Chapter 9:  Chemical Bonding and Nomenclature

 

Section 9-1:  Relationship Between Groups and Number of Valence Electrons

Section 9-2:  Electronegativity

Section 9-3:  Ionic Bonding and Ionic Nomenclature

Section 9-4:  Molecular Compounds, Covalent Bonding, and Lewis Structures

Section 9-5:  Bond Length and Resonance

Section 9-6:  Nomenclature of Binary Molecular Compounds

Section 9-7:  Introduction to Organic Nomenclature

Section 9-8:  Names of Common Acids

Chapter 9 Practice Exercises and Review Quizzes

 

 

 

 

Section 9-1:  Relationship Between Groups and Number of Valence Electrons

 

For Groups 1, 2, and 13 through 18 (except helium), all elements in a group will have exactly the same number of valence s and p electrons.  Therefore, we can quickly determine the total number of valence electrons using the following chart:

 

Group

Valence Electron Configuration

Total Valence Electrons

1

ns1

1

2

 ns2

2

13

ns2 np1

3

14

ns2 np2

4

15

ns2 np3

5

16

ns2 np4

6

17

ns2 np5

7

18

ns2 np6

8

 

Note that any d or f electrons that an element may possess will have a lower value of n than the valence electrons and, therefore, should not be included in the valence electron configuration or the total number of valence electrons.  Although most transition metals in groups 3 through 12 will have a valence electron configuration of ns2 and, therefore, a total of 2 valence electrons, there are numerous exceptions.  

 

 

 

 

 

Section 9-2:  Electronegativity

 

Electronegativity (EN) is the ability of an atom to attract electrons in a chemical bond.  Electronegativity, which is relative and has no unit, generally  increases toward the upper right corner of the periodic table table, with fluorine having the highest electronegativity value of all elements.  The following table shows the electronegativity values of select elements:

 

 

Group 1

Group 2

Groups 3-12

Group 13

Group 14

Group 15

Group 16

Group 17

Group 18

H

2.1

 

 

 

 

Li

1.0

 

B

2.0

C

2.5

N

3.0

O

3.5

F

4.0

 

Na

0.9

Mg

1.2

Al

1.5

Si

1.8

P

2.1

S

2.5

Cl

3.0

 

K

0.8

Ca

1.0

 

 

As

2.0

Se

2.4

Br

2.8

Kr

3.0

Rb

0.8

Sr

1.0

 

 

 

Te

2.1

I

2.5

Xe

2.6

Cs

0.8

Ba

0.9

 

 

 

 

 

 

Fr

0.7

 

 

 

 

 

 

 

 

 

For purposes of chemical bonding, the elements in bold print toward the lower left corner with relatively low electronegativities will generally be considered metals, while the elements in italics toward the upper right corner with relatively high electronegativities will generally be considered nonmetals.  Note that hydrogen has a relatively high electronegativity and, therefore, should be considered a nonmetal.

 

 

 

Section 9-3:  Ionic Bonding and Ionic Nomenclature

 

When a metal with a low electronegativity reacts with a nonmetal with a high electronegativity, the large difference in electronegativity (∆EN) allows for the metal to transfer one or more electrons to the nonmetal.  As a result, the metal becomes a positively-charged cation and the nonmetal becomes a negatively-charged anion.  The attraction between the cation and anion is known as ionic bonding.  

 

In the case of a Group 1 alkali metal such as lithium, a Group 2 alkaline earth metal such as calcium, or aluminum metal, the metal will lose all its valence electrons and become a cation with a noble gas electron configuration:

 

Li:  [He] 2s1 loses 1 valence electron to become Li+ :  [He]

 

Ca:  [Ar] 4s2 loses 2 valence electrons to become Ca2+ :  [Ar]

 

Al:  [Ne] 3s2 3p1 loses 3 valence electrons to become Al3+ :  [Ne]

 

Nonmetals in Group 15 such phosphorus, Group 16 such as oxygen, and Group 17 such as bromine will gain enough electrons to have an electron configuration ending in np6 that is equivalent to the electron configuration of the noble gas nearest in atomic number to the nonmetal:

 

P:  [Ne] 3s2 3p3 gains 3 electrons to become P3- :  [Ne] 3s2 3p6 or [Ar]

 

O:  [He] 2s2 2p4 gains 2 electrons to become O2- :  [He] 2s2 2p6 or [Ne]

 

Br:  [Ar] 4s2 3d10 4p5 gains 1 electron to become Br- :  [Ar] 4s2 3d10 4p6 or [Kr]

 

From the discussion above, we can conclude the following about monatomic ions:

 

Ion

Charge

Group 1 alkali metal cation

1+

Group 2 alkaline earth metal cation

2+

aluminum metal cation

3+

Group 15 nonmetal anion

3-

Group 16 nonmetal anion

2-

Group 17 nonmetal anion

1-

 

The nomenclature or naming of an ionic compound formed by combining one type of metal cation from the table above and one type of nonmetal anion from the table above will follow the format: 

 

(name of metal) +  (name of nonmetal with “ide” ending)

 

To determine the chemical formula of an ionic compound, we must choose the correct number of cations and the correct number of anions to ensure that the total positive charge of the cations equals the total negative charge of the anions, and we will typically write the cation first in the formula.  When the magnitude of the positive charge of the cation equals the magnitude of the negative charge of the anion, the formula will contain one cation and one anion.  Note, however, that the final written formula does not actually show charges:

 

sodium chloride = (combine Na+ with Cl-) = NaCl

 

magnesium sulfide = (combine Mg2+ with S2-) = MgS

 

aluminum nitride = (combine Al3+ with N3-) = AlN

 

When the magnitude of the positive charge of the cation does not equal the magnitude of the negative charge of the anion, we must use subscripts in the formula to adjust the number of ions in order to equalize the total positive and negative charges.  For example:

 

potassium oxide = (combine 2 K+ with 1 O2-) = K2O

 

aluminum fluoride = (combine 1 Al3+ with 3 F-) = AlF3

 

calcium phosphide = (combine 3 Ca2+ with 2 P3-) = Ca3P2

 

Sample Exercise 9A:

 

Write the chemical formula for each ionic compound:

 

(a) barium chloride

(b) aluminum sulfide

(c) lithium nitride

 

Solution:

 

(a) (combine 1 Ba2+ with 2 Cl-) = BaCl2

 

(b) (combine 2 Al3+ with 3 S2-) = Al2S3

 

(c) (combine 3 Li+ with 1 N3-) = Li3N

 

 

Most other metals can achieve more than one positive charge.  To distinguish among the possible cations for these other metals, the name of the metal will be followed by the magnitude of the positive charge in Roman numerals enclosed by parentheses.  For example, iron(III) = Fe3+ and tin(IV) = Sn4+.

 

Groups of bonded atoms that possess an overall charge are known as polyatomic ions.  Here is a list of common polyatomic ions:

 

1+

ammonium = NH4+

 

1-

acetate = CH3COO- or C2H3O2-

bicarbonate or hydrogen carbonate = HCO3-

hydroxide = OH-

nitrate = NO3-

 

2-

carbonate = CO32-

sulfate = SO42-

 

3-

phosphate = PO43-

 

When writing chemical formulas for ionic compounds containing polyatomic ions and a subscript greater than 1 is required after the formula of a polyatomic ion, the formula of the polyatomic ion should be enclosed in parentheses.  For example:

 

aluminum acetate = (combine 1 Al3+ with 3 CH3COO-) = Al(CH3COO)3

 

nickel(II) phosphate = (combine 3 Ni2+ with 2 PO43-) = Ni3(PO4)2

 

ammonium sulfate = (combine 2 NH4+ with 1 SO42-) = (NH4)2SO4

 

Sample Exercise 9B:

 

Write the chemical formula for each ionic compound:

 

(a) cobalt(III) carbonate

(b) strontium hydroxide

(c) ammonium nitride

 

Solution:

 

(a) (combine 2 Co3+ with 3 CO32-) = Co2(CO3)3

 

(b) (combine 1 Sr2+ with 2 OH-) = Sr(OH)2

 

(c) (combine 3 NH4+ with N3-) = (NH4)3N

 

 

If we are given the chemical formula of an ionic compound and need to write the name, we will use the format:

 

(ammonium or name of metal)

+

(name of polyatomic anion or name of nonmetal with “ide” ending)

 

Remember that in all but a few cases the name of the metal should include the magnitude of the charge in Roman numerals enclosed by parentheses, except if the metal is in Group 1, in Group 2, or aluminum. 

 

Sample Exercise 9C:

 

Write the name of each compound:

 

(a) Mn(HCO3)2

(b) Al(NO3)3

(c) (NH4)2S

 

Solution:

 

(a) Since the bicarbonate or hydrogen carbonate ion is known to have a 1- charge and there are two such ions in the formula, we can deduce that the charge on the metal must be 2+.  Therefore, the name of the compound is manganese(II) bicarbonate or manganese(II) hydrogen carbonate.

 

(b) We do not need to specify the charge of aluminum in Roman numerals enclosed by parentheses, so the name of the compound is aluminum nitrate.

 

(c) The anion S2- is simply a monatomic nonmetal anion, not to be confused with the polyatomic anion sulfate, so the name of the compound is ammonium sulfide.

 

 

On a final note, the term “molecule” should not be used to describe an individual unit of an ionic compound such as NaCl.  The more appropriate term “formula unit” should be used instead.  

 

 

 

Section 9-4:  Molecular Compounds, Covalent Bonding, and Lewis Structures

 

Molecular compounds are composed of nonmetals.  When two nonmetals are bonded, since each has a relatively high electronegativity, the small difference in electronegativity precludes the transfer of electrons that was possible in ionic compounds.  Instead, the two nonmetals will share one or more pairs of valence electrons in a covalent bond. 

 

A Lewis structure depicts each shared pair of valence electrons in a molecule using a straight line between the bonded atoms.  In addition, the unshared pairs of valence electrons in a molecule, known as a lone pairs, are depicted in a Lewis structure using one pair of dots per lone pair.  To draw a Lewis structure, we will follow the steps below as demonstrated for the molecule Br2:

 

Lewis structure Step 1:  Determine the sum of the valence electrons for each atom in the molecule or ion.  For cations, subtract the magnitude of the positive charge from the total.  For anions, add the magnitude of the negative charge to the total.  The final Lewis structure must have this total number of electrons.    

 

Br is in Group 17, so each Br atom has 7 valence electrons.  Therefore, the Lewis structure for Br2  must have a total of 2 x 7 = 14 electrons.

 

Lewis structure Step 2:  Connect each bonded atom by a single covalent bond.

 

Draw a straight line between the atoms to represent a shared pair of electrons:

 

 

Lewis structure Step 3:  Place lone pairs in separate locations adjacent to each outer atom, but never between atoms, until each outer atom has an octet of 8 valence electrons adjacent to it, thus achieving the “Octet Guideline” of having the same number of valence electrons as all noble gases, except helium.

 

Place 3 pairs of dots in separate locations adjacent to each Br atom, but not between the Br atoms, to represent 3 lone pairs of electrons on each Br:

 

The Lewis structure of Br2 is now complete with the correct total of 14 electrons and each Br atom having an octet of 8 adjacent electrons, 2 shared electrons in the covalent bond + 6 electrons in the 3 lone pairs.

 

Two bonded nonmetals with similar electronegativities are said to be joined by a nonpolar covalent bond where the electrons in the bond are essentially attracted equally to each atom and, therefore, spend roughly equal amounts of time near each of the two atoms.  When the two bonded nonmetals have a very small electronegativity difference of 0.4 or less, such as that between hydrogen (EN = 2.1) and carbon (EN = 2.5), we will generally consider the bond between the nonmetals to be a nonpolar covalent bond.  On the other hand, if the electronegativity difference between the bonded nonmetals is 0.5 or greater, we will generally consider the bond between the nonmetals to be a polar covalent bond where the electrons in the bond are attracted unequally to each atom and, therefore, spend more time closer to one atom and farther away from the other atom. 

 

Although the electronegativity of hydrogen is high enough for hydrogen to be considered a nonmetal, there is a large enough difference in electronegativity between the hydrogen and oxygen atoms in water (EN for O = 3.5, so ∆EN between H and O = 1.4) for the bonds in water to be considered polar covalent bonds.  To draw the Lewis structure of H2O, we begin by performing the first two steps above:

 

Step 1:  Each H has 1 valence electron.  O is in Group 16, so the O atom has 6 valence electrons.  Therefore, the Lewis structure of water H2O must have a total of (2 x 1) + 6 = 8 electrons.    

 

Step 2:  When a single atom of one element is bonded to two or more atoms of a second element, we will generally place the single atom in the center of the Lewis structure and spread the other atoms around the center.  As such, we will place the oxygen atom in the center and connect to each hydrogen atom with a single covalent bond:

 

 

At this point, we will skip step 3 and introduce an important exception to the “Octet Guideline” as well as the fourth step in the process of drawing Lewis structures:

 

“Octet Guideline” Exception 1:  Since the noble gas nearest in atomic number to hydrogen is helium, which has an electron configuration of 1s2 and 2 valence electrons, H will always have an incomplete octet of 2 valence electrons in Lewis structures.  As such, H must always be an outer atom joined to only one other atom by a single covalent bond.

 

Lewis structure Step 4A:  Place lone pairs in separate locations adjacent to the center atom, but never between atoms, until the correct total number of electrons has been reached in the Lewis structure.

 

Step 4A:  Place two lone pairs adjacent to the center O atom, but not between atoms, to reach the correct total of 8 electrons:

 

 

The Lewis structure of water is now complete with the correct total of 8 electrons and the O atom having an octet of 8 adjacent electrons, 4 shared electrons in the 2 covalent bonds + 4 electrons in the 2 lone pairs.

 

Below are two more notable exceptions to the “Octet Guideline”:

 

“Octet Guideline” Exception 2:  Lewis structures with boron having an incomplete octet of 6 electrons are preferable, although boron may also have an octet in some cases.

 

To draw the Lewis structure of BI3, we follow steps 1 through 3 above:

 

Step 1:  B is in Group 13, so B has 3 valence electrons.  I is in Group 17, so each I has 7 valence electrons.  Therefore, the Lewis structure must have a total of 3 + (3 x 7) = 24 electrons.

 

Step 2:  Place the single B atom in the center and spread the three I atoms around the B, then connect the B to each of the three I atoms by single covalent bonds:

 

 

Step 3:  Place lone pairs on the I atoms until each has an octet:

 

 

The Lewis structure now has the correct total of 24 electrons, but we did not perform step 4A because there were no more electrons to place on the center B atom.  Since it is preferable to leave B with an incomplete octet of 6 electrons, the Lewis structure is now complete.

 

 “Octet Guideline” Exception 3:  The center atom in a Lewis structure can exceed 8 electrons and have an “expanded octet” only if the center atom is NOT in the second row of the periodic table.

 

To draw the Lewis structure of the PCl4- anion, we follow steps 1-4A above:

 

Step 1:  P is in Group 15, so P has 5 valence electrons.  Cl is in group 17, so each Cl has 7 valence electrons.  We add one extra electron because the charge on the anion is 1-, so the Lewis structure must have a total of 5 + (4 x 7) + 1 = 34 electrons. 

 

Step 2:  Place the single P atom in the center and spread the four Cl atoms around the P, then connect the P to each of the four Cl atoms by single covalent bonds:

 

 

Step 3:  Place lone pairs on the Cl atoms until each has an octet:

 

 

Step 4A:  Place one lone pair on the center P atom to reach the correct total of 34 electrons:

 

 

The Lewis structure is now complete with the center atom P having an “expanded octet” of 10 electrons, which is acceptable because P is not in the second row of the periodic table.  Note that Lewis structures of ions are typically enclosed in brackets with the charge shown outside the bracket as a superscript. 

 

Sample Exercise 9D:

 

In some cases, noble gases can form compounds.  Draw the Lewis structure for KrF2.

 

Solution:

 

8 + (2 x 7) = 22 electrons

 

 

Kr has an “expanded octet” of 10 electrons, which is acceptable because Kr is not in the second row of the periodic table.

 

 

In addition to single covalent bonds with one pair of shared electrons between the atoms, it is also possible for two nonmetals to covalently bond with two or three shared pairs of electrons between the atoms.  For example, O2 has a double bond depicted by two straight lines between the oxygen atoms in the Lewis structure, and HCN contains a triple bond depicted by three straight lines between the carbon and nitrogen atoms in the Lewis structure. 

 

To draw the Lewis structure of O2, we can treat one of the O atoms as a center atom and the other as an outer atom as we follow steps 1 through 4A above:

 

Step 1:  2 x 6 = 12 electrons

 

Step 2:  Connect O atoms by single covalent bond:

 

 

Step 3:  Treating the oxygen atom on the right as an outer atom, place lone pairs to give the O on the right an octet:

 

 

Step 4A:  Treating the oxygen atom on the left as the center atom, place lone pairs on the O to the left until the correct total of 12 electrons is reached:

 

 

At this point, the O on the left has an incomplete octet of only 6 electrons, but we have no more electrons to add.  As such, we must introduce another step in the process of drawing Lewis structures:

 

Lewis structure Step 4B:  Move one or more lone pairs from outer atoms to create multiple bonds until the center atom has an octet.

 

Move any one of the three lone pairs on the right-hand O between the atoms to create a double bond:

 

 

The Lewis structure is now complete with each O having an octet, 4 shared electrons in the double bond + 4 electrons in the 2 lone pairs.

 

To draw the Lewis structure of BI3, it was not necessary to use Step 4A or 4B.  To draw the Lewis structure of PCl4-, it was necessary to use Step 4A but not Step 4B.  To draw the Lewis structure of O2, we used both Step 4A and Step 4B.  To draw the Lewis structure of HCN, we will need to skip Step 4A due to lack of electrons and only use Step 4B to achieve an octet on the center C atom:

 

Step 1:  1 + 4 + 5 = 10 electrons

 

Step 2:  Connect center C to H and N by single covalent bonds:

 

 

Step 3:  Since H must have an incomplete octet of 2 electrons, place lone pairs only on N to give N an octet:

 

 

At this point, C only has 4 electrons, but we must skip Step 4A because there are already 10 total electrons and there are no more electrons to add.  As such, we must proceed to Step 4B.

 

Step 4B:  Move two lone pairs from the N to create a triple bond between the C and N atoms:

 

 

The Lewis structure is now complete with the center C having an octet, 6 shared electrons in the triple bond + 2 shared electrons in the single bond.

 

 

 

Section 9-5:  Bond Length and Resonance

 

Bond length is the distance between the nuclei of two covalently bonded atoms.  As the number of shared electron pairs increases, the nuclei of the two atoms will be drawn closer together and the bond length will decrease:

 

number of bonds = bond length

 

Sample Exercise  9E:

 

Draw Lewis structures for N2 and N2H2.  Which molecule will have the shorter nitrogen-to-nitrogen bond length?  Explain briefly.

 

Solution:

 

Step 1:  N2 = 2 x 5 = 10 electrons

 

Steps 2 and 3:  Connect N atoms by single covalent bond and, treating the N on the right as an outer atom, place lone pairs to give the N on the right an octet:

 

 

Step 4A:  Treating the N on the left as the center atom, place one lone pair on the left-hand N to reach 10 total electrons:

 

 

Step 4B:  Move two lone pairs from the N on the right to create a triple bond between the N atoms:

 

 

The Lewis structure for N2 is now complete with each N having an octet, 6 shared electrons in the triple bond + 2 electrons in the lone pair.

 

Step 1:  N2H2 = (2 x 5) + (2 x 1) = 12 electrons

 

Step 2:  Each H must be an outer atom, so connect the two N atoms by a single covalent bond and then connect one H to each N by a single covalent bond:

 

 

Step 3:  Treating the N on the right as an outer atom, place lone pairs to give the N on the right an octet:

 

 

Step 4A:  Treating the N on the left as the center atom, place one lone pair on the left-hand N to reach 12 total electrons:

 

 

Step 4B:  Move one lone pair from the N on the right to create a double bond between the N atoms:

 

 

The Lewis structure for N2H2 is now complete with each N having an octet, 4 shared electrons in the double bond + 2 shared electrons in the single bond + 2 electrons in the lone pair.

 

Since the number of bonds in N2’s triple bond is greater than the number of bonds in N2H2’s double bond, N2 will have the shorter nitrogen-to-nitrogen bond length. 

 

 

We will now demonstrate the concept of resonance by drawing the Lewis structure of ozone, O3:

 

Step 1:  3 x 6 = 18 electrons

 

Steps 2 and 3: 

 

 

Step 4A:

 

 

At this point, we have the correct total of 18 electrons, but the center O has an incomplete octet of only 6 electrons.  For Step 4B, we can move a lone pair from either outer O atom to create a double bond and give the center O atom an octet.  When double or triple bonds can be formed from different outer atoms, the resulting Lewis structures are known as resonance structures and should be connected by a double-headed arrow:

 

 

If either the Lewis structure on the left or the Lewis structure on the right was alone and represented the true structure of ozone, then one of the O-to-O bond lengths would be shorter than the other due to the different number of bonds between the O atoms.  However, experimental evidence suggests that both O-to-O bond lengths are exactly equal in length and shorter than an O-to-O single bond but longer than an O-to-O double bond.  As such, the true structure of ozone cannot be either one of the Lewis structures above and instead must be an equal hybrid of the two resonance structures shown above.

 

Sample Exercise 9F:

 

Draw all the resonance structures for the carbonate ion.

 

Solution:

 

 Step 1:  CO32- = 4 + (3 x 6) + 2 = 24 electrons

 

Steps 2 and 3:

 

 

At this point, we have the correct total of 24 electrons, but the center C atom has an incomplete octet of 6 electrons.  We must skip Step 4A because there are no more electrons to add to the center C.

 

Step 4B: 

 

 

Experimental evidence suggests that all three C-to-O bond lengths are exactly equal in length and shorter than a C-to-O single bond but longer than a C-to-O double bond.  As such, the true structure of the carbonate ion cannot be any one of the three Lewis structures above and instead must be an equal hybrid of all three resonance structures shown above.

 

 

 

Section 9-6:  Nomenclature of Binary Molecular Compounds

 

The common nomenclature of binary molecular compounds composed of two different nonmetals utilizes one of the following prefixes to specify the number of atoms of the first element in the formula and a second prefix from below to specify the number of atoms of the second element in the formula:

 

Prefix

Number of Atoms

mono

1

di

2

tri

3

tetra

4

penta

5

hexa

6

 

 

Note the following:

 

1. Whereas the name of the first element in the formula is unchanged, the name of the second element in the formula will have the “ide” ending.  For example, N2O3 = dinitrogen trioxide. 

 

2.  The prefix “mono” is typically omitted if there is only one atom of the first element in the formula.  For example, BrI5 = bromine pentaiodide.

 

3.  However, the prefix “mono” is used if there is only one atom of the second element in the formula, but the prefix is typically truncated to “mon” when followed by “oxide.”  For example, CO = carbon monoxide.

 

4. The prefixes “tetra,” “penta,” and “hexa” are typically truncated by dropping the “a” when followed by “oxide.”  For example, P4O6 = tetraphosphorus hexoxide.

 

 

 

Section 9-7:  Introduction to Organic Nomenclature

 

Organic compounds contain carbon.  For some categories of organic compounds, the following prefixes are used to specify the number of carbon atoms in the chemical formula:

 

Prefix

Number of Carbon Atoms

meth

1

eth

2

prop

3

but

4

pent

5

hex

6

hept

7

oct

8

 

 

Alkanes are organic compounds composed of single-bonded carbon atoms surrounded by hydrogen atoms.  The common nomenclature of alkanes begins with one of the prefixes above to specify the number of carbon atoms and then the suffix “ane.”  To obtain the condensed chemical formula of an alkane, substitute the number of carbons for n in the formula CnH2n+2.  For example, the fuel propane has n = 3 and, thus, a formula of C3H8.

 

Alcohols are organic compounds composed of single-bonded carbon atoms surrounded by hydrogen atoms, but one of the carbons is single-bonded to an OH group.  The common nomenclature of alcohols begins with one of the prefixes above to specify the number of carbon atoms and then the suffix “anol” or “yl alcohol.”  To obtain the partially-condensed chemical formula of an alcohol, substitute the number of carbons for n in the formula CnH2n+1OH.  For example, ethanol or ethyl alcohol has n = 2 and, thus, a formula of C2H5OH.  

 

 

 

 

Section 9-8:  Names of Common Acids

 

In different aqueous acids, hydrogen ions (H+) are present in all of the acid solutions, but different anions will be present in each of the different acid solutions.  Here are the names of some common acids:

 

HCl = hydrochloric acid

 

HNO3 = nitric acid

 

H2SO4 = sulfuric acid

 

CH3COOH or HCH3COO or HC2H3O2 = acetic acid

      

 

 

Chapter 9 Practice Exercises and Review Quizzes:

 

9-1) Write the chemical formula for each ionic compound:

 

(a) ammonium phosphide

(b) chromium(III) nitrate

(c) aluminum carbonate

Click for Solution

9-1) (a) (combine 3 NH4+ with 1 P3-) = (NH4)3P

     (b) (combine 1 Cr3+ with 3 NO3-) = Cr(NO3)3

     (c) (combine 2 Al3+ with 3 CO32-) = Al2(CO3)3

 

 

9-2) Write the name of each ionic compound:

 

(a) Ca(CH3COO)2

(b) K2SO4

(c) CuOH

Click for Solution

 

9-2) (a) Ca2+ = Group 2 cation = no Roman numerals, so calcium acetate

         (b) K+ = Group 1 cation = no Roman numerals, so potassium sulfate

         (c) Since hydroxide is known to be 1-, the metal ion must be 1+, so     

           copper(I) hydroxide.

 

 

 

9-3) Draw Lewis structures for each of the following:

 

(a) boron trifluoride

(b) IBr4-

Click for Solution

 

9-3) (a) BF3 = 3 + (3 x 7) = 24 electrons

 

 

It is preferable for B to have an incomplete octet of 6 electrons.

 

     (b) 7 + (4 x  7) + 1 = 36 electrons

 

 

I has an “expanded octet” of 12 electrons, which is acceptable because I is not in the second row of the periodic table.

 

 

 

9-4) Draw Lewis structures for C2H2 and C2H4.  Which molecule will have the longer carbon-to-carbon bond length?  Explain briefly.

Click for Solution

 

9-4) C2H2 = (2 x 4) + (2 x 1) = 10 electrons:

 

 

     C2H4 = (2 x 4) + (4 x 1) = 12 electrons:

                                                                 

 

Since the number of bonds in C2H4’s double bond is less than the number of bonds in C2H2’s triple bond, C2H4 will have the longer carbon-to-carbon bond length. 

 

 

 

9-5) Draw all the resonance structures for the nitrate ion.

Click for Solution

 

9-5) NO3- = 5 + (3 x 6) + 1 = 24 electrons

 

 

 

 

 

9-6) Write the chemical formula for each binary molecular compound:

(a) sulfur hexafluoride

(b) phosphorus pentachloride

Click for Solution

 

9-6) (a) SF6

       (b) PCl5

 

 

 

9-7) Write the name of each binary molecular compound:

 

(a) N2O

(b) N2O4

Click for Solution

 

9-7) (a) dinitrogen monoxide

         (b) dinitrogen tetroxide

 

 

 

9-8) Write the chemical formula for each organic compound:

 

(a) methanol or methyl alcohol

(b) butane

Click for Solution

 

9-8) (a) CH3OH

         (b) C4H10

 

 

 

9-9) Write the name of each organic compound:

 

(a) C8H18

(b) C5H11OH

Click for Solution

 

9-9) (a) octane

       (b) pentanol or pentyl alcohol

 

 

 

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