CHEMISTRY COMMUNITY

AVerma_F19

generally carbon will always have 4 bonds in an organic compound, nitrogen will have 3+ a lone pair or 4. Oxygen would have a double bond and 2 lone pairs unless a hydrogen is drawn

AVerma_F19

The hydrogen in question must be bound to a Nitrogen, Oxygen or Fluorine. This is so that the electron density is almost completely pulled away from the hydrogen, essentially having it become an exposed proton that can bind to any lone pairs. The lone pairs, however, can come from any atom (typicall...

AVerma_F19

A chelating ligand is one that binds to a cation in more than one location, thus it is polydentate.

Some of the common polydentate ligands are

oxalate (bidentate)
ethylenediamine (bidentate)
diethylenetriamine (tridentate)
ethylenediaminetriacetate (EDTA - hexadentate)
porphyrin (tetradentate)

AVerma_F19

I'd know the common ones for the anion coordination compound naming:

iron (ferrate)
copper (cuprate)
chromium (chromate)
silver(argenate)
gold (aurate)

AVerma_F19

K+ does not influence pH since it is the conjugate base of a strong base (it is highly stable and doesn't need to take a proton).

By process of elimination, ClO- is the conjugate acid of a weak base. Thus, it is able to accept a proton.

AVerma_F19

The second name you said, potassium tetracyanonickelate (II) is the correct naming.

AVerma_F19

For (c), en is bidentate (since it forms a ring and connects to the cation 2 times. Since there are 2 en's, the en's contribute a total of 4 bonds to the TM cation. Then, the 2 chlorides provide 2 bonds that bring the coordination number to 6. For (d), if you look at the structure of EDTA, it has 2 ...

AVerma_F19

A lewis acid is an electron pair acceptor. A lewis base is an electron pair donor. A bronsted acid is a proton donor. A bronsted base is a proton acceptor. SO4 does not form H2SO4 in water; however, H2SO4 dissolved in water is a strong acid.

AVerma_F19

I believe that transition metals are a Lewis Acid, but not a Bronsted acid.

AVerma_F19

The H2O acts as a weak acid, since it is a proton donor. It's a coordinate covalent bond since both electrons are provided by the O2-.

AVerma_F19

It basically gives you the ratio between the (product of the concentration of each of the products)/(product of the concentration of each of the reactants). Using this, you can calculate the concentration of the reactants or products using the kA given.

AVerma_F19

They are able to protonate/de-protonate biological molecules like DNA, proteins, enzymes, etc. which many times inactivates their function. Additionally, many biological systems are pH sensitive, meaning that proteins will fold differently/act differently under different pHs.

AVerma_F19

Pi bonds do not affect hybridization. They are generally p orbitals that have not hybridized that are overlapping side by side.

AVerma_F19

I believe that since Iodine is larger than the Bromine, the induced-dipole/induced-dipole forces would cause the IMFs in the HI to be stronger than those in the HBr molecules.

AVerma_F19

There is no need for the orbitals to hybridize since they each have 2 unpaired electrons in the s orbital. Hybridization is observed when there are more sigma bonds than the number of unpaired valence electrons in the ground state of the central atom.

AVerma_F19

Dr. Lavelle has said that the arrows should point towards the partial negative.

AVerma_F19

So, because different atoms have different sizes, the angle is never truly/exactly 109.5 degrees unless all 4 atoms surrounding the central atom are the same. However, this phenomenon has a much smaller impact on bond angles than the presence of a lone pair, since lone pairs are untethered at one en...

AVerma_F19

You cannot, since the C-H and the C-Cl bonds are NOT equally polar. Additionally, this molecule is tetrahedral, so unless all 4 bonds are of the same polarity, the dipole moments don't cancel out.

AVerma_F19

Molecules/atoms that are bigger have more electrons, which means that the induced-dipole/induced-dipole forces would be stronger since the electron cloud can be more distorted, which creates a stronger partial charge.

AVerma_F19

I think that is exactly the kind of question that we would be asked on a test.

AVerma_F19

It would just be linear.

AVerma_F19

We haven't learned this yet in class; however, it refers to sp hybridization. Typically, a single bond is an s bond. A double bond requires the contribution of the s and p orbitals. A triple bond requires the contribution of 1 s orbital and 2 p orbitals.

AVerma_F19

They are very strong intermolecular forces that give molecules like H2O special properties.

AVerma_F19

Antioxidants are able to give a single e- to a free radical without becoming destabilized themselves.

AVerma_F19

Yes it does. They both refer to similar properties about an atom's electron-nucleus relationship.

AVerma_F19

In a perfectly covalent bond, the electrons are perfectly shared between the two atoms. However, this is never the case unless the two atoms are the same, since there will be differing electronegativities between two different element's atoms. When an atom is more electronegative than the other, it ...

AVerma_F19

Yes, you would likely have to draw/imagine the shape of a molecule and consider the dipole moments within it. For example, CH4 is symmetrical, so the dipole moments between C-H bonds cancel. However, H2O is not, and this can be seen by the strong dipoles in H-O bonds.

AVerma_F19

Yes this is correct.

AVerma_F19

It's because elements in Period 3 and higher have available space in the d shell to accommodate more electrons.

AVerma_F19

The FC charge equation gives a FC of 0 when you look at these structures. FC = # of valence e - (# of lone pair e + # of bonding/2)

AVerma_F19

I think we just need to know the general properties of all the trends.

AVerma_F19

a bond is 2 electrons.

AVerma_F19

I imagine there would be an experimental set up where we were to determine how an electron/photon will act in a certain situation as a wave/particle. Or, we could be asked to use one of the equations (De Broglie, etc) to calculate energy of a particle.

AVerma_F19

I believe you're right, since the partial charges are from covalent bonds and formal charges from elements within the compound.

AVerma_F19

I believe that they generally do. To have an FC of 0, they need an expanded octet.

AVerma_F19

Firstly, the position uncertainty that you would enter into your calculation would be 10 meters. Since the car has such a large mass, and since the speed uncertainty is on the order of 10^-39, this calculation is unhelpful since we can measure the speed of the car quite accurately. This calculation ...

AVerma_F19

The first and second electron have an energetic relationship, meaning that there is some e-/e- repulsion between the 2. When the first is removed, that repulsive force disappears, leaving the second electron more tightly bound to the nucleus. Additionally, since the nucleus has less electrons to hol...

AVerma_F19

I believe the ground state of an ion is essentially the element in its normal electron configuration (without charge or excited electrons).

AVerma_F19

You should write it from lowest to highest energy.

AVerma_F19

Yes you are correct. And going down the table means that you get more energy levels.

AVerma_F19

Yes, he hasn't covered this yet but some of the elements in Periods 3 and down can have something called an expanded octet. I think he will go over this on Friday.

AVerma_F19

Yes, they do. They have a small enough mass where theycan exhibit wave like properties via the De Broglie equation.

AVerma_F19

Δp = m Δv is the equation to relate Δv to Δp. I believe that the Δ accounts for +/-.

AVerma_F19

In class, Prof. Lavelle used a best-case scenario and maintained an "=" in his subsequent calculations; however, I don't believe it would be wrong to include the ">"

AVerma_F19

Schrodinger's equation is taught to us to demonstrate how we "define" electrons. Because of Heisenberg's uncertainty principle, we can never truly know where an electron is, and Schrodinger's equation materializes the properties of electrons as the solutions to the equation.

AVerma_F19

I believe that there is a set number of orbitals based on the subshell. s has 1 orbital, p has 3, d has 5 and f has 7. Each orbital holds 2 electrons, with an opposite spin.

AVerma_F19

Psi^2 represents something called a probability density. Probability density is DIFFERENT from probability in that probability density has a unit (volume^-1). When you take psi^2, you can multiply it by the volume in question (for example: the volume of an atom) to get the probability of finding the...

AVerma_F19

Using the equation you mentioned (KE(electron) = E(photon)-Threshold Energy), the speed of the photon is not in question. Photons (and all radiation) will travel at 3*10^8 m/s. However, depending on how much energy the photon has, the KE of the electron (not radiation) will change via velocity (sinc...

AVerma_F19

The speed of radiation does not decrease with increased frequency. c = Frequency*Wavelength, where c is the speed of light (3*10^8m/s). Since c is constant, frequency and wavelength are inversely proportional. This means that if frequency goes up, wavelength will go down, and vice versa.

AVerma_F19

Do the emission and absorption spectra of atoms always correspond?

AVerma_F19

Think of the energy levels like hitting a golf ball into the hole. If you hit the ball too lightly, it won't go into the hole. If you hit the ball too hard, it again won't go into the hole. If you miss, you have to bring the golf ball back to try and hit it again. Similarly, an electron will only mo...

AVerma_F19

Using this equation, the energy you get is the energy of 1 particle - whether it be a photon or electron. Therefore, if you wanted to find the amount of energy to remove 1 mole of electrons, you would need to calculate the energy needed for the removal of 1 electron, and then multiply by avagadros n...

AVerma_F19

I believe that 1 photon is always released per energy change. This can be observed by the fact that light of just ONE frequency can cause an electron to jump 2 energy levels.

AVerma_F19

I think that photons are considered massless packets of energy, so I'm not sure whether the same rules would apply.

AVerma_F19

I agree with the above answers; it must have something to do with the molar mass calculations.

AVerma_F19

The net ionic reaction essentially means that you only include atoms/molecules that do NOT remain in an aqueous state after the reaction has completed.

AVerma_F19

Exactly, since H2O and the compound are the only two molecules present in the solution.

AVerma_F19

I believe these would be the lowest whole number ratios that you get--essentially the empirical formula.

AVerma_F19

Technically, you would only need to do 5 problems by this Friday; however, I did many more to prepare myself for the tests, etc.

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