CHEMISTRY COMMUNITY

annaspain

It gains energy, so delta E will be positive

annaspain

yes, they must bond to hydrogen in order to form a hydrogen bond.

annaspain

The textbook is updated to the new "cyanido" naming, but "cyano" is used in lecture. Either is correct!

annaspain

Yes- sigma bonds are always single bonds. So, in a double bond there will be 1 sigma bond and 1 pi bond, and in a triple bond there will be 1 sigma bond and 2 pi bonds. I don't believe we will have to draw this, but basically in a sigma bond the orbitals overlap end-to-end and in a pi bond they over...

annaspain

The overall charge of the compound will be given outside of the bracket. From there, add the charges you already know within the compound (ex: OH has a -1 charge) and subtract from the total charge, which will leave you with the oxidation state of the transition metal.

annaspain

Ligands should always be ordered alphabetically.

annaspain

I believe Karen posted a bunch of review questions, she has a lot of helpful worksheets for peer learning sessions. You should be able to find them if you search 'Karen' in the search bar!

annaspain

The exception for ionization energy is that Nitrogen has a greater ionization energy than Oxygen, even though the trend follows that the energy increases across a period. The other exception is for electron affinity. Carbon has a greater electron affinity than Nitrogen which does not follow the tren...

annaspain

I believe we won't be required to actually draw the models, as mentioned in lecture. You will need to know the structure name and formula (AXE) though.

annaspain

No, pi bonds are formed when there's a double or triple bond. So in a double bond, one bond will be a sigma bond and the other will be a pi bond. In a triple bond, one bond will be a sigma bond and 2 would be pi bonds.

annaspain

I think either is fine, but I've noticed the separate structure seems more common for ionic compounds like salts.

annaspain

I don't believe we will need to know the exact angles in structures with lone pairs, but just need to acknowledge that those angles are smaller than they would be without the presence of the lone pairs.

annaspain

As mentioned above, the goal is to minimize formal charge. In ions with negative charges, however, usually oxygen should "carry" the negative charge rather than the central atom, which explains why using a single bond would be more stable.

annaspain

In large organic compounds like this, the structure typically follows the same order as how the formula is written. So, you will have 2 H attached to 1 C, etc. The parenthesis signify that those elements are all bonded together. So, the N will have 2 H bonded to it and then will bond to the C that h...

annaspain

Will the bond angles for a specific molecular shape always be the same for that shape regardless of what elements make it up? For example, BF3 has a trigonal planar shape and bond angles of 120 degrees each. Will another molecule with this trigonal planar shape also have bond angles of 120 degrees?

annaspain

Sulfur is not as electronegative as oxygen. Because of this, H2S is not as polar as water and has weaker intermolecular forces and therefore a lower boiling point.

annaspain

Electronegativity describes an element's ability to pull electrons closer towards its nucleus, which explains why electronegativity increases across a period (elements "want" electrons more to reach octet) and decreases down a group (atomic radius is larger so shielding causes a looser gri...

annaspain

A node shows the region where there is 0 electron density, so therefore 0 probability of finding electrons there, as mentioned in the comment above. If you're asked how many nodes are in a configuration, # nodes = (n-1). So, in an s orbital that would be 0 nodes (spherical shape), in a p orbital (n=...

annaspain

Basically, the elements in the s, d, and f blocks (except H & He) are metals. As you get towards the right side of the table, the elements are mostly nonmetals. Between the metals and non-metals, there is a diagonal pattern of metalloids (B, Si, Ge, As, Sb, Te, Po). It's easier to see with a pho...

annaspain

The photoelectric effect describes light as having particle-like properties. Because of this, increasing intensity of long-wavelength light will not cause electrons to be ejected because each individual photon is interacting with individual electrons in the metal, so if the individual photons don't ...

annaspain

There's a website called "learnbacon" that has chemistry and biology notes on it that are very helpful!

annaspain

As shown in class, certain compounds like C2 can be represented with a combination of single, double, or triple bonds. Does it matter which way we represent it as long as we show the correct total number of electrons?

annaspain

Given that l=1, that corresponds to the p-subshell, so there are 3 possible orbitals. I don't believe the fact that it is the 2p orbital will affect the amount of orbitals that could possibly be filled.

annaspain

The 2.998x10^8 value is technically more precise, however Professor Lavelle told us that it's ok to use 3x10^8 as the value for speed of light during lecture!

annaspain

Use the equation (wavelength)= (speed of light)/(frequency), so the wavelength = (3x10^8)/(2.0x10^18), which should give you 150 pm

annaspain

Are we expected to include questions from each section of the unit (in this case, sections 1A-1F) on the homework even if the topics haven't yet been covered in lecture?

annaspain

The minimum measurable wavelength is 10^-18 nm.

annaspain

In terms of the problem-solving process, I would always use standard SI units (L) to make sure your calculations are correct. You can leave your answers in standard SI units, but you could also convert them to match the given units in the problem, as long as the answer is correct!

annaspain

All objects do exhibit some sort of wavelike properties. However, as mentioned in lecture, the objects that have wavelengths smaller than 10^-18 nm (ex: 10^-34) will not be considered to express detectable wavelike behavior.

annaspain

Given that you will be answering part a with a rounded answer for sig figs, I think it's fine to use that rounded answer in part b because your final answer for part b will ultimately be rounded for sig figs as well.

annaspain

I find it much easier to make sure my work is correct when I can have clear visuals to make sure everything is balanced. I usually keep a running tally while I'm balancing my equation for each individual element, which makes the process much easier to keep track of for more complex equations.

annaspain

angelagd3l wrote:When we are rounding, how can we determine how many sig figs we need to keep in the problem?

The final answer should have the same number of sig figs as the value given in the problem itself with the LEAST amount of sig figs.

CHEMISTRY COMMUNITY

Search found 32 matches

Re: Energy Levels

Re: how can a hydrogen bond form

Re: Cyanide

Re: Identifying sigma & pi bonds

Re: Naming/ oxidation numbers/ charges

Re: How to name

Re: Practice Questions for Final

Re: Exceptions

Re: Drawing out Molecular Shape

Re: formation of pi bond

Re: Complete Lewis Structures

Re: bond angles with lone electrons

Re: 7th Edition #2E.13

Re: Lewis Structure of H2C(NH2)COOH

bond angles

Re: Boiling Points

Re: Polarizability v. Electronegativity

Re: Node

Re: Metals, metalloids, and nonmetals

Re: Photoelectric Effect - waves vs. photons

Re: Replacement for Modules

Lewis Dot Structures

Re: Homework Question 1D.23

Re: Calculating the wavelength of light

Re: Homework Question 1A.7

Homework

Re: Measurable wavelengths

Re: Changing Units

Re: Wavelike properties

Re: Number of Sig Fig

Re: Balancing reactions tips

Re: All students read this sig fig post [ENDORSED]