Counting Orbitals: Part II

As sort of a Counting Orbitals I— Appendix, I have invested in a set of painted 100 1" wooden cubes. They can be obtained from your local or online retailer in 8 colors—4 colors with 13 cubes and 4 other colors with 12. Please refer to the PowerPoint presentation (PowerPoint slides are labeled in this text as, PP2 for slide #2), PP2. The PP presentation is included as Supporting Information. We return to the simplified representation of orbitals as simple cubic blocks. 

In Figure 1 below, we see five sets of 18 cube/orbitals. With a set of three colors (4 + 9 + 5 = 18 blocks), a class of 20 (five groups of four students each) can represent orbitals up to krypton, Z = 36. In each set in Figure 1 and the left image in PP2, the blocks are arranged for the hydrogen-like orbitals with energy (not precisely to scale) on the vertical axis. The 4d and 4f subshells are not shown. 

Figure 1

As you follow along in PP2 (screenshot of the slide is below), you can see the shell and subshell labels on the hydrogen-like orbitals. The energies of all orbitals in the same shell are equivalent (degenerate), and as n increases, so does the energy, for example 1 < 2 < 3 < 4. These same shells and subshells are shifted in energies to illustrate that in a multi-electron system the subshell energies increase with the quantum number, l. The shift of subshell energies can be seen in the Shell Order, for example 3s < 3p < 3d. The Periodic Table Order is illustrated to the right, and we can see the two increasing energy trends with n and l. These balance each other in Period 4 with n =3, l = 2 (3d) and n = 4 , l = 0 (4s) where the energy of 3d is about the same as for 4s. 

Counting Orbitals

A good place to start to count bonding hybrid orbitals is with your earliest counting tool, fingers. P-block elements have four valence orbitals—one s-orbital and three p-orbitals—so, in Figure 2 (also PP3) we see labeled fingers on a hand for the four atomic orbitals. As clear as the hand before your face, the hand has clear advantages for the student and instructor. All realize that: a. there are five fingers on a hand, and b. the thumb is different from the other fingers. As only four fingers are needed for this demonstration, you see that I have taped down my left pinky finger (1), and have labeled "s" on the thumb ("the same but different" from the other fingers), "p" on the fingers (2).

 Figure 2: Image of the author's hand with orbital labels

 "Orbitals in = orbitals out" is a fundamental wavefunction mixing rule. Mixing the full set of four valence orbitals creates four hybrid orbitals (each of which can accommodate 2 electrons), and these hybrids can be used between atoms (bonding) or only on one atom (lone pair or non-bonding).  A highly successful and elegant model that uses hybrid orbitals is VSEP Repulsion Theory or just “VSEPR”. The idea is a simple one. Filled orbitals- hybrid ones in our context- keep out of each other's way, or could be said to repel each other. VSEPR's simplicity carries with it many consequences, and unpacking these consequences can be confusing to the student. This is where counting can give a simplified portal to the unfolding VSEPR's richness and its connections to hybridization, bonding and molecular shape.

The language of hybridization is the necessary start and initial source of confusion. The confusion starts when all four valence orbitals are mixed to form hybrid orbital label "sp³" (pronounced in one word "Ess-Pee-Three"). The label sp³ is confusing for what it is NOT. It IS NOT: 1) on four atoms, 2) related to the cube of anything, or 3) four orbitals that are multiplied in the wavefunction. It IS one of a set of four equivalent, symmetrical orbitals, all labeled sp³.

The persistent problem students have with hybrid orbital labels is one of counting. You will ask, "How many orbitals in an sp³ hybrid set?" And as it is very common to parrot the last number heard, they will answer, "Three". But you can short-circuit this by counting and wiggling the fingers: 1-2-3-4. Emphasizing the counting should remove exponents and multiplications from the label. It also has the advantage that each student has a version at their arm's end so they can be reminded whenever the connection bears repeating. And it bears repeating with the other hybrid orbitals, sp² and sp.

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Start With Pi (π) Bonds

"A double bond minus one gives one π-bond and a σ-bond." This amazing student mash-up of concepts has plenty of truth in it. It curiously shows that the student easily recognizes from a Lewis structure the number of π-bonds. So what that they may not know yet what a π-bond or a σ-bond look like, or they mix together parts of speech ('double' then 'one' then 'a'); they correctly count π-bonds. If we elaborate from the "Orbitals in = orbitals out" rule that when a π p-orbital on one atom bonds with the p-orbital on an adjacent atom, it cannot be used in a hybrid, and thus is "counted out" of the hybrid.  Let us take a look at Table 1 below and look back to the hand (Figure 2) to illustrate.

Table 1: Links between π-bond number, VSEPR and related properties. Also seen in PP4.

Draw the lowercase Greek letter π between the knuckles of your middle and ring fingers middle fingers as seen in PP5 . With no π-bonds, we still see 4 sp³ hybrids. On the next line of Table 1, we have one π-bond, so on the hand, pull down one p-orbital to reveal a π-bond with the remaining three sp² hybrids.  Finally, pull down another p-orbital to reveal another π-bond leaving two hybrid orbitals: sp. 

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A Few Examples Are In Order

In PP6 (screenshot below) from left to right, I lay out acetylene, ethylene, allene and methane. The triple bond of acetylene has two π-bonds, so two p-orbitals go out of the set of four, and the hybridization is sp. The double bond of ethylene has one π-bond, so one p-orbitals go out of the set of four, and the hybridization is sp². Allene is more challenging and more instructive as it has two types of hybrid bonds in a small package. Emphasizing that the p-orbital is counted in each atom of the π-bond, the outer two carbons are double-bonded, so each has one π-bond, and the hybridization is sp². The central carbon is twice double bonded, so it has one π-bond to the left and one to the right. Count two p-orbitals out, the hybridization is sp. 

PP7 (screenshot below) shows the same behavior with the nitrogen-containing compounds, hydrogen cyanide, methylene imine, hydrazoic acid and ammonia. Emphasize that hybrid orbitals can be determined by counting π-bonds, and the hybridization is named after the component atomic orbitals and the hybridized atomic orbital count IS the hybridized orbital count. Table 1 includes VSEPR angles and shapes for completeness, but in our slow-reveal, it will take us a while to address them. 

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Back to Blocks

Continuing with PP8, similar to the "Counting Orbitals I" blog, we get back to the block simplification: counting first then shapes. We begin with the set of p-block valence orbitals. There are three p-orbitals—this time directionally labeled p_z, p_y and p_x—plus one red s-orbital plus.

PP9 (screenshot below) goes over the three types of hybridization. From left to right, a) sp: two unmixed two hybridized, b) sp²: one unmixed and three hybridized, and c) sp³: no unmixed and four hybridized. 

PP10 (screenshot below) shows the mixing of the orbitals in three ways: 1) energetically, 2), by color and 3) spatially and then finally recapitulates the hand-counting. This gets us to the aforementioned elegant, simple, confusing consequences and my attempts to layer different, simplified views.  

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Energetically and Color

Though I have the set of four all together initially, the 2s orbital is lower in energy than the 2p subshell. And as you can see in PP10 immediately above, using color, you can express orbital mixing naturally and with ease. Thus, I represent the 2s with a red cube and the 2p's with yellow, just as red light is of lower energy light than yellow. Red and yellow mix to give orange which is intermediate in energy and the more yellow in color (p-orbital contribution).

In the unbound atom, the sp hybrid is both degenerate and lower in energy than the unhybridized, degenerate p_x and p_y orbitals. That is, the two sp hybrids are symmetrically enforced to be both spatially equivalent (even if they are not directionally equivalent—one pointing left and one right) and energetically averaged and equivalent. The increasing p-character from sp to sp² then to sp³ is reflected in energetic/color shifts upward/toward yellow.

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Spatial Representation

Much of the difficulty students have in dealing with hybrids is the age-old problem of representing three-dimensions on a flat surface. Obviously, seen in PP10, the linear relationship between the two sp hybrids is not difficult to show. As 3 points define a plane, three sp² hybrids can be shown on a flat surface. However, if you put the sp² hybrids in the plane, the remaining p_y orbital comes directly in and out of the page. When using simple cubes, even directionally labeled ones as I have, you lose the ability to show the contrasting "in and out of the page". However, three simple cubes can be arranged in perspective to show a trigonal planar arrangement, albeit somewhat imperfectly, and the p_y is within the page. 

It is the pesky four orbital, 3-dimensional set of the sp³ hybrids that creates the real challenge. The following may be an unfamiliar way of viewing it, but follow me. This is where the math-manipulative cubes, even more than the cubes in perspective in PP10, can help us out. More than just counting sp³ hybrid orbitals, these simple wooden cubes, with some instructions can be put in a more recognizable tetrahedral arrangement. And, because of the difficulty and 3-dimensionality, it needs repetition, so I will go over the sp³ hybrid a couple of times, before I take one pass through the other types of hybrids. 

Start with this bullet-point set, and I encourage you to follow along with in the PowerPoint, one slide for each of the following headings. 

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Create a Tetrahedron from Cubic Blocks I: Get Mental (PP11)

a. Think of space split into eight: split in half left to right, then top to bottom, finally front to back (Figure 3 below).

• Think of four smaller cubes in a square, that is that the plane they are on is split in half front to back and left to right.
• With the bottom half set, add four more cubes on top of these.
• The shared edges of the smaller cubes define the Cartesian axes x, y and z within the larger cube. 

b. Eight smaller cubes make a larger cube.

c. Half of the eight, four, still have cubic symmetry.   

• Select every other one of the smaller cubes.
  • Two from the opposite corners on the top.
  • Two from the opposing opposite corners on the bottom.
• The remaining four smaller cubes will be in a tetrahedral arrangement. 

The template for creating tetrahedral arrangement of cubic manipulative blocks is shown below in Figure 3.

Figure 3: Tetrahedron from Cubic Blocks (Also, see PP11)

Create a Tetrahedron from Cubic Blocks II: Get Manipulative (PP12)

a. Get your small cubes and have a pair of students each hold two cubic blocks: student one, color one, student two, color two in Figure above or PP11. 

• Have a piece of paper with four squares around the x-y axis. 
• Have squares on the opposite corners shaded.

b. On the paper, Student 1 uses the shaded square and Student 2 the un-shaded.  Follow along PP12.

• Each student puts their cubes on their respective squares. 
• Student 1 lifts their pair of cubes so their cube bottoms meet Student 2’s cube tops.

c. They now have a tetrahedral arrangement.

• Each cube should be touching edges of three adjacent cubes.
• The free outside corners of any three adjacent cubes make a triangular face.
• This is tricky, but if they can move in concert and hold an above-mentioned triangular face parallel to their desk, the fourth cube will point straight up.

d. This tetrahedral arrangement shows precisely the same symmetry on any of the four faces.

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Give Them Some Space

Who am I to complain about the shape of orbitals? Here I am making little cubic blocks equal every orbital, be it s, p, d, f, or any of a number of hybrids. At some point, the student needs to move from counting the orbitals to their shape and General Chemistry textbooks have so many poor figures of orbitals, particular hybrid orbitals (3). Though the orbital shapes I use are simplifications, they are better than most. 

In Figure 4 below (and on PP13) you will see a reasonably accurate probability surface of a 2p orbital and the outer part of a 2s orbital. I am not going to go into all the subtleties of their inaccuracy, but just point out that the figures have the same boundary top-to-bottom and centered on the z-axis. The hybrids have the same top-to-bottom height, but as you see with the bracket, then are not symmetrical around the z-axis. These hybrid orbital representations-

• have the correct number of planar nodes: one,
• show as p-orbital contribution increases (sp to sp² to sp³ to p), also as the p-orbital contribution increases, the more shaded, more prominent lobe decreases in size (3/4 to 2/3 to 5/8 to 1/2), and
• when I finally get back to unifying all of the types of views of hybrids, the shaded lobe of the hybrid will be represented by the hybrid orbital cube. 

Figure 4: Hybrid and Atomic Orbital (where bracket [ ] represents position and scale; taken from ChemDraw 20.1.1.125)

So, cherished readers, as I attempt to translate from my simplified shapes to cubes and finally back to fingers, what I will present to you is not a lie, but simplification. 

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Trees and Forests

One difficulty with orbitals, unfortunately one that some students never get past, is any method that clearly shows one orbital cannot show all orbitals all at once. Let me illustrate in PP14 (Figure 5 below). 

Figure 5

The initial view has individual n = 2 shell orbitals on the right of the "≡" equivalent sign (the trees). The equivalent view is on the left of the equivalent sign (the forest). Which is better the trees-view or the forest-view? I dunno. Taking a hint from the phrase, all at once, I will try to bring together in PP14, spatially and conceptually, the two equivalent sides by animating the Slide Show one orbital at a time. Activate the Animation on PP14 one click at a time and the orbitals will fly-in from the right and go to the unified view on the left, thus creating the forest one tree at a time. I am going to pursue this method in the following slides with hybrid orbitals with shapes and cubes (4). 

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 There is a lot going on here in the Forest. Wandering through it a couple of times, perhaps with a quick probe of understanding by Clicker or Socratic questioning, might help them find their way on the path.

(1) Some people can facilely hold down the left pinky finger while not moving their ring finger (usually those who play a stringed instrument). Also, some might recall a story of missing fingers told in the Roald Dahl short story “Man from the South", "Man from the South and Other Stories", Pearson, 2nd Rev. Ed., 2021, ISBN: 1405882662, or in the Alfred Hitchcock Presents television adaptation (1960, Season 5, Episode 15) starring Steve McQueen, Peter Lorre and Nellie Adams.

(2) Left-handed writers would need to label the fingernails of the right hand to maintain the thumb-to-the left orientation. However, it does make the following discussion of π-bonding less clear.