CHEMISTRY COMMUNITY

Chris_Butler_1A

That definitely makes more sense, thank you!

Chris_Butler_1A

In one of the Sapling Homework Questions, ClO 4 - is given as a molecule with resonance structures. One part of the question asked for the Oxidation number of the Chlorine and I was a little confused as to why it doesn't seem to follow the trend of Oxidation numbers. From the periodic table, we can ...

Chris_Butler_1A

Sean Wang 1J wrote:Sigma bonds consist of hybrid orbitals, pi bonds consist of unhybridized orbitals. If there was another pi bond it would also be (C 2p, O 2p).

Okay cool, thank you so much!

Chris_Butler_1A

I had a little confusion here as well. I would have thought that due to Fluorine having a higher electronegativity than Iodine that surely it would have a higher melting point. However, it seems that melting and boiling point depends on the bonds and intermolecular forces within the atom. Given Iodi...

Chris_Butler_1A

In H 2 CO, there exists a double bond between the Carbon and the Oxygen. I understand that the hybridization of both C and O is sp 2 due to their 3 areas of electron density, but I had a little confusion about how the hybridization of the double bond works. The sigma bond is (C 2sp 2 , O 2sp 2 ) whi...

Chris_Butler_1A

This was definitely a question I had as well following one of the review sessions. It seems that due to Oxygen's higher electronegativity than Chlorine, the Hydrogen is more likely to attach to the Oxygen. In conjunction with the answers above, this arrangement also helps to keep the formal charges ...

Chris_Butler_1A

Pi-bonds have 2 areas of contact (the 2 p orbitals line up side to side!). Think of a ladder. If you try to move 1 ring but not the other, the ladder (bond) will break. For sigma bonds, they have 1 point of contact (end to end), think about wringing out a towel (lol), you can twist either way witho...

Chris_Butler_1A

I was wondering if someone could help clarify why pi-bonds restrict rotation while sigma-bonds allow for it. From what I can tell, the orientation of the electron density in relation to the intermolecular axis seems to play a role. Is it that the p-orbitals seen in pi-bonds are perpendicular to the ...

Chris_Butler_1A

There was an example given with Benzene (C 6 H 6 ) where the structure was seen to have a Delocalized ring of 6 electrons, each given by the 1 unhybridized electron of each Carbon. I was wondering, in this case, does the delocalization of the electrons mean that they are not specifically "attac...

Chris_Butler_1A

Does the rule of naming in alphabetical order also apply to Anions? I know it was mentioned that in order it would go "Anion name" "(Greek prefix)" then "hydrate" but how would this work in an example with multiple anions as well as with the alphabetical order of the Li...

Chris_Butler_1A

Hydrogen bonds are stronger because of the atoms that are involved. Hydrogen bonding involves a H atom bound to a highly electronegative atom like N, O, and F. Since the hydrogen is bound to such an electronegative atom, the electronegative atom is pulling the electron density away from the hydroge...

Chris_Butler_1A

After a review session where we discussed the relative strengths of 3 separate forces--London Dispersion Forces, Dipole-Dipole, and Hydrogen Bonding--I had a little trouble understanding why Hydrogen Bonds have higher energy than Dipole-Dipole interactions. From what I understand, both kinds of inte...

Chris_Butler_1A

I makes more sense when you think of axial and equatorial bonds in terms of the 3-D arrangement of the molecule. The image I drew and attached might help you visualize its lay out. But yes, lone pairs are often in axial positions in order to minimize repulsions as you can see from many bent, pyrami...

Chris_Butler_1A

I apologize if this seems like a straightforward question, but is the only distinction between a tetrahedral and trigonal pyramidal structure that trigonal pyramidal has one lone pair in one of its axial positions? I saw some examples of both that seem to give contradictory answers. Also, by this lo...

Chris_Butler_1A

Equatorial atoms can exist on all 360 degrees of the central atom so long as they are in the same plane, and axial atoms can only occupy the space above and below or to the direct left and right of the central atom depending on how you are looking at the molecule. The relationship between these two...

Chris_Butler_1A

I was having a bit of an issue understanding the differences between axial and equatorial bonds as well as how they impact the shape of molecules. Will certain atoms or lone pairs only be in one of the two positions and thus affect the shape? Axial: On the vertical axis Equatorial: On the plane Cer...

Chris_Butler_1A

I was having a bit of an issue understanding the differences between axial and equatorial bonds as well as how they impact the shape of molecules. Will certain atoms or lone pairs only be in one of the two positions and thus affect the shape?

Chris_Butler_1A

Hi! From what I looked up online, many sources said that electron repulsion plays a role in the bond length, however failed to mention to what extent. I would assume that it would not be to a large extent because it was not reviewed in class, but I don't know for sure. From what I looked into, it s...

Chris_Butler_1A

I believe in order to distinguish between Lewis Bases and Lewis Acids in a reaction, you need to look at which molecules in the reaction are electron donors and which are electron acceptors. For example, in AlCl 4 - , which is a reaction between AlCl 3 + Cl - , the AlCl 3 is the electron acceptor ma...

Chris_Butler_1A

Another example of this I was able to find was between N2 and O2. N2, with a triple bond, sees the atoms being closer together and forming a more spherical shape while O2 has a longer double bond which creates a more rod-like shape. Due to its shape, O2 has a larger surface area and thus two molecul...

Chris_Butler_1A

For melting and boiling points lavelle talked about polarizability being the main factor. So, the common element that add to polarizability is the size of the molecule. So when you compare two elements look at the size. when I mean size atomic radius is a key trend or number of electrons. So if I a...

Chris_Butler_1A

The more bonds there are, the less lone pairs the atoms will have (i.e. two atoms joined with a triple bond have less lone pairs than two atoms joined in a single bond because more electrons are involved in the bond). Thus, molecules with single bonds have more free electrons, which repel each othe...

Chris_Butler_1A

I believe this is due to the rod-shaped structures having a larger surface area and more of the surface area being oriented to interact with each other. It might be a more simple way of looking at it but I immediately thought of stacking plates on top of one another versus placing them side by side,...

Chris_Butler_1A

Understanding that dispersion forces are what affect the boiling and melting point of molecules, I was wondering if someone could clarify how temperature interacts with these dispersion forces. Is it that with a rising temperature, electrons become more excited and thus are more likely to affect nei...

Chris_Butler_1A

Yes all molecules have London dispersion forces, and it is one of the weakest intermolecular forces. When two molecules come in close proximity they "induce" a partial negative/positive on each other instantaneously making it temporary. I'm unsure if I am correct on this but I believe it ...

Chris_Butler_1A

I had a question in regards to how electrons interact in double and triple bonds. I understand that their repelling nature results in effects such as Van der Waals forces, so would this repelling play a role in the length of double and triple bonding?

Chris_Butler_1A

I apologize if this has already been asked but I had a little confusion in regards to how Atomic Radius is calculated in ionic bonding. From what I understand, the Atomic Radius is half the distance from the center of a neighboring atom. How does this affect our ability to measure the radii of say N...

Chris_Butler_1A

I apologize if this has already been asked but because neutrons have a neutral charge, and thus they do not "pull-in" the electrons in the orbitals, can we say that they don't affect the atomic radius? For example, Carbon-12 and Carbon-13 would have the same atomic radius?

Chris_Butler_1A

I'm unsure of similar electron affinity rules for other periods but I am positive that Nobel Gases are exceptions, given that their outermost orbital is full thus making them more stable and effectively lower in electron affinity.

Chris_Butler_1A

The reason why atoms that have high electron affinity release energy is because they are so attracted to their electrons that if one of the electrons is taken away energy used to hold that atom is released. Thank you! So presumably for atoms with a low electron affinity, they absorb more energy bec...

Chris_Butler_1A

I think they release energy because they become more stable once they have attained the electron(s) needed, and stable states have less energy that unstable states. Feel free to correct me if I'm wrong :) This may be a little counterintuitive but I think my confusion is on why adding an electron, t...

Chris_Butler_1A

Electron affinity increases when it goes to the top and right side of the periodic table. Having a high electron affinity means that it is more likely to gain an electron in which it will release energy because they create bonds. Having a low electron affinity will cause the element to absorb the e...

Chris_Butler_1A

I was having a little trouble understanding what exactly a high or low electron affinity meant for an element. From what I understand, elements that desire to fill their outer orbitals will have a higher affinity but I'm not entirely sure how this relates to the absorption or release of energy. Why ...

Chris_Butler_1A

I had a bit of an issue with this problem as well. I tried to imagine wrapping each of the examples in a circle to see if they would match up (those that ended at the same side of the line or completely disconnected would not work) as well as eliminating any waves that were not consistent all the wa...

Chris_Butler_1A

Yup! Here is a summary of what equations can be used for certain particles. E=hv --> only for photon c=\lambda v --> only for photon E=\frac{1}{2}mv^{2} --> anything that has a mass and velocity (eg: electrons) so NO PHOTONS \lambda = \frac{h}{mv} --> anything that has a mass and velocity so again ...

Chris_Butler_1A

Photon does not use mass so the E=hv is only used inn that case where mass isn't part of the calculation. Additionally, when looking up the de Broglie equation, one site said it is "used to describe the wave properties of matter, specifically, the wave nature of the electron." So, with th...

Chris_Butler_1A

I was also a little confused on when to use the Rydberg equation. Thanks for the help. From the examples and the TA's I have asked, it seems like you can use either the Rydberg Equation or E = -hR/n^2 but Dr. Lavelle prefers the latter because it allows us to follow the exact process and placement ...

Chris_Butler_1A

Dr. Lavelle has said during lecture that we will mostly be dealing with UV, visible, and infrared regions, so that would be Lyman, Balmer, and Paschen series. It might also be good to know that n=4 is the Brackett series Do you think it would also be wise to know the nm measurements for the Paschen...

Chris_Butler_1A

You're right about the frequency! It's the cycles that pass through within a given time, and is also indicative of the energy, since E=hv. Energy and frequency are proportional, so the higher the frequency, the more energy. Meanwhile, intensity refers to the amount of photons in a light beam, not h...

Chris_Butler_1A

I believe the Rydberg Equation is just another way to find the emitted or absorbed light following a quantum level transition and it is derived from the empirical formula. You could use either the empirical formula or Rydberg's Equation but Professor Lavelle prefers that we use the empirical formul...

Chris_Butler_1A

I am a little confused as to when we are supposed to use the Rydberg Equation vs using the Empirical Equation for Hydrogen. Is the Rydberg Equation simply another method for finding the emitted or absorbed light following a quantum level transition? Or is it more appropriately used when we need to f...

Chris_Butler_1A

Hi! The Lyman and Balmer series just have different wavelength ranges associated with them, and unfortunately I think that's something we just have to memorize. I believe Lyman is 94-122 nm while Balmer is 410-656 nm. You could also figure out what series it is if a type of light is given, as Lyman...

Chris_Butler_1A

I was at first confused by this too, but what I found is that the intensity of light is the amount of photons present. For instance, if you increase the intensity of a light, there will be more photons which will cause the light to become brighter but that does not mean the energy capacity in each ...

Chris_Butler_1A

I think of it as the frequency is continuous for as long as there is energy. So in the system that is the energy provided, the frequency is continuous. So would you say that light acts like an assembly line where the "packages" (photons) are being continuously moved? The water analogy tha...

Chris_Butler_1A

Light intensity can be defined as photon energy per second per unit area. The intensity of light is also designated by the amplitude of the wave model. Therefore, increasing the number of photons would increase the intensity of the light, but the wavelength and frequency would remain the same since...

Chris_Butler_1A

How I thought of it was that when we get down to an extremely small scale (quantum world scale), that's when light behaves like a particle/packets of energy. But in everyday life when we're used to looking at things at a larger scale, that's when light tends to act like a wave. Remember the water a...

Chris_Butler_1A

From what Dr. Lavelle explained in the lecture, it seems that frequencies would be considered continuous for both the particle and wave models (especially helpful when using equations like v=E/h). I found the thread "Are frequencies discrete?" quite helpful if you would like a more detaile...

Chris_Butler_1A

So Dr. Lavelle mentioned that light acts both as a wave and at times as a packet of photons, thus the need for both the wave and quantum models. I was a little confused as to how exactly these properties work. Is it that light acts as either a wave or as a packet at certain times; or is it that ligh...

Chris_Butler_1A

You can use the molecular formulas to find the molar mass, but I just searched it up and got molar mass of butanone on the internet. You would only need to know the formulas for the 2-butanone and 3-methyl-3-hexanol since the other product is in excess. To approach the problem, first use the densit...

Chris_Butler_1A

Use the "Ans" key on your calculator whenever possible. We're allowed to use scientific calculators but not graphing calculators, so you should be able to store values. Also, I generally leave at least 5 numbers after each decimal point (for each step) and apply sig figs at the very end. ...

Chris_Butler_1A

From what I could find, the mantissa refers to the significant figures after a decimal point. For example, 5.382 would have a mantissa of 0.382. So, assuming we know the value of the logs, we could use the Mantissa to make sure the Significant Figures match if we were to add, subtract, or etc. so we...

Chris_Butler_1A

I could see Logarithmic Sig Figs potentially having a role when we starting getting into frequencies and wavelengths, so I imagine this next unit might give us an opportunity to practice using the Mantissa more. I just hope it is straight forward to figure out.

Chris_Butler_1A

I'm in the exact same boat, I have never seen it before. I took a screenshot from the bottom of the "Everything You Need to Know About Sig Figs" doc on his website. It seems like you need to know what the value of the log is before you begin then separate the two sides of the decimal point...

Chris_Butler_1A

In these examples of Sig Figs in Logarithms, the term Mantissa is used and unfortunately, I've never encountered this kind of problem. What exactly is the Mantissa and why does the SF of the Mantissa decide the SF in the decimal? Appreciate any help.

Chris_Butler_1A

Thank you so much, Zaid! I really appreciate it.

Chris_Butler_1A

I apologize if this question has been asked before, but is there a specific GroupMe for Lecture 1, or are there only GroupMe's for specific discussions?

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