Problem 52
Question
The following are isoelectronic species: \(\mathrm{Cl}^{-}, \mathrm{K}^{+},\) and \(\mathrm{Ca}^{2+}\) Rank them in order of increasing (a) size, (b) ionization energy, and (c) electron affinity.
Step-by-Step Solution
Verified Answer
(a) Size order: \( \mathrm{Ca}^{2+} < \mathrm{K}^{+} < \mathrm{Cl}^{-} \); (b) Ionization energy order: \( \mathrm{Cl}^{-} < \mathrm{K}^{+} < \mathrm{Ca}^{2+} \); (c) Electron affinity order: \( \mathrm{Ca}^{2+} < \mathrm{K}^{+} < \mathrm{Cl}^{-} \).
1Step 1: Understanding Isoelectronic Species
To solve this problem, we must understand that isoelectronic species have the same number of electrons. Here, \( \text{Cl}^- \), \( \text{K}^+ \), and \( \text{Ca}^{2+} \) all have 18 electrons, similar to Argon (Ar). The difference among them is the number of protons, meaning they have different nuclear charges.
2Step 2: Ranking by Atomic Size
For isoelectronic species, size depends on the nuclear charge. The greater the number of protons, the more strongly the electrons are pulled towards the nucleus, resulting in a smaller atomic size. Therefore, the order of increasing size is: \( \text{Ca}^{2+} < \text{K}^+ < \text{Cl}^- \).
3Step 3: Ranking by Ionization Energy
Ionization energy tends to increase with a stronger nuclear charge because more energy is needed to remove an electron. Thus, for isoelectronic species, the order of increasing ionization energy is: \( \text{Cl}^- < \text{K}^+ < \text{Ca}^{2+} \).
4Step 4: Ranking by Electron Affinity
Electron affinity becomes less negative (lower) as the nuclear charge increases because the added electron experiences a greater repulsion from the core electrons in the context of higher nuclear charge. Hence, the order for increasing electron affinity typically goes in the reverse order of the nuclear charge: \( \text{Ca}^{2+} < \text{K}^+ < \text{Cl}^- \).
Key Concepts
Atomic SizeIonization EnergyElectron Affinity
Atomic Size
Atomic size refers to the distance from the nucleus to the outermost electron of an atom. For isoelectronic species, which have the same number of electrons, the difference in size is influenced by the nuclear charge, or the number of protons in the nucleus. The more protons, the stronger the nuclear pull on the electrons, causing the atom to be smaller.
For example, in the isoelectronic series
For example, in the isoelectronic series
- \( \text{Cl}^- \) has 17 protons,
- \( \text{K}^+ \) has 19 protons,
- \( \text{Ca}^{2+} \) has 20 protons.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom or ion. For isoelectronic species, the ionization energy depends on the nuclear charge. A higher number of protons means a stronger attraction to the electrons, and therefore, more energy is needed to remove an electron.
In the series
In the series
- \( \text{Ca}^{2+} \)
- \( \text{K}^+ \)
- \( \text{Cl}^- \)
Electron Affinity
Electron affinity describes the energy change when an electron is added to an atom or ion. It often becomes less negative with increasing nuclear charge due to additional electron repulsion, particularly in isoelectronic species.
In our example:
In our example:
- \( \text{Cl}^- \)
- \( \text{K}^+ \)
- \( \text{Ca}^{2+} \)
Other exercises in this chapter
Problem 50
Place the following elements and ions in order of decreasing size: \(\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-},\) and \(\mathrm{Ca}^{2+}\).
View solution Problem 51
Answer each of the following questions: (a) Of the elements S, Se, and Cl, which has the largest atomic radius? (b) Which has the larger radius, Br or \(\mathrm
View solution Problem 53
Compare the elements Na, B, Al, and C with regard to the following properties: (a) Which has the largest atomic radius? (b) Which has the most negative electron
View solution Problem 54
Two elements in the second transition series ( \(Y\) through Cd) have four unpaired electrons in their \(3+\) ions. What elements fit this description?
View solution