CHEMISTRY COMMUNITY

David Chibukhchian 2G

I agree with the other answers and would add that the "n" value is found after you have balanced the electrons, so if the half-reactions have a differing number of electrons that are being transferred, you have to multiply one or both of them by some factor to make them equal. Once you do ...

David Chibukhchian 2G

One thing I also made sure to note was which half-life expressions correspond to which orders. It can get a bit confusing especially since the equations on the constant sheet don't specify this. Because of this, I memorized the half-life general format for first, second, and zero orders (like t1/2 =...

David Chibukhchian 2G

I was also wondering the same thing about these problems where we have one reactant and two products. However, in general, I believe we have to set it up as you stated where that one reactant is both the oxidizing agent as well as the reducing agent. Doing this allows us to see the oxidation reactio...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that 2nd order reactions have a half-life expression that is t1/2 = 1/k[A]0. When you have the half-life of a 2nd order reaction and its initial concentration, you can use this expression to find the rate constant. This concept was one tha...

David Chibukhchian 2G

I think that in a concentration cell, diluting the concentration of the anode or increasing the concentration of the cathode will in fact increase the potential (or E) due to the increased charge difference. The differing concentration values create a disproportionate amount of positive charge on th...

David Chibukhchian 2G

I would also add that we can use a graph to verify what we get using the initial rate method for the other orders as well! If we found that the order is 2, for example, we would graph a plot of time versus 1/[A] to make sure we get a straight line (with a slope of k, the rate constant). It's helpful...

David Chibukhchian 2G

As the other answers state, that expression helps us get the half-life expression for first, second, and zero order reactions. I would just add that it's helpful to know its function (removing natural logarithm) mostly because on the outline for the kinetics unit it says we should have a sense of ho...

David Chibukhchian 2G

I definitely agree with the other answers, I was looking at the equations sheet and noticed that the integrated rate laws and half-life expressions are all there; however, they are not labeled. It would be a really good idea to memorize which equations correspond to which orders (first, second, and ...

David Chibukhchian 2G

I definitely agree with the other answer and would just add that this is the case for Ecell as well, naught just E naught. Sometimes it's a bit confusing to remember the different ways in which E and E naught can change, so it's good to remember that coefficients will never impact their values. That...

David Chibukhchian 2G

If I'm not mistaken, it's due to the fact that there is more positive charge on the cathode side because of the higher concentration of cations. This makes it so electrons flow from the anode side to the cathode side and thus the anode gets skinnier while the cathode gets fatter. The concentration o...

David Chibukhchian 2G

I think that in general, the concentration in the anode side will be lower than that of the right side. I'm pretty sure the electrode on the left always becomes skinnier/smaller while the one on the right becomes larger. This means that the concentration on the left will go up while the one on the r...

David Chibukhchian 2G

Okay, that definitely helps! I did not realize that we should use the Nernst equation to understand this concept but now I can see the effect mathematically of varying concentrations. I also now see why E naught would not change, since it's under standard conditions. Thank you so much!

David Chibukhchian 2G

I definitely agree with the other answers and would just add that average rate, in general, is not as accurate as instantaneous rate of change as Dr. Lavelle mentioned in lecture. This is because instantaneous rate involves the use of derivatives and tangent lines, which better estimate the curve wh...

David Chibukhchian 2G

Hey guys, I had a question about one of the bullet points on outline 5. It says "Predict the effect of changes in concentration of reactants and products on the cell potential." What exactly is this referring to? Does it have to do with the relationship between E and E naught? I believe th...

David Chibukhchian 2G

I think that, in general, the way we construct cell diagrams is pretty flexible. I was really confused by this example as well, but I think that it's not the only way to express the cell diagram. Otherwise, I'm also unsure about the reasoning behind this decision. However, I definitely agree with th...

David Chibukhchian 2G

Hey guys, I had a question about one of the bullet points in outline number 5. It reads "predict whether a metal will dissolve in solution". How exactly would we go about doing this? Does it have to do with the solutions of the galvanic cell? If someone could explain what we need to know a...

David Chibukhchian 2G

If I'm not mistaken, that's basically the main function of the ion-selective electrode's thickness. Its structure decides what kind of ion is going to pass through its surface, whether it be (like you said) H+, Na+, K+. I'm assuming it has to do with ion size, but I'm not 100% sure. However, I think...

David Chibukhchian 2G

I agree with the other answer and would add that it's a good idea to remember that substances on one side are separated by a vertical line if they are in the same phase. However, if you have a half-reaction that remains in the same phase (such as Fe3+ (aq) to Fe2+ (aq)) you separate it with a comma ...

David Chibukhchian 2G

I understand the equation E = E* - (RT/Fn)lnQ when it comes to the relation between concentration and voltage; however, I am a bit confused as to what the term E* specifically means. How does it differ from the E at the left-hand side of the equation? If someone could clear that up for me that'd be ...

David Chibukhchian 2G

I was also confused about how I should go about solving this problem. However, I believe that the only way to do it is to open the table that Sapling provides and see the standard reduction potentials, observing if any of the answer choices are located inbetween the given elements. This is the metho...

David Chibukhchian 2G

If I'm not mistaken, the relationship between cell potential and delta G is delta G = -nFEcell. Thus, is delta G is negative, then the cell potential will be positive. This makes sense if we are considering that a spontaneous process occurs when delta G is negative and when the Ecell value is positi...

David Chibukhchian 2G

I definitely agree with the other answers and would add that it's important to use an R value of 0.0083145 instead of 8.3145. This is because the value of enthalpy, delta H naught, is given in kilojoules. Thus, we have to take this into account and use the appropriate R value; it was something that ...

David Chibukhchian 2G

I was also pretty confused with this section because I was unsure how it connected to the rest of the material we were learning at the time. However, I think the main point is that we can use these ideas to calculate the molar heat capacity of various molecules or particles based on their size, whic...

David Chibukhchian 2G

Yes, that is definitely the case. In addition, you can express the first law of thermodynamics delta U = q+w when you have constant pressure, not volume, by seeing that delta U = delta H - Pex*delta V. You can replace the second part of this expression with nRT because of the ideal gas law, and this...

David Chibukhchian 2G

That's basically what reversible expansion means. The external pressure is slowly decreased in small steps in order to keep it close to the internal pressure value. As a result, this leads to the maximum amount of work possible getting done. To calculate the work done during a reversible expansion, ...

David Chibukhchian 2G

Delta H naught can be thought of (for example) the delta H of a reaction when a substance is made from its elements in its standard states. For example, if we have a reaction like N2 + 3H2 -> 2NH3, then the delta H can include the naught because the substance NH3 is made out of diatomic hydrogen and...

David Chibukhchian 2G

I think that, in general, we can compare the number of gas moles on each side of a reaction to tell whether a system does work. If there are more on the right, then work was done by the system (expansion); meanwhile, if there are more on the left, we can safely assume that work was done on the syste...

David Chibukhchian 2G

Delta S total is the sum of delta S of the surroundings and delta S of the system; it must be positive in order to have spontaneity. Meanwhile, delta S naught is the entropy of a reaction where it's in terms of standard elements. To find it, you can use an appendix of molar standard entropies, sum t...

David Chibukhchian 2G

Residual entropy is basically the entropy that remains in an object when the thermal entropy (due to temperature) is eliminated. This is done by moving the temperature to 0 K. Because of this, residual entropy is due solely to the number of microstates (W) and is a result of location and orientation...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that the enthalpy value may not be given, in which case we might have to find it ourselves (for example, by using a list of standard enthalpies of formation) and then apply it to the Van't Hoff Equation. Also, be careful to convert that va...

David Chibukhchian 2G

I was also wondering this about that listed bullet point; however, I think it's just referring to the fact that a question may make use of a biological concept and we have to apply what we've learned to there. For example, ATP hydrolysis may be discussed when referring to an exothermic reaction and ...

David Chibukhchian 2G

I agree with the other answer and would just add that if the question doesn't specify whether the work of expansion is done reversibly or irreversibly, look out for the mention of the word "equilibrium". This is because reversible expansion occurs in equilibrium, where the external pressur...

David Chibukhchian 2G

Degeneracy is represented by W, and it's the number of different microstates a particle can have while maintaining the same energy. Meanwhile, entropy is the measure of disorder, represented by the letter S. We use the degeneracy to calculate the entropy through the Boltzmann equation, S = kB*ln(W)....

David Chibukhchian 2G

I was also wondering the same thing, because I understand the meaning of the graph yet I'm not completely sure about the reasoning behind it. I think the main point was to demonstrate that a system tends to move in the direction of increasing entropy. Thus, when a system is at maximum entropy, it is...

David Chibukhchian 2G

If I'm not mistaken, entropy in an isolated system cannot decrease according to the Second Law of Thermodynamics. In other words, delta S has to be greater than or equal to 0 if we are dealing with an isolated system. Topic 4I in the textbook discusses this in more detail, and I'm sure Dr. Lavelle w...

David Chibukhchian 2G

An isothermal reversible expansion can be up to any new volume, it doesn't have to be twice its volume. I think what you're referring to is an example of how much the volume can expand, but it doesn't have to be exactly twice by any means. This goes for any type of expansion; we use V2 and V1 to ind...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that the textbook offers a table in Topic 4C that shows how the molar heat capacities at constant volume and pressure vary based on atoms, linear molecules, and nonlinear molecules. It's good to note that Cp is going to be larger than Cv b...

David Chibukhchian 2G

I completely agree with the other question and would just add that the conversion from L atm to Joules is just for the constant pressure equation for work, w=-Pex*delta V. The conversion factor is 1 L atm = 101.325 J. This initially confused me because I mistakenly thought you had to make this conve...

David Chibukhchian 2G

I'm a bit unsure if you would have to use two separate equations for an irreversible reaction. However, you're correct in that an irreversible reaction expands quickly and this does not allow heat to enter the system. In this case -w does not equal q, so energy lost as work does not get replaced as ...

David Chibukhchian 2G

Basically, an irreversible reaction occurs much faster than a reversible reaction. As a result, heat is not able to slowly enter the system in steps and counteract the energy lost as work. This is the reason why reversible reactions are "isothermal" and have a constant temperature, whereas...

David Chibukhchian 2G

I definitely agree with the other answer and would just add that it's helpful to understand work in the context of volume change, since this is mainly how we think of it. When a system is compressed, work is done on the system (so its sign is positive). Meanwhile, in an expansion, it does the work (...

David Chibukhchian 2G

I definitely agree with the other answer and would just add that reversible pathways do more work than irreversible pathways because reversible ones have a greater external pressure. If we look at the work equation, we see that a higher pressure will correspond to a greater amount of work done. Thus...

David Chibukhchian 2G

We think about work in terms of volume change. Thus, heating or cooling a system does in fact lead to an energy change, but it's energy in the form of heat (it's important to remember that internal energy is made up of work, w, and heat, q). This helps us understand that whenever the volume stays th...

David Chibukhchian 2G

Another good way to remember is by thinking about compression/expansion. We talk about work in the context of volume change. When a system is compressed, its internal energy increases because work is done on the system. Meanwhile, whenever a system expands, work is done by the system (to increase th...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that a reversible process takes much more time because (for example) the external and internal pressure must be kept virtually equal during the process, so this leads to the need to alter the external pressure in very slow steps. As a resu...

David Chibukhchian 2G

Correct me if I'm wrong, but I believe the way you can deal with units when doing work calculations is by solving the problem using w=-Pex*delta V and ending up with L atm, then using the conversion factor for L atm to J to get your answer in joules (or kilojoules). The conversion is that 1 L atm is...

David Chibukhchian 2G

I think another important concept to remember is that bomb calorimeters have a bit more of a complex structure to maintain the constant volume. In addition, it may be good to keep in mind that it's called a "bomb" calorimeter because a very exothermic reaction may cause an explosion in the...

David Chibukhchian 2G

I definitely agree with the other answer and would add that standard reaction enthalpies are basically giving the reaction enthalpy when some substance is made from its elements in their standard form (like graphite for carbon, diatomic form for diatomic molecules, etc). A better way to understand i...

David Chibukhchian 2G

If I'm not mistaken, the standard enthalpy of formation is basically the standard reaction enthalpy when exactly one mole of a substance is being produced by its elements in their standard states. The important part is to make sure that the delta H for the reaction is divided by the stoichiometric c...

David Chibukhchian 2G

I definitely agree with the other answer and would just add that this process for finding the change in enthalpy is called Hess's Law, and the reason we can do it is that enthalpy is a state function. This basically means that enthalpy is a property where we only really look at the initial and final...

David Chibukhchian 2G

We can basically say that delta H equals q whenever pressure is constant. In this case, we write qp = delta H. This is because enthalpy is energy associated with a reaction (either gained or lost) when pressure is constant. That last part is what makes enthalpy a state property. When this is the cas...

David Chibukhchian 2G

Hey guys, I had a question about the following problem from Matthew Tran's Week 3 Worksheet. I attached a picture of the question below, and I was wondering if someone could give an overall guide on how to solve problems like these. The solution guide says that the answer is 4.69, but I'm honestly u...

David Chibukhchian 2G

I was also wondering if this was the basic difference between the two but I think you're right. When the calorimeter has constant pressure, it's almost like a coffee cup that is open to the atmosphere and thus volume can change (and the pressure won't vary). However, a "bomb" calorimeter h...

David Chibukhchian 2G

Okay, that definitely helps. I think it's really helpful to memorize which anions behave as bases and which cations behave as acids so it's easier to see that when solving these kinds of problems, I just wanted to make sure that this is what the outline bullet point was actually referring to. I also...

David Chibukhchian 2G

Hey guys, there's a bullet point on the outline that says "calculate the pH of a weak acid and its salt." I was just wondering what exactly this means and what kind of problems we would deal with for this topic. Would it be a reaction between a weak acid like HClO2 and a salt that contains...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that the textbook often writes just "K" instead of "Kp", but usually we will be given specific information about whether we are dealing with concentrations (and should thus use Kc) or partial pressures of gas (Kp). Corr...

David Chibukhchian 2G

Hey guys, I think I have a good understanding of the difference between a change in enthalpy vs a change in temperature. I get that an enthalpy change usually takes into account that pressure is constant. However, I want to get a better understanding of the difference. Could someone offer a quick ov...

David Chibukhchian 2G

I definitely agree with the other answers. I initially was wondering why there's this difference of steepness of the graph at various points, but I believe that the heat capacity is very significant to understanding the reason. I remember reading in the textbook that (for example) if a substance in ...

David Chibukhchian 2G

I agree with the other answer that there might be some kind of question where the total pressure may be given while the partial pressures must be found, combined with some other equilibrium calculations. However, I don't think that it's particularly too important to go in-depth about this topic. For...

David Chibukhchian 2G

You're definitely right when you say they both pertain to Kw but are different ways of thinking about or writing it. A good way to understand why is remembering what exactly Kw, or the autoprotolysis constant of water, actually is. It's the product of the hydronium and hydroxide ion concentration of...

David Chibukhchian 2G

I think that the first deprotonation is significant whereas the other deprotonations are in fact insignificant. The reason for this is (correct me if I'm wrong) because the first deprotonation increases the [H3O+] concentration so much more than the second or third deprotonations that we basically d...

David Chibukhchian 2G

Another helpful thing to remember is that the reverse of an endothermic reaction is exothermic and the reverse of an exothermic reaction is endothermic. This makes more sense later on when we learn that the delta H value of a reverse reaction is just the opposite sign of the forward reaction's delta...

David Chibukhchian 2G

I definitely agree with the other answer and would add that it's good to think about what exactly would happen if we did include liquid water (or any solvent) in the equilibrium expression. Because its concentration basically does not change, you would end up with a very similar value in both the nu...

David Chibukhchian 2G

I was also wondering along the same lines, because in the textbook I read about rotational and translational energy and am having a bit of trouble understanding what they really mean. At the same time, I want a better understanding of how exactly they differ from each other. I have a good idea of th...

David Chibukhchian 2G

I was also wondering the same thing, but I'm fairly certain that we don't have to worry too much about net ionic equations (for now at least). In general, though, I do believe you're correct that we should only dissociate the aqueous solutions and leave the solids/liquids alone. We would probably be...

David Chibukhchian 2G

Hey guys, I was wondering how we would use equilibrium concentrations to predict the solubility of a substance. The reason I ask is that this is one of the bullet points on the outline for the chemical equilibrium unit, and I'm unsure of how solubility relates to the value for K. Do we look at how m...

David Chibukhchian 2G

Another example that makes it easier to understand why we ignore water in the equilibrium expression is by thinking about a bank account filled with a million dollars. If you were to withdraw ten dollars out of that account, you would still basically have a million dollars. Thus, the change in water...

David Chibukhchian 2G

I definitely agree with the other answer and would add that it's important to keep in mind rules regarding the K value. For instance, multiplying a reaction by some factor (so that its stoichiometric coefficients are multiplied) would make the equilibrium constant be raised to the power of that numb...

David Chibukhchian 2G

I agree with the other answer and would just add that it's really helpful to understand what exactly Le Chatelier's Principle is saying. It basically states that a reaction wants to minimize the effect of any change. Keeping this in mind, we can better understand questions that describe a certain ch...

David Chibukhchian 2G

Yes, that's the equation you should use. I think it's important to remember that the Kc and Kp expressions are basically very similar, we just have to know when to distinguish between the two based on whether the given compositions are concentrations or pressures. In addition, it's helpful to know t...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that an easy way to distinguish between the two is by looking for a hydrogen atom. Amphiprotic compounds must have an H+ to donate, so if it doesn't then this automatically means it cannot be amphiprotic. However, it can still be amphoteri...

David Chibukhchian 2G

Hey guys, I had a question for one of the bullet points on the outline for this unit. It states that we have to know about "Biological examples. For example: ATP hydrolysis, osmotic pressure." Would anyone be able to give an overview regarding this topic? I don't remember Dr. Lavelle expla...

David Chibukhchian 2G

Hey guys, I was wondering if anyone could help explain why a change in pressure doesn't affect the equilibrium constant. The reason I ask this is because a compression can shift a reaction to favor the products (for example) and this would mean that there would be a higher ratio of products compared...

David Chibukhchian 2G

Yeah, so basically water isn't included in an equilibrium constant when it is in its liquid state. This is because it's in a phase where a concentration change wouldn't lead to much of a difference before and after the reaction. An example Dr. Lavelle used was imagining if you take out 10 dollars fr...

David Chibukhchian 2G

Okay, that definitely makes sense. I was a bit confused because I knew that the amount of reactants in general would change, but at the same time the actual value of K doesn't take solids into account. However, it makes sense that having more reactants would ultimately lead to increased products, ev...

David Chibukhchian 2G

Hey guys, I was wondering whether or not increasing the amount of solid reactants in a reaction will lead to an increase in the products. The reason I ask is because solids are not included in the equilibrium expression, so increasing them won't make Q less than K (for it to then counteract by produ...

David Chibukhchian 2G

I think it's just really important to always remember that adding a reactant or product at any time will ultimately not change the equilibrium constant. The reaction will always work to return to that same ratio even if you start out with or add additional reactants, so in this case I believe that K...

David Chibukhchian 2G

I definitely agree with the other answers and would just add that Lewis Structures are particularly useful for understanding K values. Whenever we have a very low equilibrium constant, this means that the reactants don't really prefer to become products and thus they are more stable in the initial f...

David Chibukhchian 2G

I agree with the other answer and would just add that this calculation is necessary if (for example) we are given equilibrium concentrations of a reaction but are asked to find the Kp value, which uses partial pressures. The ideal gas law thus allows us to convert between concentrations and partial ...

David Chibukhchian 2G

Oh okay, that definitely makes sense. I now understand that solids aren't dissolved so we can't really think about them in terms of concentration. That explains what Dr. Lavelle meant when he was explaining how leftover solid in a reaction isn't taken into account (because it remains undissolved). A...

David Chibukhchian 2G

Hey guys, I was watching today's lecture and got a bit confused when Dr. Lavelle was discussing how pure substances like solids and liquids are not supposed to be included in the expression for K. I understand that we can't really think about their concentrations changing, but why is that? I hope my...

David Chibukhchian 2G

Hey guys, I was wondering if someone could give any tips for identifying Lewis Acids and Bases whenever we are given two molecules that react. I understand their definition and can distinguish between them usually, but in some examples it seems a bit more unclear and I have trouble telling which is ...

David Chibukhchian 2G

Hey guys, I was looking over the outline for the Acids & Bases section and noticed that the last bullet point said "Explain how carbon dioxide can lead to respiratory acidosis." Would anyone be able to quickly go over what we should know about this topic? I know Dr. Lavelle discussed t...

David Chibukhchian 2G

I agree with the other answer and would just add that chelates usually have a lone pair-spacer-spacer-lone pair pattern when drawing out its structure, so this can be used to judge whether a polydentate is actually going to act as a chelate or not. I believe that all chelates have to be polydentate,...

David Chibukhchian 2G

Okay, that definitely makes sense. I think I have a better understanding now that Dr. Lavelle explained the example with Na+ when writing equations with acids and bases, I just wanted to make sure if we had to practice them for the final or not. That really helped, thank you all!

David Chibukhchian 2G

I think that molecules with only two or three regions of electron density would in fact be only "planar" or flat. However, I believe that some planar molecules would not actually be flat because they'd have lone pairs that aren't on that 2D plane. An example is square planar or t-shaped mo...

David Chibukhchian 2G

Hey guys, I was wondering whether or not we had to know how to write net ionic equations for reactions with acids and bases. It's a topic that I don't really understand well and so was wondering whether I should review it. If it is something that we should know, does anyone have any tips on how to w...

David Chibukhchian 2G

I agree with the previous answer and would just like to add that seeing a carboxyl group indicates that it's a weak acid, not a strong one, so that's definitely something to keep in mind. However, the COOH carboxyl group is stronger than a molecule with just an -OH group because the increased oxygen...

David Chibukhchian 2G

I would definitely agree with the other answer and add as a guide that most derivatives of ammonia are usually going to be weak bases as well. In addition, strong acids usually have two more oxygen atoms than hydrogen atoms, which accounts for HClO3 and HClO4 being strong acids while HClO and HClO2 ...

David Chibukhchian 2G

Okay, that definitely makes sense. I was assuming that HF doesn't dissociate completely because the bond between H and F is so short and thus strong. That makes it easier to generally understand the difference between strong and weak acids. Also, I didn't know that very concentrated solutions of HF ...

David Chibukhchian 2G

Hey guys, I was looking at a list of strong and weak acids and noticed that HF is considered to be weak. I was a bit confused because I know that HCl and HBr are both strong acids. I was wondering if anyone could explain why this particular acid is weak while other halogens that bind with hydrogen a...

David Chibukhchian 2G

Delocalized pi bonds occur in benzene, for example, when each carbon atom has one unhybridized p-orbital. Because each carbon atom is the same distance apart from each other, the unhybridized orbitals aren't really bonding in a way where we can observe separate double and single bonds in the molecul...

David Chibukhchian 2G

Okay, that definitely makes sense. I was also thinking that memorizing them would just make answering questions a lot easier. I believe Dr. Lavelle will probably specify which ones we have to learn in a future lecture so once we know which are the most important ones we can start to do so. That defi...

David Chibukhchian 2G

I remember Dr. Lavelle saying that hemoglobin is made up of four myoglobin-like molecules, so does that mean they detach from hemoglobin to form myoglobin? I'm not sure whether this is the case. If anyone has an idea that would be really helpful, I'm a bit confused as to how myoglobin forms as compa...

David Chibukhchian 2G

If I'm not mistaken, the ability to form a ring is basically referencing how chelates form and integrate the central TM cation as a part of it. In addition, I remember that Dr. Lavelle said that these ligands can rotate in this manner because they have single bonds. Single, or sigma, bonds can rotat...

David Chibukhchian 2G

I was also wondering about this topic, but I think the other answer definitely addressed it well. I believe that a good way to approach Valence Bond Theory is by just thinking about how we visualize bonds between atoms. For example, covalently bonded atoms have electrons in the shared region between...

David Chibukhchian 2G

Hey guys, I'm currently looking at Dr. Lavelle's sheet going over ligand names in coordination compounds that he provides on his website, and I'm wondering whether we should start to memorize some of the more unfamiliar ones. Is it a good idea to remember which ligands are polydentates? Or is it fin...

David Chibukhchian 2G

Yeah, that makes sense. I remember in the previous lecture Dr. Lavelle said that double bonds can't rotate because the pi bonds aren't located on the internuclear axis, so I get why the polydentate ligand would have to have single bonds. That definitely helped, thank you!

David Chibukhchian 2G

Hey guys, I was watching today's lecture and I think that I got a good understanding of what polydentates are and how they form. However, I was just wondering how exactly we would know whether or not a molecule is polydentate or not. If a question gives us the formula of a molecule, would we have to...

David Chibukhchian 2G

I believe that since cations usually have a lower electronegativity than anions, they don't exactly have extra valence electrons in the ionic bond for the anion to distort. This basically means there wouldn't really be much to polarize on a cation due to its lower electronegativity. In general, thou...

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