worked example 3.1 periodic trends: ionization energy

© 2013 Pearson Education, Inc.Fundamentals of General, Organic, and Biological Chemistry, 7eJohn McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson

Worked Example 3.1 Periodic Trends: Ionization Energy

Identify the group number of rubidium (group 1A), and find where other members of the group appear in Figure 3.1 .

Analysis

Look at the periodic trends in Figure 3.1, and predict where the ionization energy of rubidium is likely to fall on the chart.

Rubidium (Rb) is the alkali metal below potassium (K) in the periodic table. Since the alkali metals Li, Na, and K all have ionization energies near the bottom of the chart, the ionization energy of rubidium is probably similar.

Solution

Figure 3.1

Relative ionization energies (red) and electron affinities (blue) for elements in the first four rows of the periodic table.


Worked Example 3.2 Periodic Trends: Formation of Anions and Cations

Identify the group numbers of the elements, and find where members of those groups appear in Figure 3.1 .

Analysis

Which element is likely to lose an electron more easily, Mg or S?

Magnesium, a group 2A element on the left side of the periodic table, has a relatively low ionization energy, and loses an electron easily. Sulfur, a group 6A element on the right side of the table, has a higher ionization energy, and loses an electron less easily.

Solution

Figure 3.1

Relative ionization energies (red) and electron affinities (blue) for elements in the first four rows of the periodic table.


Worked Example 3.3 Electron Configurations: Octet Rule for Cations

Write the electron configuration of magnesium as described in Section 2.7 and count the number of electrons in the valence shell.

Analysis

Write the electron configuration of magnesium (Z = 12). Show how many electrons a magnesium atom must lose to form an ion with a filled shell (8 electrons), and write the configuration of the ion. Explain the reason for the ion’s charge, and write the ion’s symbol.

Magnesium has the electron configuration 1s22s22p63s2. Since the second shell contains an octet of electrons (2s22p6) and the third shell is only partially filled (3s2), magnesium can achieve a valence-shell octet by losing the 2 electrons in the 3 s subshell. The result is formation of a doubly charged cation, Mg2+, with the neon configuration:

Mg2+ 1s22s22p6 (Neon configuration, or [Ne] )

A neutral magnesium atom has 12 protons and 12 electrons. With the loss of 2 electrons, there is an excess of 2 protons, accounting for the +2 charge of the ion, Mg2+.

Solution


Worked Example 3.4 Electron Configurations: Octet Rule for Anions

Write the electron configuration of nitrogen, and identify how many more electrons are needed to reach a noble gas configuration.

Analysis

How many electrons must a nitrogen atom, Z = 7, gain to attain a noble gas configuration? Write the electron-dot and ion symbols for the ion formed.

Nitrogen, a group 5A element, has the electron configuration 1s22s22p3. The second shell contains 5 electrons (2s22p3) and needs 3 more to reach an octet. The result is formation of a triply charged anion, N3–, with 8 valence electrons, matching the neon configuration:

Solution


Worked Example 3.5 Formation of Ions: Gain/Loss of Valence Electrons

Count the number of valence electrons in each ion. For main group elements, only ions with a valence octet of electrons are likely to form.

Analysis

Which of the following ions is likely to form?

(a) S3– ( b) Si2+ ( c) Sr2+

(a) Sulfur is in group 6A, has 6 valence electrons, and needs only 2 more to reach an octet. Gaining 2 electrons gives an S2– ion with a noble gas configuration, but gaining 3 electrons does not. The S3– ion is, therefore, unlikely to form.

(b) Silicon is a nonmetal in group 4A. Like carbon, it does not form ions because it would have to gain or lose too many electrons (4) to reach a noble gas electron configuration. The Si2+ ion does not have an octet and will not form.

(c) Strontium, a metal in group 2A, has only 2 outer-shell electrons and can lose both to reach a noble gas configuration. The Sr2+ ion has an octet and, therefore, forms easily

Solution


Worked Example 3.6 Ionic Compounds: Writing Formulas

Knowing the formula and charges on the cation and anion ( Figure 3.4 ), we determine how many of each are needed to yield a neutral formula for the ionic compound.

Analysis

Write the formula for the compound formed by calcium ions and nitrate ions.

The two ions are Ca2+ and NO3–. Two nitrate ions, each with a –1 charge, will balance the +2 charge of the

calcium ion.Ca2+ Charge = 1 × (+2) = +2

2NO3– Charge = 2 × (–1) = –2

Since there are 2 ions, the nitrate formula must be enclosed in parentheses:

Ca(NO3)2 Calcium nitrate

Solution

Figure 3.4Common ions formed by elements in the first four periods.


Worked Example 3.7 Ionic Compounds: Formulas Involving Polyatomic Ions

Since magnesium is a main group metal, we can determine its ionic compound formula by identifying the charges and formulas for the anion and the cation, remembering that the overall formula must be neutral.

Analysis

Magnesium carbonate is used as an ingredient in Bufferin (buffered aspirin) tablets. Write its formula.

Look at the cation and the anion parts of the name separately. Magnesium, a group 2A element, forms the doubly positive Mg2+ cation; carbonate anion is doubly negative, CO3

2– Because the charges on the anion and cation are equal, a formula of MgCO3 will be neutral.

Solution


Worked Example 3.8 Ionic Compounds: Formulas and Ionic Charges

Using the formulas and charges for the cations and the anions (from Tables 3.2 and 3.3 ), we determine how many of each cation and anion are needed to yield a formula that is neutral.

Analysis

Sodium and calcium both form a wide variety of ionic compounds. Write formulas for the following compounds:

(a) Sodium bromide and calcium bromide

(b) Sodium sulfide and calcium sulfide

(c) Sodium phosphate and calcium phosphate


Worked Example 3.8 Ionic Compounds: Formulas and Ionic ChargesContinued

(a) Cations = Na+ and Ca2+; anion = Br–: NaBr and CaBr2

(b) Cations = Na+ and Ca2+; anion = S2–: Na2S and CaS(c) Cations = Na+ and Ca2+; anion = PO4

3–: Na3PO4 and Ca2(PO4)2

Solution


Worked Example 3.9 Naming Ionic Compounds

For main group metals, the charge is determined from the group number, and no Roman numerals are necessary. For transition metals, the charge on the metal can be determined from the total charge(s) on the anion(s).

Analysis

Name the following compounds, using Roman numerals to indicate the charges on the cations where necessary:(a) KF (b) MgCl2 (c) AuCl3 (d) Fe2O3

(a) Potassium fluoride. No Roman numeral is necessary because a group 1A metal forms only one cation.

(b) Magnesium chloride. No Roman numeral is necessary because magnesium (group 2A) forms only Mg2+

(c) Gold(III) chloride. The 3 Cl– ions require a +3 charge on the gold for a neutral formula. Since gold is a transition metal that can form other ions, the Roman numeral is necessary to specify the +3 charge.

(d) Iron(III) oxide. Because the 3 oxide anions (O2-) have a total negative charge of –6, the 2 iron cations must have a total charge of +6. Thus, each is Fe3+, and the charge on each is indicated by the Roman numeral (III).

Solution

worked example 3.1 periodic trends: ionization energy

Documents