Bohr's Atom

In the discussion of Planck’s work, I suggested that we could understand quantization as being like money - everything is a multiple of the penny. Let’s use that analogy to understand Bohr’s atom.

If an electron is a very short distance away from the nucleus, we could say that it had one cent of energy.

Of course, attraction between positive and negative don’t depend on direction, so that same one cent worth of potential energy could be anywhere on a circle around the nucleus. (Actually, it’s the surface of a sphere, although it is almost never shown as one).

If energy is quantized, then if an electron has more energy, it must be at a distance of “two cents” or of “three cents” etc. This gives us a diagram that looks like this:

Assuming that we are looking at a hydrogen atom, there is only one electron. It makes sense that the electron will be as close to the nucleus as possible, which means it will be on the first ring with the smallest amount (1 cent) of potential energy.

If we give the electron some energy (say 2 more cents), then it will have 3 cents of energy. In our picture of the atom, that means that it would need to be on the third ring. Bohr suggested that the electron would absorb the energy by “jumping” from the first level to the third. (By the way, this is called a quantum leap.)

Since the electron is still attracted to the nucleus it “wants” to go back down. More correctly, the electron would be more stable in the lower ring, so the electron will lose its excess energy and drop back down to the more stable, lower energy level. That “lost” energy is given out in the form of light.

In this way, Bohr explains the production of light. An atom is given energy (heat, electricity, chemical energy, even light), an electron absorbs that energy by moving to a higher energy level. Then the atom emits that energy in the form of light as the electron "jumps" back down.

With a more precise look at the theory, we can explain the other two issues (why only certain colors are produced and what the math of Balmer, et.al. means).

Neils Bohr and the Beginnings of Quantum Theory

Although Max Planck's work with the ultraviolet catastrophe is arguable the beginning of Quantum Physics, Neils Bohr is generally considered the father of that field.

This is because it was Bohr who was able to apply the “mathematical quirk” that Planck had found to the basic structure of matter and, in doing so, explain a number of things that were not understood. 

At the time that Bohr first proposed his theory there were several “unsolved mysteries” that we need to consider:

1. When things have a lot of energy, they give off light. This could be the glowing of embers in a fire, or the fire itself. It also could be the light coming out of a “normal” incandescent bulb or out of a fluorescent tube. It could even be the sparks given off when a piece of flint is hit with a piece of steel.

2. The light given off by an excited element is rarely, if ever, a single color of light, nor is it a smooth rainbow of colors. Instead, an excited element will give off a few (or occasionally a few dozen) very distinct, separate colors of light.

3. The work of Balmer, Paschen and Lymen found that the wavelengths of the light (visible, IR and UV) emitted by hydrogen could be found with a set of mathematical equations.

Bohr was able to synthesize an explanation that addressed all of these in a single, relatively simple idea. That idea was that:

The (Potential) Energy of Electrons is Quantized

Short? Yes. Simple? Ehhh…

Let’s pull apart what Bohr is saying:

1. The potential energy of electrons…  Because electrons are negative and the nucleus is positive they attract each other. When they are not in the same place, they have potential energy. The amount of potential energy depends on the amount of attraction and the distance. So, an electron that is further from the nucleus has more potential energy than one that is closer.

2. Energy is quantized… this means that all amounts of energy must be a multiple of some small basic amount. This is Bohr’s use of Planck’s idea.

Putting those ideas together creates a very distinct picture of the atom, that addresses and explains the three observations listed above.

What Direction Will a System Shift in Order to Reach Equilibrium

If a system is not at equilibrium, then it must either be running forward faster than backwards or backwards faster than forward. Determining which of these is a simple matter. 

We’ll use the following problem to explore this: 

The system A + 2 B ⇄ C + D has a K = 4.12. If [A] = 1.00M and all of the other concentrations are 4.12 M, the system is not at equilibrium. What direction must the system shift in order to reach equilibrium?

We know that at equilibrium, the following math must be true:

If we plug in the current values for the expression Q, we get the following:

Which is definitely NOT 4.12. 

So, which direction will it go? There are two different ways to find the answer. The first requires some mathematical logic, the other requires only careful set-up.

The Logical Way

Right now, we know that we are not at equilibrium:

But we can work out what needs to happen, again using this equation:

we know that the right side of this equation is currently too small (it's less than the value of K). In order for the fraction to be larger, the numerator of the fraction must get larger (or the denominator must get smaller...the logic works either way). The top of the fraction is comprised of the products of the reaction. So, to reach equilibrium we need more products, which means that the reaction must run to the right.

The Simple Way

We know that since K = 4.12 and Q = 1.00, that K > Q. Now all we have to do is turn the ">" sign into an arrow →. That arrow will always point in the direction that the reaction is shifting. So, in those situations where K>Q, the reaction will shift to the right (→). When K<Q, the reaction will shift to the left (←).

A word of warning

The simple way ONLY works if you put K on the left and Q on the right. If you reverse their order, you will always be exactly wrong.

A Note About Zeros

Lots of equilibrium problems start with one or more compounds missing - that is with a concentration of 0 M. That zero can mess with your math.  For instance, if you reached a point where the math looked like this:

Your math instructor would tell you that the problem is unsolvable, since dividing by zero is undefined. 

However, this is not a math class. We are not trying to solve the equation, merely to determine the direction that the system will shift. That means we can logic our way out of the mathematical trouble. 

If a system is missing one or more products, then the reaction CANNOT go backwards. (If the products don't collide, then the backward reaction can't occur.) In the same way, if the system is missing one or more reactants, it CANNOT run forward. 

That leaves us with a simple rule:

A system will ALWAYS shift toward the side that has a 0 concentration.

Determining if a System is at Equilibrium

We know that a system is at equilibrium when the equilibrium expression IS equal to the value of the equilibrium constant. That is, when K = Q.

We can turn that statement on its head to say that if K=Q then the system must be at equilibrium. That allows us to solve some simple problems like this:

The system A + 2 B → C + D has a K = 4.12. If [A] = 1.00M and all of the other concentrations are 4.12 M, is the system at equilibrium?

If it is at equilibrium, then the math below will be true.

If we substitute in our values, we see that K is NOT equal to Q

so, our system is NOT at equilibrium.

A Note about Zeros

It is fairly common to be given a system where one or more of the compounds is missing - that is, where one or more of the concentrations = 0. This can present a mathematical problem.

As an example, you may find yourself trying to determine whether the following math is true?

Your math instructor will tell you that the question is unanswerable, since the fraction on the right is undefined. But this is not a math class. We don't care what the actual value on the right is. All we need to know is that it is NOT equal to the value of K on the left. Therefore the system is NOT at equilibrium.

If you prefer logic to "undefined" math, you can think about it this way. If one of the compounds in the system (reactant or product) is missing, then it CANNOT collide with anything else and that reaction CANNOT occur at any rate. Therefore, the other reaction, no matter how slow, MUST be faster. Therefore the system is NOT at equilibrium.

Magnetic Quantum Numbers and Spatial Orientation

As discussed here, p orbitals are created when an orbital has one angular node. We used the example of the xy plane to show this, and that resulted in what is called the p_z orbital. However, we could just as easily have chosen the xz plane or the yz plane. In fact, Schrödinger’s math allows for all three of these possibilities, as shown here. 

Which of these orbitals is generated by the math depends on the value of the variable m. It is not possible to say definitively which value of m gives which orbital, because atoms don’t actually have axes, but by convention (in other words, by agreement) we say that m=-1 is the p_x orbital, m=0 is the p_y and m=1 is the p_z orbital.

We know that there are three possibilities because Schrödinger’s equation puts a numerical restriction on the possible values of m. 

Specifically, m can be any whole number from -ℓ to 0 to ℓ. 

So, if ℓ=1 (for a p orbital) then m can be -1, 0 or 1. If ℓ=3, then m has values of -3, -2, -1, 0, 1, 2, and 3. That means that on a given energy level, there can only be 1 s orbital (ℓ=0, m=0), there can be 3 p orbitals (ℓ=1, m=-1,0,1), 5 d orbitals (ℓ=2, m=-2,-1,0,1,2), etc.

Solving Problems with a System at Equilibrium

Remembering our mathematical definition of equilibrium - that a system is at equilibrium when the equilibrium expression (Q) is equal to the value of the equilibrium constant (K) - we can imagine a set of relatively simple problems.

Here’s an example:

The system A + 2 B ⇄ C + D is at equilibrium when the [A] = 1.22 M, [B] = 0.56 M, [C] = 1.00 M and [D] = 0.805 M, what is the value of the equilibrium constant?

First, we need to write the equilibrium expression:

Since we know that the reaction is at equilibrium, we know that this mathematical statement is true and we can just solve for the missing piece - in this case K.

Sometimes we need to do a little algebra:

The system 2 W + A ⇄ 2 G has a K = 1.35x10^-7. If the system is at equilibrium when [W] = 2.50 M and when [A] = 1.00 M, what is the concentration of G?

Again, we know that the system IS at equilibrium, so the equilibrium expression is true:

In this case, what we need to solve for is part of the fraction, so we’ll need to rearrange

And then take the square root

Then, we can substitute in the values we know and solve:

Angular Momentum Quantum Numbers and the Shapes of Orbitals

The angular momentum quantum number (ℓ) tells the number of angular nodes in an orbital.

Let’s pull that apart and try to make sense out of it.

Let’s start by imagining an s orbital. This is a simple sphere. We can think of this like the sound wave around a violinist that was discussed here. An s orbital looks something like this:

For the sake of this discussion, we need to impose a set of axes on this picture.

Nodes, you recall, are places where the wave has no measurable value. An angular node is a node that can be defined by an angle. These are often planes. For instance, a node could be defined as all points that are 90^o away from the z axis. That would be the xy plane. 

To see what a node on the xy plane would do to the orbital, imagine placing a band around a balloon and tightening it until it is impossibly small (like the belt that a wasp would wear). This would break the orbital into 2 parts creating a p orbital.

A second angular node would cut the orbital into 4 pieces and would produce a d orbital.

So, when ℓ=0, there are no angular nodes and the orbital is an s orbital. When ℓ=1, there is one angular node and the orbital is a p orbital. When ℓ=2, there are 2 angular nodes and the orbital is a d orbital. Likewise, ℓ=3 produces an f orbital.

There is a limitation on the value of ℓ, that is the value of n. Remember, that n tells us the number of nodes present in an orbital and that at the lowest energy level (n=1) there is no node at all. In fact we said that the number of nodes was equal to (n-1). Since ℓ is the number of angular nodes, it can never be more than the number of nodes, therefore ℓ can be any whole number from 0 up to (n-1). So if n=3, ℓ could be 0, or 1, or 2. That means that on level 3, an orbital can be an s orbital, a p orbital, or a d orbital.

Acid and Base Definitions

There are three definitions of acids and bases that you should know about. We’ll address the first two on this page in historical order. Each serves a different purpose and we’ll make sure that you understand why each definition matters as we go through them. The third definition (Lewis') is discussed here.

The Arrhenius Definition

Svante Arrhenius, a Swedish chemist, defined acids and bases for the first time in 1884. He said that:

An acid is something that, when dissolved in water, 

 would produce hydrogen ions (H⁺¹)

and

A base was something that, when dissolved in water, 

would produce hydroxide ions (OH^-1). 

Of course that doesn’t really make the definitions clear, so let’s be more concrete.

Imagine that you have a glass of water. If you add an acid (say HCl) to your water and then check at some later time (perhaps only a moment later) you will notice several changes:

1) The liquid in your glass will taste sour a bunch and
2) if you drop in a mildly reactive metal like magnesium or zinc, you will see bubbles of hydrogen gas formed.

Arrhenius related both of these properties (sour taste and ability to react with metals) to the presence of H⁺¹ ions in the solution. He did NOT explain the source of the hydrogen ions. He just used the presence of the ions to define acids.

We can understand Arrhenius's approach to bases the same way. He does not explain what a base is doing when you add a base to water, only the result.

When you put a base (say NH₃) in water, two things are observed: 

1) the solution will taste bitter and 

2) the solution will react with organic molecules to make soap-like compounds. (You will sometimes see bases described as slippery. This is actually the result of the solution making your skin into soap.)

Pros and Cons of the Arrhenius definition:
Pro: Arrhenius' definition is simple and clear.
Pro: It has real-world application (lemons are sour because they contain acid, pure chocolate is bitter because it contains bases)
Con: Acids and Bases are only defined in terms of their behavior with water


The Brønsted/Lowry Definition

Johannes Brønsted and Thomas Lowry proposed essentially identical theories at nearly the same time and, as such, their names are both attached to the theory. (This is a composite photo, they were competitors, not friends or colleagues.) Both men were attempting to interpret Arrhenius’ definitions and to explain where the hydrogen and hydroxide ions came from.

Before we can do that we need to address one small but crucial item. There is no such thing as a hydrogen ion in aqueous solution. Period. So, what was Arrhenius finding (and tasting)? The simple answer is hydronium (H₃O⁺¹). Hydronium is just a water molecule with an extra H⁺¹ attached, so Arrhenius was pretty close to right.

To understand the Brønsted/Lowry definition of an acid, it is easiest to look at a reaction. Since Arrhenius defined his acids when added to water, let’s start there. We’ll put some HCl into water, like this:

HCl + H2O → 

Arrhenius tells us that this will produce hydrogen ions (although we now know that it makes hydronium ions), so let’s put that in:

HCl + H2O → H3O+1 + 

Now, we know that matter is neither created or destroyed, so we can figure out what else needs to be on the right side.

HCl + H2O → H3O+1 + Cl-1

Brønsted and Lowry noticed that this reaction actually only involved the movement of one ion. Specifically, a hydrogen moved from the chlorine (leaving behind its electron) to the water. Of course, you remember that a hydrogen atom is made of a single proton and electron, so they were noticing that the HCl was losing a proton (an H+1 ion) which was being transferred to the water molecule to create hydronium. They landed on this definition of an acid:

An acid is a proton donor

A few things to notice:

a) We have ions with charges in this reaction. That’s something we’ve only ever seen in net-ionic equations before

b) The total charge on the right is the same as the total charge on the left (this is another aspect of conservation that we haven’t stressed about before)

Now let's look at a basic reaction, specifically we’ll drop some ammonia (NH3) into water. We know, according to Arrhenius that this reaction will produce hydroxide, so we can start with:

NH3 + H2O → OH-1 +

Remembering the conservation of matter, we can determine the other product:

NH3 + H2O → OH-1 + NH4+1 

Once again, the reaction involves ONLY the movement of an H+1 ion, in this case from the water to the ammonia. That left Brønsted and Lowry with the definition of a base:

A base is a proton acceptor 

Defining acids and bases this way has a HUGE consequence - the action of an acid (giving a proton) requires the presence of a base (taking the proton).

If we then remember that reactions can run in both directions, we end up with acids and bases on BOTH sides of the equation: 

There are three additional consequences of this definition

a) Reactions don’t have to involve water 

b) Some things (like water) can behave as either an acid or as a base

c) The existence of Conjugate Pairs

Let’s tackle those one at a time

a) According to Brønsted and Lowry, these reactions involve the transfer of an H+1 from one thing to another, but water is not specifically part of their definition, so we could have a reaction between HCl and NH3 which would look like this:

b) If you look at the reactions between HCl and water and between NH3 and water below, you’ll notice that water is listed as a base in one and an acid in the other. 

Compounds and ions that can react as either an acid or a base are called amphoteric, and there is more information here.

c) If you look closely at the two reactions above you may notice that in all cases the base on one side of the reaction is ALMOST the same as the acid on the other side. In fact, not only are the similar, they differ by a SINGLE H+1 ion. Here are the pairs in the two reactions above so that you can clearly see what I mean: 

Pairs of this type (one acid and one base that differ by a single H⁺¹) are called conjugate pairs. The language is used in a number of ways:

HCl and Cl^-1 are a conjugate pair

Cl^-1 is the conjugate base of HCl

HCl is the conjugate acid of Cl^-1  

Pros and Cons of the Brønsted/Lowry definition:
Pro: Explains what is going on at the molecular level

Pro: Works with or without water

Pro: Helps define conjugate pairs

Con: Not easily applied to "real-life"

Electron Configuration - Positive Ions

Now that you have a sense of electron configurations and how electrons are arranged in the orbitals of an atom, it is time to look at ions, specifically positive ions. Some positive ions can be quite simple to understand, while others are a bit messier. We’ll start with the easy ones.

Easy positive ions

Remember that a set of orbitals with the same energy is most stable when it is empty, half-filled or filled. To understand negative ions, we will focus on the last of those three possibilities - when the orbitals are filled and in particular how it applies to the p block of the periodic table.

The p block can hold up to 6 electrons (3 orbitals, 2 electrons in each). The result of that is a column of remarkably stable elements - the noble gases.

Elements that are just beyond those elements often lose electrons to other atoms, and in the process achieve the same electron configuration as a noble gas. For instance, sodium (#11) has an electron configuration of 1s2 2s2 2p6 3s1, which is just one electron more than that of neon (1s2 2s2 2p6). If another atom steals an electron from sodium, the electron configuration becomes the same as that of stable neon. 

Vocabulary alert:

We say that the Na+1 ion is isoelectronic with neon — that is, it has the same electron configuration. That does NOT mean that they are the same. Different numbers of protons in the nucleus and a different overall charge cause them to behave differently.

In the same way, magnesium atoms often lose 2 electrons to become Mg+2 ions which are isoelectronic with neon, and aluminum atoms often lose 3 electrons to become Al+3 ions.

As an aside, it is never technically correct to say that an element gives away an electron (although you will hear this all the time.) No positive nucleus has ever given away negative electrons. However, it turns out that the metals on the left side of the periodic table have fairly low ionization potential. That means that it is relatively easy for another atom to steal an electron or two from them. As a result, they are often found as positive ions.

More complex positive ions

Elements that are further away from the noble gases may also lose electrons to other atoms making positive ions. However, the further an element is from the noble gases, the less likely it is that another atom will take enough electrons to make the ion isoelectronic with the previous noble gas.

At that point, it becomes difficult (or impossible at this point in your chemistry education) to predict how many electrons will be lost, and therefore to predict the charge on the ion. 

We can still write the electron configuration of the ion once we know it’s charge however. It is, unfortunately, not as simple as taking off the right number of electrons. There’s a catch.

Electrons are always removed from the highest numbered orbital first.

Let’s use iron as our working example. Iron (#26) is found as an atom in nature, but more commonly as either a +2 or +3 ion. The +2 ion has lost two electrons so that it now has only 24 electrons and the +3 ion has lost three electrons and now has 23 electrons.

Let’s see how that works. Iron has an electron configuration of [Ar]4s2 3d6. If the atom loses two electrons, it is tempting to think that the configuration would be [Ar]4s2 3d4, but that is WRONG!

We need to follow the rule above, so the two electrons are removed from the 4s orbital, not the 3d orbitals.

The Fe+3 ion loses the 2 electrons in the 4s orbital and then 1 from the 3d (since the 4s orbital is then empty). So the electron configuration of Fe+3 is [Ar]3d5

Things get even more complicated when the element is in the p block. As an example, let’s look at tin (#50).

Tin is commonly found in nature with either a +2 or +4 charge (a loss of 2 or 4 electrons). The atom has an electron configuration of [Kr]5s24d105p2. When tin loses two electrons, they are taken from the 5p orbital, leaving [Kr]5s24d10. When two additional electrons are taken, they are NOT taken from the 4d. Instead, following the rule above, they are taken from the 5s orbital, leaving the Sn+4 ion with a configuration of [Kr]4d10.