Question

(a) The average distance from the nucleus of a \(3 s\) electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a \(3 s\) electron from the chlorine atom, as compared with a \(2 p\) electron? Explain.

Short answer

The \(3s\) orbital has lower energy than the \(3p\) orbital as it is closer to the nucleus, having more negative potential energy. However, due to the lower principal quantum number, the ionization energy for the \(3s\) electron in chlorine is higher than predicted. Thus, it requires more energy to remove a \(3s\) electron than a \(2p\) electron.

Step-by-step solution

Understand the concept of atomic orbitals and the potential energy of electrons

Atomic orbitals are regions in an atom where the probability of finding an electron is the highest. The energy of an electron in an atom depends on both the principal quantum number (n) and the angular momentum quantum number (l). The potential energy of an electron depends upon its distance from the nucleus. The closer an electron is to the nucleus, the more negative its potential energy and therefore, the lower its energy. On the other hand, the farther away an electron is from the nucleus, the more positive its potential energy and the higher its energy.

Comparing the energies of \(3s\) and \(3p\) electrons

Given that the average distance from the nucleus of a \(3s\) electron in a chlorine atom is smaller than that of a \(3p\) electron, we can conclude that the potential energy of a \(3s\) electron is more negative than that of a \(3p\) electron. Since the potential energy of an electron is related to its total energy, this implies that the \(3s\) orbital has lower energy than the \(3p\) orbital.

Comparing the energies required to remove a \(3s\) electron versus a \(2p\) electron

When removing an electron from an atom, the energy required to do so is called ionization energy. The ionization energy depends on the distance of the electron from the nucleus and the shielding by other electrons. Since the \(3s\) electron is farther from the nucleus and experiences more shielding than a \(2p\) electron, it would be expected that the ionization energy to remove the \(3s\) electron would be lower compared to that of the \(2p\) electron. However, because the \(3s\) electron is closer to the nucleus than the \(3p\) electron and has lower principal quantum number, the ionization energy for the \(3s\) electron in chlorine is higher than predicted, so it requires more energy to remove a \(3s\) electron than a \(2p\) electron.

Key concepts

Principal Quantum Number

The principal quantum number, denoted as "n", is an essential aspect when discussing atomic orbitals. It determines the size and energy level of an electron's orbital. Imagine it as the number that signifies which 'shell' or 'layer' of the atom an electron resides in. Each subsequent larger value of "n" represents a shell that is further out from the nucleus,
and these shells can hold more electrons.

• The value of "n" can be any positive whole number, starting from 1.
• This number primarily indicates the energy of an electron; higher "n" means higher energy and an electron that's further from the nucleus.
• For example, in a chlorine atom, electrons in the 3rd shell (n=3) have higher energy than those in the 2nd shell (n=2).

The principal quantum number also helps determine the maximum number of electrons that can be held in any shell using the formula \(2n^2\).

This means in the third shell, it can hold up to 18 electrons in total. This concept plays a crucial role when considering the overall potential energy and stability of electrons within an atom. Remember, a higher principal quantum number means the electron is less tightly bound to the nucleus.

Angular Momentum Quantum Number

The angular momentum quantum number, represented as "l", is critical in understanding the shape and type of electron orbitals. It can be thought of as describing the 'subshells' within each principal shell determined by the principal quantum number "n".
The values of "l" range from 0 to \(n-1\) and determine the orbital's shape and angular momentum.

• If "n" = 3, the possible values for "l" can be 0, 1, and 2, corresponding to s, p, and d orbitals, respectively.
• The "s" orbitals are spherical, "p" orbitals are dumbbell-shaped, and "d" orbitals have more complex shapes.

This number explains how electrons are arranged spatially and impacts the atom’s electronic configuration.

So, while the principal quantum number gives us information on energy levels, the angular momentum quantum number provides us with details about an electron's path. Both are essential for determining how electrons interact with each other and the nucleus. Understanding these interactions helps clarify why different orbitals have different energy levels.

Potential Energy of Electrons

When discussing electrons and their movement within an atom, the potential energy of electrons is vital. This energy depends on the electron's position relative to the nucleus.
In simpler terms, it's the energy electrons have down to their distance away from the nucleus.

• Electrons closer to the nucleus have more negative potential energy, making them more stable.
• Electrons that are farther away possess higher potential energy, which translates to higher energy levels or less stability.

In the context of the chlorine atom from the exercise, a "3s" electron, which is closer to the nucleus, has a more negative potential energy compared to a "3p" electron.
This aspect makes it lower in energy overall, due to its proximity to the nucleus.

The balancing act between kinetic and potential energy in electrons is what provides an atom its characteristic stability and behavior. Lower potential energy means an electron is less likely to be ejected from the atom without substantial external energy input.

Ionization Energy

Ionization energy refers to the energy required to remove an electron from an atom or ion. This concept is crucial in determining an element’s chemical properties. Essentially, it is the energy 'cost' required to extract one electron out of its orbit.

• Electrons that are closer to the nucleus, as seen with lower quantum numbers, usually have higher ionization energies.
• Conversely, electrons in outer shells require less energy for removal due to less nuclear attraction and more electron shielding.

In the given exercise, the "3s" electron in a chlorine atom, despite being in a higher shell number than "2p", is quite close to the nucleus in comparison to a "3p" electron.
This results in a paradox where logic suggests it should be easier to remove than a "2p" electron, but because it's closer, it requires more energy.

This highlights one of the semi-counterintuitive principles in atomic chemistry where spatial arrangements and interactions within sublevels and shells significantly influence ionization energy levels.