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Chapter 8: Problem 141

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

Short Answer

Expert verified
a. \(\mathrm{O}^{2-} > \mathrm{O}^{-} > \mathrm{O}\) b. \(\mathrm{Fe}^{2+} > \mathrm{Ni}^{2+} > \mathrm{Zn}^{2+}\) c. \(\mathrm{Cl}^{-} > \mathrm{K}^{+} > \mathrm{Ca}^{2+}\)

Step by step solution

01

(Set a: O, O-, O虏-)

For this set, we need to compare the size between a neutral atom and its ions with different charges. As we increase the negative charge on an atom, the number of electrons increases, thus causing a decrease in the effective nuclear charge (a lower positive force from the protons) which results in a larger atomic radius. So, among \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\), we can arrange them in decreasing size as follows: \(\mathrm{O}^{2-} > \mathrm{O}^{-} > \mathrm{O}\).
02

(Set b: Fe虏鈦�, Ni虏鈦�, Zn虏鈦�)

For this set, we see that all ions have the same positive charge (+2). Given that they belong to the same period, their size depends on the number of protons. As we move from left to right across a period, atomic radius generally decreases due to an increase in the number of protons and thus an increase in effective nuclear charge. We have \(\mathrm{Fe}^{2+}\) (26 protons), \(\mathrm{Ni}^{2+}\) (28 protons), and \(\mathrm{Zn}^{2+}\) (30 protons). Thus, we can arrange them in decreasing size as follows: \(\mathrm{Fe}^{2+} > \mathrm{Ni}^{2+} > \mathrm{Zn}^{2+}\).
03

(Set c: Ca虏鈦�, K鈦�, Cl鈦�)

In this set, we need to compare ions with different charges. The key to evaluate their size is to compare their effective nuclear charge. When comparing \(\mathrm{Ca}^{2+}\) (20 protons, 18 electrons) and \(\mathrm{K}^{+}\) (19 protons, 18 electrons), both have the same number of electrons but \(\mathrm{Ca}^{2+}\) has more protons which increases the effective nuclear charge and thus has a smaller radius. When comparing \(\mathrm{K}^{+}\) and \(\mathrm{Cl}^{-}\), \(\mathrm{K}^{+}\) has 1 less electron than \(\mathrm{Cl}^{-}\) but they both have one less proton than their respective numbers of electrons. In this situation, \(\mathrm{Cl}^{-}\) has more electron-electron repulsion, which leads to a larger atomic radius. Thus, we can arrange them in decreasing size as follows: \(\mathrm{Cl}^{-} > \mathrm{K}^{+} > \mathrm{Ca}^{2+}\). In summary, we have: a. \(\mathrm{O}^{2-} > \mathrm{O}^{-} > \mathrm{O}\) b. \(\mathrm{Fe}^{2+} > \mathrm{Ni}^{2+} > \mathrm{Zn}^{2+}\) c. \(\mathrm{Cl}^{-} > \mathrm{K}^{+} > \mathrm{Ca}^{2+}\)

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Key Concepts

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

Effective Nuclear Charge
The effective nuclear charge (ENC) plays a key role in determining the atomic radius of an element or ion. It is the net positive charge experienced by electrons in an atom. Essentially, while the nucleus attracts electrons via the protons it contains, this attraction is countered by electron-electron repulsion. As a result, electrons don't feel the entire positive charge of the nucleus, but rather the 'effective' part of it.

Hence, the more protons an atom has, the stronger the effective nuclear charge, drawing electrons closer to the nucleus and reducing the atomic radius. For example, when comparing ions such as \(O\), \(O^{-}\), and \(O^{2-}\), an increase in negative charge reduces the ENC felt by the outer electrons. This is because added electrons increase repulsion and spread out more, resulting in a larger atomic radius.
Isoelectronic Species
Isoelectronic species are atoms or ions that have the same number of electrons. Despite having the same electronic configuration, their sizes may differ depending on their effective nuclear charge. This arises because the different species may have a varied number of protons.

For instance, \(\ca^{2+}\), \(\k^{+}\), and \(\cl^{-}\) are isoelectronic, each having 18 electrons. However, their sizes differ due to the number of protons: \(\cl^{-}\) with 17 protons, \(\k^{+}\) with 19 protons, and \(\ca^{2+}\) with 20 protons. \(\ca^{2+}\) experiences a stronger pull towards the nucleus because of its 20 protons, giving it a smaller atomic size compared to the others.
Atomic Structure
The atomic structure explains why atoms exist in different sizes, partly due to electron arrangements and nuclear charge. An atom's structure is centred on its nucleus, containing protons and neutrons, surrounded by electrons in orbitals. The distance between the outer electrons and nucleus ultimately determines the atom's radius.

As electron shells are filled, new electrons tend to repel each other, and this repulsion can increase the atomic size. However, simultaneously more protons can contract the electron cloud, reducing the size due to an increased effective nuclear charge. Hence, as we move from \(\fe^{2+}\) to \(\zn^{2+}\) in the periodic table, the effective charge increases, shrinking the atomic radius despite the same ionic charge.
Ionic Size Comparison
When comparing the sizes of ions, several factors come into play, including electron count and charge. The size of ions can vary significantly even if they start as a single atom, depending on the gain or loss of electrons.

Take the example of neutral \(\o\), \(\o^{-}\), and \(\o^{2-}\): gaining electrons reduces the ENC per electron due to increased electron-electron repulsions, resulting in larger ions. On the other hand, losing electrons, as in the case with cations like \(\fe^{2+}\), typically reduces size because the ENC is stronger, pulling the electrons closer. Consider comparing \(\cl^{-}\) and \(\k^{+}\); although both contain the same number of electrons, \(\k^{+}\) is smaller due to increased effective nuclear attraction, showing the complexity of ionic size trends.

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Most popular questions from this chapter

Oxidation of the cyanide ion produces the stable cyanate ion, \(\mathrm{OCN}^{-}\). The fulminate ion, \(\mathrm{CNO}^{-}\), on the other hand, is very unstable. Fulminate salts explode when struck; \(\mathrm{Hg}(\mathrm{CNO})_{2}\) is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{N}\) is the central atom in \(\mathrm{CNO}^{-} .\) )

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: Given the following bond lengths, $$ \begin{array}{llll} \mathrm{N}-\mathrm{N} & 167 \mathrm{pm} & \mathrm{N}=\mathrm{O} & 115 \mathrm{pm} \\ \mathrm{N}=\mathrm{N} & 120 \mathrm{pm} & \mathrm{N}-\mathrm{O} & 147 \mathrm{pm} \\ \mathrm{N} \equiv \mathrm{N} & 110 \mathrm{pm} & & \end{array} $$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

Indicate the bond polarity (show the partial positive and partial negative ends) in the following bonds. a. \(\mathrm{C}-\mathrm{O}\) d. \(\mathrm{Br}-\mathrm{Te}\) b. \(\mathrm{P}-\mathrm{H}\) e. \(\mathrm{Se}-\mathrm{S}\) c. \(\mathrm{H}-\mathrm{Cl}\)

Write electron configurations for the most stable ion formed by each of the elements Te, \(\mathrm{Cl}, \mathrm{Sr}\), and \(\mathrm{Li}\) (when in stable ionic compounds).
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