Question

Write equations that show the process for (a) the first two ionization energies of lead and (b) the fourth ionization energy of zirconium.

Short answer

The first two ionization energies of lead (Pb) are: 1. \(Pb(g) \rightarrow Pb^{+}(g) + e^{-}\) 2. \(Pb^{+}(g) \rightarrow Pb^{2+}(g) + e^{-}\) The fourth ionization energy of zirconium (Zr) is: \(Zr^{3+}(g) \rightarrow Zr^{4+}(g) + e^{-}\)

Step-by-step solution

Identify the element and its symbol

Lead (Pb) is a heavy metal with the atomic number 82. We will be concerned with the process of removing the first two electrons from the lead atom.

Write equation for the first ionization energy

The first ionization energy of lead (Pb) represents the energy required to remove the first electron from a neutral lead atom. The reaction can be written by converting the neutral atom to its cation: Pb(g) → Pb+(g) + e^-

Write equation for the second ionization energy

The second ionization energy of lead (Pb) represents the energy required to remove the second electron from the lead ion with a charge of +1. The reaction can be written by converting the ion with a charge of +1 to its cation with a charge of +2: Pb+(g) → Pb+2(g) + e^- (b) Writing equation for the fourth ionization energy of zirconium (Zr)

Identify the element and its symbol

Zirconium (Zr) is a transition metal with the atomic number 40. We will be concerned with the process of removing the fourth electron from the zirconium atom.

Write equation for the fourth ionization energy

The fourth ionization energy is the energy required to remove the fourth electron; hence, we will go through the reactions that lead us to the ion with a charge of +3 in order to remove the fourth electron: First ionization energy: Zr(g) → Zr+(g) + e^- Second ionization energy: Zr+(g) → Zr+2(g) + e^- Third ionization energy: Zr+2(g) → Zr+3(g) + e^- Now, we will write the equation for the fourth ionization energy: Zr+3(g) → Zr+4(g) + e^- To recap, the equation for the first two ionization energies of lead (Pb) are: 1. Pb(g) → Pb+(g) + e^- 2. Pb+(g) → Pb+2(g) + e^- The equation for the fourth ionization energy of zirconium (Zr) is: Zr+3(g) → Zr+4(g) + e^-

Key concepts

Chemical Equations

Understanding chemical equations is essential when studying the reactions involved in ionization energy processes. A chemical equation is a symbolic representation of a chemical reaction where the reactants are displayed on the left side and the products on the right, separated by an arrow that indicates the direction of the reaction. For instance, when an atom loses an electron to form a cation, a single arrow is used because this process is not reversible under normal conditions.

In our exercise dealing with the ionization energies of lead (Pb) and zirconium (Zr), we witness the step-by-step removal of electrons. Each step can be represented by a separate chemical equation. For lead, the first, Pb(g) → Pb+(g) + e^-, and the second, Pb+(g) → Pb²⁺(g) + e^-, ionization energies depict the transformation from a neutral atom to a more positively charged ion, step by step. Likewise, in the case of zirconium, the equation for the fourth ionization is Zr³⁺(g) → Zr⁴⁺(g) + e^-. Each equation represents a discrete event where an electron is detached, which is central in understanding the ionization process.

Atomic Number

The atomic number of an element is a fundamental concept in chemistry as it determines the element's position in the periodic table and its chemical properties. The atomic number, often denoted by the symbol 'Z', is the number of protons found in the nucleus of an atom. As electrons are negatively charged and protons are positively charged, the atomic number also indicates how many electrons are present in a neutral atom.

In our exercise, you have come across lead and zirconium with atomic numbers 82 and 40, respectively. This atomic number not only helps us to locate them on the periodic table but it also guides us when forming and determining ionization energies. The atomic number is directly related to the distribution of electrons around the nucleus, which affects how tightly the electrons are held. It is this 'hold' that ionization energy seeks to break to remove an electron. Thus, atomic number is intrinsic in predicting the energy required for ionization.

Transition Metals

Transition metals are a group of elements located in the central block (groups 3-12) of the periodic table, distinguished by their ability to form positive ions with incomplete 'd' subshells. Zirconium (Zr) is one such transition metal, known for its high melting point and strong resistance to corrosion. Transition metals have complex electron arrangements, which can result in varying oxidation states, lending them unique chemical behaviors.

The ionization energy in transition metals, like zirconium, is particularly intriguing because these metals often exhibit multiple ionization energies corresponding to the removal of electrons from both the s and d orbitals. The fourth ionization energy in the exercise is a higher energy process because it may involve removing an electron that is closer to the nucleus or engaged in stabilizing 'd' orbital interactions. By studying the step-by-step ionization of transition metals, students gain insights into electron configurations and the energetic considerations of forming cations, which are integral in fields like metallurgy and catalysis.