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

(a) The 3p orbital is higher in energy because 3s electrons are closer to the nucleus on average, experiencing a stronger attractive force from protons in the nucleus. (b) It would require more energy to remove a 3s electron from the chlorine atom, compared to a 2p electron, since 3s electrons are closer to the nucleus and have a stronger electrostatic force acting on them.

Step-by-step solution

Identify the energy levels of orbitals

According to the Aufbau principle, electrons fill the orbitals in ascending order of energy. The energy levels of the orbitals can be determined by considering the (n + ℓ) rule, where n is the principal quantum number and ℓ is the azimuthal quantum number. In this case, n for 3s is 3 and ℓ is 0, whereas n for 3p is also 3 but ℓ is 1.

Compare the energies of 3s and 3p orbitals

Since the (n + ℓ) values for the 3s and 3p orbitals are the same (both equal to 3), we need to consider the principal quantum number (n) to determine which orbital has higher energy. Higher n values indicate higher energy levels. In this case, 3s and 3p orbitals have the same principal quantum number, so we cannot definitively state which has more energy simply based on n. However, the question states that the average distance from the nucleus for a 3s electron is smaller than for a 3p electron. This means 3s electrons are closer to the nucleus on average, and thus experience a stronger attractive force from positively charged protons in the nucleus, which results in 3s electrons having lower energy than 3p electrons. Answer to (a): The 3p orbital is higher in energy.

Determine energy requirement for electron removal

When an electron is removed from an atom, it requires energy; this process is known as ionization. The energy required for ionization depends on the electrostatic force between the electron and the nucleus. The stronger the force, the more energy is required for ionization. Electrons with a lower energy level are closer to the nucleus and have stronger electrostatic forces acting on them. Answer to (b): It would require more energy to remove a 3s electron from the chlorine atom, as compared with a 2p electron, because 3s electrons are closer to the nucleus and have a stronger electrostatic force acting on them. Explanation: As the question mentioned, the average distance from the nucleus for a 3s electron is smaller than that of a 3p electron. This implies that the electrostatic force between the nucleus and a 3s electron is stronger, making the 3s electron more strongly bound to the nucleus. Consequently, it would require more energy to remove this electron compared to a 2p electron, which has a weaker electrostatic force acting on it since it is further from the nucleus.

Key concepts

Aufbau Principle

The Aufbau principle is a fundamental cornerstone when it comes to understanding electron configuration in atoms. Aufbau, a German word meaning 'building up', accurately conveys the essence of this principle—electrons fill atomic orbitals starting from the lowest available energy levels before moving to higher ones.

To visualize this, imagine a set of stairs where each step represents an orbital with a specific energy level. Just as you would start climbing from the bottom, electrons start 'climbing' from the lowest step. However, not all steps—or orbitals—are created equal. Some are closer to the nucleus and have a lower energy, and others are further away with higher energy. Using the rules of the Aufbau principle, we can predict the electronic structure of an atom, which is crucial in determining its chemical behavior.

For students tackling chemistry homework, grasping the Aufbau principle provides the groundwork for solving more complex problems involving electron arrangements and predicting atomic properties.

Orbital Energy Levels

Orbital energy levels are the various stages at which electrons exist within an atom. These levels depend on two quantum numbers: n, the principal quantum number which defines the main energy level or shell, and l, the azimuthal quantum number which defines the shape of the orbital within a shell.

When comparing energy levels between orbitals, one can use the (n + l) rule; orbitals with a lower (n + l) value are filled first. If two orbitals have the same (n + l) value, the electron fills the orbital with a lower n. Understanding this can seem tricky, but once you realize it's like an address system, with n as the city and l as the street, it becomes more clear.

The energy difference between orbitals influences how tightly electrons are bound to the nucleus. Electrons in lower energy orbitals are more strongly attracted to the nucleus and are more difficult to remove—think of it as being closer to the planet's core in a powerful gravitational field.

Ionization Energy

Ionization energy refers to the amount of energy necessary to remove an electron from an atom or ion in its gaseous state. This concept is akin to untying a tightly bound knot—the closer and more intimately the electron 'knot' is tied to the nucleus, the more energy you will need to free it.

The energy required is directly related to how strongly electrons are attracted to the atom's nucleus. Electrons in orbitals closer to the nucleus, such as 3s electrons, are more tightly held due to a strong electrostatic force of attraction. In contrast, electrons in higher orbitals can be removed with relatively less energy.

During homework exercises, students might wonder why some elements are more reactive than others. Ionization energy offers a clue: atoms with low ionization energies lose electrons easily, making them more reactive. Conversely, elements with high ionization energies do not give up their electrons readily and are less likely to react. This concept not only helps students with their exercises but also deepens their understanding of element reactivity and trends across the periodic table.