CHEMISTRY COMMUNITY

kpang_4H

Pi bonds have to be atomic orbitals,not hybrid orbitals

kpang_4H

For hybridization, the 2 is not needed, but it is if asking about specific orbitals

kpang_4H

Any polydentate ligand is a chelating ligand

kpang_4H

Formal charge is valence minus number of lines and dots, and you can get a negative or a postiive answer when you do that

kpang_4H

Fe3+ has more polarizing power because it is more positively charged and smaller.

kpang_4H

It sounds like you are doing fine in the class. I would say that reading over some of the textbook and attending the workshops are very helpful in reviewing and applying what we learned in lecture.

kpang_4H

Is there going to be a final review session held by the UAs like there was for the midterm?

kpang_4H

Can someone explain the difference between conjugate, Bronsted, and Lewis acids and bases?

kpang_4H

Are pH=-log(H+) and pH=-log(OH-) the only equations we need to know, regarding pH, for the final?

kpang_4H

What exactly is chelation and what are some examples of chelating ligands?

kpang_4H

305376058 wrote:Typically, yes, because the hydrogen is attracted to atoms N, O, and F, that have stronger partial negative charges.

I would also add that the N O F atoms must have a lone pair for hydrogen to bond.

kpang_4H

What is the point of using EDTA^4- to remove metals from solutions?

kpang_4H

Bilal Pandit 3F wrote:They are used when the ligand has a prefix itself, so the ligand would have a di, tri, etc, so instead of saying didi, you would just say dis.

How do you know which is the ligand?

kpang_4H

How do you determine the bond angles for a seesaw shaped molecule?

kpang_4H

Why cisplatin is used for chemotherapy and not transplatin?

kpang_4H

The electronegativity difference is big between the N and the H and also the O and the H which creates dipoles.

kpang_4H

My iq drops as well, but i think just try to not think about the time and just try your best!

kpang_4H

I think vsepr and molecular shape is where we start for test 2, and intermolecular forces

kpang_4H

If no other intermolecular forces are different between the two, larger ones should have stronger attractions because of the london forces.

kpang_4H

Equatorial ones are on the x-axis or horizontal plane like an equator and axial ones are the vertical ones,

kpang_4H

I am pretty sure you can go to your TA to review the test!

kpang_4H

Oxygen has a 4th electron which makes a pair in the orbital which has more electron repulsion due to that paired electron, giving it a lower ionization energy.

kpang_4H

Because those have high electronegativity which makes a strong dipole when bonded to the H.

kpang_4H

I dont think there is a relationship, I think larger atoms are generally more polarizable.

kpang_4H

There definitely wont be any discussion thursday or friday but not sure about tues-wed

kpang_4H

I think it is talking more about attractive energy. So the more negative the potential energy is, the stronger the attractive force is.

kpang_4H

Pi bonds are side by side overlap of p-orbitals, sigma bonds are other types of overlapping orbitals.

kpang_4H

I think the idea is that bonds are not binary. For example, like gender, bonds lie on a spectrum of ionic and covalent character. Therefore, they are not always equally sharing (covalent) or completely giving up electrons (ionic), so they have some characteristics of both.

kpang_4H

The exceptions for ionization energy are the one mentioned with O and N and it also applies to the elements directly below, which are P and S.

kpang_4H

Yes, noble gases have ionization energies and electron affinities. Since noble gases are very very stable (inert/won't react), they do not want to gain or lose an electron. Therefore, they will have very high ionization energies and very low electron affinities.

kpang_4H

Any element from row 3 and beyond can have expanded octet because when n=3, l=0,1,2 which means elements in the third row have access to a 3d orbital, but none of the third row elements have any electrons in that orbital yet, so they can accommodate more electrons. Elements in the fourth row have a ...

kpang_4H

I don't think we count valence electrons for d-block elements but they can have multiple oxidation states.

kpang_4H

The 3rd row and beyond can have expanded octet, more than 8 electrons. H, He, Li, Be will always have an incomplete octet. The 13th column or the Boron column can have an incomplete octet.

kpang_4H

sbottomley3a wrote:How will the difficulty of these problems compare to the difficulty of the midterm?

At the review session, he said the problems just as hard or harder than midterm questions ideally.

kpang_4H

Since some people already responded to your question on what shield is, I'll just say how it affects ionization energy. Basically, you will see a trend of decreasing ionization energy down the periodic table and a trend of increasing ionization energy across the period.

kpang_4H

Are we expected to know how to calculate these values in order to find electronegativity differences?

kpang_4H

They only need 2 electrons to be satisfied, so if lithium loses an electron it will be stable.

kpang_4H

How can you tell if an atom is in its ground state or an excited state by looking at its electron configuration? For example, Li: [Ar] 1s^2 2p^1

kpang_4H

In ionic compounds, the anions transfer electrons to cations, which are lacking, so that both atoms achieve a noble gas like configuration. As for covalent bonds, the electrons are simply shared, meaning that electrons are not given into the other atom unlike in ionic bonds.

kpang_4H

1 Hz = 1 cycle/second

kpang_4H

the equation you used was right, but you may have messed up when you plugged in the values. I did 10g/194.19g.mol^-1 divided by .240 L and got .215M.

kpang_4H

Arianna Perea 3H wrote:does the metal eject electron if the energy and work function are the same?

As long as the energy of the photon is greater or equal to the work function, the metal will eject the electron. Any excess energy is regarded as kinetic energy.

kpang_4H

Could someone explain to be how you use the work function to find kinetic energy of an electron and the energy needed to remove an electron. Specifically in the case of these two questions from the video module assessments. "Light hits a sodium metal surface and the velocity of the ejected ele...

kpang_4H

You would add it up, so the uncertainty value would be 6.

kpang_4H

The number you are referring to is Planck's constant(h) which relates the energy in one photon of electromagnetic radiation to its frequency.

kpang_4H

The photoelectric experiment consisted of shining light onto a metal surface to measure the energy needed to remove electrons from different metals. Kinetic energy is equal to energy of the photon minus the the energy to remove the electron (work function/threshold energy). This equation illustrates...

kpang_4H

You can just divide the speed of light by the given velocity using c=hv

kpang_4H

De Broglie's equations applies as long as the particle in question has p, momentum(mv), and has wavelike properties. Depending on the question, you can use the equation to find mass, velocity, or wavelength.

kpang_4H

Diffraction results from constructive and destructive interference. Constructive waves are in phase, which means their crests and troughs line up with one another to result in a wave with a larger amplitude. Destructive waves are out of phase, so the resulting wave's amplitude is lower than the init...

kpang_4H

The equation for the speed of light is c = frequency x wavelength. If you remember, c is a constant, 2.998 X 10^8 m/s.

kpang_4H

Hertz is the SI unit for frequency, and it tells you cycles/second. When anything is denoted by ^-1, it puts the unit goes under 1. For example, s^-1 is 1/s, and g.mol^-1 is the equivalent of g/mol.

kpang_4H

I noticed that in the practice problems, you can just write the elements in the same order the mass percentage compositions or mass of the elements are shown.

kpang_4H

If you can, try to look back at your old notes. If not, watching Khan Academy and other available videos is super helpful. Just make sure that afterwards you are able to apply your knowledge by practicing textbook problems, especially in sections where you find you are more unsure.

kpang_4H

https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/EVERYTHING_YOU_WANTED_TO_KNOW_ABOUT_SF.pdf This is the link on Lavelle's website that explains how sig figs work. There is no set number of decimal places or sig fig for the whole set; after reading over the pdf, use sig figs at the e...

kpang_4H

Based on the lecture, base SI units we need to know are grams (mass), meters (length), and seconds (time). It's also helpful to note certain prefixes, such as: Giga (G), mega (M), kilo (k), centi (c), milli (m), micro, and nano (n), and the degree each indicates.

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