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Chapter 7: Problem 46

Place the following ions in order of decreasing size: \(\mathbf{K}^{+}\) \(\mathrm{Cl}^{-}, \mathrm{S}^{2-},\) and \(\mathrm{Ca}^{2+}\).

Short Answer

Expert verified
The order is \(\text{S}^{2-}, \text{Cl}^-, \text{K}^+, \text{Ca}^{2+}\).

Step by step solution

01

Understand Ionic Size

Ionic size, or ionic radius, refers to the size of an ion compared to its neutral atom. Generally, when an atom loses electrons to become a cation, it becomes smaller. Conversely, when an atom gains electrons to become an anion, it becomes larger.
02

Analyze the Ion Charges

The ions given are \(\text{K}^+\), \(\text{Cl}^-\), \(\text{S}^{2-}\), and \(\text{Ca}^{2+}\). \(\text{Ca}^{2+}\) has lost two electrons, and \(\text{K}^+\) has lost one. Meanwhile, \(\text{Cl}^-\) has gained one electron and \(\text{S}^{2-}\) has gained two electrons.
03

Compare Same-Period Ions First

\(\text{K}^+\) and \(\text{Ca}^{2+}\) are both cations from the same period (Period 4) in the periodic table. The \(\text{K}^+\) ion, having lost one electron, is larger than \(\text{Ca}^{2+}\), which has lost two electrons.
04

Compare Sulfide and Chloride Ions

\(\text{Cl}^-\) and \(\text{S}^{2-}\) are anions. Sulfur and chlorine atoms are in the same period (Period 3), with sulfur being larger because it is to the left in the period. Adding electrons increases size, so \(\text{S}^{2-}\) is larger than \(\text{Cl}^-\).
05

Order All Ions by Size

Combining insights, \(\text{S}^{2-}\) is the largest (most electrons added), followed by \(\text{Cl}^-\), then \(\text{K}^+\) (only one electron lost), and finally \(\text{Ca}^{2+}\) being the smallest (two electrons lost).

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Key Concepts

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

Cation
A cation is an ion that carries a positive charge. This positive charge results from an atom losing one or more electrons. Electrons are negatively charged, so when they are removed, the ion becomes positively charged.
In an atom, electrons are present in shells around the nucleus. When electrons are removed, the remaining electrons experience a stronger pull from the nucleus. This causes the radius of the ion to decrease, making cations smaller than their neutral atoms.
For example:
  • When a potassium atom (\(\text{K}\)) loses an electron, it becomes a cation (\(\text{K}^+\)), decreasing in size.
  • Similarly, calcium (\(\text{Ca}\)) becomes \(\text{Ca}^{2+}\), even smaller as it loses two electrons compared to potassium's one.
Anion
An anion is an ion with a negative charge. This occurs when an atom gains extra electrons. More electrons increase the negative charge, causing the ion to expand.
When additional electrons are added, they repel each other and occupy more space, enlarging the ion's size. Anions are thus larger than the neutral atom from which they derive.
Consider:
  • Chlorine atoms (\(\text{Cl}\)) gain an electron to become \(\text{Cl}^-\), an anion that is larger than the neutral chlorine atom.
  • Similarly, sulfur (\(\text{S}\)) becomes \(\text{S}^{2-}\) by gaining two electrons, increasing its size more than \(\text{Cl}^-\).
Periodic Table
The periodic table is a vital tool in understanding the properties of elements. It organizes elements by increasing atomic number and recurring chemical properties, which helps predict the behavior of ions.
Each row in the periodic table is called a period. Elements in the same period have the same number of atomic orbitals. Moving from left to right, atomic size generally decreases as more protons in the nucleus attract the electrons more strongly.
Some key insights include:
  • Cations like \(\text{K}^+\) and \(\text{Ca}^{2+}\) sit within the same period and lose different numbers of electrons, affecting their ionic size.
  • Anions like \(\text{Cl}^-\) and \(\text{S}^{2-}\) also display different ionic sizes based on their position in a period.
Electron Configuration
Electron configuration describes the distribution of electrons in an atom's orbitals. It outlines an atom's structure, crucial for understanding ionic behavior and size changes.
When atoms form cations, they lose electrons from their outermost shell. For anions, additional electrons are added to this outer shell. This alteration changes the electron configuration, leading to variations in ionic radius.
For example:
  • Potassium (\(\text{K}\)) normally has the electron configuration of \([ \text{Ar} ] 4s^1\). When it becomes \(\text{K}^+\), it loses the \(4s\) electron, shrinking in size.
  • Chlorine (\(\text{Cl}\)) with configuration \([ \text{Ne} ] 3s^2 3p^5\) gains one electron to become \(\text{Cl}^-\), filling the 'p' orbital, thus becoming larger.

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Most popular questions from this chapter

Nickel(II) formate \(\left[\mathrm{Ni}\left(\mathrm{HCO}_{2}\right)_{2}\right]\) is widely used as a catalyst precursor and to make metallic nickel. It can be prepared in the general chemistry laboratory by treating nickel(II) acetate with formic acid (HCO,H). \(\mathrm{Ni}\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2}(\mathrm{aq})+2 \mathrm{HCO}_{2} \mathrm{H}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Ni}\left(\mathrm{HCO}_{2}\right)_{2}(\mathrm{aq})+2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) $$ Green crystalline \(\mathrm{Ni}\left(\mathrm{HCO}_{2}\right)_{2}\) is precipitated after adding ethanol to the solution. (a) What is the theoretical yield of nickel(II) formate from 0.500 g of nickel(II) acetate and excess formic acid? (b) Is nickel(II) formate paramagnetic or diamagnetic? If it is paramagnetic, how many unpaired electrons would you expect? (c) If nickel(II) formate is heated to \(300^{\circ} \mathrm{C}\) in the absence of air for 30 minutes, the salt decomposes to form pure nickel powder. What mass of nickel powder should be produced by heating 253 mg of nickel(II) formate? Are nickel atoms paramagnetic?

Slater's rules are a simple way to estimate the effective nuclear charge experienced by an electron. In this approach, the "shielding constant," \(\mathrm{S}\), is calculated. The effective nuclear charge is then the difference between S and the atomic number, \(Z\). (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) $$ Z^{*}=Z-\mathrm{S} $$ The shielding constant, \(\mathrm{S}\), is calculated using the following rules: 1\. The electrons of an atom are grouped as follows: (1s) \((2 s, 2 p)(3 s, 3 p)\) (3d) \((4 s, 4 p)\) (4d), and so on. 2\. Electrons in higher groups (to the right) do not shield those in the lower groups. 3\. For \(n s\) and \(n p\) valence electrons (a) Electrons in the same \(n s, n p\) group contribute 0.35 (for \(1 s 0.30\) works better). (b) Electrons in the \(n-1\) group contribute 0.85 (c) Electrons in the \(n-2\) group (and lower) contribute 1.00 4\. For \(n d\) and \(n f\) electrons, electrons in the same \(n d\) or \(n f\) group contribute \(0.35,\) and those in groups to the left contribute 1.00 As an example, let us calculate \(Z^{*}\) for the outermost electron of oxygen: $$ \begin{array}{c} \mathrm{S}=(2 \times 0.85)+(5 \times 0.35)=3.45 \\ Z^{*}=8-3.45=4.55 \end{array} $$ Here is a calculation for a \(d\) electron in \(\mathrm{Ni}\) : $$ Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55 $$ and for an s electron in \(\mathrm{Ni}\) : $$ \begin{aligned} Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] & \\ -[1 \times 0.35] &=4.05 \end{aligned} $$ (Here \(3 s, 3 p,\) and \(3 d\) electrons are in the \((n-1)\) groups.) (a) Calculate \(Z^{*}\) for \(F\) and \(N e\). Relate the \(Z^{*}\) values for O, F, and Ne to their relative atomic radii and ionization energies. (b) Calculate \(Z^{* *}\) for one of the \(3 d\) electrons of \(\mathrm{Mn}\), and compare this with \(Z^{*}\) for one of the \(4 s\) electrons of the element. Do the \(Z^{*}\) values give us some insight into the ionization of Mn to give the cation?

A possible excited state for the H atom has an electron in a \(4 p\) orbital. List all possible sets of quantum numbers \(\left(n, \ell, m_{\ell}, m_{\mathrm{s}}\right)\) for this electron.

What is the maximum number of electrons that can be identified with each of the following sets of quantum numbers? In some cases, the answer may be "none." In such cases, explain why "none" is the correct answer. (a) \(n=3\) (b) \(n=3\) and \(\ell=2\) (c) \(n=4, \ell=1, m_{\ell}=-1,\) and \(m_{s}=+1 / 2\) (d) \(n=5, \ell=0, m_{\ell}=-1, m_{\mathrm{s}}=+1 / 2\)

Sodium metal reacts readily with chlorine gas to give sodium chloride. $$ \mathrm{Na}(\mathrm{s})+1 / 2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{NaCl}(\mathrm{s}) $$ (a) What is the reducing agent in this reaction? What property of the element contributes to its ability to act as a reducing agent? (b) What is the oxidizing agent in this reaction? What property of the element contributes to its ability to act as an oxidizing agent? (c) Why does the reaction produce \(\mathrm{NaCl}\) and not a compound such as \(\mathrm{Na}_{2} \mathrm{Cl}\) or \(\mathrm{NaCl}_{2} ?\)
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