CHEMISTRY COMMUNITY

Francis_Nguyen3D

You use the suffix -ate at the end of the metal name when the entire complex has a negative charge. In this problem the complex has a 2- charge which explains the utilization of the suffix -ate.

Francis_Nguyen3D

en= ethylenediamine
edta=Ethylenediaminetetraacetic acid
These are the abbreviated forms of these compounds when they are a ligand.

Francis_Nguyen3D

The coordination number for:
a) 4
b) 2
c) 6
d) 6

The coordination number for c would come from the fact that there are four bonds coming from (en) and two from cl
For d, there are six bonds which is implied through the name of the complex.

Francis_Nguyen3D

In regards to neutral ligands, are we just expected to memorize which compounds are neutral or is there a way to determine if they are neutral?

Francis_Nguyen3D

When there is a single bond there is 1 sigma bond and 0 pi bonds. When there is a double bond, there is 1 sigma bond and 1 pi bond. When there is a triple bond, there is 1 sigma bond and 2 pi bonds. Hope this helps! :)

Francis_Nguyen3D

Yes, regions of electron density do match up with the number of hybridized orbitals. For problem 2F.13, it asks for the hybrid orbitals of each carbon atom. So when you draw the lewis structure of CH2CHCN, there should be three regions of electron density around two of the three carbon atoms and two...

Francis_Nguyen3D

A larger atom would result in a weaker bond because the nucleus of the atom doesn't have a strong hold of the outer most electrons. I believe when we refer to the size of the atom, we discuss its molecular mass, atomic number, and atomic radius interchangeably but when discussing bonding, I think at...

Francis_Nguyen3D

The smaller cation has higher polarizing power because of its electronegativity. Because of its small size and the fact that it wants electrons, the cation is able to hold onto electrons better. So overall, the smaller the atomic/ionic radius, the more polarizing power the atom/cation has. (charge a...

Francis_Nguyen3D

I believe that is correct!
regions of electron density around the central atom= number of hybrid orbitals

Francis_Nguyen3D

As electronegativity decreases, polarizability increases.

Francis_Nguyen3D

Hello, the bold and dashed lines within the lewis structure represent the 3D position of an atom within the molecule. When it is a bold line, that means that the atom is more towards the foreground. When it is a dashed line, that means that the atom is more towards the background. I believe you can ...

Francis_Nguyen3D

If this problem is from the 7th edition textbook, the correct answer is [Ar]3d^10 because the question asks for the electron configuration of Cu+ not Cu^2+.

Francis_Nguyen3D

I believe that it would be best to use the Lewis Structure with the lowest energy and with octet exceptions because it would give a more accurate representation of the shape of the molecule. If we were to ignore formal charges and octet exceptions, we might miss a crucial double bond that could help...

Francis_Nguyen3D

The first four elements also have octet exceptions where they do not obtain an octet of electrons. This is due to the fact that the first four elements are stable/feel complete with the amount of electrons they currently have. Another exception would consist of elements such as Boron and Aluminum wh...

Francis_Nguyen3D

I believe you can determine whether a molecule is color or nonpolar based on its charge distribution. When you draw the lewis structure, if there is symmetric distribution of charge, then the molecule would be nonpolar. If you draw the lewis structure and there is an uneven distribution of charge, t...

Francis_Nguyen3D

I believe the reason why Hydrogen Bonds are stronger than ion-dipole interactions is because Hydrogen bonds involve greater differences in electronegativity. This can then create a large electric dipole of positive and negative charges.

Francis_Nguyen3D

I know this is a pretty simple question but, how do we indicate resonance structures? Can we just draw the structures next to one another or is there a more formal indication? I think one way to indicate resonance structures is to draw out one lewis structure. Then from there see if 1. the structur...

Francis_Nguyen3D

I believe the exceptions we have learned so far consists of: 1. Electron deficiency- which is when there is less than an octet, therefore causing a fourth bond to be made. For example: BF3, since Boron is short one bond to complete its octet, it would seek out another F atom to form BF4-.This would ...

Francis_Nguyen3D

Bond lengths could be the same because of resonance. Resonance molecules do not have one true lewis structure, but rather a variety of structures. Because of this, the resonance lewis structure of a given molecule would be a hybrid of all the possible structures that could be created. This would mak...

Francis_Nguyen3D

Hello,
when you do the electron configuration diagram for 3d^8 three orbitals would be filled out of the five orbitals available. Two orbitals would have only one electron in them-making them unpaired.
https://www.adichemistry.com/jee/qb/coordination-chemistry/1/q1-3.png

Francis_Nguyen3D

Hello,
the S orbital is not degenerate because its l value= 0 therefore making the m1 value=0. This makes the s orbital have 0 degeneracy.

Francis_Nguyen3D

Hello,
the central atom is usually the atom with the lowest ionization energy.

Francis_Nguyen3D

Hello,

this was posted on the Chem 14A website under announcements : Midterm covers: Fundamentals, Quantum, Bonding to end of 3.11 (6 Ed.) and to end of 2C (7 Ed.)

Francis_Nguyen3D

Hello,
this would be considered as an excited state because it does not follow Hund's Rule. The second electron should have filled in the next orbital rather than pairing up in the first one, if it followed this rule.

Francis_Nguyen3D

It looks like you're correct. Dr. Lavelle has 1.45 on his list of solution manual errors.

Here is the link:https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Solution_Manual_Errors_6Ed.pdf

Hope this helps!

Francis_Nguyen3D

Use the equation Energy= hc/(wavelength). Then use the energy of the photon you got from part b, which should be 2.25 x 10^-17J. Rearrange the equation so that you are solving for wavelength so it would be: wavelength= hc/energy Then plug in 6.626 x 10^-34 for h and 3.0 x 10^8 for c and divide by 2....

Francis_Nguyen3D

For this problem use the equation: Energy= hc/(wavelength). Then use the energy of the photon you got from part b, which should be 2.25 x 10^-17J. Rearrange the equation so that you are solving for wavelength so it would be: wavelength= hc/energy Then plug in 6.626 x 10^-34 for h and 3.0 x 10^8 for ...

Francis_Nguyen3D

I believe Dr. Lavelle said that the cut off would be 10^-15 and anything smaller than that is undetectable.

Francis_Nguyen3D

The number of photons in a light beam affects the brightness of the whole beam, whereas the frequency of the light controls the energy of each individual photon. In the Photoelectric effect electrons are ejected from a metal surface based on the amount of energy absorbed. So, a certain threshold has...

Francis_Nguyen3D

Planck proposed the idea that light and other forms of electromagnetic waves were released in certain packets of energy. This became known as "Quanta" because quanta by definition, is a discrete amount of energy proportional between its magnitude and frequency. So, this is known as Planck'...

Francis_Nguyen3D

So for a theoretical yield, it is the amount(grams) of product that you receive through calculations considering that the conditions of the experiment are perfect. One way to think about theoretical yield is the amount you calculated on paper. As for the actual yield, it is the amount(grams) of prod...

Francis_Nguyen3D

For calculations, use the given molar mass on the periodic table without rounding. Round based on the amount of sig figs at the end of your calculation.

Francis_Nguyen3D

Typically, it is best not to round the atomic weights of the elements before using it for calculations because your final answer would not be as accurate as it could be. Usually, you would round at the end of your calculation depending on the amount of significant figures there are. You determine th...

Francis_Nguyen3D

When given the name of the compound, I usually like to write out the symbols of the elements. So in the case of this problem, start off with Magnesium. Its symbol would be Mg and then move onto Sulfate. Since Sulfate is not on the periodic table, try utilizing a common ions chart where you will find...

Francis_Nguyen3D

(1) : Balance N first because it appears the less within the equation. 2C10H15N + O2 --> CO2 + H2O + CH4N2O (2): Balance H, there are 30 H on the left side and a total of 6 H on the right, leaving the (4 H) alone, that would be 30-4= 26/2= 13 as your coefficient for H2O 2C10H15N + O2 --> CO2 + 13H2O...

CHEMISTRY COMMUNITY

Search found 35 matches

Re: When to use -ate

Re: "(en)" and "(edta)"?

Re: 6th Edition 17.37

Neutral Ligands

Re: 2F.17 7th ed: Pi bond

Re: Electron Density and Hybridization

Re: Atom Size and Bond Strength

Re: higher polarizability

Re: Regions of electron density

Re: Electronegativity and Polarity

Re: Bold/Dashed Lines in Bonds

Re: Homework Focus 2A

Re: Lowest Energy Lewis Structures

Re: octet exceptions

Re: VSEPR Model

Re: Hydrogen Bonds vs. Ion-Dipole Bonds

Re: Resonance Structures

Re: List of octect exceptions

Re: bond length

Re: Number of Unpaired Electrons

Re: Degeneracy

Re: central atoms

Re: Midterm

Re: Ground vs Excited State

Re: Exercise 1.45, 6th Edition

Re: Wavelength Calculation

Re: 7th Edition 1B.15

Re: Wave-Like Properties Value

Re: electromagnetic radiation

Re: Packets of Energy

Re: Percent Yield [ENDORSED]

Re: Rounding molar mass on the periodic table

Re: Periodic Table

Re: Question E.9: Epsom Salts

Re: Question H21 (6th Edition)