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For each of the following sets of atoms and ions, arrange the members in order of increasing size: (a) \(\mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se}\); (b) \(\mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\), (c) \(\mathrm{Ca}, \mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}\) (d) \(\mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}\)

Short Answer

Expert verified
(a) Se < Se²⁻ < Te²⁻; (b) Fe³⁺ < Co³⁺ < Fe²⁺; (c) Ti⁴⁺ < Sc³⁺ < Ca; (d) Be²⁺ < Na⁺ < Ne.

Step by step solution

01

(a) Se²⁻, Te²⁻, Se - Analyzing their properties

In this set, we have a neutral atom (Se) and two anions (Se²⁻ and Te²⁻). In general, anions are larger than their parent atoms because they have gained extra electrons, which increase the electron-electron repulsion. Meanwhile, within the same group in the periodic table, atomic and ionic size increase down the group due to the increasing number of electron shells.
02

(a) Se²⁻, Te²⁻, Se - Arranging in order of increasing size

Knowing that anions are larger than their parent atoms and that size increases down a group, we can arrange them in increasing order of size: Se < Se²⁻ < Te²⁻.
03

(b) Co³⁺, Fe²⁺, Fe³⁺ - Analyzing their properties

In this set, we have three cations (Co³⁺, Fe²⁺, and Fe³⁺). Cations are smaller than their parent atoms because they have lost electrons, reducing electron-electron repulsion and increasing the effective nuclear charge, thus pulling the remaining electrons closer to the nucleus. Among cations, higher charged ones are smaller due to the greater effective nuclear charge they experience.
04

(b) Co³⁺, Fe²⁺, Fe³⁺ - Arranging in order of increasing size

Following the reasoning that cations with higher charge are smaller and that Fe³⁺ and Co³⁺ belong to the same period in the periodic table, we can arrange them in increasing order of size: Fe³⁺ < Co³⁺ < Fe²⁺.
05

(c) Ca, Ti⁴⁺, Sc³⁺ - Analyzing their properties

In this set, we have a neutral atom (Ca) and two cations (Ti⁴⁺ and Sc³⁺). Again, we generalize that cations are smaller than their parent atoms. Additionally, within the same period, atomic and ionic size decrease as we move across the period due to the increasing effective nuclear charge without adding more electron shells.
06

(c) Ca, Ti⁴⁺, Sc³⁺ - Arranging in order of increasing size

Considering that cations are smaller than their parent atoms, and size decreases across a period, we can arrange them in increasing order of size: Ti⁴⁺ < Sc³⁺ < Ca.
07

(d) Be²⁺, Na⁺, Ne - Analyzing their properties

In this set, we have a neutral atom (Ne) and two cations (Be²⁺ and Na⁺). As mentioned before, cations are smaller than their parent atoms. Furthermore, when moving across a period, atomic and ionic size decrease due to the increasing effective nuclear charge.
08

(d) Be²⁺, Na⁺, Ne - Arranging in order of increasing size

Taking into account the previously mentioned generalizations and periodic trends, we can arrange them in increasing order of size: Be²⁺ < Na⁺ < Ne.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cations vs Anions
Atoms and ions come in various sizes, and understanding the difference between cations and anions is crucial in predicting them. Cations are positively charged ions that form when an atom loses one or more electrons. This loss decreases the electron-electron repulsion and increases the effective pull of the nucleus on the remaining electrons, causing cations to be smaller than their parent atoms.
  • Cations are smaller than their parent atoms due to the reduced electron repulsion.
  • Positive charge increases the effective nuclear charge on each electron.
Anions, on the other hand, are negatively charged ions created when an atom gains electrons. Adding extra electrons increases repulsion among the electrons, causing the anion to be larger than the neutral atom.
  • Anions are larger than their neutral atoms due to extra electron repulsion.
  • More electrons lead to a larger spread to minimize repulsion.
Understanding this difference helps in arranging ions and atoms by size, with general rules being anions are larger and cations are smaller compared to their parents.
Periodic Table Trends
The periodic table is organized in such a way that it reflects trends in atomic and ionic sizes, among other properties. These trends are predictable patterns that can be used to compare the sizes of different atoms and ions. Trends Down a Group: As you move down a group in the periodic table, the atomic size generally increases. This increase is due to the addition of electron shells, which outweighs the effect of the slightly increased nuclear charge.
  • More electron shells are added as you go down a group.
  • The greater distance from the nucleus makes the atom bigger.
Trends Across a Period: As you go across a period from left to right, atomic and ionic sizes tend to decrease. This decrease is a result of the increasing effective nuclear charge, which pulls electrons closer to the nucleus despite the electrons being added to the same shell.
  • Effective nuclear charge increases, reducing atomic size.
  • More protons pull electrons closer, shrinking the atomic radius.
Using these trends allows chemists to predict the relative sizes of atoms and ions across the table.
Effective Nuclear Charge
Effective nuclear charge refers to the net positive charge experienced by electrons in an atom. This concept is paramount in understanding why some atoms are smaller or larger within a period or group. The effective nuclear charge is the actual nuclear charge minus the effect of electron shielding or repulsion from inner electrons. Key Points:
  • Protons in the nucleus exert an attraction on electrons.
  • This attraction is partially offset by other electrons, especially the inner shells.
A higher effective nuclear charge means electrons are pulled closer to the nucleus, making the atom or ion smaller.
  • An increase in this charge across a period pulls outer electrons closer.
  • The greater the charge, the tighter the electrons are held.
Thus, effective nuclear charge is a vital factor in predicting and explaining periodic trends in size and other properties.
Electron Shells and Repulsion
Electron shells and repulsion play a critical role in determining the size of an atom or ion. Electrons repel each other because of their negative charge, and this repulsion affects how electron shells are arranged and how large an atom or ion becomes. Electron Shells: Electrons are organized into shells around the nucleus based on energy levels.
  • Each shell can hold a certain number of electrons.
  • As more shells are added, the size of the atom increases.
Electron Repulsion: The repulsion between electrons increases the effective size of the electron cloud.
  • More electrons result in more repulsion, leading to a larger size.
  • This repulsion explains why ions with extra electrons (anions) are larger.
Understanding the balance between electron shells and their repulsion helps explain the relative sizes of different ions and atoms. The intricate dance of these factors determines the spatial arrangement and ultimate size of atomic structures.

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Most popular questions from this chapter

In Chapter 1 we learned that silicon is the second most abundant element in Earth's crust, accounting for more than one-fourth of the mass of the crust (Figure 1.6). Yet we see that silicon is not among the elements that have been known since ancient times (Figure \(7.2\) ), whereas iron, which accounts for less than \(5 \%\) of Earth's crust, has been known since prehistoric times. Given silicon's abundance how do you account for its relatively late discovery?

Arrange the following atoms in order of increasing effective nuclear charge experienced by the electrons in the \(n=3\) electron shell: \(K, M g, P, R h\), and \(T\). Explain the basis for your order.

Consider the gas-phase transfer of an electron from a sodium atom to a chlorine atom: $$ \mathrm{Na}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(g) $$ (a) Write this reaction as the sum of two reactions, one that relates to an ionization energy and one that relates to an electron affinity. (b) Use the result from part (a), data in this chapter, and Hess's law to calculate the enthalpy of the above reaction. Is the reaction exothermic or endothermic? (c) The reaction between sodium metal and chlorine gas is highly exothermic and produces \(\mathrm{NaCl}(\mathrm{s})\), whose structure was discussed in Section 2.7. Comment on this observation relative to the calculated enthalpy for the aforementioned gas-phase reaction.

Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in Kr? Which will be closer to the nucleus? Explain.

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?\) Explain.

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