(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.

Atomic structure forms the foundation for understanding how matter is constructed at the smallest scale. Atoms are composed of a central nucleus, made up of protons and neutrons, surrounded by electrons that orbit in different energy levels or shells. These electrons exist in regions called orbitals, which can be visualized as clouds of probability where an electron is likely to be found.

The key to grasping atomic structure lies in understanding these electron orbitals. Each orbital can hold a specific number of electrons and is associated with a particular energy level. The 's' orbitals are the simplest, having a spherical shape, while 'p', 'd', and 'f' orbitals are more complex. The 's' orbital is always the first orbital in any given shell because it is the closest to the nucleus and has the lowest energy. As electrons fill up these orbitals, they follow a specific order known as the Aufbau principle, which aids in predicting the electronic configuration of an atom.

Understanding atomic structure is crucial when considering properties such as reactivity and bonding. The configuration of electrons determines how atoms will interact with each other, and it lays the groundwork for chemical reactions.

Electron removal energy, also known as ionization energy, is the amount of energy required to remove an electron from an atom or ion. This energy is a key indicator of an element's reactivity and is influenced by several factors, including the distance of the electron from the nucleus and the amount of charge on the nucleus. Generally, the closer an electron is to the nucleus, the more energy is needed to overcome the attraction and remove the electron.

The ionization energy tends to increase as you move across a period in the periodic table because the nuclear charge increases, pulling the electrons in closer. Conversely, ionization energy decreases as you move down a group because the electrons being removed are in higher energy levels, further away from the nucleus, and more shielded by inner electron shells. This concept is important when comparing the ease with which different electrons can be removed from an atom and can help predict the atom's chemical behavior.

The attraction between electrons and the nucleus is one of the central aspects of atomic structure and chemistry. This attraction is due to the opposite charges of electrons (negative) and protons (positive) in the nucleus. Coulomb's Law describes the force of this attraction: the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Electron-nucleus attraction influences how tightly an electron is held by the atom, which in turn affects the atom's ionization energy. A strong attraction means more energy will be required to remove an electron. The concept becomes particularly interesting when comparing electrons in different orbitals, as in the exercise. Electrons in orbitals that are closer to the nucleus, like the 3s compared to the 3p, experience a stronger attraction, meaning they usually have lower energy and are harder to remove, translating to higher ionization energy for these electrons. This principle underlies many of the trends observed in the periodic table, including atomic radii, ionization energies, and electronegativity.