CHEMISTRY COMMUNITY

TheresaDsilva4A

Will we have to know how to solve questions using equilibrium tables (example 6D4 in the textbook)? I just wanted to check since Dr. Lavelle didn't cover these types of questions during the lecture. Thank you so much!

TheresaDsilva4A

di- and tri- are used when there are more than one of the same ligand attached to the central metal ion. bis-, tris- and tetrakis- are used to signify the presence of more than one polydentate ligand. For instance, if you have two oxalato ligands in a coordination compound, it would be written as -b...

TheresaDsilva4A

I believe the integer in front of the hybridization signifies the principal energy level n at which the hybrid bond is formed. E.g. n= 2 for 2sp^3
Hope this helps!

TheresaDsilva4A

1. Whenever a "p" is placed before the letter (e.g. pKa or pKb), it signifies that you have to take the -log of the corresponding value (pKa = -logKa, pKb = -logKb, pKw = -logKw). 2. Ka is the acid dissociation constant and measures the extent to which an acid breaks down into its componen...

TheresaDsilva4A

I believe the -OH and -O bases formed from alkali and alkaline earth metals are typically classified as strong bases. Hope this helps!

TheresaDsilva4A

When I was looking through the solution manual, I noticed that water was often written as OH2 instead of H2O in the formulas of coordination compounds. Can someone explain why this is the case? Will we be penalized on an exam if we write it one way as opposed to the other? Thank you so much!

TheresaDsilva4A

Does anyone know what information we have to specifically know about these chemotherapy drugs? Is it just that they're an example of a coordination compound? I was absent during this lecture, so I'd appreciate any information! Thank you so much :)

TheresaDsilva4A

I think you have to label each bond individually for double/triple bonds (e.g. write pi above one of the double bonds and sigma above the other). However, the order with which you label pi/sigma for multiple bonds does not matter as long as you have the correct number of each type of bond. And yes, ...

TheresaDsilva4A

The pKa value is one method that is utilized to indicate the strength of an acid. It is calculated by taking the negative of the log base 10 of the acid dissociation constant, Ka (thus the p before the Ka). As the acid dissociation constant Ka increases, pKa decreases. Thus lower values of pKa are a...

TheresaDsilva4A

I believe it depends on whether an acid is strong or weak. Strong acids (e.g. HCl, HBr) are completely disassociated in solution, so they are written as a sum of their ionic components. However, weak acids (e.g. HF) typically retain their identity/not disassociate, so they are not broken into their ...

TheresaDsilva4A

Here are the different molecular shapes and their bond angles as defined by VSEPR theory: (2 regions of electron density) Linear: 180 degrees (3 regions of electron density) Trigonal Planar: 120 degrees, Bent: <120 degrees (4 regions of electron density) Tetrahedral: 109.5 degrees, Trigonal Pyramida...

TheresaDsilva4A

Yes! D-orbitals are often involved in hybridization in elements belonging to Period 3 or later because they can accommodate five or more electron pairs. As a result of the expanded valence shell, they require d-orbitals in their hybridized orbitals to account for all of the regions of electron densi...

TheresaDsilva4A

I think the difficulty of this class depends on your prior experience with chemistry. If you took AP chemistry in high school, I would tend to believe that many of the concepts in 14A are a review. Personally, this class is moderately challenging for me because I only took an honors chem class in hi...

TheresaDsilva4A

I have a follow-up question: How can we tell that dipole moment vectors cancel when they are at angles to one another (e.g. in a molecule with trigonal planar geometry)? Do we have to break up the vectors into their individual components?

TheresaDsilva4A

I think the format of the final will be similar in format to the midterm, but slightly longer (3 hours long vs 2 hours long). I believe it is cumulative, so any material covered during the quarter is fair game. Hope this helps! :)

TheresaDsilva4A

Multiple bonds are treated as single regions of high electron density because these bonds always act as a single unit. Thus, any resonance structure of a particular molecule will have the same shape as determined by the VSEPR model. Hope this helps!

TheresaDsilva4A

Yes, we can! A nonzero dipole moment implies that electrons are shared unequally by bonded atoms, resulting in the formation of partial charges.

TheresaDsilva4A

To add on the previous answer, lone pair repulsions are typically stronger than atom-atom repulsions because lone pairs are less restrained than bonds and take up more space!

TheresaDsilva4A

Bond order is determined by the type of bond (single, double, triple etc.) found between a pair of atoms. So, in order to determine the bond order, I would try drawing the lowest-energy Lewis structure for the molecule. For example, in the carbon monoxide molecule, the C-O bond order is 3 because ca...

TheresaDsilva4A

No, you do not need to know the bond order when finding the bond angle. Relating to the VSEPR model, I think the main concept regarding bond order that we have to be familiar with is that a multiple bond is treated as a SINGLE region of high electron density when determining the molecular shape.

TheresaDsilva4A

VSEPR stands for the Valence Shell Electron Pair Repulsion Model. It is utilized to determine/explain the experimentally observed shapes of molecules by taking into account the various lone and bonding pairs of the molecule. Hope this helps!

TheresaDsilva4A

Does anyone know if we will have to commit this information to memory for Test 2/Final exam? Will we be given this reference sheet during the exams? Thank you!

TheresaDsilva4A

Can someone explain how hydrogen bonding affects the polarity of a molecule? I'm trying to understand the connection between these two concepts. Thank you so much!

TheresaDsilva4A

No :( I think the first time we will view our grades will be when they are posted on myUCLA. I'm not sure if they will be handing back the midterm in discussion section...

TheresaDsilva4A

Just to add on to the previous replies, typically smaller cations and larger anions bonded together to increase the covalent character of a molecule. A smaller cation is able to more strongly attract electrons from the anion while a larger anion more easily shares electrons due to a weaker effective...

TheresaDsilva4A

I think each TA will grade one question to standardize grading across the board. So, multiple TAs?

TheresaDsilva4A

Based on the homework problems, an element that is commonly associated with an expanded octet (hypervalence) is the noble gas Xenon. Boron and Aluminum are two elements that often exist in compounds with an incomplete octet. Hope this helps!

TheresaDsilva4A

Elements that are able to hold an expanded octet occur from Period 3 onwards because they have empty d-orbitals in their valence shell. They do not always have to hold more than 8 electrons. Typically, an expanded octet will consist of 10 or 12 electrons, but it may even be greater in particular cas...

TheresaDsilva4A

Just to add on, when a molecule has resonance structures, it's important to calculate the formal charge distribution of each structure. The resonance structure with the lowest formal charge makes the dominant contribution to the resonance hybrid because it is most energetically favorable (as you all...

TheresaDsilva4A

Just to add on, ionic bonds can also have covalent characteristics. The strong positive charge from a cation can induce distortion in the electron cloud of an anion resulting in electrons occupying the region between the two nuceli of the bonded atoms. This "sharing" of electrons between a...

TheresaDsilva4A

A Lewis base is defined as a species that donates an electron pair (has lone pair electrons) while a Lewis acid is defined as a species that accepts an electron pair (has vacant orbitals). The acceptance of an electron pair by the Lewis acid from the Lewis base enables the formation of a coordinate ...

TheresaDsilva4A

Elements with low ionization energy tend to form cations because it is easier (requires less energy) to remove electrons from their orbitals. Thus, the formation of a positively charged ion (cation) is favored. The characteristic of having low ionization energy also enables these elements (metals) t...

TheresaDsilva4A

Just to add on to the previous replies, the cations and anions of an ionic compound are bonded together in a crystalline formation (I think of a cube). So, that's basically rows of alternating cations (+) and anions (-) attached to each other. Hope this helps you get a better picture of how ionic bo...

TheresaDsilva4A

Can someone explain how trends in effective nuclear charge explain the trends in electron affinity? I'm kind of confused because electron affinity does not have the same "periodic nature" as the other properties of elements. Thank you!

TheresaDsilva4A

I spoke with an LA after class about this specific topic. He explained that the electrons of the NO3- ion are evenly distributed amongst the three bonds. Thus, in reality, there is no distinguishable difference between the double/single bonds. This is the reasoning behind the experimental 1.24-angst...

TheresaDsilva4A

Based on physical laws, it is typically more energetically favorable for compounds formed through covalent/ionic bonding to have a lower energy state than the elements that make them up. Since ionic/covalent compounds have a lower energy state than their constituents, they are often much more stable...

TheresaDsilva4A

Is there a clear trend for electron affinity across a period/down a group (as for atomic/ionic radii)? Do you have any helpful tips for remembering the trend/exceptions to the rule? I struggled with this topic in AP chem, so I'd really appreciate any insight :) Thank you so much!

TheresaDsilva4A

1. E = hv calculates the energy contained by an individual photon of light 2. E = 1/2 mv^2 is the formula for the kinetic energy of any object with a mass. Since photons are assumed to be massless, you cannot use this formula to calculate their energy. Oftentimes, the formula for kinetic energy is u...

TheresaDsilva4A

If the 3d orbital is filled after the 4s orbital, then why does it have less energy? Wouldn't electrons tend to fill subshells with lower energy first? Can someone please explain why this is the case? Thank you!

TheresaDsilva4A

1. By Heisenberg's Uncertainty Principle: (uncertainty in momentum)(uncertainty in position) >/= (h)/(4 pi) 2. The given values in this problem allow you to calculate the uncertainty in momentum. Then, using Heisenberg's Uncertainty Principle, you must solve for the uncertainty in the position. 3. u...

TheresaDsilva4A

In my AP Chem class, I learned to write the 4s subshell before the 3d subshell when writing an electron configuration. However, during the lecture and in the examples/problems from the textbook, the 3d orbital is written prior to the 4s orbital. (I also remember Dr. Lavelle saying that this is the p...

TheresaDsilva4A

In my AP Chem class, I learned to write the 4s subshell before the 3d subshell when writing an electron configuration. However, during the lecture and in the examples/problems from the textbook, the 3d orbital is written prior to the 4s orbital. (I also remember Dr. Lavelle saying that this is the p...

TheresaDsilva4A

I thought that the excited electrons are released from the hydrogen atom in the case of atomic spectra? Can you please clarify what you mean when you say that the electron is still "attached"? Thank you!

TheresaDsilva4A

The symbol N subscript A stands for the number of photons in 1 mol of photons (so, by definition, Avogadro's number of 6.02 * 10^23 photons). The example asks to find the total amount of energy in 1 mol of photons. So, we have to multiply the energy of a single photon by the total number of photons ...

TheresaDsilva4A

Can someone explain how black body radiation is related to the photoelectric effect?

TheresaDsilva4A

The final units of your answer depend on what you are solving for. The principal energy levels (n1 and n2) do not have units. However, if you are solving for the wavelength using the Rydberg equation, the final units of your answer must be meters (or some other measurement of length). For reference,...

TheresaDsilva4A

Is there a particular equation relating intensity to other variables? I'm still a little confused why intensity plays no role at all in the photoelectric effect. Thanks!

TheresaDsilva4A

I think you need to refer to the diagram of the scale shown below the problem in order to find the solution. If you look at the picture, there are 9 blue gallium atoms on the left side of the scale. Looking at your periodic table, the molar mass of astatine (210 g) is approximately three times great...

TheresaDsilva4A

Well, first you'd use the calculated molar mass of KCl (74.5513 g/mol) to convert the mass of KCl into moles of KCl: 5.51 g KCl * 1 mol KCl/74.5513 g KCl = 0.739 mol KCl Then, the problem states that water is added to a final volume of 125 mL. Basically this means that water was poured into the cont...

TheresaDsilva4A

On a side note, I'd like to point out that the equation was incorrectly written in the original question. There is no subscript 2 for the oxygen gas on the product side of the equation. Nitrogen monoxide NOT nitrogen dioxide is formed! The entire correct equation is written below: 4NH[3] + 5O[2]--> ...

TheresaDsilva4A

Well, in the balanced equation 4NH3 + 5O2 --> 4NO + 6H20, the ratio NH3 to O2 = 4:5. If you divide both sides of this ratio by 4 you get 1: 1.25. That is why you have to multiply the number of moles of NH3 by 1.25 to get the required moles of O2.

TheresaDsilva4A

I compared the experimental mole ratios of NH3 and O2 to the ratios of the stoichiometric coefficients given in the problem. As you probably calculated: 21.4 g NH3 = 1.2566 mol NH3 42.5 g O2 = 1.328125 mol O2 In the reactant side of the given chemical equation, the ratio of NH3: O2 = 4:5. Based on t...

TheresaDsilva4A

In the last sentence, I think you meant to say add the molar mass of Calcium (39.94g/mol) to the molar mass of a Sulfide ion (32.05 g/mol) in order to find the solution.

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