CHEMISTRY COMMUNITY

Anna De Schutter - 1A

No the lone pairs on the double bonded O won't bind with the metal because that would give O three bonds (one double with C and one single with the metal) and one lone pair, and thus a formal charge of : FC=6-2-3=+1 which is less stable than it was when O was just bound to the C with a double bond a...

Anna De Schutter - 1A

Thank you so much! :)

Anna De Schutter - 1A

Thank you! :)

Anna De Schutter - 1A

Hi!

I know that if a transition metal is bound to 4 ligands, its shape could either be square planar or tetrahedral. How do I distinguish between these two shapes? And what would be the shape of cisplatin, tetrahedral or square planar?

Thank you!! :)
Anna De Schutter - section 1A

Anna De Schutter - 1A

Thank you Dr. Lavelle! :)

Anna De Schutter - 1A

Hi! I'm a bit confused about the answers given in the solutions manual regarding problem 17.33. For part a, we consider HN(CH2CH2NH2)2 and the solutions manual says it is tridentate. For part b however (CO3 2-), the solutions manual says it can be monodentate or bidentate. I understand why CO3 2- ca...

Anna De Schutter - 1A

Hi! For last Friday's lecture I wrote down in my notes that: Cl- is "good on its own" in a solution as shown by the equation HCl --> H+ + Cl- I'm a bit confused by this though, how can we know Cl- won't affect the reaction through this equation? Is this also why Cl- doesn't affect the pH o...

Anna De Schutter - 1A

I had my discussion section today, and my TA explained to me how to find the charge of a ligand :) She told us that we needed to calculate the formal charge of the atom of the ligand bound to the transition metal. So for example for the coordination compound [Co(en)2(Cl)2)]*, we need to calculate th...

Anna De Schutter - 1A

Also, if it is true that double bonds don't count as ligands, why can (CO3)2- be a monodentate or a bidentate (it has one double bonded O and two single bonded O), but oxalate (C2O4)2- can only be a bidentate (it has two double bonded O's and two single bonded O's)? Can't (C2O4)2- also be a monodent...

Anna De Schutter - 1A

Thank you!! :)

Anna De Schutter - 1A

Hi! In problem 17.29, we need to determine the oxidation number of each metal. In the solutions manual we are given the formula: (#metal atoms)(oxidation number of the metal) + summation sign(# each ligand)(charge of each ligand)=charge of the ion I was wondering, how to we determine the charge of e...

Anna De Schutter - 1A

Hi! I'm a bit confused on how to do problem 17.33. For 17.33 a for example, I thought HN(CH2CH2NH2)2 would be tridentate (which it is!) because there are three available lone pairs, one on each nitrogen. But then for 17.33 c, where we are given H2O, the solutions manual says it is monodentate. I jus...

Anna De Schutter - 1A

That makes a lot of sense, thank you! :)

Anna De Schutter - 1A

Yeah I looked at the solutions manual error document on the website but it didn't show an error for 4.13 so I'm confused as to what I'm doing wrong :( Do you see any errors in my reasoning? Thank you!! :)

Anna De Schutter - 1A

Hi! For the (CH3)2Be molecule, the shape is said to be tetrahedral around the carbon atoms in the solutions manual. Could I also say that the shape of the whole molecule is linear because we have Be in the middle and a CH3 on either side? Or would that be wrong? Also, for the SnCl2 molecule, I drew ...

Anna De Schutter - 1A

Hi! Isn't the Lewis structure for IO3- an I with two O around it with double bonds and one O with a single bond instead of the one shown in the solutions manual (where I is attached to three O's, each with a single bond)? The structure shown in the solutions manual would give the formal charges: I =...

Anna De Schutter - 1A

Oki, thank you! :)

Anna De Schutter - 1A

Thank you!! :)

Anna De Schutter - 1A

Hi! In homework 4.3 we are asked to draw the structure of CH2F2. I was going to draw the Lewis structure the way we did it before (with a 90 degrees angle between each three consecutive atoms), but in the solutions manual, the structure drawn shows on H atom as being in front of the other (with the ...

Anna De Schutter - 1A

Hi!

Last Wednesday in class we said that water doesn't have a linear shape because of its two lone pairs of electrons. What is the name of a molecule that has the notation AX2E2, such as H2O?

Thank you!
Anna De Schutter - section 1A

Anna De Schutter - 1A

Hi! In class we talked about how the interaction potential energy for induced dipole - induced dipole interactions is proportional to (polarizability 1 * polarizability 2)/distance^6 We also said that the polarizability depends on the size of the atom and the number of electrons, with an increase in...

Anna De Schutter - 1A

Hi! Is there a specific hw problem you had in mind that showed this? The only reason I can think of why we would say less than a certain bond angle instead of the actual value is because of the presence of lone pair electrons. We said in class, repulsion strength: lone pair - lone pair > lone pair -...

Anna De Schutter - 1A

Hi! Dr. Lavelle also talked about during last Friday's lecture when he said that the reason Xe can interact with other atoms even if it is a noble gas is because it is a very large atom. As a result, Xenon doesn't hold on to its electrons very well and sharing electrons with other atoms, such as F i...

Anna De Schutter - 1A

Hi!

I also had the same problem as you, but then Dr. Lavelle referred me to one of his posts which was very helpful!

Here it is:
viewtopic.php?f=33&t=31699&p=100750

Hope this helps! :)
Anna De Schutter - section 1A

Anna De Schutter - 1A

Another way to look at it is that N is less electronegative than O which is another reason why N3- has a higher polarizability than O2- because the polarizability of an anion increases as its electronegativity decreases. What I don't really understand is why do smaller cations have a higher polarizi...

Anna De Schutter - 1A

Hi!

I'm a bit confused, how are He, Li and Be exceptions to the octet rule? If they don't aim to have 8 valence electrons, what do they do then?

Thank you!
Anna De Schutter - section 1A

Anna De Schutter - 1A

Would we be given a chart that tells us the electronegativity of each element in order to solve this question? Or is there another way to know that O is more electronegative than Cl?
Thank you! :)

Anna De Schutter - 1A

And when you say separation of charges in the molecule do you mean where the electrons are located?
Thank you! :)

Anna De Schutter - 1A

I'm a bit confused, what does the term dipole mean? And is it different from a dipole moment?

Thank you so much! :)
Anna De Schutter - section 1A

Anna De Schutter - 1A

Thank you! :)

Anna De Schutter - 1A

In class we said that when we form an ionic compound, we transfer the electrons from a metal to a nonmetal. I just want to make sure are the metalloids: boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po) and then everything to the right of it is a ...

Anna De Schutter - 1A

I had the same question, because when I calculated the formal charges of chlorine and oxygen in the solution's manual configuration I got: Cl: 7valence electrons - 5 lone pair electrons - 0.5*2 bonding electrons = +1 O: 6valence electrons -6 lone pair electrons - 0.5*2bonding electrons = -0.5 Wherea...

Anna De Schutter - 1A

To add on to that, the reason chlorine would hypothetically gain one electron or lose 7 electrons is so that it has the electron configuration of a noble gas. if chlorine gains one electron, it would have the electron configuration of argon: [Ar] if chlorine loses 7 electrons, it would have the elec...

Anna De Schutter - 1A

I think that another way you can determine which element has the lowest ionization energy between two elements that are in two different rows or groups is to use the concept of diagonal relationship which states that elements diagonal to each other (one period down and one group to the right) have v...

Anna De Schutter - 1A

Hi! To answer your first question (how to do the Lewis structure for CCl4), the way I do it is I first count the number of valence electrons present in the molecule. So in CCl4, carbon has 4 valence electrons and chlorine has 7 valence electrons. Because there are four Cl in the molecule CCl4, the t...

Anna De Schutter - 1A

Thank you that helps a lot! :)

Anna De Schutter - 1A

Hi! During class (Wednesday of week 5) we talked about main group metals and how they form cations by losing their s and p valence electrons. As a result, they adopt the electron configuration of the preceding inert-gas atom. I'm a bit confused, what does main group metals mean? Moreover, Dr. Lavell...

Anna De Schutter - 1A

Thank you!! :)

Anna De Schutter - 1A

Hi! Problem 2.55 says: "Give the notation for the valence-shell configuration (including the outermost d-electrons)". I'm a bit confused, what does valence-shell configuration mean? For example, for part b it asks us to give the valence-shell configuration of group 15 elements and I was ju...

Anna De Schutter - 1A

But why is it 5d^4 6s^2 and not 5d^5 6s^1? I thought it was more stable to have d^5 than d^4 (as well as d^10 instead of d^9)? Also are we expected to know how to include the f-block in electron configurations? I'm not sure but I thought we had to know how to do electron configurations of the s-bloc...

Anna De Schutter - 1A

That makes a lot of sense! Thank you so much! :)

Anna De Schutter - 1A

Hi! Problem 1.57 says: "Lines in the Balmer series of the hydrogen spectrum are observed at 656.3, 486.1, 434.0, and 410.2 nm. What is the wavelength of the next line in the series?" I thought because in the Balmer series we are starting at n=2, that the next line in the series would be n=...

Anna De Schutter - 1A

Hi! The way I answered that question is by saying that the photoelectric experiment shifted our perception from seeing light behaving only as a wave to seeing it as composed of packets of energy called photons as well. Indeed, the photoelectric experiment didn't match the wavelike properties of ligh...

Anna De Schutter - 1A

Thank you Beverly! :)

Anna De Schutter - 1A

I'm still a bit confused, like if we take away one electron, the second electron could still be on the outermost shell right? Like for oxygen for example, there are 6 electrons in n=2 so wouldn't removing one electron mean that the second electron we remove is still on the outermost shell (n=2)? Why...

Anna De Schutter - 1A

Hi! I'm not too sure if it was called shell attraction, but how I understood it is that across a period, the atomic radius decreases. This is because across a period, we still remain in the same shell (n=1 for example, or n=2, etc.), yet the number of protons increases. Thus, the electrons are more ...

Anna De Schutter - 1A

Hi! Today in lecture we discussed how removing the second electron from an element is always harder. Indeed, the 2nd ionization energy of an element is always higher than the first one. I just didn't really understand the reason behind this statement. Why is it harder to remove an electron from an e...

Anna De Schutter - 1A

Thank you Gianna and Kimberly! :)

Anna De Schutter - 1A

While I was doing the homework problems, I came across some questions where I wasn't sure which quantum number they were asking me to find. So I just wanted to make sure: - when they ask for the number of orbitals, does that mean the quantum number ml? - when they ask for the number of subshells, do...

Anna De Schutter - 1A

I don't think it indicates different properties of the electrons configuration. Dr. Lavelle just used half arrows, but he said it didn't matter which one you used.

Anna De Schutter - 1A

I agree, we don't really have a number to which we can compare the uncertainty of momentum to see if it's realistic, but it's easy to compare the uncertainty of velocity to the speed of light to see if it is realistic.

Anna De Schutter - section 1A

Anna De Schutter - 1A

I agree that for the example we did in class, we could say it was an impossible outcome because the the speed of an electron can never be faster than the speed of light since it has a mass. Maybe to determine if the uncertainty of position is reasonable we could have compared it to the diameter of t...

Anna De Schutter - 1A

Ek=0m/s in this question (part b) because the question is asking us to find "how much energy is required to remove the electron from the metal surface". This implies that we are looking for the threshold energy (also called the work function) in this question and the threshold energy means...

Anna De Schutter - 1A

Thank you! :) And yes you're right, n2>n1, I wrote it the wrong way, my bad!

Anna De Schutter - 1A

Thank you!!

Anna De Schutter - 1A

Hi! For quiz 2 if it asks us for example to calculate the frequency of light emitted by an electron that goes from n=4 to n=2 can I use the Rydberg equation: frequency = R [(1/n1^1) - (1/n2^1)] or should I go through these calculations: delta E = E(final) - E(initial) = E2-E4= (-1/4*h*R) + (1/16*h*R...

Anna De Schutter - 1A

Hi! I've been struggling a bit with understanding how the solutions manual presents 1.15. So the problem states: "In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6nm. Determine the values of n for the initial and final energy levels of the electron during the emission ...

Anna De Schutter - 1A

Hi! So when we did question 1.25 a, we found that the energy emitted with one sodium atom is 3.37*10^-19J. In this question we are asking how much energy is emitted with 5.00mg of those sodium atoms. To solve this problem, we need to find how many sodium atoms there are in 5.00mg so that we can mult...

Anna De Schutter - 1A

Hi! The way I went about this problem is I looked at the wavelengths I found and then I associated each with an event. In class I believe we mentioned these sample wavelengths for each component of the spectrum of light: Radio 10cm - Microwave 1mm - IR 1 micrometer - Visible 700-400nm - UV 100nm - x...

Anna De Schutter - 1A

Hi! The information given to us in this question is that no electron is emitted until the frequency of radiation reaches 2.50*10^16 Hz. We are trying to find the amount of energy required to remove the electron from the surface, thus the threshold energy (also known as work function) required to rem...

Anna De Schutter - 1A

I'm not too sure that there would exist a case where you would need to use both equations. However, if for example you are giving n(final) and M(final) as well as M(initial), you could use M=n/V to find V(final) and then the equation M(initial)*V(initial)=M(final)*V(final) to find V(initial). You co...

Anna De Schutter - 1A

When you are trying to determine the limiting reactant of the reaction, you find the initial amount of moles for each reactant (either given to you or you convert masses to moles using the molar mass) and then you choose one reactant (let's call it reactant 1) and depending on the molar ratios (give...

Anna De Schutter - 1A

I believe that if the problem had asked to find the empirical formula, we would have also found C49H78N6O12 because as said by Dr. Lavelle just above, the empirical formula and molecular formula are the same in this problem because the ratios of the molecular formula (C49H78N6O12) are irreducible. M...

Anna De Schutter - 1A

Hi Madeleine! To come back to your question about G5 part c I think this is one way to look at it: According to the key concept: When diluting a solution, moles of solute remain the same, we can write: n(initial)=n(final) or M(initial)*V(initial)=M(final)*V(final) In part c of problem G5 we want to ...

Anna De Schutter - 1A

Hi Brynne! From what I understood during my discussion section points will not be taken off if we don't use the correct number of significant figures; however, you shouldn't give too many numbers after the decimal point either (like 10 for example). If I remember correctly my teaching assistant told...

Anna De Schutter - 1A

Hi! I've noticed in the solutions manual that units are used for every value when doing calculations. For example if they want to calculate the number of moles of C knowing that the mass of C is 67.49g (problem F15) they write: (67.49g)/(12.01g/mol)=5.619mol I was just wondering would it be okay to ...

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