Question

Arrange these elements in order of decreasing first ionization energy: \(\mathrm{Cl}\) , \(\mathrm{S}, \mathrm{Sn}, \mathrm{Pb} .\)

Short answer

The elements in order of decreasing first ionization energy are: Cl > S > Sn > Pb.

Step-by-step solution

Understanding the concept of ionization energy

First ionization energy is the energy required to remove the most loosely bound electron from a neutral atom in its ground state. The ionization energy generally increases across a period from left to right and decreases down a group in the periodic table.

Locate the elements in the periodic table

Determine the position of each element in the periodic table. This will help in comparing their ionization energies, as the elements are Cl (group 17, period 3), S (group 16, period 3), Sn (group 14, period 5), and Pb (group 14, period 6).

Apply the trends in ionization energy

Using the periodic table trends, we can say that generally, Cl will have the highest first ionization energy because it is furthest to the right and top. Sulfur (S) comes next, as it is to the left of chlorine in the same period. Tin (Sn) will have a lower ionization energy than both Cl and S because it is further to the left and in a lower period. Lead (Pb) will have the lowest ionization energy because it is below Sn in the same group.

Arrange the elements in decreasing ionization energy

Considering the established trends, the elements in order of decreasing first ionization energy are: Cl > S > Sn > Pb.

Key concepts

Periodic Table Trends

Understanding the periodic table trends is crucial when studying the properties of elements, including their ionization energies. The first ionization energy refers to the energy needed to remove an electron from an isolated, gaseous atom. Within the periodic table, this energy generally increases as you move from left to right across a period, due to higher nuclear charges attracting electrons more strongly. Conversely, ionization energy tends to decrease as you descend down a group, as electrons are further from the nucleus and more shielded by other electrons, reducing the nuclear attraction.

When attempting to arrange elements in order of decreasing first ionization energy, it's essential to remember these trends. They are the key to predicting how easily an atom will give up an electron, which is fundamental in chemical reactivity and bonding.

Electron Removal

The process of electron removal from an atom is termed ionization, and it is a delicate interplay between the pull from the nucleus and the repulsion among electrons. The ease of removing an electron from an atom is determined by several factors, including the atomic number and the electron configuration. Electrons in atoms are arranged in shells, and those in the outermost shell (valence electrons) are the easiest to remove. The energy required to take away this first electron is indicative of an element’s reactivity and is particularly important in chemical reactions involving electron transfer, such as oxidation-reduction reactions.

To conceptualize this process, imagine a tug of war between the nucleus and the electron. The first ionization energy measures how hard the nucleus 'tugs' on the electron, and differs from one element to another based on the balance of forces involved.

Elemental Comparison

Comparing elements based on their first ionization energies can offer insight into their chemical nature. Highly electropositive elements, often found in the lower left of the periodic table, have low ionization energies and tend to lose electrons readily, forming cations. On the other hand, elements with high ionization energies are usually located in the upper right portion, are less willing to part with their electrons, and may readily accept electrons during chemical reactions to become anions.

For instance, in the case of chlorine (Cl), sulfur (S), tin (Sn), and lead (Pb), the trend from highest to lowest ionization energy typically follows their positions in the periodic table. Chlorine, being farther up and to the right, will have a higher ionization energy compared to sulfur, which is just left of it in the same period. Tin and lead, being farther down and to the left, will have significantly lower ionization energies. This knowledge is valuable when predicting the behavior of elements and their propensity to form certain types of ions or bonds.

Atomic Structure

The influence of atomic structure on an element’s properties cannot be overstated. The number of protons in the nucleus, the arrangement of electrons, and the size of the atom – all determine how strongly the outermost electrons are held. In elements where the nuclear charge is effectively more concentrated (due to fewer electron shells), the force on valence electrons is stronger, resulting in higher ionization energies.

The structure extends beyond just a single shell of electrons. Inner electrons can shield the outer ones from the full effect of the nuclear charge, a relationship termed electron shielding or screening effect. For example, although lead has more protons than sulfur, the additional shells in lead increase electron shielding, making it easier for an electron to be removed, and thereby, lead has a lower ionization energy than sulfur. Grasping atomic structure offers a valuable window into the forces at play within an atom that affect its behavior.