General Chemistry Review – Valence Electrons of The First Row Elements

1. The Most Critical Graph Ever Created?


Which statement would contain the most information in the fewest words if all scientific knowledge were to be lost in a cataclysm and only one sentence were to be passed down to the next generation of creatures? All things, according to the atomic hypothesis, are composed of atoms, which are tiny particles that are in a state of perpetual motion, attracting each other at a distance but repelling when pressed together. If you apply a little imagination and critical thinking, you will discover that this single sentence contains a wealth of information about the world.

–Richard P. Feynman

Obviously, a chemist like myself will appreciate a quote that emphasises the centrality of chemistry to scientific knowledge.  You can always rely on a barber to find an encouraging quotation about the significance of haircuts.

Aside from this proviso, if I had to choose a single graph that could survive a cataclysm for the next generation of creatures, I would choose this one.

What is happening here?

The x-axis lists each element of the periodic table in ascending atomic number order.

The y-axis represents the amount of energy required to ionise each neutral element to a +1 charge ("first ionisation energy").  In other words, "how straightforward is it to remove an electron from each element?”.

This extremely simple plot contains a wealth of information about atomic structure, but the concept itself is not difficult to grasp: "how much energy does it take to rip an electron off a neutral element?”

Two essential points to note:

First, observe the general trend: the amount of energy required to pull an electron away from an atom decreases as the size of an element increases. The further away an electron (planet) is from a nucleus (sun), the weaker the attractive force between them will be (as measured by Newton's law of gravitation in one case and Coulomb's law in the other). In fact, this "planetary" analogy served as the foundation for the Bohr model of the atom.

Second, observe the periodic trend: certain elements (He, Ne, Ar, Kr, Xe) have unusually high ionisation energies, followed by those with unusually low ionisation energies (Li, Na, K, Rb, and Cs).

2. Periodic Filling of Electronic Energy Levels (aka "Orbitals") Explains the Graph.

Observe how naturally interpreting the above diagram leads to a discussion of orbitals.

Bohr was the first to make the connection between the periodicity of atomic properties and the periodic filling of electronic energy levels (1923).

In a 1913 model of the atom, it was assumed that electrons orbited the nucleus at progressively greater distances, similar to the planets in their ever-expanding celestial spheres around the sun. While Bohr's "electronic energy levels" are now referred to as "orbitals," the analogy ends there.  Bohr did not anticipate the peculiar behaviour of electrons relative to planets. Nor did anyone else in 1913. [Note 1]

What do I mean by "weirdly-behaved"? If one knows the position and velocity of Venus, for instance, one can calculate its positions at distant future times (such as the transit of Venus predicted for 10 June at 3:48 a.m. UTC).

Heisenberg demonstrated that there are limits to the precision with which one can determine the position and momentum of a subatomic particle, such as an electron. As a result, our knowledge of the precise positions of electrons is imprecise; they must be expressed as probability functions. What we call "orbitals" are actually three-dimensional shapes with a 95% chance of containing an electron with a given set of quantum numbers. [Note 2]

3. n, l, and m Are The Three Quantum Numbers That Define Orbitals.

The three essential terms in a particular form of Schroedinger's wave equation define the properties of these orbitals.

The principal quantum number n (1, 2, 3) is sometimes referred to as the "electron shell" because it relates to the distance of electrons from the nucleus.

l (also known as the azimuthal quantum number; however, it is not essential to know this name). For a given value of n, the range of possible values for l is between 0 and (n–1). Therefore, when n = 1, l = 0. When n equals 2, l can take on the values 1 or 0.

The value of l determines the orbital's shape. For l = 0, the orbital is spherical; we refer to this orbital as a s orbital.

For l = 1, which is only possible for n = 2 or greater, the orbital has a dumbbell-like shape and is known as a p-orbital. The higher observed values are l = 2 (d orbitals) and l = 3 (f orbitals), which are fascinating in their own right, but which we will skip in true organic chemistry fashion.

m, which is the quantum number for magnetism. The possible values of m depend on the value of l and range from –l to +l, inclusive of zero. Consequently, for l = 1 (the p-orbital), m can take on the values –1, 0, or +1. The physical interpretation of this is the three possible orientations along the x, y, and z axes for the p-orbital. m can take on five different values for d-orbitals (l = 2) and seven different values for f-orbitals (l =3).

In addition to n, l, and m, there is a fourth quantum number known as electronic spin, which can take on values of either +12 or –12 for electrons.

Lastly, the final bullet point:

In accordance with Wolfgang Pauli's exclusion principle, no two electrons can have the same set of quantum numbers (i.e., they cannot have the same quantum state). (Electrons belong to the family of fermions, which share a property analogous to "one seatbelt per occupant").  It's not a clown car, but rather an atom!

This means that electrons in atoms "build up" in energy level in a well-defined pattern, beginning with 1s (which can hold two electrons with opposite spins), 2s (which also holds two electrons), 2p (which, due to the three possible values of m, can hold 3 2 = 6 electrons), 3s (2 electrons), etc.

4. A Tour of the Electronic Structures of the Initial Eleven Elements

We could continue filling electrons, but that is not the purpose of this post, which is merely to review the atomic configurations of the first 11 elements with reference to their initial ionisation energies.

So, shall we explore the electronic configurations of the first eleven elements?

5. Hydrogen and Helium: The first shell

Consider the (first) energies of ionisation for hydrogen and helium:

Hydrogen: 1312 kJ/mol

Helium: 2372 kJ/mol

Hydrogen has a single electron in the 1s orbital with a spin of either +1/2 or -1/2 (they have the same energy, except in the presence of a strong magnetic field). 

The electronic configuration can be represented in two ways:

1 represents the shell number (principal quantum number n), s represents the shape of the orbital (other values are p, d, and f), and the superscript represents the number of electrons in that orbital.

alternatively, with a box depicting the orientation of electrons in each orbital (the arrow's direction indicating the spin). For a half-filled orbital, the direction of the spin is arbitrary; I will draw it as "up" in this and all subsequent examples, but it is not incorrect to draw it pointing down instead.

Two electrons in the 1s orbital of helium have opposing spins (+1/2 and –1/2), as depicted below.

Almost twice as much energy is required to remove an electron from helium as from hydrogen (2372 kJ/mol versus 1312 kJ/mol).

Why is the ionisation energy higher? According to the Coulomb equation, each electron in the 1s orbital of helium now has two protons from the nucleus pulling on it rather than one, resulting in a stronger attractive force.

Note that it is not exactly twice as much; we can explain the lower value as a result of electron repulsion in the 1s orbital.

6. Valence Electrons of Lithium and Beryllium in the 2s Shell

With the 1s orbital already occupied, the third element (lithium) must have its electron placed in the 2s orbital, which is not only further away from the nucleus but also partially shielded from the attractive force of the nucleus by the pair of electrons in the 1s orbital. This shielding causes the electron in lithium's 2s orbital to "feel" only a nuclear charge of (3 - 2) = 1 from the nucleus.

This electron is therefore exceptionally easy to remove, requiring only 520 kJ/mol (compared to 2372 kJ/mol for helium!). Lithium metal readily participates in chemical reactions; for instance, it reacts slowly with water.  On the contrary, helium has never been observed combining with another element.

In contrast, the two electrons in lithium's 1s shell are never involved in chemical reactions. These electrons can be referred to as "inner shell" electrons, as opposed to the "outer shell" or "valence" electron in the 2s orbital. 

Since all the interesting chemistry occurs with the electrons in the 2s orbital (and not with the electrons in the "inert", closed, 1s shell), it is convenient to represent lithium's electronic configuration as [He] 2s1, which indicates that lithium has the electronic configuration of helium plus one valence electron in the 2s orbital.

In the fourth element, beryllium, the 2s orbital is completely occupied by an electron pair with opposite spin. Note the increased ionisation energy (+379 kJ/mol for lithium) because the extra proton in the nucleus exerts a stronger pull on the electrons.

Notably, despite the fact that beryllium has a filled 2s orbital, it does not behave like a noble gas. For instance, beryllium metal readily reacts with oxygen to form an oxide layer.  In contrast to the energy gap between 1s and 2s, the energy gap between the 2s orbital and the next-highest energy level (2p) orbital is relatively small, such that when both orbitals are filled, they can accommodate a "octet" of valence electrons.

7. The 2p Shell: Boron, Carbon, and Nitrogen Valence Electrons

Now that the 2s orbital is full, additional electrons must be placed in the 2p orbital, which has an even higher energy level.

Each set of p orbitals can hold six electrons, whereas each s orbital can only hold two. Each level of p orbitals (2p, 3p, 4p, etc.) is composed of three individual dumbbell-shaped p orbitals that are aligned at right angles to each other along the x, y, and z axes, according to our interpretation.

Here is how the electronic configuration of boron appears. (Note: we could also choose to label the three 2p orbitals as 2px, 2py, and 2pz; however, for our purposes, all of these orbitals have equivalent energies, so it is unnecessary to do so here). 

Noting that the ionisation energy of boron is 99 kJ/mol lower than that of beryllium, we can conclude that the 2p orbital is slightly more distant from the nucleus than the 2s orbital. [Note 3]. 

The ionisation energy rises precipitously from boron to carbon.... (+ 286 kJ/mol)

...then carbon to nitrogen (+316 kJ/mol):

8. A Sudden Dip At Oxygen

The ionisation energy of oxygen suddenly decreases by –89 kJ/mol:


Have you ever boarded a bus only to discover there are no more empty seats and you must (horrors!) sit next to a complete stranger?

Since each of the 2p orbitals (px, py, pz) is currently occupied by a single electron, any additional electrons must pair up. This decrease in ionisation energy reflects an increase in electron-electron repulsion in a double-occupied orbital, making it (slightly) easier for the electron to escape.

9. Fluorine's Electronic Configuration

From oxygen to fluorine, the ionisation energy again increases (+368 kJ/mol):

10. A Maximum At Neon

Then, at neon, a second local maximum of ionisation energy (+399 kJ/mol from fluorine) is reached:

Here, the 2s and 2p orbitals (acting as a "octet") have reached their maximum capacity, and each electron in the octet will "feel" the maximum effective charge from the nucleus. In neon, each of the eight electrons in the valence "octet" feels a net force of +8 from the nucleus (ten protons minus two electrons in the 1s shell) and a certain amount of electron-electron repulsion in the filled 2s and 2p orbitals.

Neon is a local maximum where both the Coulombic attractive term and ionisation energy are maximised.  This is the origin of the well-known "octet rule" in which atoms are said to "seek out full octets" and so on.

11. Sodium in its 3s orbital

To illustrate the dramatic difference in behaviour between sodium and neon, let's go one atom further.

There is a significant difference between the energies of the n=2 shell orbitals (2s and 2p) and the n=3 shell orbitals (3s and 3p).

The 3s orbital is further from the nucleus, and the 10 electrons in lower-energy orbitals shield it from the attractive force of the nucleus. We state that the "effective nuclear charge" experienced by an electron in the 3s orbital of sodium is merely 1. (i.e. 11 – 10 = 1).

The ionisation energy of sodium is 496 kJ/mol, which is 1,584 kJ/mol lower than that of neon. Sodium readily gives up its valence electron, even violently reacting with water, whereas no other element has ever been observed to combine with neon.

We'll stop here, but you can imagine where the story goes from here. [Note 4]

12. Endnote: A Puzzle. How About A Simple Molecule Such As CH4? 

So here's a question. If the atomic configurations of carbon and hydrogen are [He]2s22p2 and 1s1, respectively, what might the structure of the simplest hydrocarbon (CH4) resemble?

Wouldn't we anticipate seeing C-H bonds along the px, py, and pz axes (bond angles of 90°) and then a fourth C-H bond in the 2s orbital as far from the other electrons as possible (135°)?

Wouldn't we expect different bond lengths for C-H bonds attached to p orbitals (further from the nucleus) versus those attached to s orbitals?

Wouldn't we anticipate CH4 to have a (small) dipole moment?

Instead, the following has been observed regarding methane:

All C-H bonds possess the same bond length (1.09 Angstrom).

All angles of C–H bonds are identical. All H-C-H bond angles are 109.5 degrees, and the hydrogens are arranged around carbon in a perfect tetrahedron.

There is no dipole moment.

What's up? How could this possibly be true, given our current understanding of the s and p orbitals? How can we explain the fact that the distinctions between the 2s and 2p orbitals have disappeared entirely?

So what is happening? This topic will be discussed in the next post.

How Do We Know That Methane Is Tetrahedral?

Again, I appreciate Matt's assistance with this post. Employ Matt as a tutor! 

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