General orrganic and biological chemistry structures off liffe 5th by karen timberlake06 worked examples

Sample Problem 6.1 Ions
a. Write the symbol and name for the ion that has 7 protons and 10 electrons.
b. Write the symbol and name for the ion that has 20 protons and 18 electrons.

Solution
a. The element with seven protons is nitrogen. In an ion of nitrogen with 10 electrons, the ionic charge would be
3–, [(7+) + (10–) = 3–]. The ion, written as N3–, is the nitride ion.
b. The element with 20 protons is calcium. In an ion of calcium with 18 electrons, the ionic charge would be 2+,
[(20+) + (18–) = 2+]. The ion, written as Ca2+, is the calcium ion.

Study Check 6.1
How many protons and electrons are in each of the following ions?
a. S2– b.
Al3+

Answer
a. 16 protons, 18 electrons b.

13 protons, 10 electrons

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.2 Writing Symbols for Ions
Consider the elements aluminum and oxygen.
a. Identify each as a metal or a nonmetal.
b. State the number of valence electrons for each.
c. State the number of electrons that must be lost or gained for each to achieve an octet.

d. Write the symbol, including its ionic charge, and name for each resulting ion.

Solution

Study Check 6.2
Write the symbols for the ions formed by potassium and sulfur.

Answer
K+ and S2–

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.3 Writing Formulas from Ionic Charges
Write the symbols for the ions, and the correct formula for the ionic compound formed when lithium and
nitrogen react.

Solution
Lithium, which is a metal in Group 1A (1), forms Li+; nitrogen, which is a nonmetal in Group 5A (15), forms N3–.
The charge of 3– is balanced by three Li+ ions.
3(1+) + 1(3–) = 0
Writing the cation (positive ion) first and the anion (negative ion) second gives the formula Li3N.

Study Check 6.3
Write the symbols for the ions, and the correct formula for the ionic compound that would form when calcium
and oxygen react.

Answer
Ca2+, O2–, CaO

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.4 Naming Ionic Compounds
Write the name for the ionic compound Mg3N2.

Solution

Step 1

Identify the cation and anion. The cation, Mg2+, is from Group 2A (2), and the anion, N3–, is from
Group 5A (15).

Step 2

Name the cation by its element name. The cation Mg2+ is magnesium.

Step 3

Name the anion by using the first syllable of its element name followed by ide. The anion N3– is nitride.

Step 4

Write the name for the cation first and the name for the anion second. Mg3N2 magnesium nitride

Study Check 6.4
Name the compound Ga2S3.

Answer
gallium sulfide

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.5 Naming Ionic Compounds with Variable
Charge Metal Ions
Antifouling paint contains Cu2O, which prevents the growth of barnacles and algae on the bottoms of boats. What
is the name of Cu2O?

Solution

Step 1

Determine the charge of the cation from the anion.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 6.5 Naming Ionic Compounds with Variable
Charge Metal Ions
Continued
Step 2

Name the cation by its element name and use a Roman numeral in parentheses for the charge. copper(I)

Step 3

Name the anion by using the first syllable of its element name followed by ide. oxide

Step 4

Write the name for the cation first and the name for the anion second. copper(I) oxide

Study Check 6.5
Write the name for the compound with the formula Mn2S3.

Answer
manganese(III) sulfide

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.6 Writing Formulas for Ionic Compounds
Write the correct formula for iron(III) chloride.

Solution

Step 1

Identify the cation and anion. Iron(III) chloride consists of a metal and a nonmetal, which means that it
is an ionic compound.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.6 Writing Formulas for Ionic Compounds
Continued
Step 2

Balance the charges.

Step 3

Write the formula, cation first, using subscripts from the charge balance. FeCl3

Study Check 6.6
Write the correct formula for chromium(III) oxide.

Answer
Cr2O3

General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.7 Writing Formulas Containing Polyatomic Ions
An antacid called Amphojel contains aluminum hydroxide, which treats acid indigestion and heartburn. Write the
formula for aluminum hydroxide.

Solution
Step 1

Identify the cation and polyatomic ion (anion).
Cation
Polyatomic ion (anion)
aluminum
hydroxide
3+
Al
OH–

Step 2

Balance the charges.

Step 3

Write the formula, cation first, using the subscripts from charge balance. The formula for the compound
is written by enclosing the formula of the hydroxide ion, OH–, in parentheses and writing the subscript 3
outside

theChemistry:
right parenthesis.
© 2016 Pearson Education, Inc.
General, Organic, and
Biological
Structures of Life, 5/e
Karen C. Timberlake


Sample Problem 6.7 Writing Formulas Containing Polyatomic Ions
Continued

Study Check 6.7
Write the formula for a compound containing ammonium ions and phosphate ions.

Answer
(NH4)3PO4

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.8 Naming Compounds Containing Polyatomic Ions
Name the following ionic compounds:
a. Cu(NO2)2 b.
KClO3

Solution

Study Check 6.8
What is the name of Co3(PO4)2?

Answer
cobalt(II) phosphate

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.9 Naming Molecular Compounds
Name the molecular compound NCl3.

Solution

Step 1

Name the first nonmetal by its element name. In NCl3, the first nonmetal (N) is nitrogen.

Step 2

Name the second nonmetal by using the first syllable of its element name followed by ide. The second
nonmetal (Cl) is named chloride.

Step 3

Add prefixes to indicate the number of atoms (subscripts). Because there is one nitrogen atom, no prefix

is needed. The subscript 3 for the Cl atoms is shown as the prefix tri. The name of NCl3 is nitrogen
trichloride.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.9 Naming Molecular Compounds
Continued

Study Check 6.9
Write the name for each of the following molecular compounds:
a. SiBr4 b.
Br2O

Answer
a. silicon tetrabromide

b.

dibromine oxide

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 6.10 Writing Formulas for Molecular Compounds
Write the formula for the molecular compound diboron trioxide.

Solution

Step 1

Write the symbols in the order of the elements in the name.

Step 2

Write any prefixes as subscripts. The prefix di in diboron indicates that there are two atoms of boron,
shown as a subscript 2 in the formula. The prefix tri in trioxide indicates that there are three atoms of
oxygen, shown as a subscript 3 in the formula.
B 2O 3

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.10 Writing Formulas for Molecular Compounds
Continued

Study Check 6.10
Write the formula for the molecular compound iodine pentafluoride.

Answer
IF5

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.11 Naming Ionic and Molecular Compounds
Identify each of the following compounds as ionic or molecular and give its name:
a. K3P
b. NiSO4 c. SO3

Solution
a. K3P, consisting of a metal and a nonmetal, is an ionic compound. As a representative element in Group 1A (1), K
forms the potassium ion, K+. Phosphorus, as a representative element in Group 5A (15), forms a phosphide ion,
P3–. Writing the name of the cation followed by the name of the anion gives the name potassium phosphide.
b. NiSO4, consisting of a cation of a transition element and a polyatomic ion SO 42–, is an ionic compound. As a
transition element, Ni forms more than one type of ion. In this formula, the 2– charge of SO 42– is balanced by one
nickel ion, Ni2+. In the name, a Roman numeral written after the metal name, nickel(II), specifies the 2+ charge.
The anion SO42– is a polyatomic ion named sulfate. The compound is named nickel(II) sulfate.
c. SO3 consists of two nonmetals, which indicates that it is a molecular compound. The first element S is sulfur (no
prefix is needed). The second element O, oxide, has subscript 3, which requires a prefix tri in the name. The
compound is named sulfur trioxide.

Study Check 6.11
What is the name of Fe(NO3)3?

Answer
iron(III) nitrate
General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.12 Drawing Lewis Structures
Draw the lewis structure for PCl3, phosphorus trichloride, used commercially to prepare insecticides and flame
retardants.

Solution
Step 1

Determine the arrangement of atoms. In PCl3, the central atom is P because there is only one P atom.
Cl

Step 2

P
Cl

Cl

Determine the total number of valence electrons. We use the group number to determine the number of
valence electrons for each of the atoms in the molecule.

Step 3

Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be
represented
by a bond

line.
© 2016 Pearson Education, Inc.
General, Organic, and
Biological Chemistry:
Structures
of Life, 5/e
Karen C. Timberlake


Sample Problem 6.12 Drawing Lewis Structures
Continued
Step 4

Place the remaining electrons using single or multiple bonds to complete the octets. Six electrons
(3 × 2 e–) are used to bond the central P atom to three Cl atoms. Twenty valence electrons are left.
26 valence e– – 6 bonding e– = 20 e– remaining
We use the remaining 20 electrons as lone pairs, which are placed around theouter Cl atoms and on the
P atom, such that all the atoms have octets.

Study Check 6.12
Draw the Lewis structure for Cl2O.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 6.13 Drawing Lewis Structures for Polyatomic Ions
Sodium chlorite, NaClO2, is a component of mouthwashes, toothpastes, and contact lens cleaning solutions. Draw
the Lewis structure for the chlorite ion, ClO2–.

Solution
Step 1

Determine the arrangement of atoms. For the polyatomic ion ClO 2–, the central atom is Cl because there is
only one Cl atom. For a polyatomic ion, the atoms and electrons are placed in brackets, and the charge is
written outside to the upper right.
[O Cl O]–

Step 2

Determine the total number of valence electrons. We use the group numbers to determine the number of
valence electrons for each of the atoms in the ion. Because the ion has a negative charge, one more electron
is added to the valence electrons.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.13 Drawing Lewis Structures for Polyatomic Ions
Continued
Step 3

Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be
represented by a line, which indicates a single bond.

Step 4

Place the remaining electrons using single or multiple bonds to complete the octets. Four electrons are used
to bond the O atoms to the central Cl atom, which leaves 16 valence electrons. Of these, 12 electrons are
drawn as lone pairs to complete the octets of the O atoms.

The remaining four electrons are placed as two lone pairs on the central Cl atom.

Study Check 6.13
Draw the Lewis structure for the polyatomic ion NH2–.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.14 Drawing Lewis Structures with Multiple Bonds
Draw the Lewis structure for carbon dioxide, CO2, in which the central atom is C.

Solution
Step 1

Determine the arrangement of atoms. O C O

Step 2

Determine the total number of valence electrons. Using the group number to determine valence electrons,
the carbon atom has four valence electrons, and each oxygen atom has six valence electrons, which gives a
total of 16 valence electrons.

Step 3

Attach each bonded atom to the central atom with a pair of electrons.
O : C : O or O—C—O

Step 4

Place the remaining electrons to complete octets. Because we used four valence electrons to attach the C
atom to two O atoms, there are 12 valence electrons remaining.
16 valence e– – 4 bonding e– = 12 e– remaining

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.14 Drawing Lewis Structures with Multiple Bonds
Continued
The 12 remaining electrons are placed as six lone pairs of electrons on the outside O atoms. However, this
does not complete the octet of the C atom.

To obtain an octet, the C atom must share lone pairs of electrons from each of the O atoms. When two
bonding pairs occur between atoms, it is known as a double bond.

Study Check 6.14

Draw the Lewis structure for HCN, which has a triple bond.

Answer

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.15 Bond Polarity
Using electronegativity values, classify each of the following bonds as nonpolar covalent, polar covalent, or ionic:
O—K, Cl—As, N—N, P—Br

Solution
For each bond, we obtain the electronegativity values and calculate the difference in electronegativity.

Study Check 6.15
Using electronegativity values, classify each of the following bonds as nonpolar covalent, polar covalent, or ionic:
a. P—Cl
b. Br—Br
c. Na—O

Answer
a. polar covalent (0.9)

b. nonpolar covalent (0.0) c. ionic (2.6)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.16 Shapes of Molecules
Predict the shape of a molecule of SiCl4.

Solution

Step 1

Draw the Lewis structure. Using 32 e–, we draw the bonds and lone pairs for the Lewis structure of SiCl4.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 6.16 Shapes of Molecules
Continued
Step 2

Arrange the electron groups around the central atom to minimize repulsion. To minimize repulsion, the
electron-group geometry would be tetrahedral.

Step 3

Use the atoms bonded to the central atom to determine the shape. Because the central Si atom has four
bonded pairs and no lone pairs of electrons, the SiCl4 molecule has a tetrahedral shape.

Study Check 6.16
Predict the shape of SCl2.

Answer
The central atom S has four electron groups: two bonded atoms and two lone pairs of electrons. The shape of SCl 2 is
bent, 109°.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.