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?

Short answer

(a) The 3\(p\) orbital is higher in energy because the 3s electron is closer to the nucleus and experiences a stronger attraction. (b) It requires less energy to remove a 3s electron from the chlorine atom, compared to a 2p electron, due to the 3s electron being further from the nucleus and experiencing a weaker attractive force.

Step-by-step solution

(a) Determine the higher energy orbital

As the average distance from the nucleus for a 3s electron is smaller than that for a 3\(p\) electron, it means the 3s electron is closer to the nucleus. Since electrons experience a stronger attraction to the nucleus when they are closer, the 3s electrons will have lower energy than the 3\(p\) electrons. Hence, the 3\(p\) orbital is higher in energy.

(b) Compare the energy required to remove 3s and 2p electrons

We are now comparing the energy required to remove an electron from a 3s orbital to that of a 2p orbital. Based on their principal quantum numbers (3 for 3s and 2 for 2p), we can infer that the 3s electron is, on average, further from the nucleus than the 2p electron. Since the 3s electron is further from the nucleus, it experiences a weaker attractive force and requires less energy to be removed from the atom as compared to a 2p electron. Thus, it requires less energy to remove a 3s electron from the chlorine atom, compared to a 2p electron.

Key concepts

Atomic Orbitals

Atomic orbitals are regions in an atom where there is a high probability of finding electrons. These orbitals have different shapes and energy levels, determined by their quantum numbers. Generally, electrons will occupy the orbitals in a way that minimizes the energy of the atom. This concept explains why electrons close to the nucleus are lower in energy—they feel a stronger attraction due to the positive charge of the protons in the nucleus.

There are different types of orbitals—s, p, d, and f. The 's' orbitals are spherical, while 'p' orbitals have a dumbbell shape. As seen in the exercise, a 3s electron is closer to the nucleus on average than a 3p electron, which leads us to conclude that a 3s orbital is generally lower in energy than a 3p orbital. Understanding the shape and energy differences of these orbitals can help students predict electronic configurations and chemical properties.

Principal Quantum Number

The principal quantum number, symbolized by 'n', is an integer that determines the overall size and energy of an atomic orbital. It essentially denotes the 'shell' or energy level in which an orbital resides. The larger the principal quantum number, the further the shell is from the nucleus and the higher the energy of the electrons within that orbital.

For example, a 2p electron is in the second shell, and a 3s electron is in the third shell. Though both are in p and s orbitals respectively, the different principal quantum numbers indicate that the 3s electron is further away from the nucleus, therefore in a higher principal energy level. The concept is crucial in understanding why certain electrons are easier to remove from an atom and how this affects ionization energy, as explained in the given exercise.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or ion in its gaseous state. It is a vital concept in understanding chemical reactivity and trends across the periodic table. Generally, the closer an electron is to the nucleus, the more energy it will take to remove it due to the stronger electrostatic forces of attraction.

As the exercise illustrates, a 3s electron, which occupies an orbital with a higher principal quantum number, has lower ionization energy compared to a 2p electron because it is, on average, further from the nucleus. Recognizing these relationships helps students predict the reactivity of elements and provides insights into the periodic trends in ionization energy, such as why ionization energy tends to increase from left to right across a period.