another hint towards the steric nature of chemical pressure: '$\ce{Sr}$ is isovalent to $\ce{Ca}$ but has a larger ionic radius (1.31 Å and 0.99 Å for $\ce{Sr^2+}$ and $\ce{Ca^2+}, respectively) and thus allows for the effect of chemical pressure to be studied without changing the nominal Mn valence.' DOI: 10.1063/1.4994089.

'Lattice forces can be substantial. Just by substituting the metal ion of the inert hosts, the effect of the chemical difference in pressure can be equivalent to several kbar of external pressure.' Not sure what this implies in terms of units, or if there really are any for chemical pressure. 'Of course, the concept of internal pressure is  very crude and constitutes only a first approximation to the complex problem of guest-host interactions.' Hauser et al, 2002, link

Not sure if this is helping at all, or the comments further obfuscate the matter. A long summary: The external pressure you apply has a different effect on the surface vs deep in the solid. Think of chemical pressure as a force inside the solid that you compare to the effective external pressure at that point inside the solid. Change surrounding atoms, the force alters. If it's as if it adds to the effective external pressure, this is positve chemical pressure. If the model effective external pressure inside is less than external pressure at the surface, this is negative chemical pressure.

@LinearChristmas Wow, that's a lot of comments. Maybe you could put them in a structured form and post a clear and definitive answer. For now I see a bunch of facts, but no underlying theory or model. By definition, pressure is a physical quantity  numerically equal to the force acting per unit surface area perpendicular to this surface. Assuming this force is the Coulombic one, then what is the surface? What is an approximate magnitude/order/units of chemical pressure? How to measure it? Is it always anisotropic? Can it be predicted (I saw a couple of papers mentioning DFT does that)?

I'll have another look tomorrow. In the mean time, I think it's indeed a good idea to transfer the comments to chat and delete them here.

External and internal pressure
To study the effect of pressure on properties of a solid, is equivalently to learn how changes in volume transform physical parameters. For external pressure at constant temperature, this relationship manifests through compressibility $\kappa$.
$$\kappa =-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$$
An alteration in volume will modify the energy of a system. The relationship is internal pressure, something chemists first encounter when discussing real gases.
$$\pi_T=\left(\frac{\partial U}{\partial V}\right)_T$$
In solids, one might discover different notations. For example[1],
$$\alpha =\left(\frac{\partial W}{\partial V}\right)_T \sim \frac{\mathrm{d}W}{\mathrm{d}V} $$
where $W$ is potential energy, and $\alpha $ is called energy–volume correlation. Disregarding possible sorption, varying external pressure is the purest option of probing internal pressure.[2] Unfortunately, this might not always be feasible.[2][3] Most actinides, for instance, are highly radioactive, thus limiting their study to very specialised laboratories.[3] Many actinides are too scarce to perform necessary large scale measurements.[3] Here is where chemical pressure comes in.
Chemical pressure
Instead of external pressure or thermodynamic internal pressure, we now discuss chemical pressure. Much akin to external pressure, scientists take interest in how energy is dependent on volume. Chemical pressure comprises of substituting in or adding chemical entities to a solid network.[2] [4] [5] [6]

External pressure presses down on the solid. Volume is modified via a compressibility relationship.[1] A more direct or pure method.[2]
Chemical pressure is induced by changing cell volumes via the introducing of new chemical entities, such as atoms or compounds, in the lattice. An indirect, less pure approach.[2]

Obviously, chemical doping modifies a system in other ways besides volume alone.[2] [5] [7] By choosing an element with similar electronic configuration, these deviations from ideality are minimised.[2] [7] Another advantage of chemical pressure over extrernal pressure is that the former may be positive and negative.[2] [8] Positive pressure implies that pressure from chemical substitution is directed inwards, and adds to external pressure. Negative pressure pushes the solid apart, hence acting against external pressure.[2]
Immunisation against terminological confusion
Condensed matter science, like other fields of study, is entitled to its own nomenclature. For chemists specialising elsewhere, this might seem a superfluous obfuscation. This is especially true for cutting-edge chemistry where terminology is fresh and perhaps not entirely settled. For this reason, synonyms are widespread, and overlap with other terms may occur.

External pressure is also called physical pressure[4] to establish a physical pressure – chemical pressure dichotomy.
Chemical pressure is often enclosed between quotation marks. More confusingly, internal pressure is used as an alternative to chemical pressure.[5] Perhaps this is to emphasise an external pressure – internal pressure dichotomy. Or, it could be, that since chemical pressure is used as a tool to measure internal pressure, the concepts have blurred lines because the distinction seemed unnecessary.

Units and experiments
Internal chemical pressure will probably have concurring units with external physical pressure. Pascals ($\ce{Pa}$) when operating in the SI framework. Very few studies explicitly mention this fact[5] because more often the changes in some concrete parameter are discussed[1] [9]. Because of what chemical pressure constitutes, we do not apply a chemical pressure of $10\ \mathrm{kPa}$. We might, however, bring about (or apply) a chemical difference in pressure ... equivalent to several $\mathrm{kbar}$ of external pressure[5]. For solids where external pressure experiments are currently unthinkable, this equivalence cannot be discussed. But the goal is not to compare chemical pressure to external pressure anyway; as before, we wish to learn more about how volume affects (potential) energy.
As far as I am aware, chemical pressure cannot be measured directly. Chemical pressure is induced inside a solid material by substituting or introducing new chemical entities. Where would one put the barometer? If it is possible to carry out external pressure measurements alongside chemical doping, we may compare one to the other. To emphasise, the effects need not match. In these and other cases, volume changes in cells are estimated or measured, and their effect on the system as well.
/(short overview/list of experiments in progress, next week)/
In my opinion, this subsection deserves a question on its own.
TL; DR
/(in progress, next week)/
Additional reading
For an overview of highly-correlated electron systems in metallic solids, I recommend C. Marini. Pressure-induced metallization process in Strongly Correlated Electron Systems [Online]; G. Stefani, G. Altarelli, P. Postorino, Eds. Università degli Studi Roma TRE: Rome, Italy, 2009–2010. Archived link: bit.ly/2yPJcCo
References and bibliography

[1] G. Borelius. 'Internal Pressure in Solids and Liquids'. Physica Scripta, 1970, 1, 2-3, pp 141–147. DOI: 10.1088/0031-8949/1/2-3/011
[2] J-M. Fournier. 'Chemical pressure in actinide systems'. Physica B: Condensed Matter, 1993, 190, 1, pp 50–54. DOI: 10.1016/0921-4526(93)90441-8
[3] U. Benedict. 'Properties of Actinide Metals Under High Pressure'. Journal de Physique Colloques, 1984, 45, C8, pp C8-145–C8-148. DOI: 10.1051/jphyscol:1984826
[4] D. M. Friedman. New Research on YBCO Superconductors; Y. Itoh, Ed. Nova Science Publishers: 2008. Chapter 1: Recent Advances of NMR and NQR Studies of $\ce{YBa2Cu3O_{7-δ}}$ and $\ce{YBa2Cu4O8}$, pp 25–69 Subsection 6: 'Chemical Pressure, Physical Pressure, and Site Disorder', pp 59–59. ISBN: 978-1-60456-084-8
[5] A. Hauser, A. Nahid, S. Delahaye, A. Sadki, S. Schenker, R. Sieber, M. Zerara. 'Chemical Pressure'. CHIMIA International Journal for Chemistry, 2002, 56, 12, pp 685–689. DOI: 10.2533/000942902777679858
[6] R. S. Liu, C. H. Shen, T. S. Chan, R. Gundakaram, S. F. Hu, J. G. Lin, C. Y. Huang. 'Chemical Pressure Induced Phase Transition in Single and  Double Perovskites with Magnetoresistance Effect'. Tamkang Journal of Science and Engineering, 2002, 5, 1, pp 59–61. Archived link: bit.ly/2fvxrsv
[7] A. Huon, D. Lee, A. Herklotz, M. R. Fitzsimmons, H. N. Lee, S. J. May. 'Effect of chemical pressure on the electronic phase transition in $\ce{Ca_{1-x}Sr_xMn7O12}$ films'. APL Materials, 2017, 096105, pp 096105-1–096105-7. DOI: 10.1063/1.4994089
[8] Masatomo Uehara, Tsuyoshi Amano, Sachiko Takano, Tatsuya K[o]ri, Takahiro Yamazaki, Yoshihide Kimishima. 'Chemical pressure effect on the superconductor $\ce{MgCNi3}$'. Physica C, 2006, 440, 1–2, pp 6–9. DOI: 10.1016/j.physc.2006.03.017
[9] M.Tropeano, C.Fanciulli, F.Canepa, M.R.Cimberle, C.Ferdeghini, G.Lamura, A.Martinelli, M.Putti, M.Vignolo, A.Palenzona. 'Effect of the chemical pressure on superconductivity and SDW in undoped and $15$% $\ce{F}$ doped $\ce{La_{1-y}Y_yFeAsO}$ compounds'. Physical Review B, 2009, 79, 174523, pp 174523-1–174523-6. DOI: 10.1103/PhysRevB.79.174523

Instead of MathJax for non-mathematical formulae use HTML entities with escaping brackets: \[1\]. Years in the publication list should also not use maths, but either HTML or markup **bold** or bold. Please don't use link shorteners, this is completely unnecessary and might introduce dead links when going out of service. They also pose a potential security risk.

Thanks for the pointers. I'll switch what you mentioned later today. As for link shorteners here: I haven't technically used them. The link is archived and does not include a redirect. The bit.ly link is only for those who, for some reason, cannot open hyperlinks, hence the short copy.

Apologies for delay... Hopefully February is the magic month where I can complete this answer.

First off, "chemical pressure" (CP) is often portrayed as an empirical concept or effect for linking chemical composition and physical properties in the first place, and not a physical quantity -- even though it is one. That might be a reason for certain confusion.
Doing some research I encountered "CP" term being used in other fields with substantially different meanings:

Chemical engineering: a pressure created by the reactants inside an apparatus;
Medicine: a negative stimuli of narcotic drugs; also effect of hormones on sexual behaviour;
Biology: a synonym for osmotic pressure;
Ecology: a suppression caused by prolonged exposure to insecticides;
Industry: "Chemical Pressure Company, Inc." (USA) manufactured polymers in 1970-80s.

One of the first reliable literature sources where "CP" has been coined as a truly chemical term used these days, was an exceptionally well-written review "A brief review of a study of cohesion and chemical attraction" [1]. To the mid-1920s van der Waals concept of "incompressible" atoms has been criticized, and transitory collisions in gases were compared to the "united" atoms in solids and liquids. Here are the key points (emphasis mine):

When a molecule is composed of more than one atom, different intrinsic internal pressures must evidently exist on different sides of each atom ... For instance in liquid bromine ...,evidently the two atoms in a molecule are bound together much more firmly than in the mere cohesion of the liquid. Hence much greater pressure (and therefore much greater compression)
  must exist between the atoms of a given molecule than exists between two different molecules. The first-named pressure, the greater one, may be called chemical intrinsic pressure; the second, the smaller one, may be called cohesive intrinsic pressure. They are both probably due in this case (as in others) to the same forces being exerted to different extents. In such
  cases the atomic fields of force must be distorted, the atomic structure being far more compressed on one side than on all the other sides. [...]
Evidently the main influences which work in any physicochemical atomic contacts may be represented by the following equation of two terms indicating compressing agencies, with two terms indicating distending agencies:
  $$p + \Pi = \Pi_p + P_\Theta \tag{1}$$
  in which $p$ represents external pressure, $\Pi$ the sum of all possible intrinsic compressing effects; $\Pi_p$ the intrinsic distending or "repulsive" pressure, and $P_\Theta$, thermal pressure. This equation is expressed as between pressures rather than forces because the first and the last terms may then be directly measured; for $p$ may be found by an ordinary pressure gauge, and $P_\Theta$ equals $T_\alpha / \beta$ where $\alpha$ and $\beta$ are coefficients of expansion and compression respectively. [...]
Common observation of the phenomena of cohesion and chemical combination shows that both $\Pi$ and $\Pi_p$, must decrease as the distance of the particles (or the volume of the system) increases. Thus we may write for a solid or liquid with monatomic molecules:
  $$p + \Pi_0 \left( \frac{v_0}{v} \right)^m = \Pi_p \left( \frac{v_0}{v} \right)^n + T_\alpha / \beta \tag{2}$$
  while admitting that the exponents $m$ and $n$ are not necessarily constant.
  $n$ must be larger than $m$, otherwise the system could not adjust itself to
  equilibrium, and a slight increase of external pressure would cause the solid
  or liquid to collapse. [...]
Heats of evaporation indicate that the exponent m is probably not far from 2, as van der Waals assumed, and the other exponent ($n$) may likewise be evaluated (in a somewhat complicated way) with reasonable probability. [...]
... chemical combination, in polyatomic elements or compounds, involves much more complicated treatment. ... the compressibility of a binary compound
  may be roughly pictured by the following equation, in which $x$ is the
  fraction of each atom subjected only to cohesive pressure $\Pi_1$ and $(1 - x)$ is the fraction subjected to the intense chemical pressure $\Pi_2$.
  $$\beta \approx \frac{x}{n_1 - m} \left( \frac{x}{\Pi_1} + \frac{x-1}{\Pi_2} \right) \tag{3}$$
... the importance of the idea which postulates the existence of different pressures on different sides of a single atom is of great importance. ... whereas in ethyl ether the cohesive pressure is probably of the order of 2000 atmospheres, the intrinsic chemical pressure existing between atoms of carbon and hydrogen or carbon and oxygen (different in each case) must be much over 100,000 atmospheres. Precisely the same sort of difference, although perhaps of less degree, must exist within anistropic crystals, even when they are built up of only one kind of atom (e.g., zinc). [...]
... every substance must exist as such by means of a balance of pressures: external pressure plus affinity pressure equals intrinsic distending pressure plus thermal pressure.
Chemical affinity pressure must differ from cohesive pressure mainly in its intensity and in the fact that the chemical union of two atoms must involve great atomic distortion on one side of each, not equalled by the effect of the cohesive pressure around the periphery of the molecule.
  Simple cases involving the above considerations have been subjected to
  mathematical analysis, with reasonably satisfactory results. Very high internal pressures in non-volatile solids are indicated, having the order of the breaking strength of very rigid substances like glass. These pressures are consistent with the heats of chemical combination and the heats of evaporation.

Important conclusion is also that CP is defined as an actual physical quantity and should be treated accordingly. So, the same pressure units (Pa, bar, atm etc.) are applicable. Also, elemental substitutions create local stresses in a crystal structure and therefore might substantially affect CP.
The next milestone probably was a short review from late 1960s "Das „Druck-Abstands-Paradoxon”" [2] (English "The "pressure-distance paradox""), which attributed chemical pressure as a counterpart to the external physical pressure, influencing interatomic distances, coordination numbers, phase transformations,  densities and unit cell dimensions in various systems: graphite - diamond, $\ce{BN}$, $\ce{SiO2}$, $\ce{ZnS}$, $\ce{ZnSe}$ and $\ce{AgI}$.
Since 1980s CP is widely used as a concept for the description of chemical substitution and phase transition in solids. It was shown that chemical pressure coefficient ($\epsilon$, [Pa]) can and should be used to discover the symmetry of the solid phase and dilution coefficient (viz. $x$ from eq. 3, here denoted as $t$) by applying the counterbalance condition $P + \epsilon t = 0$ ($P$ is physical pressure) to the phase diagram [3] (emphasis mine):

The formulation of guide-lines for the chemical pressure ($P^*$) concept represents a difficult task. Generally chemical pressure manifests itself as any response of the crystal lattice to a given change in substitution. A reasonable requirement is that all qualitative features of the pressure, temperature ($P, \mathrm{T}$) phase diagram for a given solid solvent ($T$) should be retained in the composition, temperature ($t, \mathrm{T}$) phase diagram for its solid solution systems, and that only the phase boundaries are shifted by the substitution. It is convenient to consider the $P, \mathrm{T}$
  phase diagrams for different substitution levels ($t$) of a solid solution phase as made up of (roughly) additive contributions from a chemical component
  (which can be positive or negative) and an external (positive) pressure. ... The shift of the $P, \mathrm{T}$ phase boundaries as a function of $t$ can be used as a quantitative measure of the chemical pressure.

Interestingly, from the thermodynamics point of view, chemical potential $\mu_i$ is defined as a subset of CP [4, p. 206]; so CP can also be quantitatively expressed in units of heat/energy.
During the past decade several advancements in simulation of CP by quantum mechanical approach (DFT-CP, Hirshfeld surface analysis) [5] and EXAFS analysis [6] have been made.
References:

Richards, T. W. Transactions of the Faraday Society 1928, 24, 111 DOI: 10.1039/tf9282400111.
Kleber, W. Krist. Techn. 1967, 2 (1), 13–14 DOI: 10.1002/crat.19670020103. (in German)
Ziȩba, A.; Zach, R.; Fjellvåg, H.; Kjekshus, A. Journal of Physics and Chemistry of Solids 1987, 48 (1), 79–89 DOI: 10.1016/0022-3697(87)90144-2.
Weinhold, F. Classical and Geometrical Theory of Chemical and Phase Thermodynamics; Wiley: Hoboken, N.J, 2009. ISBN 978-0-470-40236-8.
Berns, V. M.; Engelkemier, J.; Guo, Y.; Kilduff, B. J.; Fredrickson, D. C. J. Chem. Theory Comput. 2014, 10 (8), 3380–3392 DOI: 10.1021/ct500246b.
Mukherjee, S.; Nag, A.; Kocevski, V.; Santra, P. K.; Balasubramanian, M.; Chattopadhyay, S.; Shibata, T.; Schaefers, F.; Rusz, J.; Gerard, C.; Eriksson, O.; Segre, C. U.; Sarma, D. D. Physical Review B 2014, 89 (22) DOI: 10.1103/PhysRevB.89.224105.

Daniel Fredrickson's formalism of DFT-CP is probably the most useful and applicable description or formalism of chemical pressure. Your Ref 5 is a good use of it. Here is the other paper that gives a lot of detail on it. https://doi.org/10.1021/ja300685j

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Why don't ligands affect the oxidation state of the central metal ion?
If $\ce{Fe^3+}$ were complexed with 6 $\ce{H2O}$ ligands, the resulting complex ion would be $\ce{[Fe(H2O)6]^3+}$, and the central metal ion would still have the same oxidation state of +3 as it did before complexation. However, didn't the $\ce{Fe^3+}$ ion receive 12 electrons (2 from each water molecule) in forming the complex? Is this not a kind of reduction?
Also, is the positive charge of the complex ion still localized on the metal ion, or is it spread over the entirety of the complex?

The oxidation state of the metal is a formal count not a real charge.

@Zhe I acknowledge that, and I had asked a separate subquestion regarding the real charge of the complex. But whenever something is reduced, their oxidation state goes down, no? So my main question is essentially why $\ce{Fe^3+}$ isn't reduced by being coordinated by ligands.

Even if the end result is a reduction, it doesn't matter because the oxidation state is just computed from a formula that doesn't account for that. The difference is that water is approximated as a neutral ligand whereas something like a chloride is approximated as an anionic ligand. The former does not affect the oxidation state of the metal ion whereas the latter does. There's nothing to be confused about here because we just created a definition, and then, we applied it. Definitions aren't reality.

You don't become any richer when the banks lend you money. Same thing here, with metal ion instead of you and electrons instead of money.

If you take a glimpse at the molecular orbital scheme, you will notice that the electrons donated by the water molecules still reside in orbitals that are largely ligand-centred (and have a relatively low energy). On the other hand, iron(III) also has five d electron itself; these reside in the $\mathrm{t_{2g}}$ and $\mathrm{e_g}^*$ orbitals, which are largely metal-centred (and have a relatively high energy).
This is true for most complexes. The metal tends to have rather high-lying orbitals while the ligands tend to donate rather low-lying ones to the overall complex. As long as this order is retained, there is no chemical redox behaviour going on.
For some systems, however, the orbitals are either so similar or even crossed-over that suddenly a metal orbital will become lower-lying than a ligand orbital. In this case, an electron that used to reside in a ligand orbital will relax into a metal orbital — a redox process. You can for example think of the reaction of iodide with copper(II) in this manner: copper(II) will immediately get reduced to copper(I) by iodide ions. While this is a unidirectional process, some systems are also known to bind ligands in a redox fashion reversively, meaning that the electrons will be exchanged both ways. The binding mode of oxygen to haemoglobin is understood in this manner.
Finally, sometimes there is a change in oxidation state that is merely formal. Tetracarbonylhydridoiron(0) $\ce{[Fe(CO)4H]}$ is such a case. It can be protonated to formally give tetracarbonyldihydridoiron(II) $\ce{[Fe(CO)4H2]+}$ or deprotonated to give tetracarbonylferrate(–II) $\ce{[Fe(CO)4]-}$. Both changes in oxidation state are merely formal and chemically all that is happening is a protonation/deprotonation.

Another point here is that ligands are labile. The $\ce{H2O}$ molecules aren't "glued" to the complex but "swap" in and out. The $\ce{Fe^{3+}}$ can also have ligands other than $\ce{H2O}$ if there any are in the solution. This ligand swapping is why it is called a "complex."

@MaxW As I tried to address when referring to the redox binding mode of oxygen to haemoglobin, lability does not mean that no redox process is involved.

Another thing: Even if the ligands are firmly glued to the central metal and no swapping occurs at any reasonable timescale (eg $\ce{[Fe(CN)6]^4-}$) it is called a complex.

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Thermochromism in PbI2
I noticed while heating some PbI2 powder that it reversibly changes from a yellow to a dark red. However, I was unable to find anything on this thermochromism.
There are related cases in TlI and AgI but there it is mostly about occupying the free sites in the crystal structure or if Ag is present there is also the possibility of Ag coming close to Ag (so-called argentophilia). This, however, doesn't seem to be the case for Pb(II).
It is also no thermal charge transfer, for Hg(II) + I(-) a thermal charge transfer to Hg(I) + I(0) was often mentioned and it makes sense there but Pb(I) is not really useful here.
The structure is also different forming layers. I read they have weak VdW-bonds so perhaps those break in the process of heating it? If impurities are introduced there is the possibility of forming a perovskite structure but I doubt this would be reversible. So this has to be a smaller shift in the structure.
PbI2 has quite many different room temperature and high-temperature polytypes, so perhaps it's really a change in the crystal structure? I couldn't really find much on this because I don't know how the colors change for the different polytypes.
Does anyone have an idea what might be the dominating effect for a reformation in the structure, like an actual effect that might take place here?
You see often it is the case in these solid-state thermochromic compounds, that the high-temperature form seems to be the best situation, but cooling it down allows weaker forces like this argentophilia, for example, to distort the structure a bit. Like TlI is more or less an orthorhombically distorted NaCl in its low-temperature form.
Given the fact however that Pb(II) still has a lone-pair that can be positioned quite well in the gap between the layers I don't see a real demand for any change at all.

Does the colour change happen only in air, or does it also happen in an inert atmosphere? And how hot do you have to get it for it to change colour? Presumably less than its melting point of 402 °C? Rather than thinking about structural changes, there's a real chance the colour change is due simply to semiconductor electronic properties; maybe with enough heating you can populate enough electronic states closer to the valence band to provide significant absorption of lower energy visible light.

That might be another option, thank you for the suggestion. I can't really put it under nitrogen or argon here and for the temperature I'd have to check at what point it changes. But is was around 200-250°C I guess. I think the transition is at around 300°C so it could be one of the other polytypes or as you said the semiconductor properties.

Just found something:
So the most dominant thermal change, and we are talking here about annealing single crystals for several days to around 150°C seems to be a polytypic change from the 2H-Type to the 12R-Type. Or, as there are several hexagonal and rhomboedric phases let's just assume a change from H to R. You may know polymorphs, like allotropes for example, diamond vs graphite or different phosphorus allotropes. And polytypes are structures that usually appear in layers where the overall composition and connections are pretty much the same but the stacking order of the layer is different. I found a sketch how the layers have to change:

1 E.Salje, B.Palosz, B.Wruck, Structural and Magnetic Phase Transitions in Minerals, Advances in Physical Geochemistry, vol 7, Springer, New York, p.231, 1988.
It was often reported that the change from H to R seems to be thermally while the change from R to H takes a long time. As the crystals were grown from sodium based gels traces of sodium could be found and the higher the sodium content the easier and faster they turned back to H. I didn't grow any crystals but precipitated it using NaI, so I probably have a high sodium content which acts as an impurity to fastly change the structure back, which appeared to me as being reversibly thermochromic.
I don't know whether this also causes the color change but many other color changes, like in TlI depend on minor distortions in the crystal structure as well so it could be plausible.


text:
Gas phase during water sampling
I have a question concerning simultaneous sampling of water and gas phase.
The situation is a pipe (diameter somewhat around 15 cm) where a mixture of water and gas is flowing through (two-phase flow). Assuming that the flow is separated and the gas phase is moving on top, what will happen if I open a sampling tap at the bottom of the pipe? The flow is induced by a pump and the pressure in the system is above ambient (20 bar).
My assumption is, that I only sample the water and the gas phase on top will not leave the pipe unless I allow the gas phase to "connect" with the tap by extracting a big amount of fluid.
The problem is maybe comparable to a bottle of sparkling water, where I drill a small hole at the bottom. Will I only sample the fluid phase or will I sample a mixture of both?
I hope somebody has an answer. Thanks in advance!

In your sparkling water bottle example, assuming it is capped tight, some water would leak out and somewhat of a vacuum would form. Would that be the case in the flowing system or is there some pressure associated with (i.e. causing) the flow of the fluids? In other words, is this an open flow system, or sealed, or under some pressure, etc. Some clarification of the physical setup with respect to this issue would be helpful.

Thanks. Added the conditions of the physical setup. Concerning the vacuum: wouldn't the gas phase on top just expand as the volume for the gas phase increases?

Of course it would.

That's a tricky question.
Given your setup, it follows that you will only be sampling water (not simply "fluid", mind you; fluid is a common term for both gas and liquid). But gases might be more or less soluble in liquids. Chances are that your water is saturated with some gas (say, $\ce{CO2}$) and pressurized with the same gas in the pure form. Then, as soon as you take your sample and put it in a sealed (or worse yet, unsealed) container, it will release some gas, and so the composition of the sample will become different from that of the liquid in the pipe.
Whether or not this issue is relevant to your situation, is up to you to assess.

Thanks for the answer and the clarification about "fluid". As a non-native I was not aware of this difference. The current sampling setup is taking account of the gas release which is occuring due to sampling by a gas counting meter (it's the volume I am most interested in). I was just asking myself if the sample is representative for the whole system, which I can now deny.

Well, if you take account of the released gases, then your sample is OK as long as we are talking of the liquid. Then again, the very intention manifested in counting the gas implies that you're interested in it, hence you might want to know the gas/liquid ratio, which you won't be able to know by sampling the liquid alone.


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Why is the bond dissociation energy of C-H bond higher than that of a N-H bond?
Looking at a chart of BDEs (bond dissociation energies) regarding elements bonded with hydrogen, the general trend seems to be that the BDE increases as we go to the top and to the right. This can be explained by the fact that the elements to the right have stronger attraction for the covalent bonds, making homolytic cleavage harder. Likewise, those elements near the top have shorter radii, which means that their attraction is larger too.
However, there is a single anomaly in the $\ce{N-H}$ and $\ce{C-H}$ bonds, where the BDE of $\ce{C-H}$ is higher than $\ce{N-H}$. Why is this?

Think about - lone pair electrons require greater stabilization (more s-character in the orbital) than bonding pair electrons.

It is experimental data and one could only say according to data they have. You have to be careful in comparing bond strengths even in diatomic molecules, but especially when the bond is in a polyatomic molecule. In the case of polyatomic molecules, the bond strengths are even more scattered depending upon the structure of the rest of the molecule. A somewhat more sensitive parameter of bond strength is actually the bond length (short bonds -> stronger bonds & long bonds -> weaker bonds) but data on the bond lengths in polyatomic molecules is much more scarce than bond energies because the

con't from above: "detailed molecular structures are more difficult to obtain.The bond energies in CH and NH are essentially the same, but even so, the electronic states of the diatomics may be different so that comparisons may be mismatched. So I think we cannot clearly tell why it is so."

https://pubs.acs.org/doi/abs/10.1021/ed077p1062

If you look at an orbital diagram of nitrogen you will see that each p-orbital has one electron in it, if you want to add a second electron to any one of them, it makes the orbital higher energy. Atoms would generally like to be in the lowest possible energy state, and half or completely filled shells are far more low energy, and thus, more stable, this is most clearly seen with the electron configuration of copper and chromium.
Cu: [Ar] 4s1 3d10
Cr: [Ar] 4s1 3d5
Nitrogen has a half filled 2p sub level, so it has to move to a substantially higher energy state to add a fourth electron. If it gets one in a bond with Hydrogen, Nitrogen finds it energetically cheaper to dissolve the bond and return to its base configuration. The reverse applies to Carbon. Because it now is able to get another electron from Hydrogen, it now has a half filled 2p sub-level and is in a fairly stable configuration that it would like to keep.
Most of this comes from Chemistry by Maciel, Traficante, and Lavallee. The book was published in 1978, so I am sorry if anything is dated.

But this comparison isn't fair because you are looking at a single bond, possibly in some molecule. Clearly, in the usual molecule, like methane or ammonia, the 2p subshell would technically be fully populated. But even if you were to take this into consideration, your explanation is still flawed because bonding involves hybrid orbitals and/or molecular orbitals. So it is incorrect to still talk about the atomic orbitals.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Concept of Limiting Current
I am investigating the current output from an electrochemical cell. The electrochemical cell can be broken down into several stages, for example:

Diffusion of reactants across a capillary
Diffusion in aqueous electrolyte
Adsorption and desorption at the electrode
Current generated due to anodic and cathodic though the Butler-Volmer Equation

From the readings I have done so far, current will be generated at each of these stages due to transfer of electrons. However, most authors will introduce the concept of limiting current at the rate determining step and neglect the remainder current generated from other stages.
If so, is it possible to obtain the total current output by summing the limiting current in each individual stages of the entire chemical reaction? Or focusing on solely the limiting current of the rate determining step will give a good approximation?

I think you have the wrong idea. The notion that "current will be generated at each of these stages due to transfer of electrons" is wrong. Rather current will be limited at each step. Also current and voltage are dependent. As the current flowing through a cell goes up, the voltage goes down.

@MaxW So, does it mean that the current measured when connecting the 2 ends of the electrode across a resistor is not the limiting current? What are the procedures needed to calculate this current then? Taking the net battery voltage (Energy given by Nernst equation - cathodic and anodic overpotential) divided by the value of the resistor and internal resistance of the cell?


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Why oxidation number of oxygen in OF2(oxygen difluoride) is +2?
I mean oxygen is electronegative element it should have negative oxidation number
Also how it is able to make bond with fluorine which itself is electronegative?
 Why the oxidation number of oxygen is +2.?

related https://chemistry.stackexchange.com/questions/59465/can-we-oxidize-oxygen

Oxygen is an electronegative element -True.
But oxygen is not the most electronegative element. It has an electro negativity of 3.5 on Pauling scale. That means if oxygen combines with an element which is more electronegative than it will surely possess a positive oxidation state.
Fluorine being the most electronegative element (electro negativity of 4.0 on Pauling scale) will in any case (except in fluorine gas) have an oxidation state of -1. So that's the reason why oxygen has a +2 oxidation state in OF2.
Dioxygen difluoride (O2F2) is another fluoride of oxygen in which oxygen has an oxidation state of +1.
EDIT: Fluorine in its gaseous form (F2) has zero oxidation state.

Fluorine gas itself has F in oxidation state zero.

Yep! I forgot about that. I'll make an edit.

Fluorine is more electronegative than oxygen. Thus it would attract electrons towards itself during complete transfer. It would result in a +2 charge on oxygen. Thus oxygen has +2 oxidation no.


text:
Derivation of Henderson–Hasselbalch equation
While deriving the Henderson–Hasselbalch equation why don't we consider the hydrolysis reaction?
For example if we consider an acidic buffer of acetic acid and sodium acetate, why do we not consider the hydrolysis of the acetate ion furnished by sodium acetate?

Unclear what you mean here. Isn't the whole point that you are accounting for the interaction of the acid and conjugate base with water via the equilibrium?

Just because it's negligible.
Well, I tried to make some calculations. I was too lazy to make them using all this internal stackexchange math system, so I hope you'll understand my handwriting.
I also need to say what conditions I used: activity=equilibrium concentration; I didn't care much about hydroxonium, because it doesn't change anything; I used 0.1M solutions of both acetate and acid just because I wanted to.
The conclusion is: really, if you have 0.1M solutions, a change in 10-10M will change absolutely nothing, so why suffer and not neglect this?
P. S. All calculations made in Excel 2013 :)

Do you mind using MathJax? Images are not easily searchable?

@PrittBalagopal I probably will, but not in the middle of working day :)


text:
Determining the stability of four organic compounds given below

Answer given is a but how on earth is it possible because here one C Atom has 5 covalent bonds. Infact the aromatic character is decreased.  I think c is correct because of resonance structure.  But why the anser is a?  Can anyone explain please?

For halogen is in meta position ? So -I effect of NO2 effecting?

The answer is (a), but the double bonds in it are drawn the wrong way (one carbon seems pentavalent, another trivalent). Mistakes happen.

The main idea is that you don't want to end up with a (formal) positive charge next to a strongly electron-withdrawing group (EWG). No resonance structure of (a) (especially if it were drawn correctly) can have that, whereas in (b) that's already the case in the depicted structure (NO2 being the EWG), and in (c) you get that by shifting the two pi bonds in sequence. Not sure about (d); did they mean to put double bonds somewhere, or is it really this aliphatic cycle we're talking about? It's always sad to see such inaccurate and shabbily put together exam questions... how will students learn?

Ok I got the idea.  It's because of hyperconjugation (mainly)  and then about like charges on resonating structures or normal structures will decrease the stability
And so the order is
A>C>D>B
Or A>D>C>B


text:
Reason for the greater energy of the eclipsed conformer of alkanes?
My textbook states:

In the staggered form of ethane, the electron clouds of C-H bonds are as far apart as possible, thus there are minimum repulsive forces, minimum energy and maximum stability. On the other hand, when the staggered form changes into the eclipsed form, the C-H bond clouds come closer to each other resulting in increased electron cloud repulsion. To check the increased repulsive forces, molecule will have to possess more energy. And therefore it is less stable

My question pertains to the statement in bold text. Am I correct in thinking that in the eclipsed conformer the electron clouds repel more, this repulsion would (sort of) want to break up the molecule so to fight this breaking tendency, the molecule has to possess greater energy to control these new repulsive forces?
If I am wrong, what does this statement mean?

I think you have the right idea. To move from the staggered to the eclipsed conformation, you have to put in additional energy because otherwise the repulsive forces are enough to keep the molecule in the staggered form.

Just take two electrons. If you want to push them closer together you need to input some energy and the system overall will have higher energy. There is a preference for the two electrons to be farther apart.
This situation is not any different if we look at pairs of electrons in bonds instead of individual electrons.

The way I think of the two electron system is that the potential energy content is higher once we push them together is that that is the energy needed to set up the closer configuration on bringing the charges from infinite separation to the closer separation, which is the classical picture of potential energy of a system. While your answer does give insight, do you essentially mean that the molecule tries to push the electrons together so has greater energy? Which is what I asked, my question remains, am I correct in thinking the way I did?

It's a little hard to parse what you mean. The classical picture doesn't matter because electron-electron repulsion is still defined in the same way. The electrons are overall closer in the eclipsed versus the staggered conformation. Not sure why you need to invoke breaking up the molecule. Increased repulsion increases energy.


text:
Is it possible to have a gas so dense a solid or liquid can be formed on top?
any ideas? I’ve been wondering about this for a while.
Thanks

I ‘any ideas’ your question? Please elaborate

Li weighs 0.534g/ml, Ar already 0.536g/ml at it's critical point, Kr 0.9.
So: Yes.


text:
Why does like dissolve like?
Polar solvents love polar solutes to be dissolved in it and non polar with non polar. Often said as like dissolves like.
Okay, polar loving polar can be understood with help of the facts: same polar nature, same kind of interactions etc.
But how will you explain the application of this rule with organic or non-polar solvents?
Solomons’ and Fryle’s Organic Chemistry says that there are some unfeasible entropy changes when polar solutes are added in non polar solvents and vice versa.
Why so?

"Like dissolves like" is a thumb rule and generalisation. There are many cases where it does not apply, for polar and unpolar compounds.

I do not consider this a duplicate of the question linked. The linked question asks why non-polar compounds dissolve in non-polar solvents. This question asks about why polar solvents cannot easily dissolve non-polar compounds and vice-versa.

The reason behind this is the hydrophobic effect. Everyone has seen it if they pour a spoonful of vegetable oil into a pot of water, e.g. to cook pasta. As long as nothing is disturbing the vegetable oil, it will collect itself together in one big bubble rather than form many small bubbles.
Polar solvents will always be arranged in a way that positively polarised areas are close to negatively polarised areas in the neighbouring solvent molecule. Hydrogen bonds — especially in water — are nothing but the extension of this concept to even more polarisation. Polar molecules, as you stated, will fit together with this scheme well. Unpolar compounds mixed in do not. As soon as you add an unpolar compound to a polar solvent, you are creating a type of artificial boundary and only the areas of the solvent molecules that are neither particularly positive or negative will be happy to be in the near vicinity. That means that you have a much higher ordering of the solvent molecules where they hit an unpolar compound, because one direction is basically doomed to be neither positive nor negative.
This ‘solvent wall’ will form no matter whether the unpolar island is large or small. However, it will also have almost the same thickness, no matter how big the contained part is. Therefore, if multiple unpolar molecules clump together, the overall number of polar molecules constrained in that wall is lower — an entropic gain. Thus, unpolar compounds tend to either not dissolve or precipitate out of polar solvent solutions.
This also works in the opposite direction. A polar molecule will prefer, for energetic reasons, to have a polar neighbour. However, an unpolar solvent cannot provide that polar environment. A neighbouring undissolved polar molecule, however, can. Here again, the polar compounds rather stay clustered together as it allows the central molecules to be more disordered while only a small layer on the border must take care to interact as well as possible with the unpolar solvent.
Note that this answer assumes some kind of black/white dichotomy. In reality, most compounds are somewhere on a scale from absolutely unpolar to very polar and are able to adapt to a wide range of organic solvents. Conversely, some organic solvents have the reputation of dissolving almost anything organic; most notably dichloromethane. However, the solubilities may greatly vary.

Nothing's absolutely unpolar.

@Mithoron:  Can you elaborate on that?  I would have thought that highly symmetric molecules like carbon dioxide, methane, benzene, etc. couldn't have a permanent dipole moment due to their symmetry.  Or are you talking about London dispersion forces?

@MichaelSeifert I said here that it's wrong like hundred times. Why for crying out loud everyone cares about stupid dipoles when they have only slight effect on properties of compounds? Polarity is property of solvent.

@Mithoron Well, how polar is liquid hydrogen going to be? (I know it’s an extreme example). But even if that has some polarity, most organic chemists just define the polarity of something like pentane to be zero units on an arbitrary scale.

More polar then liquid helium.

I would be quite dubious in ascribing much validity to answers which use the word "unpolar", since that word indicates either a lack of familiarity with chemical terminology or a lack of respect for it.
The way I consider solvation (for two pure chemical compounds) is that in their pure state the interactions are between molecules of themselves. For two compounds, A & B, call them A2A and B2B interactions. These interactions can be either weak or strong (and this can be determined using Thermodynamic experimentation). When considering solubility, the added term is the A2B interactions.  When A2A and B2B interactions are weak, then even weak A2B interactions should result in substantial interactions (and hence substantial mixing). When A2A and B2B interactions are both strong, A2B interactions must also be strong to get a homogeneous equilibrium phase. So we should look at these interactions relative to the strongest.  If A2A interactions are strong (say NaCl) and B2B interactions are weak (say n-hexane) then we'd predict that the A2B interactions are also going to be (very) weak, and that the solubility of NaCl in hexane is going to be small (negligible, actually). So in this case the strong interaction in the (non)solute (NaCl) dominates. But instead of NaCl, consider C20H42 mixed into n-hexane. We'd predict weak A2A and B2B interactions and so, since we know all matter interacts electromagnetically, we'd assume A2B interactions are going to be relatively significant, and so the icosane will be soluble in the hexane. But what about icosane in water? Here we'd have weak A2A interactions and strong B2B interactions. What about A2B? Well, looking at the molecules, there is no obvious specific interactions which could add to the interaction energy, so I'd assume that the water isn't going to substantially interact with the icosane, and the solubility will be small. You'll note that it is the stronger interaction, whether it is in the solvent or the (potential) solute which determines the 'baseline'. The hetero-interaction (A2B) must be similar in magnitude. The easiest way for that to occur is for both A and B to have similar polarity. Hence, like dissolves like. (with lots of exceptions for specific interactions which "like" doesn't cover (h-bonding, polarizability, steric effects, etc.) [Added clarification: A2A is an abbreviation of "a molecule of A interacting with another molecule of A" or "A-to-A interactions"]

Can you add paragraphs? Such a wall of text is hard to read.


text:
Bomb Calorimeter vs Cup Calorimeter
In the textbook,it is said that bomb calorimeter measures the constant volume heat ,while the cup calorimeter measure Constant pressure Heat.I was wondering, why cup calorimeter is considered constant pressure,even though it also does not allow a gas to expand (it has a fixed volume).So if not then that means cup calorimeter allows some air to get in so that pressure is constant ?(I.e the system is not closed )?

Even though the pressure remains constant, heat can get in (or out) of the system. From the cup calorimeter, we are trying to find $\Delta H$ which is equal to the heat evolved from the system at constant pressure. We measure this heat and our job is done.
$$\Delta H=q_p$$
If we want to calculate $\Delta U$, we just to find the heat evolved from the bomb calorimeter, which can be done by measuring the change in calorimeter's final and initial temperature given that the heat capacity of the calorimeter is known.
$$\Delta U=q_v=nS\Delta T$$


text:
Find the Relative Error in the Mass of the Product
I have the following question,

A student was given the assignment of $4.00$ $\pu{mL}$ of $0.50$ $\pu{M}$ $\ce{KI}$ with $16.00$ $\pu{mL}$
  of $0.50$ $\pu{M}$ $\ce{Pb(NO3)2}$ in a stoichiometry study using the method of
  continuous variations. If the student used only $3.50$ $\pu{mL}$ of $0.50$ $\pu{M}$ $\ce{KI}$,
  what would be the relative error in the mass of the product?

I need to find the relative error, I know that
Relative Error = Absolute Error/Actual Value. I also know that Absolute Error = Expected Value - Actual Value.
I believe that the way to solve this is to find the moles of $\ce{KI}$ at $4.00$ $\pu{mL}$ and add that with the moles of $\ce{Pb(NO3)2}$ at $16.00$ $\pu{mL}$. Both at $0.50$ $\ce{M}$. This would be the Expected Value?
Then I need to find the moles of $\ce{KI}$ at $3.50$ $\pu{mL}$ and then add this with the moles of $\ce{Pb(NO3)2}$ at $16.00$ $\pu{mL}$. This would be the Actual Value?

More of a long comment than answer.
Well you are all wrong.
Absolute Error = Actual Value - Expected Value
That way if you have too little product the error is negative, and if you have too much product the value is positive.
Relative Error would typically be expressed as a percentage.
$ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Expected Value}} \times 100\%$.
First when you mix $\ce{KI}$ and $\ce{Pb(NO_3)_2}$ what happens?

Hint - What is the formula for the chemical reaction?

PB(NO3)2 + 2 KI = 2 KNO3 + PBI2 is the reaction balanced. So to find the moles of 4.00mL of .50 M Kl I would do, moles = Molarity * Liters, so .50*4.00 = 2.0 or do I need to make 4.00mL to liters first?

Do a "dimensional analysis." 1 mL = 0.001 liters, 1.0 molar = 1 mole/liter. So 1 ml of a 1.00 molar solution is 0.001 moles. So how many moles of Pb(NO3)2 and how many of KI? Which is the limiting reagent? How much (grams) of PbI2 should be made?

Dimensional analysis see https://en.wikipedia.org/wiki/Dimensional_analysis#The_factor-label_method_for_converting_units


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Why is there a bulge at the top of my volumetric flask?
Sometimes volumetric flasks are designed with a bulge in the neck right below the stopper opening. What is the design intent of that bulge?

I'd assume just an artifact of joining the ground glass joint to the neck of the flask.

It's also there to help you grip better. This way there's a stop if it starts to slide in your hand.

From ASTM manual [1, p. 204], emphasis mine:

Volumetric flasks should be diluted to just below the line, mixed thoroughly, then allowed to stand for 4 to 24 h to allow any heat from an acid-water reaction to dissipate. [...] Thorough mixing consists of at least 20 complete inversions of the volumetric flask, allowing enough time on each for the air bubble to completely traverse the length of the vessel. Some borosilicate volumetric flasks contain a bulge in the neck above the line as an aid to proper mixing.

Volumetric pipetts also have a large bulge in the middle, but it serves a different purpose – to keep the $\rm length : volume$ ratio minimal and still maintain quick liquid collection with high precision.
Reference

Dulski, T. R. A Manual for the Chemical Analysis of Metals; ASTM manual series; ASTM: West Conshohocken, PA, 1996. ISBN 978-0-8031-2066-2.


text:
Acidic Environment and Rust
I'm trying to understand why in an acidic environment, rust increases. The explanations I have seen so far basically have the following,

"In acidic environments the cathode for the rusting of iron changes to
  $O_{2(g)}+4H^+ _{(aq)} + 4e^- \rightarrow 2H_2 O_{(l)}$. The anode
  reaction remains $Fe_{(s)} \rightarrow Fe^{2+} + 2e^-$. The  cell
  value increases as the cathodic process is lower in the
  electrochemical series, producing a more effective electrochemical
  cell, accelerating corrosion rate."

Though I don't understand why in acidic environments the cathode reaction changes? These are some of the explanations I've read

In normal oxygenated conditions hydronium ions increase the electrode potential of the reduction of oxygen making it a better oxidant.
In acidic environments the reduction of oxygen proceeds more rapidly. (This doesn't seem to really explain it)

Question
Why in acidic environments does the cathode reaction for the rusting of iron change?

related https://chemistry.stackexchange.com/questions/35245/why-is-acidic-hydrogen-peroxide-a-potent-oxidizer


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Why is an isothermal process not unique in nature?
To be clear, I have got two doubts,

One, about the non-uniqueness of an isothermal process &
Two, about the uniqueness of an adiabatic process.

The first doubt
The Wikipedia page of isothermal process states that:

For an adiabatic process, in which no heat flows into or out of the gas because its container is well insulated, Q = 0. If there is also no work done, i.e. a free expansion, there is no change in internal energy. For an ideal gas, this means that the process is also isothermal.[4] Thus, specifying that a process is isothermal is not sufficient to specify a unique process.

Now I am confused about the meaning of the last line.

Thus, specifying that a process is isothermal is not sufficient to specify a unique process.

I am not able to understand the significance of that last statement. Does it mean that any two states, say A & B, can be connected by more than one isothermal processes?
The second doubt
Similarly, I had read in my textbook that an adiabatic process is unique in nature. That is, between any two states only one adiabatic process can exist. To say the same thing in other words,
if two thermodynamic states are connected by a reversible adiabatic process, then the same two states cant be connected by an irreversible adiabatic process
Kindly help me understand these statements with the help of some PV/VT/PT diagrams.
Any help would be deeply appreciated.

Please clarify what you are actually asking. You claim to be confused about the uniqueness of an isothermal process in the line between the quotes but later claim to be unsure about the uniqueness of an adiabatic process. So which one is it?

Sir/Madam Jan,  I have tried to make my question clearer, kindly put the question off hold and help me. Thanking you in advance...

@user134175 I still think this is two separate questions so it should be posted as two separate questions. Edit this one down to one of your doubts and post the second as a separate question. I will vote to reopen this post if you do that because I think they are worthwhile questions.

(1) An isothermal process is one during which the system temperature $T=constant$. For an ideal satisfying $pV=RT$ this means that $pV=constant$ irrespective of how you achieve it. You can go up and down, back and forth on the same isothermal, but it is still an isothermal process having the same temperature. So it is not a unique process between two fixed points.
(2) The uniqueness of the adiabatic process is true only for a simple system, that is one whose state can be described by temperature and one additional "mechanical" parameter, e.g., for a gas $p = p(T,V)$, or a magnet $M=M(T,B)$, etc. It is definitely not true for more complicated systems, for example a gas with magnetic properties $p=p(T,V,B)$. Instead what is true is that through every equilibrium point there is a unique adiabatic surface (not a curve) determined by a pair of variables say $T,V$ or $V,B$ or $T,B$ such that any curve lying in that surface is an adiabatic process. That such adiabatic surfaces exist is the basis of one of the mathematical proofs of the existence of the entropy function and this quite famous proof is due to Caratheodory.


text:
Do weak acid/weak base neutralisation reactions go to completion?
I'm wondering if reactions that involve a weak acid and a weak base go to completion. For example, say we have equal amounts and equal volumes of acetic acid and ammonia being mixed together. Would this neutralisation reaction go to completion or would it reach a state of equilibrium?
$\ce{NH_3 + CH_3COOH <=> NH_4^+ + CH_3COO^-}$
Instinctively, I'm thinking that the reaction would be an equilibrium, since proton transfer should be able to occur between the products ($\ce{NH_4^+ + CH_3COO^-}$) to reform the reactants. Can anyone clarify if this is true?

Yes, there is an equilibrium.

Weak acids and bases do indeed remain in an equilibrium state with a certain concentration of the free acid and the free base alongside a certain concentration of the deprotonated acid anion and the protonated base cation.
In general, you can consider any acid-base reaction to be essentially the following:
$$\ce{HA <=> H+ + A-}\tag{1}$$
Which in turn means that the $\mathrm pK_\mathrm a$ value is defined as in $(2)$:
$$K_\mathrm a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{2}$$
Reactions $(1)$ and $(2)$ are usually written out in water with hydronium in place of a naked proton but the same principle applies. Of course, a base reaction can be considered essentially the reverse, meaning that the acid constant typically given is actually $\mathrm pK_\mathrm a (\ce{HB+})$ rather than anything directly corresponding to the free base. If we now take the reaction of a (weak) acid and a
(weak) base, we get equation $(3)$ and the equilibrium constant as given in $(4)$:
$$\begin{align}\ce{HA + B &<=> A- + HB+}\tag{3}\\[0.7em]
K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\end{align}$$
We can now perform simple mathematics with $(4)$ to arrive at the modified equation $(5)$ as below:
$$\begin{align}K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\\[0.7em]
&= \frac{[\ce{A-}][\ce{HB+}][\ce{H+}]}{[\ce{HA}][\ce{B}][\ce{H+}]}\\[0.7em]
&= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{HB+}]}{[\ce{B}][\ce{H+}]}\\[0.7em]
K &= \frac{K_\mathrm a(\ce{A})}{K_\mathrm a (\ce{HB+})}\tag{5}\end{align}$$
We have thus arrived at a way to determine the equilibrium constant of the proton transfer reaction just from the acidity constants of both participants. If both acids weak and have similar $\mathrm pK_\mathrm a$ values — as ammonium and acetic acid do — the equilibrium constant will be close to $1$ and thus both sides of the equation will have similar concentrations.
Using $(5)$, you can also predict a ‘degree of completion’ for any acid-base reaction.

I'll add another bit of terminology, the equivalence point. So given a solution of acetic acid you add an equivalent amount of ammonia, a base. However because both are weak the final pH need not be 7.00 (neutral).

@HernanMiraola That is indeed a typo (or a brainfart, I’m not sure).


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The electron configuration and electron density distribution in singlet oxygen
The molecular orbital schemes for singlet ($\mathrm{^1\Delta_g}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.

Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.
It is pretty clear that the dioxygen molecule, if just observing the position of the nuclei, must be $D_\mathrm{\infty h}$ symmetric. This implies that both $\unicode{x3c0}^*$ bonds are symmetry equivalent. Therefore, it seems logical that in the $\mathrm{^3\Sigma_g^-}$ triplet state the two orthogonal $\unicode{x3c0}^*$ bonds are each populated by a single electron, leading to an overall rotation-symmetric linear molecule.
In the $\mathrm{^1\Delta_g}$ singlet state, this degeneracy seems to be destroyed, just by looking at the MO scheme. We now seem to have one populated $\unicode{x3c0}^*$ orbital and one unpopulated one, going by my understanding of this scheme.
What does this imply for the electron density distribution of $\mathrm{^1\Delta_g}$ singlet oxygen? Is it valid to assume that two (arbitrary) opposite sides have a higher electron density than the two other orthogonal sides? How can this be interpreted?

Probably not a full answer, but hoping to point readers in a useful direction.
If we denote the two $\pi^*$ orbitals as $\pi_x^*$ and $\pi_y^*$ respectively, then the spatial wavefunctions for each state can be written as follows (all lower orbitals are ignored):
$$\begin{align}
{}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) + \pi_x^*(2) \pi_y^*(1)] \\
{}^3\Sigma_\mathrm g^- &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) - \pi_x^*(2) \pi_y^*(1)] \\
{}^1\!\Delta_\mathrm g &: \begin{cases}\pi_x^*(1) \pi_x^*(2) \\ \pi_y^*(1) \pi_y^*(2) \end{cases}
\end{align}$$
For the ${}^1\!\Delta_\mathrm g$ case, one would expect a spatial degeneracy of 2, because the letter $\Delta$ in the term symbol indicates that the quantum number $|\Lambda| = 2$. Therefore, there is one state with $\Lambda = +2$, and one state with $\Lambda = -2$. On the other hand, the $\Sigma$ term has $|\Lambda| = 0$ and hence is spatially non-degenerate (the $^3\Sigma$ term is triply degenerate due to spin).
(Digression: $\Lambda$ represents the projection of the angular momentum along the internuclear axis. As such, there is no further projection quantum number ("$M_\Lambda$") that can take values $-2, -1, 0, +1, +2$. In comparison, in atoms, $\ell$ indicates the total angular momentum and $m_\ell$ the projection of this angular momentum onto the $z$-axis. For more information about term symbols of diatomic molecules, I suggest reading this link. There are many other sources on the Internet, but many are sloppy with notation, which only leads to confusion down the road.)
So, the MO diagram above - which ties the ${}^1\!\Delta_\mathrm g$ state to only one wavefunction - is incomplete. The ${}^1\!\Delta_\mathrm g$ state corresponds to two possible wavefunctions, which are symmetry-equivalent; there is no "preference" for the $x$-axis over the $y$-axis.
If you want to go further than what I've written, then I'd point you to the case of the boron atom which I asked about earlier, which is exactly analogous to this. The point about boron is that it has one electron in the 2p subshell. Does this go into the $\mathrm p_x$, $\mathrm p_y$, or $\mathrm p_z$ orbital? In the case of dioxygen, do the paired electrons go into the $\pi_x^*$ or $\pi_y^*$ orbital?
As you said, it does not make sense for us to assign the electron(s) to any of the orbitals in particular, as that leads to an asymmetric electron density distribution. However, I must admit that I don't fully understand the CASSCF calculation there, and I never quite got round to following up on that question, so you'll have to ask somebody else about that.

Am I distilling the following take-home message correctly: Assuming we split the space around the bond axis using the planes $x=z$ and $x=-y$ into four equal spaces, it is wrong to imagine that two (arbitrary, but opposite) of these spaces have a higher electron density than the other.

Yes, that's essentially it.

$$\begin{align}
{}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[|\pi_x^(1) \pi_y^(2)| \color{red}{-} |\pi_x^(2) \pi_y^(1)|] \
{}^3\Sigma_\mathrm g^- &: 2^{-1/2}[|\pi_x^(1) \pi_y^(2)| \color{red}{+} |\pi_x^(2) \pi_y^(1)|]
\end{align}$$ Acc to doi: 10.1021/cr60272a004

@Martin-マーチン if you're talking about the equations on p 397, those are the full spatial+spin wavefunctions; what I wrote above is only the spatial part. If you expand the determinants and factorise out the spin part, you get the minus sign for the spatial part of the triplet state (I didn't bother checking the singlet state). (I added the word spatial to hopefully clarify.)


text:
How do chemists verify certain phenomena like 1,2-hydride shifts, 1,2-methyl shifts and ring expansions?
I've been learning about organic chemistry independently on a ship without an instructor or university resources like a lab setting for the past year. I was wondering how do chemists verify that their mechanisms are correct?
For example, take 3-methylbutene and add the reagent HCl. From the 1,2-hydride shift we get 3-chloro-3-methylbutane. For a 1,2-methyl shift take 3,3-dimethylbutene with the same reagent HCl. We know that from the mechanism we get 2-chloro-2,3-dimethylbutane. But how do you verify that these are the major products?

Still at it? Congratulations, keep up the good work!

NMR is the usual way to prove the structure obtained (quantitatively), and then you start switching substituents. Deuterate the educt at some position, and check where the deuterium is in the product, switch methyl for ethyl, etc.

I wanted to put this as a comment but I don't have enough rep :p
It happens the other way round, first the products are obtained, then the mechanism is proposed.
The point is, according to me that once it has been verified which product is major and which minor, chemists may try to come up with a suitable explanation for the product obtained, on paper. They may try some things, come up with possible mechanisms and sometimes even phenomena like hydride shift, methyl shift etc. Once they come up with such an explanation, they try to apply it to other related problems and upon reaching the right solution every time, they can say this is what might have happened to a high degree of certainty.
So you see, chemistry is also sometimes glorified problem solving :p


text:
Identifying a Positive or Neutral Ligand Inside Coordination Sphere?
There are some coordination compounds like $\ce{Na2[Fe(CN)5NO]}$, where ligands like $\ce{NO}$ are used which exist as both $\ce{NO+}$ (nitrosonium ion) and $\ce{NO^.}$ (neutral nitrosyl). How to identify which ligand is being used in such complexes?
A similar question was also on Quora but none of the answers were satisfactory.
Please don't answer with respect to Raman/infrared using stretching frequencies etc. since all that is out of scope at the level I'm currently studying.

Long story short, you can never tell.

Why not ? If such a question is asked in some examination as an MCQ , what then?

Then you resort to some Raman/infrared frequencies stuff.

Possible duplicate of Back bonding in coordination complex

@Mithoron Proposed dublicate is a different question altogether. OP is just using this compound as an example to understand the concept in general while the other question is specifically dealing with only this compound.

Books have been written on the topic of whether a certain nitrosyl ligand is the nitrosyl cation, the nitrosyl radical or — you forgot this one — the nitrosyl anion $\ce{NO-}$. This ligand is termed a non-innocent ligand, meaning we cannot just look at it and tell what it is.
You asked for an answer that does not require IR or Raman spectroscopy. Well sadly, that means one has to resort to something even more exotic such as Mössbauer spectroscopy and ESR techniques. If you get lucky, a crystal structure may also help. Because the only chances we have to somewhat determine what we are dealing with are answers to the questions:

what is the $\ce{NO}$ bond length?
what is the $\ce{M-N-O}$ angle?
what is the spectroscopic oxidation state of the central metal (not the formal one)?
what is the count of unpaired spins?

One of the most obvious data points is the bond angle. A nitrosyl cation will typically only coordinate in a practically linear matter with an angle of $\approx 180^\circ$ while the nitrosyl anion will typically coordinate in a ‘bent’ manner with an angle of $\approx 120^\circ$. However, we also have the nitrosyl radical which has been shown to adopt both bond angles so we just lost our key piece of evidence.
Even using ESR to determine the spin state (i.e. number of unpaired electrons) can be inconclusive. Taking the related example of the binding of oxygen to haemoglobin, the current consensus accepts a structure that does not immediately conform to the observed spin of $S = 0$. The answer here is antiferromagnetic coupling that keeps to spins antialigned even though they could also not couple and observe Hund’s rule.
Thus, without a plethora of experiments you cannot determine in which state non-innocent ligands such as nitrosyl, dioxygen (which can be $\ce{O2, O2^.-}$ or $\ce{O2^2-}$ etc. bond to the metal. You may want to look at the related question on the electronic structure of $\ce{[Fe(H2O)5(NO)]^2+}$ for which I went through the trouble of actually looking up literature.
Because the oxidation state of iron and nitrosyl are so unclear, a curly bracket notation has been suggested. In this case, we would be dealing with $\ce{\{FeNO\}^6}$ (if I did the maths correctly).
Thankfully, you can always name these compounds by IUPAC nomenclature. Because IUPAC nomenclature desires to allow naming even when the actual electronic structure is unknown, it has defined nitrosyl to always be a neutral three-electron donor for nomenclature purposes. Thus, we would be able to call the complex you are interested in disodium pentacyanidonitrosyliron(III).


text:
Is the boundary of an atom well defined?
That question might seem weird, but that's what I have been imagining lately; going through Bohr's theory on the hydrogen atom. He clearly defines the radius of an orbit with an exact mathematical expression. But what do they mean when they say any atom (say $\ce{Na}$) has a radius $r$. Are we to consider atoms as solid, rigid spheres? If that's the case, what about water?
If all the atoms are defined as solid, rigid, spherical balls, how come an aggregation of so many solid objects be a liquid? How can it flow? So, in a nutshell what is the boundary of an atom made up of, or even this: does it have a defined boundary?
This troubles me a lot. I think maybe the modern atomic model has dealt with this problem. I haven't read it in detail, and I would be highly thankful to any references.

You ask too many questions at once. Stop using Bohr's theory. Atoms are neither solid nor rigid, and don't have any boundary at all. Then again, a collection of solid spheres may pretty well act as a liquid. Ever heard about quicksand?

Any modern chemistry textbook (=less than fifty years old) only uses Bohr as an introduction to orbital theory, except perhaps for the very first year of chemistry in school (8th grade or so, fourteen-year-olds). Exceptions prove to be frauds. ;-)

@Karl I'm in high school, and still we're being taught the Bohr theory, even worse we are made to solve a huge number of problems with the Bohr theory..and when I ask my teachers about quantum mech, they ward me off by saying I'm too young to know! I can't seem to get the hang of the schrodinger wave equation as it's beyond the level of calc I know..

@YourAverageEuler Nobody would bother school kids with wave functions, it also works without. Get yourself next years schoolbook if you are really eager. And  you can do a lot of instructive things with Bohr, go along with it! In first and second year school chemistry, you should learn things that prepare you for more, not the final "truth". Also Schrödingers model only works quantitativley for single hydrogen atoms. Still it indispensable to be very familliar with it to understand anything further on.

@Karl Correction: Hydrogen-like atoms. $\ce{He+}$ and $\ce{Li^2+}$ also work ;)

https://chemistry.stackexchange.com/questions/59096/atomic-radius-definition-disproved-bohrs-model-to-schroedingers-wave-equatio

No, the boundary of an atom is not well defined. According Quantum mechanics the electron density around an atoms decreases exponentially, thus getting asymptotically close to zero.
Considering atoms as rigid spheres is only an approximation. The criteria used to decide on the actual value of the atomic radius is arbitrary. You can for example choose the expectation value $\langle\hat r\rangle$ of the electronic wavefunction. For the Hydrogren atom, this results in the bohr radius $a_0$. But you might as well choose the maximum of the electron density ($\frac{3}{2}a_0$ for $\ce{H}$), or something based on experiments.
Now the aggregate state is something completely different and not directly  related to atomic radius and whether they are rigid sphere or not. Solid means the molecules/atoms of a substance are fixed in their positions. They can vibrate, but do not change their average position with respect to each other. In gases molecules may move freely. And a liquid is something in between, you have some order if you look at a small group of molecules, but not on a larger scale as for solids.

As was already mentioned, the boundary of an atom is everything but well-defined. Quantum mechanics tell us this very well: the probability of finding an electron at the distance $r$ from the nucleus is proportional to $\mathrm e^{-r}$, meaning that it will never hit zero.[1] The lack of a zero-point means that there is no cutoff after which we can clearly say the atom is over. Instead, when performing quantum calculations an arbitrary cutoff value is chosen. The boundary within which the probability of finding the electron is larger than the cutoff value is then plotted as the ‘extent of the orbital’.
However, even if atoms were rigid spheres that does not mean your macroscopic postulate (that rigid spheres cannot form a liquid) holds true because you are misestimating the difference between microscopic and macroscopic properties. I present you the following example:
We will both agree that sand is made up of rigid solid particles even if they are not perfect spheres. Yet, sand shows a few properties that almost make it look liquid if you look from far away. For example, it can ‘flow down’ a slope much in the same way as water could. And also if you try to pile it into a pyramid, it will flow outwards to a certain extent as water (or better: mercury) drops would. You might extend this analogy to marbles which are clearly spherical (usually). You cannot stack them on a blank surface as they will roll away much like you cannot stack water molecules because they will flow to the side.
The main difference between my macroscopic examples and actual molecules is the type and extent of attractive forces between them. While the rough texture of sand can be said to give it some kind of a surface tension (like water and mercury), the attractive forces between the particles are still low and the main force acting on them is gravity (of the Earth, not of other grains of sand). The same thing is true for the marbles example which even lack surface tension and are only attracted to the Earth by gravity. Actual molecules have much stronger intermolecular forces that dominate over the external gravity force source in the microscopic world. That contributes to the final edge of a difference between many grains of sand/many marbles and a sample of water.
All I have discussed above is, however, completely independent of the question whether the spheres that make up atoms are rigid and solid or not.

Seriously? Bohr's Atomic Model? It was known by 1920 that that model failed (badly) to explain observational data. That is, it hasn't been taken seriously for nearly 100 years. Perhaps you should invoke the 4 humors and reason from there. (The only atom the the theory did a pretty good job for is hydrogen. One down, 91 elements to go....)  There are lots of different models which are useful for different purposes. One thing we know about all of them, is they're all wrong. So, since the "right" model isn't known, picking a "best" model requires you to know what your purpose is, what the context is. The issue is deciding which model is most useful, not which model is correct. That turns out to depend on context. What I was taught as an undergraduate chemistry major (almost 50 years ago) was that an electron bound in an atom (or molecule, or ion, or radical) is best thought of as a probability wave. This means that the location of the electron is "smeared out" in space and that for any distance, r, from the nucleus (center), there is a non-zero probability of the electron being at that distance. This means, to answer your question, that atoms do not have "hard surfaces". The reason why some Physical Chemistry textbooks still cover the Bohr Model is because it helps the student transition between the physics of everyday objects to the non-intuitive world of quantum mechanics.  The best book for you depends on what your mathematical knowledge is, how much effort you want to spend, and how detailed you want to be. It is a mature science, and many popular books have been written about it. Concepts such as "radius" are mostly nonsensical at the quantum scale. For instance, the radius of an atom depends on how you measure it. Consider the simple case of shooting electrons at it. Then shooting protons at it. Then shooting neutrons at it. Then microwaves. Then (visible) light waves. Then x-rays. Each will give you a different radius. In some ways it is similar to the question of how long is a coastline. If you measure it with a 100 meter cable, you'll get one answer (ignoring differences in sea level, tides, wind, rain, etc.), if you measure it with a meter stick, you'll get a larger value. Measure it centimeter by centimeter, and it will be even larger, and just imagine if you could measure it atom-by-atom, it would be very difficult to determine where the liquid began and where the land ended. Coastline is a macroscopic concept, with little meaning at the micrometer scale (let alone the nanometer scale). Very much like your concept of radius. (Although to be fair, we "steal" the word and use it in various contexts which often have similarities to its macroscopic use.)

Wall of text which is terrible to read. Making fun of OP for the first 4 lines, despite the Bohr model beeing a pretty good start into chemistry, then a lot of things that doesn't answer the question at all (why does the measurement using electrons yield a different result than with protons? Just saying it does doesn't explain anything) and the comparison with the coastline doesn't work well. Yes you get different results depending on the length of your measuring stick, but the problem with atoms is a "soft boundary"

Chill out @alphonse, I'm only in high school, just a little curious kid you know? I am very much aware of the drawbacks of the Bohr model. I was just curious to know what was the banter on radius all about. How to electrons "enter" an atom? Which made me think, hold on! Just now I read about the atoms having radii (which logically means they have a boundary) and now you say, electrons can enter it? Is the' wall' of an atom electron permeable? I know thats too much but still, curiosity never dies


text:
Ionic interaction in crystal field theory
How correct is the assumption of ionic interaction between the ligands and the central metal atom in cft when many ligands do not even possess charge over them?
Besides, I also referred to the conditions required for formation of ionic bond. They are mostly formed when electronegativity difference between the two atoms is around 2.0 on the Pauling scale. The EN difference between ligands and metals is less than that mentioned above in most of the cases. So on what assumptions does the idea of ionic interaction hold?

Neutral ligands are considered as dipoles and the negative side of the dipole repel the electrons of the metal causing the d orbital splitting

Crystal field theory does not really assume ionic interactions between the ligand and the central metal. It assumes that the ligands can be replaced in a thought experiment by point-shaped negatively charged particles that approach the central metal from certain directions causing the orbital energies to change.
The idea of seeing them as negatively charged has its merits. All ligands (formally) ligate through the action of a lone pair, whether this lone pair contributes to a charge, a formal charge or not. Note that positively charged ligands can occur; the most famous being $\ce{NO+}$. The electron pair that ligates will, however, always carry some local negative charge just because electrons are negatively charged particles.
This should not really be part of the question but I’ll mention and answer it anyway. For one, you seem to be assuming that ionic and covalent bonds are two different things. That is not true. All bonds between different elements can be understood to have a partially covalent and a partially ionic character; the main difference is which character is stronger. And the $2.0$ difference on the Pauling scale is part of the Pirate Code: more of a rough guideline than an actual rule. You can find counterexamples in both directions: predominantly ionic bonds with less and predominantly covalent bonds with more electronegativity difference. However, as I mentioned this is not relevant to crystal field theory because you should not be assuming an ionic bond between the ligand and the metal.


text:
Does adding prefixes to SI units make them no longer SI units?
Our chemistry professor said that "mg for mass, g/m3 for density, ms for time are all SI units".
I want to make sure now, even if we added a prefix to the unit, will it still be an SI unit?

Have a look here.

Yes, when prefixes are used with coherent SI units, the resulting units are still SI units; however, they are no longer coherent.

The metre per second, symbol m/s, is the coherent SI unit of speed. The kilometre per second, km/s, the centimetre per second, cm/s, and the millimetre per second, mm/s, are also SI units, but they are not coherent SI units.

BIPM: The International System of Units (SI)

Worth mentioning the exception to the rule: kg is the base unit of mass, not grams.  See also https://www.unc.edu/~rowlett/units/cgsmks.html: the CGS system (centimeter, gram, second) used to be the base units (with force in dynes for example), but now almost everyone uses MKS (with force in Newtons).

The main advantage of Metric/SI is that everything makes sense and is so easy to work with. Which is why it infuriates me that the base unit of mass is kg. WHY? Why would you ruin an otherwise perfect system? Like, in a system all about logical unit choices, you go and do that. Sigh.


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Affinity to mobile or stationary phase during column chromatography
I have carried a column chromatography and obtained two compounds. The first one is liquid (ester) and the second one is solid (nitro) yet I know that the second compound is less polar than the first one. Why does the result differ from the knowledge about polarity (the more polar the molecule is, the slower the compound)?

I am very strongly inclined to believe that you incorrectly predicted the polarities. Just from the comparison ester versus nitro, I want to assume that the nitrocompound is more polar. Please give us structures or we cannot have the slightest of clues how to answer. Voting to close as unclear.

@Jan I had o-dinitrobenzene and ethyl benzoate

I don’t know in which universe you would consider a dinitrobenzene less polar than an ethyl ester but this universe is not the correct one.
As your experimental data (the $R_\mathrm f$ values of the two eluted compounds) clearly show, the dinitrobenzene is more polar than ethyl benzoate.


text:
Concentration of a solution of NaOH
I'm about to try to make my own soap for the first time. In order to create my recipe, I'm using this calculator.
As I'm using a lye solution, I have to fill the box %w/w of the solution. The calculator says:

If you are using a ready made lye solution, check the box above and
  enter the enter the strength of your solution. The solution strength
  needs to be measured as a percentage of lye weight to water weight (%
  w/w).

I'm a bit confused here, I would have thought the w/w ratio was the ratio of the mass of NaOH in the solution to the total mass of the solution (and not the water mass used to disolve NaOH). So my first question is: what is really that w/w ratio?
My second question is how to interpret what I read on the lye solution (1L) I've bought. It's written:

Sodium hydroxyde - N°CAS: 1310-73-2: 30%

What does this 30% refer to? Is this a ratio of volumes? Of mass? Is this the w/w ratio I want to know?
Thanks for helping me clarify this!

CAS: 1310-73-2 is a chemical ID number for sodium hydroxide. The 30% would be 30% sodium hydroxide by weight of solution. So 100 grams of solution would contain 30 grams of sodium hydroxide. // No idea what the doofus lye calculator is doing.

Whatever that lye calculator is doing it is either wrong by design or internally correct and using a very obscure definition that no other chemist would understand.
Percentage solutions are always (barring calculators programmed by laymen) defined as weight of the solute divided by weight of the solution times 100. So a $\pu{30\%}$ solution of $\ce{NaOH}$ means $\pu{30g}\ \ce{NaOH}/\pu{100g}\ \text{solution}$.


text:
Clarification on the $\mathrm{S_N2}$ mechanism
The question is about the specifics of the $\mathrm{S_N2}$ mechanism (nucleophilic bi-molecular substitution).

(Source: Wikipedia)
As we know that in this reaction the nucelophile ($\ce{Nu-}$) approaches the molecule. The main step (According to the molecular orbital theory) of the reaction is that the the nucleophile bonds with its lone pair in the anti-bonding molecular orbital (ABMO) of the central carbon atom. As this happens the leaving group ($\ce{Lg}$) is kicked off by gradual weakening of the $\ce{C-Lg}$ bond. Now the thing that bothers me here is:
Why does carbon prefer to add electrons in its ABMO? As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order whereas, while electrons are pushed into the ABMO it clearly decreases the bond order, which directly indicates the instability of bond formation.
The main question being, what makes the reaction favourable? What is the 'driving force'? Why does it take place when there is a decrease in bond order?

sigh You know the mechanism and don't know why reactions happen? You put enough energy and bonds break.

I reckon knowing a reaction mechanism and understanding it step -by-step are quite different things  @Mithoron

@YourAverageEuler Not really. Unless you understood the steps you cannot say you know the mechanism.

"what makes the reaction favourable? What is the 'driving force' Why does it take place when there is a decrease in bond order?" It's almost always like this with reactions. You input activation energy to get unstable transition state which leads to products

Why does carbon prefer to add electrons in its ABMO?

Well, preferring is a bit of a strong word. Orbital interactions that are productive towards forming and breaking bonds can only happen between a filled and an empty orbital. Therefore, you should always identify a HOMO (highest occupied molecular orbital) and a LUMO (lowest unoccupied molecular orbital), compare their energies and thereby decide whether they are able to react.
Here, the lowest unoccupied orbital of the carbon compound is typically the $\ce{C-Lg}$ antibonding $\unicode{x3c3}^*$ orbital — because that is the antibonding orbital belonging to the leaving group (I am aware that this is circular reasoning). Its energy is also somewhat favourable with respect to the energy of the filled lone pair orbital of the nucleophile meaning the two can react well.

As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order …

No. Why?

electrons are pushed into the ABMO it clearly decreases the bond order …

To specify: decreases the bond order *of the $\ce{C-Lg}$ bond.

which directly indicates the instability of bond formation.

No, it doesn’t. It only indicates that the $\ce{C-Lg}$ bond is being broken and the leaving group is getting ready to leave.

Why does it take place when there is a decrease in bond order?

You fail to see that while the bond order of the $\ce{C-Lg}$ bond is steadily decreasing the bond order of the $\ce{C-Nu}$ bond is steadily increasing. At any point in time on the reaction coordinate, the sum of both bond orders is $1$.

The main question being, what makes the reaction favourable? What is the 'driving force'?

Two things: the formation of a more stable $\ce{C-X}$ bond (meaning that $\ce{C-Nu}$ is a stronger bond than $\ce{C-Lg}$) and the liberation of a better stabilised anionic species $\ce{Lg-}$ as opposed to $\ce{Nu-}$.
Actually, in many $\mathrm{S_N2}$ reactions only one of those two factors is the driving force. And while we tend to write $\ce{Nu-}$ and $\ce{Lg-}$, we should always remember that these compounds need not be anions. For example, $\ce{H2O}$ is a very good leaving group (after alcohols have been protonated) but not an anion. The same is true for $\ce{N2}$ (from a diazonium salt). And equally, water or ammonia can be nucleophiles although they are neutral compounds.
So it can well be that the reaction is only driven by the liberation of a neutral compound and thereby the removal of charges (as would be the case in the $\mathrm{S_N2}$ reaction of iodide and a protonated alcohol when $\ce{HI}$ is used to generate iodoalkanes). It might also be that the reaction is driven only by the stronger $\ce{C-Nu}$ bond — or by the breaking of the weaker $\ce{C-Lg}$ bond as would be the case in nucleophilic substitutions with tosylate ($\ce{Me-C6H4-SO3-}$) or mesylate ($\ce{MeSO3-}$) as leaving groups.
There is not one same truth for all $\mathrm{S_N2}$ reactions.