2.6: Orbitals, Electron Clouds, Probabilities, and Energies (2023)

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    Our current working model of the atom is based on quantum mechanics that incorporate the ideas of quantized energy levels, the wave properties of electrons, and the uncertainties associated with electron location and momentum. If we know their energies, which we do, then the best we can do is to calculate a probability distribution that describes the likelihood of where a specific electron might be found, if we were to look for it. If we were to find it, we would know next to nothing about its energy, which implies we would not know where it would be in the next moment. We refer to these probability distributions by the anachronistic, misleading, and Bohrian term orbitals. Why misleading? Because to a normal person, the term orbital implies that the electron actually has a defined and observable orbit, something that is simply impossible to know (can you explain why?)

    Another common and often useful way to describe where the electron is in an atom is to talk about the electron probability density or electron density for short. In this terminology, electron density represents the probability of an electron being within a particular volume of space; the higher the probability the more likely it is to be in a particular region at a particular moment. Of course you can’t really tell if the electron is in that region at any particular moment because if you did you would have no idea of where the electron would be in the next moment.

    Erwin Schrödinger (1887–1961) developed, and Max Born (1882–1970) extended, a mathematical description of the behavior of electrons in atoms. Schrödinger used the idea of electrons as waves and described each atom in an element by a mathematical wave function using the famous Schrödinger equation (\(H \Psi=E \Psi\)). We assume that you have absolutely no idea what either \(H \Psi\) or \(E \Psi\) are but don’t worry—you don’t really need to. The solutions to the Schrödinger equation are a set of equations (wave functions) that describe the energies and probabilities of finding electrons in a region of space. They can be described in terms of a set of quantum numbers; recall that Bohr’s model also invoked the idea of quantum numbers. One way to think about this is that almost every aspect of an electron within an atom or a molecule is quantized, which means that only defined values are allowed for its energy, probability distribution, orientation, and spin. It is far beyond the scope of this book to present the mathematical and physical basis for these calculations, so we won’t pretend to try. However, we can use the results of these calculations to provide a model for the arrangements of electrons in an atom using orbitals, which are mathematical descriptions of the probability of finding electrons in space and determining their energies. Another way of thinking about the electron energy levels is that they are the energies needed to remove that electron from the atom or to move an electron to a “higher” orbital. Conversely, this is the same amount of energy released when an electron moves from a higher energy to a lower energy orbital. Thinking back to spectroscopy, these energies are also related to the wavelengths of light that an atom will absorb or release. Let us take a look at some orbitals, their quantum numbers, energies, shapes, and how we can used them to explain atomic behavior.

    Examining Atomic Structure Using Light: On the Road to Quantum Numbers

    J.J. Thompson’s studies (remember them?) suggested that all atoms contained electrons. We can use the same basic strategy in a more sophisticated way to begin to explore the organization of electrons in particular atoms. This approach involves measuring the amount of energy it takes to remove electrons from atoms. This is known as the element’s ionization energy which in turn relates directly back to the photoelectric effect.

    All atoms are by definition electrically neutral, which means they contain equal numbers of positively- and negatively-charged particles (protons and electrons). We cannot remove a proton from an atom without changing the identity of the element because the number of protons is how we define elements, but it is possible to add or remove an electron, leaving the atom’s nucleus unchanged. When an electron is removed or added to an atom the result is that the atom has a net charge. Atoms (or molecules) with a net charge are known as ions, and this process (atom/molecule to ion) is called ionization. A positively charged ion (called a cation) results when we remove an electron; a negatively charged ion (called an anion) results when we add an electron. Remember that this added or removed electron becomes part of, or is removed from, the atom’s electron system.

    Now consider the amount of energy required to remove an electron. Clearly energy is required to move the electron away from the nucleus that attracts it. We are perturbing a stable system that exists at a potential energy minimum – that is the attractive and repulsive forces are equal at this point. We might naively predict that the energy required to move an electron away from an atom will be the same for each element. We can test this assumption experimentally by measuring what is called the ionization potential. In such an experiment, we would determine the amount of energy (in kilojoules per mole of molecules) required to remove an electron from an atom. Let us consider the situation for hydrogen (\(\mathrm{H}\)). We can write the ionization reaction as: \[\mathrm{H} \text { (gas) }+\text { energy } \rightarrow \mathrm{H}^{+} \text {(gas) }+\mathrm{e}^{-}.\][15]

    What we discover is that it takes \(1312 \mathrm{~kJ}\) to remove a mole of electrons from a mole of hydrogen atoms. As we move to the next element, helium (He) with two electrons, we find that the energy required to remove an electron from helium is \(2373 \mathrm{~kJ/mol}\), which is almost twice that required to remove an electron from hydrogen!

    Let us return to our model of the atom. Each electron in an atom is attracted to all the protons, which are located in essentially the same place, the nucleus, and at the same time the electrons repel each other. The potential energy of the system is modeled by an equation where the potential energy is proportional to the product of the charges divided by the distance between them. Therefore the energy to remove an electron from an atom should depend on the net positive charge on the nucleus that is attracting the electron and the electron’s average distance from the nucleus. Because it is more difficult to remove an electron from a helium atom than from a hydrogen atom, our tentative conclusion is that the electrons in helium must be attracted more strongly to the nucleus. In fact this makes sense: the helium nucleus contains two protons, and each electron is attracted by both protons, making them more difficult to remove. They are not attracted exactly twice as strongly because there are also some repulsive forces between the two electrons.

    The size of an atom depends on the size of its electron cloud, which depends on the balance between the attractions between the protons and electrons, making it smaller, and the repulsions between electrons, which makes the electron cloud larger.[16]The system is most stable when the repulsions balance the attractions, giving the lowest potential energy. If the electrons in helium are attracted more strongly to the nucleus, we might predict that the size of the helium atom would be smaller than that of hydrogen. There are several different ways to measure the size of an atom and they do indeed indicate that helium is smaller than hydrogen. Here we have yet another counterintuitive idea: apparently, as atoms get heavier (more protons and neutrons), their volume gets smaller!

    Given that

    1. helium has a higher ionization energy than hydrogen and
    2. that helium atoms are smaller than hydrogen atoms, we infer that the electrons in helium are attracted more strongly to the nucleus than the single electron in hydrogen.

    Let us see if this trend continues as we move to the next heaviest element, lithium (\(\mathrm{Li}\)). Its ionization energy is \(520 \mathrm{~kJ/mol}\). Oh, no! This is much lower than either hydrogen (\(1312 \mathrm{~kJ/mol}\)) or helium (\(2373 \mathrm{~kJ/mol}\)). So what do we conclude? First, it is much easier (that is, requires less energy) to remove an electron from \(\mathrm{Li}\) than from either \(\mathrm{H}\) or \(\mathrm{He}\). This means that the most easily removed electron in \(\mathrm{Li}\) is somehow different than are the most easily removed electrons of either \(\mathrm{H}\) or \(\mathrm{He}\). Following our previous logic we deduce that the “most easily removable” electron in \(\mathrm{Li}\) must be further away (most of the time) from the nucleus, which means we would predict that a \(\mathrm{Li}\) atom has a larger radius than either \(\mathrm{H}\) or \(\mathrm{He}\) atoms. So what do we predict for the next element, beryllium (\(\mathrm{Be}\))? We might guess that it is smaller than lithium and has a larger ionization energy because the electrons are attracted more strongly by the four positive charges in the nucleus. Again, this is the case. The ionization energy of \(\mathrm{Be}\) is \(899 \mathrm{~kJ/mol}\), larger than \(\mathrm{Li}\), but much smaller than that of either \(\mathrm{H}\) or \(\mathrm{He}\). Following this trend the atomic radius of \(\mathrm{Be}\) is smaller than \(\mathrm{Li}\) but larger than \(\mathrm{H}\) or \(\mathrm{He}\). We could continue this way, empirically measuring ionization energies for each element (see figure), but how do we make sense of the pattern observed, with its irregular repeating character that implies complications to a simple model of atomic structure?


    Questions to Answer

    • Why are helium atoms smaller than hydrogen atoms?
    • What factors govern the size of an atom? List all that you can. Which factors are the most important?

    Questions to Ponder

    • What would a graph of the potential energy of a hydrogen atom look like as a function of distance of the electron from the proton?
    • What would a graph of the kinetic energy of an electron in a hydrogen atom look like as a function of distance of the electron from the nucleus?
    • What would a graph of the total energy of a hydrogen atom look like as a function of distance of the electron from the proton?


    What is the probability of finding an electron between orbitals? ›

    Orbitals are commonly represented by the boundary surfaces that encloses the region where there is a 90-95 % probability of finding the electron.

    What is the probability of finding the electron answer? ›

    - From the above discussion it's clear that an electron orbital is most commonly defined as the radius of the sphere that encloses 95 % of the total electron probability and the probability of finding an electron in an orbital is approximately 95%.

    How many electrons can fit in the first 3 shells of the electron cloud? ›

    The first shell (closest to the nucleus) can hold two electrons. The second shell can hold 8 electrons. The third shell can hold 32 electrons. Within the shells, electrons are further grouped into subshells of four different types, identified as s, p, d, and f in order of increasing energy.

    How many electrons can fit in the electron cloud? ›

    Each orbital can hold no more than two electrons. So, each s sublevel can have two electrons, each p sublevel can hold six electrons, etc.

    What is the probability of finding a 2p electron within a nodal plane? ›

    Nodal Planes:

    Nodal planes are planes that are assumed by scientists, so they do not actually exist. They are simply used to simply the study and visualization of the distribution of electrons. In nodal planes, the probability of finding an electron is zero.

    What do orbitals do with probability? ›

    In atomic theory and quantum mechanics, an atomic orbital (/ɒrbədl/) is a function describing the location and wave-like behavior of an electron in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus.

    What is the 2 8 8 rule in chemistry? ›

    Elements react in ways to obtain a full outer shell, whether by sharing, losing or gaining electrons (see our videos on bonding and types of compounds). • The 1st, 2nd, and 3rd shells can hold 2, 8 and 8 electrons respectively. Once a shell is filled up, any further electrons must fill in a new higher level shell.

    What is the rule for electron shells? ›

    Each shell can contain only a fixed number of electrons: the first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.

    What is the 2 8 8 18 rule in chemistry? ›

    It is an arrangement of electrons in various shells, sub-shells and orbitals in an atom. It is written as 2, 8, 8, 18, 18, 32. It is written as nlx ( where n indicates the principal quantum number), l indicates the azimuthal quantum number or sub-shell, and x is the number of electrons.

    How many electrons can the first energy level in the electron cloud hold? ›

    The first principal energy level contains only an s sublevel; therefore, it can hold a maximum of two electrons. Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons.

    How do you count electron clouds? ›

    # of Bonds: Regardless of having a multiple bond, chemical bond(s) between two nuclei is considered as one charge cloud. # of Lone Pairs: Count the number of lone pair electrons. Total # of Charge Cloud: Add the # of bonds and # of lone pairs.

    How many electrons can each energy level fit? ›

    Questions and Answers
    Energy Level (Principal Quantum Number)Shell LetterElectron Capacity
    2 more rows

    Why is the probability of finding the electron in an orbital never 100%? ›

    100 % cannot be used since that would put everything into an orbital and therefore the concept of an orbital would lose its usefulness.

    Why is the probability of finding an electron never 100%? ›

    The answer is at any distance from the nucleus, the probability density of finding an electron is never zero. It will always have some finite value, so it is not possible to draw a boundary surface diagram which encloses a region with 100% probability density.

    What is the probability of finding an electron in p orbital? ›

    px orbital is a dumbbell-shaped orbital along the x-axis. So, the probability of finding the electron in px orbital is maximum on the x-axis and zero at the nucleus.

    What is the probability of finding an electron everywhere? ›

    The diagrams cannot, however, show the entire region where an electron can be found, since according to quantum mechanics there is a non-zero probability of finding the electron anywhere in space.


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