CHEMISTRY COMMUNITY

Sebastian Lee 1L

For 1E, I thought increasing the concentration of Ag+ on the cathode side would increase the cell potential. In the Nernst equation for a concentration cell where Q=[anode]/[cathode], increasing the cathode concentration will decrease Q which makes the term more negative which means a more positive ...

Sebastian Lee 1L

The frequency factor tells us about the reaction mechanism and angles in which the reactants collide. It kind of indicates the ratio of collisions that occur at the correct angle to form products. This is useful in the Arrhenius equation lnk=lnA-\frac{E_{A}}{RT} when you want to find the activation ...

Sebastian Lee 1L

I assume you are referring to the derivation of the Arrhenius equation in which ln\frac{k2}{k1}=\frac{A_{E}}{R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}] . It doesn't matter what T1 or T2 are as long as they correspond to the correct k1 and k2. So if you want to find what the rate constant is at a higher tem...

Sebastian Lee 1L

My TA emailed us and said she could hand them back sometime Friday in person for anyone who would want it. I would wait to see if your TA says anything about it.

Sebastian Lee 1L

I think this has to do with the unique rate and the differential rate law. Remember that the rate of change of reactants and products relative to each other depends on their stoichiometric coefficients. For example in the reaction 2A\rightarrow B , the reactant A is used up twice as fast as the prod...

Sebastian Lee 1L

It doesn't really matter if you use kJ or J BUT you must use the same units in your equation. For example, if $\Delta H$ is given in kJ/mol but $\Delta S$ is given in J mol^-1 K^-1, you HAVE to convert either kJ to J or vice versa to have consistent units.

Sebastian Lee 1L

Just to expand on the previous comment, you can generalize the rate constant's units as \frac{1}{M^{n-1}s} where n is the order of the reaction. So for a zero order reaction, the units would be M/s. For first order, it's just s^-1. For second order, it's \frac{1}{Ms} , etc. The rate constant's units...

Sebastian Lee 1L

Thank you Lyndon for all of the work, time, and dedication you have done in helping all of us learn, succeed, and grow in these past two quarters. You have been an absolutely wonderful, smart, and fun UA and your review sessions have helped all of us improve our understanding and skills. I know you ...

Sebastian Lee 1L

There actually is a set where the concentration of B changes but A and C don't. Look at experiments 1 and 3. For this, you would divide the final over initial concentration of B and divide the final over initial reaction rates. You should find that B changes by a factor of about 2.4 while the rate c...

Sebastian Lee 1L

Under standard conditions, the combustion of gas is most likely spontaneous since you would usually be releasing heat (-delta H) and increasing entropy (+delta S). However, I wouldn't say that combustion is always spontaneous, especially depending on non-standard conditions. For example, if you have...

Sebastian Lee 1L

For a galvanic cell to work and provide a spontaneous current, you do NOT need both (or any) half-reactions to have a positive standard reduction potential. Remember that for these problems you want to find the E cell of the OVERALL reaction. If the cathode has a higher reduction potential than the ...

Sebastian Lee 1L

When working with a pH electrode meter you have the following equation: E=E^{\circ}-\frac{RT}{F}ln[H^{+}] . This is derived from the Nernst equation where in this case, Q is just the concentration of H+ because in buffers, [HA] and [A-] are the same. From here, you know that you are given the cell p...

Sebastian Lee 1L

\Delta G^{\circ} refers to the change in Gibbs free energy of a reaction under STANDARD conditions (298K and 1 atm/barr or 1M). \Delta G is unique to a reaction under certain conditions that are NOT STANDARD like a different temperature or concentrations. The equation of \Delta G=\Delta G^{\circ}+R...

Sebastian Lee 1L

To study the reaction given, you have to use both oxidation and reduction half-reactions to end up with AgBr (s) from Ag+ and Br-. If your Ag+ is being reduced to form Ag (s), you know that there must be another oxidation half-reaction that uses Br- to form your desired AgBr(s) (not just Br). If you...

Sebastian Lee 1L

So from the cell diagram, we know that the anode on the left is the oxidation half-reaction and the cathode on the right is the reduction half-reaction. For the oxidation half-reaction we have two species they give you, H 2 (g) and HCl (aq). In this case, since HCl is aqueous, you can think of it as...

Sebastian Lee 1L

Remember that you can only use one method to find the E cell . If you use the equation E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode} , you do NOT flip any signs and only use the reduction potentials as given in the chart. In your example, you would find the reduction potential of NO 3 - (0....

Sebastian Lee 1L

The rate-determining step (aka the slow step) can be identified using the rate law given. For example, in class we were given a rate law of Rate = k [NO 2 ] 2 . In this case, we would know that the rate-determining step is the one that leads to that rate law, which in this case is the bimolecular co...

Sebastian Lee 1L

In this case, the given reaction is the overall cell reaction and doesn't seem to have any change in oxidation numbers for reactants and products. However, the oxidation and reduction half-reactions will. This might not be the best way to determine the half-reactions, but I honestly just looked at t...

Sebastian Lee 1L

I think what he was trying to say was that when plotted, the ln or log of a variable will appear linear when graphed, but in reality, the values are exponentially related. So even though the value for the logQ may be only 1 higher, the actual value for Q will be 10 times as much. If log is 2 higher,...

Sebastian Lee 1L

Hmm, I'm not sure but the solutions manual includes NO not O. The cell diagram you should get is Hg(l)|Hg₂²⁺(aq)||NO₃^-(aq), H+(aq)|NO(g)|Pt(s). This should make sense since the anode is where Hg is oxidized into Hg₂²⁺ and the N in NO₃^- is reduced from +5 to +2 in NO.

Sebastian Lee 1L

Why is E^{\circ} = 0? In what scenarios is E^{\circ} = 0? In this problem, there are actually two reasons why E^{\circ} is 0. Remember that the standard cell potential for a galvanic cell is equal to the standard reduction potential of the cathode minus the standard reduction potential of the anode...

Sebastian Lee 1L

n should be the moles of electrons that are transferred in the redox reaction. In the problem, you are given the cell diagram and can figure out the oxidation (anode) and reduction (cathode) reactions and how many electrons are involved in each (should be the same when balanced). For part a, you hav...

Sebastian Lee 1L

The reason why the left side (anode) doesn't have the platinum electrode is because Hg (l) is a special case in that it can also act as a conductor. In this case, a wire could be inserted into the mercury which would be able to become oxidized, losing two electrons and sending a current through the ...

Sebastian Lee 1L

You can refer to the equation q=mC\Delta T to think about this problem. From this equation, we know that a greater q means a greater change in temperature and that a larger mass means a lower change in temperature. As the moles of HCl and NaOH react, the formation of water releases energy as heat (q...

Sebastian Lee 1L

\Delta G^{\circ} refers to the Gibbs free energy of a reaction at the STANDARD state whereas \Delta G refers to the Gibbs free energy under other non-standard conditions. For example, you might want to find \Delta G when the reaction is occurring at a different temperature T or when it's not at equ...

Sebastian Lee 1L

3B: You can refer to the equation q=mC\Delta T to think about this problem. From this equation, we know that a greater q means a greater change in temperature and that a larger mass means a lower change in temperature. As the moles of HCl and NaOH react, the formation of water releases energy as hea...

Sebastian Lee 1L

I believe the i represents current. In a galvanic cell, you have the maximum potential difference as the current approaches 0. This should make sense since if you have more current, then the difference in potentials will decrease at a faster rate since the charges are slowly becoming neutralized (he...

Sebastian Lee 1L

In these cases, you would have two separate half-reactions with the same reactant for both the oxidation half-reaction and reduction half-reaction. In the case of Br2 that you gave, the oxidation half-reaction would be Br_{2} (l)+12OH^{-}(aq)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2...

Sebastian Lee 1L

In this problem, you are solving for the change in gibbs free energy of the reaction by adding/subtracting standard gibbs free energy of formation of the products and reactants. The equation is \Delta G_{rxn}^{\circ} = \sum \Delta G^{\circ}_{f}(products)-\sum \Delta G^{\circ}_{f}(reactan...

Sebastian Lee 1L

Entropy can be thought of as the possible number of positions or orientations a molecule can occupy (these are called microstates). Basically, if there is more space to occupy, more possible configurations, more thermal movement, etc. then there is more entropy. (Honestly you can still think of it a...

Sebastian Lee 1L

If you are using moles (n), then C will be the molar heat capacity. Make sure that in problems, if you're only given the molar heat capacity, that you change any mass into moles. Specific heat capacity will only work for mass (g).

Sebastian Lee 1L

The standard entropy of a reaction will generally increase (+delta S) if you go from solid to liquid/gas or liquid to gas because you are going from more ordered, confined molecules to a lot more space and more possible states for the molecules. There's more space and movement in a gas than a solid ...

Sebastian Lee 1L

Yep! On the equation sheet on the cover of the test there should be the equations for the molar heat capacities of monatomic ideal gases at constant pressure and constant volume. I believe that Cp = (5/2)R and Cv = (3/2)R.

Sebastian Lee 1L

Typically, if you are given a mass of molecules in a reaction, you will use the specific heat capacity because q=mC_{s}\Delta T where m is mass (in grams) and C is the specific heat. However, if you are only given the molar heat capacity you will need to convert the mass to moles so that it satisfie...

Sebastian Lee 1L

Adding an inert gas will increase the pressure of a system. However, it does NOT affect the equilibrium of a reaction because there is no change in concentration which means Q does not change. Adding an inert gas will not shift a reaction towards products or reactants.

Sebastian Lee 1L

Knowing that q=-w when the change in internal energy is 0 just helps you solve problems that may require work (but you're only given heat) or vice versa. You know that the work done by a system expanding isothermally is the opposite sign of the heat coming into the system. Note that the internal ene...

Sebastian Lee 1L

For ideal gases, you might want to know that for an atom Cp=(5/2)R and Cv=(3/2)R. I believe the first one is given on the equation sheet.

Sebastian Lee 1L

I think what your TA meant is that substances in their standard state in their most stable form have a standard enthalpy of formation of 0. The most stable form of an element is one that only contains that element in its natural state. For example, oxygen's most stable form is O2 gas so the enthalpy...

Sebastian Lee 1L

All this means is that in the Boltzmann Equation S=k_{B}lnW , if you have a large error in W (degeneracy), you're not going to have much of an error in S (entropy). So you might have a large difference in values you get for W, but once you take the natural log and multiply by the Boltzmann constant,...

Sebastian Lee 1L

I also had a question on this problem. I get how to do the problem using the entropy equations when temperature changes. But why is it that you can cool the water vapor back down to 85 Centigrade and still have it be in the vapor state? I assume it's in the vapor state since the molar heat capacity ...

Sebastian Lee 1L

I believe you need to convert the temperature to Kelvin. This is because the units for entropy are Joules per Kelvin. So $\Delta S=\frac{q}{T}$ will only give you the right units for entropy if you use Kelvin not Celsius.

Sebastian Lee 1L

So for this problem you're actually given the enthalpies of combustion not the enthalpies of formation (the 3rd method we discussed in class). Basically what the enthalpy of combustion means is the amount of heat given off/absorbed when a mole of that molecule reacts in a combustion reaction. For ex...

Sebastian Lee 1L

In this problem, you need to find equations for both the heat change of the ice and the water to find the final temperature. The heat change (q) of the ice involves both the enthalpy of fusion (what you're confused about) and the heat needed to increase the temperature of the melted ice. \Delta H_{f...

Sebastian Lee 1L

Yep! The change in enthalpy is +765J since $q=\Delta H$ under constant pressure.

Sebastian Lee 1L

Just as an extra note: If the system were to be compressed, work would be done on the system by the surroundings. In this case, the sign for work would be positive since it is gaining energy from the surroundings. You'd still use the same equation of w=-P\Delta V ; it's just that if compressed, delt...

Sebastian Lee 1L

Sometimes a calorimeter can be considered a surrounding but usually the problem will make that clear. For example, if you look at problem 4D.3, the calorimeter absorbs heat as a result of the exothermic reaction between gases. In that case, you would consider q (heat) as the transfer in energy betwe...

Sebastian Lee 1L

An isolated system has no transfer of matter or energy.
From the equation $\Delta U=q+w$ , it should make sense that the change in internal energy of an isolated system will be 0 because there is no heat transfer (energy) and no work being done (also energy).

Sebastian Lee 1L

A reversible expansion is basically one that takes place in really small, slow steps. Conversely, an irreversible expansion is one that happens very quickly. I'll give examples of both. So for both processes, you can think of a beaker in which a piston is compressing a gas at say 2 atm. In an irreve...

Sebastian Lee 1L

Here's a link to a rough visual I created: https://ibb.co/XC7dP1d
Hopefully it can clarify the last part of my post :)

Sebastian Lee 1L

To answer your first question: in this case, the system that does more work will have the more negative work (or a greater absolute value). Remember that our equation for work resembles w=-P\Delta V . If the volume changes more, then more work is done and the value is more negative. We make the sign...

Sebastian Lee 1L

Enthalpy doesn't necessarily mean a change in temperature; it's just a measure of the heat energy being released/absorbed. For example, when water is undergoing a phase change to become vapor, there is a positive \Delta H value because heat is being applied for vaporization BUT the temperature will ...

Sebastian Lee 1L

I doubt the solutions would be posted anywhere. I don't believe they're on the chem 14a website.

Sebastian Lee 1L

I believe Lavelle was talking about how one can combine a weak acid and a salt solution containing its conjugate base to make a buffer. (The same applies with a weak base and its salt with the conjugate acid). A buffer is a solution that can resist changes in pH and balance the pH if an acid or base...

Sebastian Lee 1L

Yeah, I think this is just an error in the solutions manual. The values you put in the quadratic equation are correct based on the ICE table and Kc. The solutions manual says that the values for x are 9.2 and .071 BUT you should've gotten 8.7 and .074. In this way, solving for the equilibrium concen...

Sebastian Lee 1L

Again, in addition to the small (<10 -3 ) approximation, you can find the percent ionization to test if you can use the approximation. If the percent ionization is less than 5%, your approximation is acceptable. Remember that to find percent ionization, you take the equilibrium concentration of the ...

Sebastian Lee 1L

If you took Lavelle for 14A, we did cover the transition metals acting as acids in salts briefly. Remember that transition metals like aluminum can form coordination compounds with ligands including water when in solution. There are 6 available bonding sites in an octahedral arrangement where the al...

Sebastian Lee 1L

Like others have said, acid and base reactions are not always at equilibrium. However, it is important to note that when we do calculations for pH, pK a , and pK b , these values will be based on equilibrium concentrations. This is especially important when considering weak acids/bases' dissociation...

Sebastian Lee 1L

To add on to previous comments, since weak bases do not fully dissociate into their respective ions, you will have concentrations of both reactant and product at equilibrium. This is the basis for K a which is equal to the concentration of the dissociated products over the remaining concentration of...

Sebastian Lee 1L

I used the information in table 5G.2 to help figure the problem out. If you look at that table, you'll find the equilibrium constant for the reverse reaction ( N_{2}O_{4} (g) \rightleftharpoons 2NO_{2} (g) ) at 298 K. From here you should be able to figure out K for the given reactio...

Sebastian Lee 1L

Yes, it is indeed possible to have pH outside of the range from 0 to 14. For example, if you had a solution of 10M HCl (a strong acid), you would find the pH = -log(H+) to be -1. Of course, this would be a tremendously strong acid solution. Likewise for very strong bases, if you had a solution of 10...

Sebastian Lee 1L

To do this problem, you would use the same concept that we used when finding the concentration of H 3 O + and OH - at the normal room temperature of 25C. Remember that if the solution is neutral, that means that both concentrations of H 3 O + and OH - are equal. The problem gives us that K w is 2.1 ...

Sebastian Lee 1L

when a question simply asks for "the equilibrium constant" or "K" does that mean related to concentration or pressure? and why is this so easily interchangeable? I believe that when a problem asks you to find the equilibrium constant, unless it specifies to find Kc or Kp, you sh...

Sebastian Lee 1L

So, I know that solids and liquids are not supposed to be included in Kc but I was wondering if Aq is excluded as well. Isn’t Aq liquid too? Also, I still don’t really understand why these things are excluded from the expression so can someone please explain to me why. Aq is included in the equilib...

Sebastian Lee 1L

In lecture, Dr. Lavelle had on the slides that if Q<K, then [R]>[P] and the forward reaction is favored. I understand that when Q is less than K that means the ratio of products over reactants is less, signifying that there are more reactants (and less products) at that time than there would be at e...

Sebastian Lee 1L

Could someone please explain how to get part a? I keep getting 13.9 instead of 6.9 Alright I think I know where you went wrong. When you're finding the reaction quotient Q c for an equilibrium reaction with concentration you MUST make sure that you are using the correct concentrations. In the probl...

Sebastian Lee 1L

For heterogeneous equilibria, you will use the same general equation to find the equilibrium constant as for homogeneous equilibria: K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b} . However, in heterogeneous equilibria, you will NOT include any solids or pure liquids in the formula. I believe this is because t...

Sebastian Lee 1L

I believe what is meant by dynamic equilibrium is that when a reaction is in equilibrium, there still are reactions occurring. It is not a static system. When a reaction is at equilibrium, concentrations of reactant and product are the same because the reactants are forming products at the same rate...

Sebastian Lee 1L

I think bond order just means is it single, double, or triple bond. So the bond order for Cl₂ is 1 since it's a single bond.

Sebastian Lee 1L

please quote my question so i get a notification, thanks in advance! i know there is something wrong with my thought process, but i cannot figure it out. my thought process is that if a base is weaker, less oh- will be dissociated out. taking the -log of that smaller number will result in a smaller...

Sebastian Lee 1L

Yep, HOF is the stronger acid! When the acids dissociate and donate their proton, you will be left with the anions, FO - and ClO - . Since F is more electronegative it will be able to better delocalize the negative charge of the Oxygen by pulling the electrons closer to it. This delocalization/induc...

Sebastian Lee 1L

For reference: 8b) A newly designed laser pointer with a certain frequency is pointed at a sodium metal surface. An electron is ejected from the metal surface with wavelength 1.10 nm. What is the frequency of the light from the laser pointer? The work function of sodium is 150.6 kJ∙mol-1. I'll just ...

Sebastian Lee 1L

Recall that hydrogen bonds can form when a hydrogen that is bonded to a Nitrogen, Oxygen, or Fluorine is strongly attracted to a Nitrogen, Oxygen, or Fluorine with a lone pair. In problems where it asks you to find how many hydrogen bonding sites there are, you should look for the hydrogens bonded t...

Sebastian Lee 1L

HOCl and HOI are both oxoacids and act quite differently, because of the oxygen involved in their structure. Since the electrons are withdrawn (or pulled due to the high electronegativity of oxygen and chlorine), they are delocalized and are more prone to get rid of the H+ proton than HOI. But when...

Sebastian Lee 1L

I think we should know conjugate acids/bases because they are related to the strength of acids/bases and how salts can or cannot affect pH of a solution. The conjugate base of an acid is basically the product once the acid donates the proton (assuming Bronsted acid). For example, the conjugate base ...

Sebastian Lee 1L

The interaction potential energy equation says that the energy/strength of an intermolecular force is directly proportional to the polarizability of the two molecules and inversely proportional to the distance between them. You don't need to do any calculations with this but you should be able to co...

Sebastian Lee 1L

The conjugate base of a strong acid will not affect the pH because it is weak and stable. For example, if you have a Cl- anion in a salt, that is the conjugate base of HCl, a strong acid. Remember that strong acids will completely dissociate into ions so for HCl: HCl\rightarrow H^{+} + Cl^{-} . Cl- ...

Sebastian Lee 1L

To add onto Benjamin's post: https://slideplayer.com/slide/6819688/23/images/9/Acidity+of+Aqueous+Transition+Metal+Ions.jpg For problem e, the Aluminum cation will form those 6 bonds to water forming a coordination compound. Metal cations will act as acids because the strong positive charge of the m...

Sebastian Lee 1L

Actually, the sig figs for pH are a little different than normal. Essentially, if you are calculating the pH from a given concentration, the number of sig figs in the concentration is the number of sig figs you put AFTER the decimal point in the pH. For example, if I was given a 3.14 * 10 -2 M solut...

Sebastian Lee 1L

For the calculations we will be doing in 14A, this is the value for K w . All this means is that in water at standard temperature and pressure, the concentration of hydronium ions [H 3 O + ] times the concentration of hydroxide ions [OH - ] is 1 x 10 -14 . This tells us that pH + pOH is 14. However,...

Sebastian Lee 1L

What Dr. Lavelle was talking about was how much an acid will dissociate depending on the acidity of the solution it is in. Remember that for Bronsted acids, the acid in solution can dissociate into hydronium ions (H 3 O + ) and the conjugate base (A - ). For example, HF \rightleftharpoons H^{+} + F^...

Sebastian Lee 1L

From my understanding, the second solution in part b is diluted relative to the desired solution. Therefore, to find the pH of the diluted acid, you need to find the new concentration/molarity. Using the equation M_{1}V_{1} = M_{2}V_{2} where M1 is .250, V1 is .2000, and V2 is .2500, you can find th...

Sebastian Lee 1L

In coordination compounds, a polydentate ligand is a single molecule that can bind to the same central transition metal at multiple sites. This means that it likely will have multiple atoms with lone pairs that can create coordinate covalent bonds with the metal. For this type of question, it's help...

Sebastian Lee 1L

Seeing that there are assigned homework problems that are similar to that, I would be prepared to answer those if they do appear on the final.

Sebastian Lee 1L

For the final, you should know the ligand names listed in the document Dr. Lavelle has provided: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf I believe the names in the textbook are the new IUPAC ones, but we only need to know how to name coordina...

Sebastian Lee 1L

The roman numeral indicates the positive charge of the transition metal cation. This notation is often used since transition metals can take on multiple oxidation states. For example, to distinguish Fe 2+ and Fe 3+ you would write Iron (II) or Iron (III). If we have the coordination compound [Co (N...

Sebastian Lee 1L

We know that nonmetal oxides will typically form acidic compounds (ie. H 2 SO 4 or HNO 3 ) and that most metals oxides will form basic compounds (ie. CaO or NaO 2 ). However, amphoteric compounds are ones that can have either acid or base character depending on the reaction. Typically, elements near...

Sebastian Lee 1L

The roman numeral indicates the positive charge of the transition metal cation. This notation is often used since transition metals can take on multiple oxidation states. For example, to distinguish Fe 2+ and Fe 3+ you would write Iron (II) or Iron (III). If we have the coordination compound [Co (NH...

Sebastian Lee 1L

The definition of an Arrhenius acid is a molecule that forms a hydronium ion (H 3 O + ) in the presence of water. An Arrhenius base forms a hydroxide ion (OH - ) in water. This definition applies to most Bronsted acids/bases. A Bronsted acid is a molecule that is a proton donor, meaning that it will...

Sebastian Lee 1L

I believe this is referring to the fact that at any moment in time, electron distribution in a molecule can be slightly unequal. Electrons may be "hanging out" on one side of a molecule at an instant point in time. This creates a temporary/instantaneous dipole moment because of this unequa...

Sebastian Lee 1L

Also, extra stuff not regarding the original question: The VSEPR shape of SO 2 would be bent with a bond angle of slightly less than 120 degrees. This is because there are 3 regions of electron density (trigonal planar derived) but one of them is a lone pair which repels a little more than a bonded ...

Sebastian Lee 1L

The Lewis Structure for SO₂ is a central Sulfur double bonded to each of the Oxygen atoms. A double bond consists of one sigma bond and one pi bond so in total you would have two sigma bonds and two pi bonds (one sigma and pi for one bonded Oxygen and another sigma and pi for the other).

Sebastian Lee 1L

I suppose you should know that since there is likely a double bond in the cis and trans isomers, there is a pi bond between p orbitals which prevents the atoms from rotating. This ensures that the cis can't turn into a trans without some kind of chemical reaction.

Sebastian Lee 1L

Dipole-dipole interaction occurs when you have two polar molecules, each with a permanent dipole moment. The negative partial charge of one dipole is attracted to the positive partial charge of the other. For example, if I had H 2 O and NH 3 , the partial positive of the hydrogen would be attracted ...

Sebastian Lee 1L

When atomic size increases, polarizability ( \alpha ) increases because the electrons are held on less tightly by the nucleus. High polarizability creates the stronger intermolecular forces described by the larger interaction potential energy. The distance between molecules is important but I think ...

Sebastian Lee 1L

I know that hydrogen bonding can occur when a hydrogen that is polar covalently bonded to a N,O, or F has a highly positive partial charge that it is attracted to a lone pair of a different molecule's electronegative atom. Can this other electronegative atom with the lone pair be any atom or does it...

Sebastian Lee 1L

Yes. These have appeared in readings and HW.
You should know linear, trigonal planar, bent (<120 degrees), tetrahedral, trigonal pyramidal, bent (<109.5 degrees), trigonal bipyramidal, seesaw, T-shaped, linear (but with 3 lone pairs), octahedral, square pyramidal, and square planar.

Sebastian Lee 1L

You have to understand the VSEPR shape of the molecule to determine its polarity. AsF 5 has a trigonal bipyramidal shape since it has 5 regions of electron density and no lone pairs. In a trigonal bipyramidal shape, you have 2 axial bonded atoms and 3 equatorial bonded atoms (2 are straight up and d...

Sebastian Lee 1L

I believe that the hydrogen bond (-20 kJ/mol) is indeed stronger than the ion-dipole force (-15 kJ/mol).

Sebastian Lee 1L

I believe you would still treat the lone electron as one region of electron density. So it would act similar to a lone pair but have a lower repulsion than a lone pair would.

Sebastian Lee 1L

The HCH bond in CH 2 2- has the smallest bond angle because it has a bent molecular shape. You are right in saying that the molecule will be based off a tetrahedral shape since it has 4 regions of electron density (2 bonding pairs and 2 lone pairs). Since it has 2 lone pairs, you would replace 2 ato...

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