General Chemistry Topics with Explanations

The worst way to learn periodic trends (or anything else in the PS portion of the MCAT) is to memorize them. This is because memorization without understanding will only allow you to solve problems that are asked in the same format that you have memorized. However, the MCAT is infamous for forcing you to take information that you've learned in your college classes and apply it to new scenarios with which you have no previous experience.

The periodic trends that you should know for the MCAT (or DAT, OAT, PCAT) are atomic radius, ionization energy, electron affinity, and electronegativity. We will consider each one in turn in a second post, but first, we must consider a fifth trend upon which the other four all depend, which is the effective nuclear charge, Z.

Effective Nuclear Charge (Z or Zeff): This trend is probably the hardest for students to understand, but once you do understand it, all of the other trends will seem very logical and intuitive to you. Z is basically a book-keeping method, where we cancel out each core electron with one of the protons in the nucleus. (Core electrons are the ones in the inner shells; in other words, they are the non-valence electrons.) We do not cancel the valence electrons. (Remember that valence electrons are the ones in the outer shell.) When we have completed this exercise, we will find that, for a neutral atom, we are left with the same number of uncancelled protons in the nucleus as we have valence electrons. The number of uncancelled protons is equal to the value of Z, and it will be the same as the group number for a neutral atom.

The reason why we only cancel out core electrons is that these electrons are considered to be shielding the valence electrons. This means that they block the valence electrons from easily having access to the nucleus. Remember that the nucleus is positive, and all electrons are negative. Since unlike charges attract, the valence electrons are attracted to the nucleus. But the problem is that there are some core electrons in the way, and they too are negatively charged. Since like charges repel, core electrons will repel the valence electrons, somewhat countering their attracting to the nucleus.

Let's look at some actual atoms to make this concept clearer.

Calculating Z for Li: We'll start with lithium, which has three protons and three electrons as a neutral atom. If you begin to write the electron configuration of lithium, where do the electrons go? Well, the first two go into the 1s orbital. These are two core electrons. The third goes into the 2s orbital, and it's a valence electron. Doing our book-keeping, we find that we have cancelled two protons for our two core electrons, and this leaves us one proton left over. Thus, the value of Z for Li is approximately 1.

Calculating Z for Be: The next element after Li is beryllium, which has four protons and four electrons as a neutral atom. Its configuration is 1s2, 2s2, which means that again we have two core electrons as with Li. However, we now have two valence electrons, and when we have finished cancelling our core electrons with two of the protons, we now have two protons left over. Thus, the value of Z for Be is approximately 2. (Note that in doing this, we are assuming that electrons in the same shell do not shield one another. That is, we are assuming that the two valence electrons in Be do not shield one another. Technically that is not exactly true, but for our purpose, the effect of electron-electron repulsion by electrons in the same shell can be ignored.)

Calculating Z for B: After Be comes boron, which has five protons and five electrons. Its configuration is 1s2, 2s2, 2p1. As we have already seen, we have two core electrons from the 1s-orbital, which we can cancel with two of the five protons. This leaves us with three protons left over, and three valence electrons. (Again, we are assuming that the 2s electrons do not shield the 2p electron, but technically this is not completely true.) So now our Z value is 3.

Calculating Z for the rest of the elements in row 2: As we continue to fill the 2p orbitals, one electron at a time, we will find that the Z value continues to increase by one as we move from left to right. Here are the Z values for the rest of the elements in row 2:

Element Z-value
C.............4
N.............5
O.............6
F.............7
Ne............8

Calculating Z for the Group I Metals: Now, let us consider what happens to Z when we go down a group. We'll use Group I, the alkali metals. We've already seen that Li has a value of 1 for Z. What about Na, right below Li? Well, sodium has 11 protons and 11 electrons as a neutral atom. Its configuration is 1s2, 2s2, 2p6, 3s1. All of the electrons in the second shell are now core electrons; only the one electron in the 3s orbital is a valence electron. If we cancel out the ten core electrons with ten of the protons, we are left with one proton, just as with Li. So the value of Z for Na is also approximately 1, just as it is for Li. What about K, element 19? Its configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 as a neutral atom. Just as before, there is only one valence electron. The other 18 electrons are all part of the core. So they each cancel a proton, leaving us with 1 proton left over, and a Z value of about 1. Clearly, then, ALL of the alkali metals have a Z value of approximately 1. This same pattern will hold true for each of the other main groups in the periodic table.

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Here is a summary of the trends for Z.

As we travel from left to right across the periodic table, Z increases. This occurs because we are adding more protons as we move from left to right, but we are not adding more shielding core electrons.

As we move down a group from top to bottom, Z stays approximately equal. This occurs because we are not changing the net number of protons left over after we have cancelled out all of the shielding core electrons.

If you have read and understood the previous post about Z, you will now find that the other four trends we mentioned are all quite intuitive. Let's consider them one at a time.

Atomic Radius: Remember that the size of the nucleus of an atom is extremely small relative to the size of the entire atom. So when we talk about an atom's size, what we are really considering is the size of its electron cloud.

Radius of Neutral Atoms: As we move from left to right across the periodic table, we find that Z is increasing. So what exactly does this mean? Well, as Z increases, the net positive pull of the nucleus increases. (Z is, after all, the number of protons that are not needed to cancel out with the core electrons.) As we increase the net positive charge of the nucleus, we will also increase its ability to pull on the electron cloud, even the valence electrons. (This should make sense to you from Coulomb's Law, also; if you increase the magnitude of Q, you also increase the magnitude of F between the two charges.) So as Z increases from left to right, those nuclei will pull their electrons in more and more tightly to the nucleus, and this will decrease the overall size of the atom. Thus, the smallest atoms are found on the right side of the table, and the largest are on the left.

As we go down a group, Z doesn't change very much. But what does change? Well, we are adding a new principle quantum number each time we move down a row, which means that we've added more core electrons. So even though we aren't changing the net pull of the nucleus, we ARE changing the distance between the nucleus and the valence electrons. In fact, we are greatly increasing it. (From Coulomb's Law, we know that increasing the distance between two charges decreases the force between them.) So as we move down the periodic table, we will continue to add more layers of electrons without significantly changing Z, making the atom grow larger and larger.

Radius of Cations: You may be asked on your test to compare the radius of a cation with that of a neutral atom. The most important thing that removing an electron does is that it decreases electron-electron repulsion within a shell. Recall from the previous post that we are assuming that valence electrons don't repel one another, but in reality, they actually do a little bit. So when you remove an electron to form a cation, you don't change your value of Z, but you do decrease repulsion among the remaining valence electrons, which allows the nucleus to pull them in even tighter. Thus, cations are always smaller than the neutral atom.

This effect is especially pronounced if you remove an electron from a Group I element, since you lose an entire shell when you do this. Thus, Li+ cation is MUCH smaller than neutral Li.

Radius of Anions: The effect of adding an electron to a neutral atom is the opposite of the effect of removing one. Again, you are not changing Z, but you ARE increasing repulsion among the valence electrons, and so the net effect is to increase the size of the atom. If you add an electron to a group VIII element, this effect is especially pronounced, since you will be adding another whole quantum level to the atom.

Ionization Energy (IE): IE is the energy required to remove an electron from a lone atom in the gaseous state. Note that the atom MUST be in the gas phase, and it MUST be alone (i.e., not bonded to anything else).

As we have seen, the value of Z increases from left to right as we go across the periodic table. This means that the atoms are able to hold on to their valence electrons more and more tightly. Thus, it gets harder to remove an electron as the value of Z increases, and IE increases from left to right.

Going down a group, Z doesn't change much. But, the size of the atom increases greatly, and because of this, so does the distance between the nucleus and the valence electrons. That decreases the magnitude of the force of attraction between the nucleus and the outermost electrons. Thus, IE decreases going down a group from top to bottom.

You may also get asked on your test to compare first and second IE values. In general, it always takes more energy to remove a second electron than it does to remove the first. This is because you are trying to remove an electron from an already positively charged species, and make it have a positive charge of even higher magnitude. So second IE is always greater than first IE, third IE is greater than second IE, and so on.

Electron Affinity (EA) and Electronegativity (EN): These two concepts are similar to one another, and they follow the same trends. EA measures the energy released when a lone gaseous atom accepts an electron. It is the opposite of IE, and as with IE, the atom must not be in a bond, and it must be in the gas phase. Note that EA values are given as the absolute values of the energy released. (That is, we usually report exothermic reactions, where energy is being released, as having a negative energy value. However, EA is always reported as the absolute value of that negative energy magnitude, and so EA values are always positive.) EN measures the ability of an atom in a bond with another atom to pull the bond's electron density toward itself.

As we travel from left to right across the periodic table, and Z increases, the ability of the atom to accept an electron will also increase. This is because the pull of the nucleus on the valence electrons increases from left to right. Thus, both EA and EN increase from left to right.

As we go down a group, Z doesn't change much, but the valence electrons get farther and farther from the nucleus. Because of this, the atom is less and less able to "handle" having another electron added to it. Thus, both EA and EN will decrease as we go from top to bottom down the group.

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Summary of Periodic Trends

Effective nuclear charge increases from left to right, and does not significantly change from top to bottom. The left to right trend occurs because the number of core electrons remains constant, while the number of net protons left over after cancelling out the core electrons increases. The top to bottom trend occurs because the net number of protons left over after cancelling out the core electrons remains constant.

Atomic radius decreases from left to right, and increases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z.

Ionization energy increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.

Electron affinity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.

Electronegativity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.

Students commonly confuse intramolecular interactions (bonds) with intermolecular interactions (molecular attractions). This post will explain the types of bonds that hold an individual molecule or crystal together, as well as the molecular attractions that bind molecules to their neighbors.

Intramolecular Interactions (Bonds)

Interactions that are inTRAmolecular are within a single molecule or ion pair. We normally call these associations bonds. The bonding interactions within molecules are relatively strong, especially in comparison to the attractive interactions among molecules. Although bonds are often considered to fall into one of two discrete categories, covalent and ionic, it is actually best to consider bonding as a continuum, with perfectly covalent bonds and perfectly ionic bonds as the two extremes.

Nonpolar Covalent Bonds:
Although we often consider C-H bonds to be "nonpolar" in organic chemistry, in reality, perfectly nonpolar covalent bonds with completely equal sharing of bond electrons between both atoms only occur in homonuclear diatomic molecules. Homonuclear means that both of the atoms being bonded are identical (homo = same); diatomic means that there are only two atoms. So examples of perfectly nonpolar covalent bonds would include H2, O2, N2, and the halogens (F2, Cl2, Br2, and I2). For your test, you should memorize that in nature, these seven elements occur in pairs.

Ionic Bonds:
Sometimes two elements have such a large difference in electronegativity (see post 3 for an explanation of electronegativity) that one atom is able to grab an electron completely away from another. This generally happens when highly electronegative elements from the extreme right side of the periodic table (ex. halogens) are mixed with highly electropositive elements from the extreme left side of the periodic table (ex. alkali metals). Each element forms an ion, with the metal becoming a positively charged cation and the nonmetal becoming a negatively charged anion. The two ions are held together by electrostatic attraction, which is described by Coulomb's Law.

Polar Covalent Bonds:
Most other bonds between two different elements that have more moderate differences in electronegativity fall somewhere between the above two extremes. Some, such as a C-H bond, lie closer to the nonpolar covalent end of the spectrum. Others, such as a Mg-C bond, lie closer to the ionic end of the spectrum. Polar covalent bonds have covalent character because the bonding electrons are shared between the two atoms, but they also have ionic character because the bonding electrons are not shared equally. Rather, the more electronegative element is able to pull extra electron density to itself, and the more electropositive element is left slightly electron-deficient. This forms a bond dipole, with a partial negative charge on the electronegative atom and a partial positive charge on the electropositive one. Polar covalent bonds can either be protic or aprotic (explained below in Post 10).

All pre-health test students should memorize this list of strong acids and strong bases. Any acid not on this list should be considered to be a weak acid. Most other bases are also weak bases, but realize that there are also some strong organic bases. I have listed some examples.

Strong Acids:
-H2SO4
-HNO3
-HCl
-HBr
-HI
-HClO3
-HClO4
*note that HF is NOT a strong acid, and its omission from the list is not an accident!!!
**only the first proton on H2SO4 is strong (i.e., completely dissociates); the second one is weak.

Strong Bases:
-Group I elements with hydroxide (ex. NaOH, KOH)
-Group II elements with hydroxide [ex. Ca(OH)2]
-some organic bases, including alkoxide ions, sodium hydride, Grignard reagents, and LDA

Here is SilvrGrey330's extremely clever mnemonic to remember salt solubility rules.

C A S H n Gia

Read it as "Cashin' Gia"...how to remember that? well the story is...im a pimp...and gia is my hoe, and i need to get my cash from her. hence...Cashing from gia.

C is clorates, A is acetates, S is sulfates, H is halogens, n is Nitrates, and Gia is Group I A metals. ---> THESE ARE ALL SOLUBLE, XCEPT

for S: Ca, Ba, Sr.....just remember the tv network CBS
for H: Ca, Ba, Sr + Happy...whats happy? Hb Ag Pb ...mercury, silvr and lead...add a py to the end and all the first letters spell HAPPY

and if its not part of CASHnGIA...its insoluble.

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Please note that salt solubility is a complex equilibrium, and it isn't an either-or phenomenon like a simplified set of rules makes it out to be. Books may differ on how detailed they get about the rules for predicting it, and also on where they draw the line between calling a salt "soluble" versus calling it "insoluble." Keep in mind that the solubility rules are guides to help you predict the relative solubilities of different salts, and take them all with a grain of (ahem!) salt.

Calculating logs without a calculator is not as difficult as you might expect it to be. First, you should be familiar with the following log rules:

log (xy) = log x + log y
log (x/y) = log x - log y
log (x)^y = y(log x)

You can use these rules to derive the Henderson-Hasselbalch equation from the Ka expression for an acid:

HA -> A- + H+

Ka = [A-][H+]/[HA]

Taking the -log of both sides gives us:

-log Ka = -log {[A-][H+]/[HA]}

The left side is the definition of pKa:

pKa = -log {[A-][H+]/[HA]}

The right side can be re-written using the log rules:

pKa = -log [A-] - log [H+] - (-log [HA])

The second right side term is the definition of pH:

pKa = -log [A-] + pH + log [HA]

Solving for pH:

pH = pKa + log [A-] - log [HA]

Using the log rules in reverse gives us the equation as it is normally written:

pH = pKa + log [A-]/[HA]

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You should also know the common base 10 logs. For example:

log 1/100 = -2
log 1/10 = -1
log 1 = 0
log 10 = 1
log 100 = 2
and one more that is very helpful for pH estimations: -log 0.5 = 0.3

So let's say that I'm working on a pH problem. I find out that my [H+] is equal to 3.6 x 10^-5. What is the pH? We don't have a calculator on any of the pre-health tests, so we need to be clever about this. I don't know what the -log of 3.6 x 10^-5 is. But here's what I do know:

-log (1 x 10^-5) = 5
-log (10 x 10^-5 = 1 x 10^-4) = 4

So, the -log (3.6 x 10^-5) must be some number between 4 and 5. That is usually sufficient to solve the problem. Your answer choices might be 3.4, 4.4, 5.4, and 6.4, and you will of course pick 4.4.

If necessary, we can do even better than this. Say two of our answer choices are 4.2 and 4.5. How do we pick between them? (Note that this situation should not come up on your test, but just in case it does, you'll know what to do!)

Here's where knowing that -log 0.5 = 0.3 comes in useful. We know that -log (5 x 10^-5) must be 4.3, because we already memorized that -log (0.5) = -log (5 x 10^-1) = 0.3. Since we want the -log (3.6 x 10^-5), it must be a number that is close to 4.3, but a bit higher (because 3.6 is between 1 and 5, then the pH must be between 4.3 and 5). So we can eliminate 4.2, and 4.5 must be the correct answer. (The actual pH value is 4.44.)

Deviations Due to Raising Pressure: Ideal gas behavior is most closely approximated at low pressure. There are two factors that cause real gas behavior to deviate from ideal gas behavior as pressure is increased. One is the attraction that exists among gas particles, which becomes significant at medium pressure; and the other is the volume of the gas particles themselves, which is significant at high pressure. The ideal gas law assumes that these two factors are insignificant, which is generally true at low pressures. As you start raising the pressure to a medium level, the gas particles begin to attract one another. This causes them to "stick" together and contracts the volume of the gas overall below what would be expected if no attraction were occurring. As you continue to raise the pressure, the volumes of the individual gas particles begin to occupy a significant portion of the volume of the container. This causes the volume of the gas to be larger than expected, because the ideal gas law does not take into account the volumes of the gas molecules. (According to the ideal gas law, the molecules have no volume, as well as no intermolecular interactions.)

Deviations Due to Lowering Temperature: Ideal gas behavior is most closely approximated at high temperatures. Remember that temperature is a measure of the average kinetic energy of the molecules in the gas. As the temperature is decreased, the molecules move more and more slowly. This makes the intermolecular attractions among them more significant. If the temperature is decreased enough, the gas will reach its vaporization point (boiling point), and turn into a liquid (or possibly a solid if the pressure is below the triple point pressure). As explained above, intermolecular interactions cause the gas's volume to be smaller than what is predicted by the ideal gas law.

Correcting for Deviations in Pressure and Temperature: The ideal gas law works fairly well as long as the pressure is kept low and the temperature is kept high. However, under conditions of high enough pressure and low enough temperature, it breaks down. The van der Waals equation of state is one example of a modified gas law equation that attempts to correct for the effects of particle volume and intermolecular attractions. It looks similar to the ideal gas law, but it has two extra terms:

van der Waals equation of state: [P + (n^2)(a)/(V^2)](V-nb) = nRT

The a-term is to compensate for attractive forces among the molecules, and it will be lowest for nonpolar molecules and higher for polar ones. The b-term compensates for molecular volume, and it will be lowest for small molecules and higher for larger ones. Note that if a = b = 0, the van der Waals equation reduces down to the ideal gas law:

[P + (n^2)(0)/(V^2)][V-n(0)] = nRT
[P][V] = nRT

Question: Is it ok to calculate the emf just as if the cell were galvanic, and then simply switch to a negative value?

Yes, that will work as long as you write one half-reaction as a reduction and the other as an oxidation. Many times both half-reactions will be written as reductions, so check where the electrons are in the equations. If they are on the left side of the equation, it's a reduction half-reaction, and if they are on the right side of the equation, it's an oxidation half-reaction. You must always have one of each in any redox reaction, because if something is getting reduced than something else must be getting oxidized.

Question: for standard cell potential: which formula is correct: Eo = Ereduction + Eoxidation? Or Eo(cell) = Eo(cathode) - Eo(anode)? How can this problem be solved?

Consider the following electrode potentials:

Cu2+ + 2e- --> Cu(s) Eo = +0.34 V
2H20 --> O2 + 4H+ + 4e- Eo = -1.23 V

What is the Eo(cell) for the reaction shown in the following equation?

2Cu2+ + 2H20 --> 2Cu(s) + O2 + 4H+

A) -0.89 V
B) +0.55 V
C) +1.57 V
D) + 1.91 V

My advice to you is to not use either version of the formula, as it will only confuse you. (Guess I didn't need to tell you that, right? ) Instead, what you should do is attack any electrochemistry problem by first considering the following:

What is the form in which the two half-cell reactions are written? Generally, but not always, you will see both half-reactions written as reduction potentials. You can determine this because you will see that the electrons are being added on the left side of the equation. Remember, when you add electrons, it's a reduction. Conversely, if you lose electrons, it's an oxidation, and you will see the electrons on the right side of the equation.

If both half-reactions are written as reductions, you must turn one of the them backward into an oxidation half-reaction. (If something is being reduced, something else better be getting oxidized because those electrons have to come from somewhere.) The tough part is deciding which half-reaction to turn backward. If you have a galvanic cell (most common case), you will want to turn the half-reaction with the smaller reaction potential backward. (Don't forget that a number like -1.2 would be smaller than one like -0.3). You do this because galvanic cells will always have a positive Ecell (they have to, because they are spontaneous, and the equation that relates G with Ecell has a minus sign in it: G=-nFEcell). If the cell is electrolytic, then it is not spontaneous, Ecell should be negative (making G positive) and you will turn the half-reaction with the larger reaction potential backward instead. Remember, in either case, when you turn the half-reaction backward, you change the sign of that half-reaction's reaction potential.

Ok, so now let's consider your specific example. The first half-reaction (the Cu one) is a reduction half-reaction. Again, I know that because the electrons are being added on the left side; Cu is gaining electrons. The second one, in contrast, is an oxidation half-reaction. The electrons are on the right side; O is losing electrons. Good, so in this case, nothing needs to be turned around, because we already have one reduction and one oxidation. You can see that this cell is an electrolytic one as written. (Add the two reaction potentials together, and you are going to get a negative number for Ecell) Looking at the third equation that combines both half-reactions, you can see that it is written in the same fashion as the two half-reactions above it: Cu is getting reduced, and O is getting oxidized. Thus, again, you have an electrolytic cell, nothing needs to be turned backward because one partner is oxidizing and one is reducing, and you merely need to add the two reaction potentials together to get your -0.89.

Really, there isn't any need to do any math on this problem. As soon as you recognize that you have an electrolytic cell, there is only one possible right answer for this question, because the others are all positive for Ecell.

Remember that you do not have to multiply the values of the reaction potentials by the number of moles of electrons. The reason why is that reaction potential is an intrinsic property; it is independent of the amount of material you have. That is, one gram of Cu has the same reaction potential as 1 kg of Cu does. (Temperature and density are other examples of intrinsic properties.) The properties where you do have to multiply by the number of moles are extrinsic properties; those do depend on the amount of material. For example, when you use Hess's law, it matters greatly how much material you have because the more compound you have, the more heat it will give off. (Energy and volume are some other examples of extrinsic properties.)

Intermolecular Interactions (Molecular Associations)

Interactions that are inTERmolecular are among more than one molecule of a substance. These are the forces that attract the molecules of a liquid or solid to each other. Intermolecular interactions occur among nonpolar molecules as well as among polar ones, but polar interactions are much stronger than nonpolar ones are.

Polar Molecules:
Polar molecules contain one or more dipoles. (See Post 4 for an explanation of dipoles.) They fall into two categories: polar protic, and polar aprotic. Polar protic molecules have protons (hydrogens) attached to electronegative atoms, namely F, O, or N. Polar aprotic molecules still contain dipoles, but there are no protons attached to F, O, or N. It is essential that you understand the distinction between protic and aprotic polar molecules in order to properly understand the organic chemistry substitution mechanisms.

Polar Aprotic Molecules:
Polar aprotic molecules have intermolecular interactions known as dipole-dipole interactions. The positive end of the dipole on one molecule is attracted to the negative end of the dipole on another molecule. This is an electrostatic interaction, and it is governed by Coulomb's Law. Coulomb's Law tells us that either a larger charge separation (leading to a larger molecular dipole) or a smaller distance between the molecules will strengthen the dipole-dipole interactions between them. There are many examples of polar aprotic molecules, including ketones, aldehydes, esters, tertiary amines and amides, HCl, HBr, HI, cyanides, nitro groups, and DMSO (dimethyl sulfoxide). Polar aprotic solvents destabilize nucleophiles, making them suitable solvents for Sn2 reactions in organic chemistry.

Polar Protic Molecules:
Polar protic molecules have intermolecular interactions known as hydrogen bonds. In spite of their name, note that hydrogen bonds are not true bonds. Rather, hydrogen bonds are an especially strong type of dipole-dipole interaction. The positive end of a dipole on one molecule (the H) is sometimes so strongly attracted to the negative end of another molecule's dipole (the F, O, or N) that the H will dissociate from the first molecule and bind to the second one. You are already familiar with this process for water, which autoionizes to form H3O+ and OH-:

2 (H2O) <-> (OH)- + (H3O)+

Water is the most common example of a polar protic molecule, but some other examples include alcohols, carboxylic acids, primary and secondary amines and amides, and hydrogen fluoride. Polar protic solvents stabilize carbocations, making them suitable solvents for Sn1 reactions in organic chemistry.

Nonpolar Molecules:
Molecules that do not contain dipoles are called nonpolar molecules. They are held together by intermolecular interactions that are called London dispersion forces or van der Waals interactions. These interactions are due to temporary, induced dipoles. Where do these dipoles come from? To answer this, you must realize that the electrons in a bond or in an atom are not stationary. Rather, these electrons are constantly in motion. Sometimes, by sheer chance, they are not equally distributed on the atom or in the bond, and this inequality of charge forms a temporary dipole. It is temporary because soon thereafter, as the electrons continue to move, they distribute more equally again, and so the dipole goes away. But in the meantime, that short-lived dipole has affected its neighbors, and it has caused similar dipoles to form in them as well. This is why these dipoles are induced. So you can imagine a bunch of molecules, held together by small dipoles that are constantly forming and unforming. But there are always some dipoles present at any given time, and that is what holds the liquid or solid together.

Note that all substances contain dispersion forces, but these forces are only greatly significant for nonpolar elements and molecules that lack dipole-dipole interactions. In polar compounds, dispersion forces can be neglected.

Vapor pressure is the part of the total pressure that is accounted for by the vapor. Ideally, this means that if the total pressure is 12 atm, and the air is 25% vapor, then the vapor pressure is 25% of 12 atm = 3 atm. In a closed system, the amount of vapor in the gas phase will be set by the saturation vapor pressure. The central idea here is that of equilibrium.

Equilibrium is often thought of as the point where movement stops. In chemistry, a better way to think of it is that the rate in one direction and the rate in the other direction are equal and opposite, so there is no net movement.

In the case of the saturation vapor pressure, this means that for every molecule in the liquid that gets enough energy to break free of the liquid and enter the gas phase, there is a molecule already in the gas phase that strikes the surface of the liquid, and lacks the energy necessary to bounce back, hence remaining in the liquid. There is no net change in the ratio of molecules in the gas phase to molecules in the liquid phase.

For pressure in general, one must continue to think of this in terms of the single molecule colliding against others. If the pressure is high, then it will be harder (i.e., requires more energy) for a molecule to break free of the liquid and become a vapor, because the gas phase is crowded with other molecules. Under low pressure, there's more space, and hence it is easier (i.e., requires less energy) for a molecule to break free of the liquid and become a vapor.

When you boil water, the water will get hot enough to vaporize, but no hotter--because if it got hotter than needed to vaporize, it would have already evaporated. So if the boiling point of water is 100 degrees, and you have boiling water, then you know that the water is 100 degrees--any more and it would vaporize, and any less and it wouldn't be boiling. When the pressure is low, it takes less energy for a molecule to jump into the empty space above the liquid, meaning that the temperature required to make the water boil is less. In a pressure cooker, the water molecule needs more energy to get into the crowded space above the liquid, so the temperature required to make it boil gets higher.

Pressure is lower at higher altitude for a similar reason: imagine you are under a rug, and then imagine you are under ten rugs. The more layers of carpet on top of you, the higher the pressure is. The atmosphere is no different. There is no air in space, and the only reason that earth has an atmosphere is because gravity pulls the air down to earth. The lower your altitude is, the greater the thickness of the atmosphere above you, and the greater the pressure of the air weighing down on you. Conversely, the higher you go, the closer you are to space, and the less air there is above you weighing down on you.

Since the pressure at high altitude is low, water boils at a lower temperature. That means that the pot of boiling water at the top of Mt. Whitney is significantly cooler than the pot of boiling water at the bottom of Death Valley, and cooler still than the boiling water in the pressure cooker. The pressure cooker makes it possible for the water to remain a liquid at higher temperatures, so you can make boiling water hotter than normally possible (the water can't vaporize because of the high pressure). Hence, due to the temperature differences, it would take longer to boil an egg at the top of a mountain, and less time to boil an egg in a pressure cooker.

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This post was written by Nutmeg.

Electrons always go from the anode to the cathode. ALWAYS, no matter what kind of cell you have. This is because the cathode is defined as the reducing electrode, while the anode is being oxidized. Again, this is always the case, regardless of what kind of cell it is. What changes for galvanic vs. electrolytic cells is the SIGN of the electrodes. In a spontaneous (galvanic) cell, the negative electrons go from a negatively charged anode down their electrical gradient to a positively charged cathode. Makes sense, right? Negative charges would like to go to a positively charged electrode, given their druthers. On the other hand, in an electrolytic cell, you are forcing the negative electrons to go backward, against their electrical gradient, to a negative electrode. Thus, in electrolytic cells, the anode is now positive while the cathode is negative.

As for what flows to the anode, that would be the current. Recall that current is defined by physicists as the direction that imaginary positive charges would travel (though of course it is the electrons we are concerned with in chemistry, not the current). It is also possible if you were reading about a galvanic cell that the problem might mention ions from the salt bridge moving to the cathode and anode. (Since the electrodes are in separate cells for a galvanic cell, you need the salt bridge to complete the circuit or you won't get any electron flow.) What happens? Well, when the electrons move from the anode, they leave behind positive charges. So negative charges from the salt bridge (anions) travel to the anode cell to counterbalance them. Conversely, the electrons going to the cathode side would build up a net negative charge if positive ions (cations) from the salt bridge didn't travel to the cathode side to balance them. So salt bridge anions go to the anode side, and cations go to the cathode side.