3.4: Ionic Bonding: Anion Formation, Symbolism, and Nomenclature
- Page ID
- 214226
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Define anion.
- Explain how neutral atoms ionize to form anions.
- Determine the charges achieved when main group non-metals ionize.
- Symbolize and name main group anions.
Anions
Anions are negatively-charged ions that are most often formed when non-metals, which are found on the right side of the periodic table, gain valence electrons.
For example, consider sulfur (S).
An atom of sulfur contains 16 protons, because its atomic number is 16, and 16 electrons, in order to be net-neutral. For reasons identical to those described in the previous section, the neutron count for this particular sulfur atom is irrelevant to the analysis that will be performed in the following paragraphs and will not be discussed further.
Sulfur's electron configuration, which was first determined in Section 2.6, is shown below.
1s22s22p63s23p4
Sulfur has 6 valence electrons, as determined either by totaling the electrons found within the orbitals in the highest occupied energy level, which, in this case, includes both the 3s and 3p orbitals, or by identifying the "A/B System" group number for the column in which the element is found.
In order to be stable, a particle must possess an octet, or eight, fully-paired valence electrons. Sulfur would need to gain 2 electrons in order to achieve an octet configuration. Unlike in the calcium example presented in the previous section, gaining this number of electrons is possible, as doing so would not exceed the maximum gain-limit of three electrons. Therefore, sulfur will gain two additional valence electrons.
Consider the impact that this will have on sulfur's electron configuration. The rules presented in Section 2.6 dictate that if two additional electrons are added, they must both be placed in the 3p orbital, in order to fill that orbital to its maximum capacity.
1s22s22p63s23p(4+2)
Simplification of the 3p orbital's superscript results in a new electron configuration, as shown below.
1s22s22p63s23p6
In this new electron configuration, the number of valence electrons contained in the corresponding particle are still determined by totaling the electrons found within the the 3s and 3p orbitals.
3s23p6
The superscripts associated with these orbitals now sum to 8. Therefore, this new particle has 8 valence electrons and has achieved, through the process described above, a highly-stable octet configuration.
Note that the number of protons was unchanged in the process described above, so this new particle still contains 16 protons. Since the identity of an element is defined by the number of protons that it contains, due to the association between atomic number and proton-count, this new particle is still a form of sulfur. However, since this new particle was formed through the gain of two valence electrons, it now has 18 (16 + 2) electrons. Therefore, this new particle is no longer a sulfur (S) atom which, as stated above, contains 16 electrons.
This particle is now an ion, a charged particle, as a result of the imbalance between the number of positively-charged protons (+16) and negatively-charged electrons (–18) that it possesses. This particle contains two more negatively-charged electrons than positively-charged protons; therefore, it bears an overall net –2 charge. Additionally, since this ion is negatively-charged, it is classified as an anion.
Again, the notation used to symbolize this particle must incorporate the charge of the ion as a superscript on the corresponding elemental symbol, in order to distinguish an elemental symbol, which represents a neutral atom, from an ion symbol, which represents a charged particle. As stated above, this ion is still a form of sulfur (S). Because this particle is now an ion, its specific charge (–2) must be included as a superscript, and an ion symbol of "S–2" results.
Finally, this particle is referred to as a "sulfide ion." As was the case in the calcium example presented in the previous section, the word "ion" is now a necessary part of this particle's name. However, the name of this particle has changed in one additional way, relative to the name of the element from which it was formed ("sulfur"). Specifically, the suffix, or ending, of the element's name has been replaced with "-ide." This suffix is generally used by chemists to denote chemicals that bear negative charges. Therefore, the names of anions can be easily distinguished from the names of cations, as the suffixes of the former change to "-ide," and the endings of the latter remain unmodified.
Finally, note that all non-metals will form anions, as indicated previously, because they will only be able to achieve octet configurations by gaining additional valence electrons. The resultant ions will, therefore, always contain more negatively-charged electrons than positively-charged protons and will bear a net-negative overall charge.
Consider each of the following neutral elements.
- Chlorine
- Nitrogen
For each,
- describe its ionization process,
- determine the charge that will result upon its ionization,
- provide the ion symbol for the resultant ion, and
- provide the ion name for the resultant ion.
- Answer a
- An atom of chlorine contains 17 protons and 17 electrons, 7 of which can be classified as valence electrons, as determined either by using the electron configuration written below, or by recognizing that chlorine is located in Group 7A on the periodic table.
1s22s22p63s23p5
Chlorine would need to gain 1 electron in order to achieve an octet configuration. Gaining this number of electrons is possible, as doing so would not exceed the maximum gain-limit of three electrons. Therefore, chlorine will gain one additional valence electron.
If one additional valence electron is added, it must be placed in the 3p orbital, in order to fill that orbital to its maximum capacity.1s22s22p63s23p(5+1)
Simplification of the 3p orbital's superscript results in a new electron configuration, as shown below.1s22s22p63s23p6
In this new electron configuration, the number of valence electrons contained in the corresponding particle are still determined by totaling the electrons found within the the 3s and 3p orbitals.3s23p6
The superscripts associated with these orbitals now sum to 8. Therefore, this new particle has 8 valence electrons and has achieved, through the process described above, a highly-stable octet configuration.
The number of protons was unchanged in the process described above, so this new particle still contains 13 protons. Since the identity of an element is defined by the number of protons that it contains, this new particle is still a form of chlorine. However, since this new particle was formed through the gain of one valence electron, it now has 18 (17 + 1) electrons. Therefore, this new particle is no longer a chlorine (Cl) atom which, as stated above, contains 17 electrons.
This particle is now an ion, a charged particle, as a result of the imbalance between the number of positively-charged protons (+17) and negatively-charged electrons (–18) that it possesses. This particle now contains one more negatively-charged electron than positively-charged protons; therefore, it bears an overall net –1 charge. Since this ion is negatively-charged, it is classified as an anion.
An ion symbol incorporates the charge of the ion as a superscript on the corresponding elemental symbol. As stated above, this ion is still a form of chlorine (Cl). However, because this particle is now an ion, its specific charge (–1) must be included as a superscript, and an ion symbol of "Cl–1" results. Finally, this particle is referred to as a "chloride ion." The suffix of the element's name is replaced with "-ide," because this ion is an anion. - Answer b
- An atom of nitrogen contains 7 protons and 7 electrons, 5 of which can be classified as valence electrons, as determined either by using the electron configuration written below, or by recognizing that nitrogen is located in Group 5A on the periodic table.
1s22s22p3
Nitrogen would need to gain 3 electrons in order to achieve an octet configuration. Gaining this number of electrons is possible, as doing so would meet, but not exceed, the maximum gain-limit of three electrons. Therefore, nitrogen will gain three additional valence electrons.
If three additional valence electrons are added, they must all be placed in the 2p orbital, in order to fill that orbital to its maximum capacity.1s22s22p(3+3)
Simplification of the 2p orbital's superscript results in a new electron configuration, as shown below.1s22s22p6
In this new electron configuration, the number of valence electrons contained in the corresponding particle are still determined by totaling the electrons found within the the 2s and 2p orbitals.2s22p6
The superscripts associated with these orbitals now sum to 8. Therefore, this new particle has 8 valence electrons and has achieved, through the process described above, a highly-stable octet configuration.
The number of protons was again unchanged in the process described above, so this new particle still contains 7 protons and is, therefore, still a form of nitrogen. However, since this new particle was formed through the gain of three valence electrons, it now has 10 (7 + 3) electrons. Therefore, this new particle is no longer a nitrogen (N) atom which, as stated above, contains 7 electrons.
This particle is now an ion, as a result of the imbalance between the number of positively-charged protons (+7) and negatively-charged electrons (–10) that it possesses. This particle now contains three more negatively-charged electrons than positively-charged protons; therefore, it bears an overall net –3 charge. Since this ion is negatively-charged, it is classified as an anion.
Because this particle is now an ion, its specific charge (–3) must be included as a superscript on the corresponding elemental symbol, and an ion symbol of "N–3" results. Finally, this particle is referred to as a "nitride ion." The suffix of the element's name is again replaced with "-ide," because this ion is an anion.