Chemistry Exam 2 Study Guide
Ionization Energy- minimum energy required to remove an electron from an atom in a gas phase, resulting in an ion (chemical species with net charge
POSITIVE charge means it is a CATION
A. Example: Na (g) —> Na+ (g) + e
B. Therefore, sodium has an ionization energy of 495.8 kJ/mol (first ionization energy of sodium).
C. IE1 (Na) which corresponds to the removal of the most loosely held electrons In general, as Zeff increases, ionization energy also increases Higher atomic radius = lower Zeff = lower ionization energy Lower energy = closer to the radius = more stable
NOTE: removing a paired electron is easier because of the repulsion forces between 2 electrons If you want to learn more check out What are the four layers of the atmosphere?
NOTE: Removing electrons can
IE1 values for main group Don't forget about the age old question of What is the effect of marriage on wages for women?
elements (kJ/mol) are to the right
It is possible to remove Don't forget about the age old question of What is cellular respiration & fermentation?
additional electrons in
subsequent ionizations, giving
IE1, IE2, etc.
It takes more energy to remove the 2nd/3rd/4th electrons because it’s harder to remove core electrons than valence electrons
IE1 (Mg) > IE1 (Na) because Mg is to the right so if it has a greater Zeff and more difficult to remove (496 kJ/mol < 738 kJ/mol)
IE2 (Na) > IE2 (Mg) because the second ionization of Mg removes a valence electron where the second ionization of Na removes a core electron If you want to learn more check out What is the steady-state distribution?
Electron affinity (EA) is the energy
released when an atom in the gas
phase accepts an electron
Cl (g) + e- —> Cl- (g)
Like ionization energy, electron affinity
increases from left to right across a
period as Zeff increases
NOTE: It’s easier to add an electron to an empty orbital than to add an electron into an full orbital
More than 1 electron may be added to an atom.
O (g) + e^- —> O^ - (g) (EA1 = -141 kJ/mol)
O^ - + e^ - —> O^ 2- (g) (EA2 = -741 kJ/mol)
While many 1st electron affinities are positive, subsequent EA’s are ALWAYS NEGATIVE. Considerable forces must be used to overpower compulsion energy
EA1 (Si) > EA1 (Al) because Si is more to the right therefore it has a greater Zeff
EA1 (Si) > EA1 (P) because even though P is more to the right it involves putting it in a 3p orbital. The energy it costs of PAIRING outweighs advantage of adding electrons to atoms with easier Zeff
A. Metals tend to be shiny, lustrous, malleable, ductile, and
good conductors of heat and electricity
B. They have low IE (because they have CATIONS and not
C. Many of the periodic trends of elements can be
explained using Coulombs Law which states that the
force (F) between two charged objects (Q1 and Q2) is
directly proportional to the product of the two charges
and inversely proportional to the distance (d) between
the objects squared
Ions of main group element
A. Species with identical electron configurations to the noble gas to the right are called isoelectronic
B. Common monatomic ions are arranged by their positions in the periodic table C. Exceptions: Mercury is actually a polyatomic ion (Hg2^2+)
Electron Configuration of Ions
A. Write the original configuration
B. Add/remove to make appropriate number of electrons
C. Note: in CATIONS, electrons are removed from the occupied orbitals with the highest value of n, but in ANIONS, electrons are added to the empty/partially filled orbitals with the lowest value of n
A. Low electron affinity- hard to accept electrons
B. High electron affinity- easy to accept electrons
C. Low ionization energy- easy to lose electrons
D. High ionization energy- hard to lose electrons
E. Noble gases have lower ionization energies
Ions of d-block elements
A. Ions of d block elements are formed by removing electrons first from the shell with the highest value of n. (When you're removing electrons from the final configuration, you’re making it more stable by lessening the repulsion forces.) B. Always go to fill s first, then p.
C. Exception: when you're trying to fill the p blocks
Worked example 4.8: write the electron configuration for the following ions of d block elements
A. Zn^2+ —> [Ar] 3d^10
B. Mn ^2+ —> [Ar] 3d^5
C. Cr ^3+ —> [Ar] 3d^5
Ionic radius- the radius of a cation or an anion (radius- distance between valence shell and nucleus)
A. When an atom loses an electron to become a CATION, its radius decreases due in part to a reduction on electron-electron repulsions in the valence shell B. A significant decreases in radius occurs when ALL of an atoms valence electrons are removed
C. When an atom gains one or more electrons and becomes an ANION, its radius increases due to increased electron-electron repulsions D. When referring to radius, a cation < neutral atom < anion
A. An isoelectronic series is a series of two or more species that have identical electron configurations but different nuclear charges
B. Ionic radius and attraction force are difference
C. Example: O^2- and F^- have the same electron configuration but different attraction forces
NOTE: Stronger attraction forces make ionic radius shrink
Worked example 4.9- know how to rank them in order of increasing radius
A compound is a substance composed of two or more elements combined in a specific ratio and held together by chemical bonds, such as salt (NaCl) or water (H20)
Lewis dot symbols
A. When atoms form compounds, their valence electrons actually interact B. A LDS consists of the elements symbol with dots (do not pair until needed) C. Examples
1. Boron —> 1s2/2s2/2p1 = 3 valence electrons
2. Carbon—> 4 valence electrons
3. Nitrogen—> 5 valence electrons
D. The atoms combine in order to achieve a more stable electron configuration E. Maximum stability results when a chemical specifies is isoelectronic with a noble gas
For main group metals such as Na the number of dots is the number of electrons that are lost
For nonmetals in the second period the number of unpaired dots is the number of bonds the atom can form
Ions may also be represented by Lewis dot symbols (O^2-) which can be represented by writing the element with correct number of electrons then brackets with the charge on the outside
Ionic Compound Bonding- refers to the electrostatic attraction that holds oppositely charged ions together in an ionic compound
A. Metals- low ionization energies lose electrons very easily
B. Nonmetals- high electron affinity accepts electrons very easily
C. Ionic Compounds
1. Lewis Dot Symbol
2. Na (dot) —> Na+ + e
A. A 3-dimensional array of oppositely-charged ions is called a lattice
B. Lattice energy is the amount of energy required to convert a mole of ionic solid to its constituent ions in the gas phase
C. The magnitude of lattice energy is a measure of an ionic compounds stability D. Lattice energy depends on magnitudes of the charge and on the distance between them
E. If they have the same charges, the only thing they have that differs one from another is the distance between them
F. *Know how to arrange ionic compounds in order of increasing/decreasing lattice energy (opposite of atomic radius)*
G. NOTE: big atomic radius means low lattice energy
H. The formation of ionic bonds RELEASES a large amount of energy
The resulting electrically
neutral compound, sodium
chloride, is represented with
the chemical formula NaCl
Naming Ions and Ionic Compounds
A. A monoatomic cation is named by adding the word ION to the name of the element
B. A monoatomic anion is named by changing the ending of the elements name to IDE, such as oxide, carbide, sulfide
C. Some metals can form cations of more than one possible charge (especially true for elements of D-block)
1. Fe2+ : ferrous ion [Fe(II)]
2. Fe3+ : ferric ion [Fe(III)]
3. Mn2+ : manganese (II) ion
4. Mn3+ : manganese (III) ion
5. Mn4+ : manganese (IV) ion
Formulas of Ionic Compounds
A. Ionic compounds are electronegatively neutral
1. In order for ionic compounds to be electrically neutral,
the sum of the charges of the cation and anion in each
formula must be zero
2. Example: aluminum oxide. 2(+3) + 3(-2) = 0
B. To name ions and ionic compounds:
1. Name the cation by omitting the word ion and using a roman numeral if the cation can have more than one charge
2. Name the anion by adding the word IDE
a. NaBr —> Sodium Bromide
b. CaO —> Calcium Oxide
c. Mg3N2 —> Manganese Nitride
d. Fe2S3 —> Iron(III) Sulfide
C. NOTE: the subscript of the anion is the charge of the cation
D. NOTE: the subscript of the cation is the charge of the anion
Covalent Bonding and Molecules
A. When compounds form between elements with similar properties with similar properties, electrons are not transferred from one electron to another but instead are shared in order to give each atoms a noble gas configuration B. This approach is known as the Lewis Theory of Bonding, named for its proponent, Gilbert Lewis
C. Lewis Theory depicts bond formation in H2 (right)
D. A molecule may be an element or a compound
E. Different samples of a given compound always contain the
same ratio, known as the law of definite proportions
F. A molecule is a combination of at least two atoms in
specific arrangement held together by chemical forces
G. A molecule may be an element or a compound
H. Different samples of a given compound always
contain the same ratio. This is known as the law
of definite proportions.
I. If two elements can form two or more different
compounds, the law of multiple proportions tells
us that when the masses of two elements when
combined with each other to form more than one
compound are in a ratio of small whole numbers
J. The mass ratio of oxygen to carbon dioxide is 2.66:1, and the ratio of oxygen to carbon in carbon monoxide is 1.33:1
K. The ratio of two such mass rations can be expressed as small whole numbers L. Diatomic molecules contain two atoms and may either be heteronuclear or homonuclear
(a) would be homonuclear
diatomic but it has more than
two so it is not diatomic. Imagine
the same colors but with only
two atoms bonded.
(b) is polyatomic
(c) is heteronuclear diatomic
A. A chemical formula denotes the composition of the substance B. A molecular formula shows the exact number of atoms in each element in a molecule
C. Some elements have two or more distant forms known as allotropes, such as oxygen (O2) and ozone (O3) are allotropes of oxygen
D. A structural formula shows not only the elemental composition but also the general arrangements
E. An empirical formula uses whole-number ratios of elements to get the smallest bit of information that we can get from observation. While the molecular formulas tell us the actual number of atoms (the true formula), the empirical formula gives the simplest formula. Sometimes the true formula IS the empirical formula, like with H2O.
1. Molecular formula: N2H2
2. Empirical formula: NH2
3. Look at the worked example 5.6
Naming Molecular Compounds
A. Remember: binary molecular compounds are substances that consist of just two different elements
1. Name the first element that appears in the formula
2. Name the second element that appears in the formula, change ending to IDE
C. Greek prefixes are used to denote the number of atoms of each element 1. mono- is usually omitted for the first element
2. For ease of pronunciation, we usually eliminate the last letter of a prefix that ends in “o” or “a” when naming an oxide
3. Worked example
a. NF3 ???? nitrogen trifluoride
b. N2O4 ???? dinitrogen tetroxide
D. Compounds containing hydrogen
1. The names of molecular compounds containing hydrogen do not usually conform to the systematic nomenclature guidelines
2. Many are called by the common, nonsystematic names or by names that do not indicate explicitly the number of H atoms present
a. B2H6 ???? Diborane
b. SiH4 ???? Silane
c. NH3 ???? Ammonia
d. PH3 ???? Phosphine
e. H2O ???? water
f. H2S ???? hydrogen sulfide
E. One definition of an acid is a substance that produces hydrogen ions (H+) when dissolved in water (HCl is an example of a binary compound that is an acid when dissolved in water)
1. To name these acids:
a. Remove the –gen ending from hydrogen
b. Change the –ide ending to on the second element to –ic
(Hydrogen chloride ???? hydrochloric acid)
2. A compound must contain at least one ionizable hydrogen atom to be an acid upon dissolving
A. Our nomenclature discussion so far as focused on inorganic compounds, generally defined as those without carbon
B. Organic compounds contain carbon and hydrogen, sometimes in combination with other atoms
C. Hydrocarbons contain only carbon and hydrogen
D. The simplest hydrocarbons are called alkanes
E. Many organic compounds contain groups of atoms known as functional groups, which often determine a molecule’s reactivity
Covalent Bonding in Ionic Species
A. Polyatomic ions consist of a combination of two or more atoms B. Formulas are determined following the same rule as for ionic compounds containing only monatomic ions: ions must combine in a ratio that gives a neutral formula overall
C. Oxoanions are polyatomic anions that contain one or more oxygen atoms and one atom (the central atom) of another element
D. Starting with the oxoanions that end in –ate, we can name these ions: 1. The ion with one MORE O atom atom than the –ate ion is called the per…ate ion. (ie ClO3- = a chlorate ion, so CLO4- = a perchlorate ion) 2. The ion with one LESS O atom than the -ate ion is called the –ite ion. (ie ClO2- is the chlorite ion)
3. The in with TWO FEWER O atom than the -ate ion is called the hypo…ite ion. (ie ClO- is the hypochlorite ion)
1. percholate ClO4
2. chlorate ClO3-
3. chlorite ClO2-
4. hypochlorite ClO
5. nitrate NO3-
6. nitrite NO2-
7. phosphate PO43-
8. phosphite PO33-
9. sulfate SO42
10. sulfite SO32-
F. Oxoacids, when dissolved in water, produce hydrogen ions and the corresponding oxoanions
1. An–ate ion is called ……. ic acid (HClO3 ???? chloric acid)
2. An –ite ion is called …….. ous acid (HClO2 ???? chlorous acid) 3. Prefixes in oxoanions names are retained in naming oxoacids
4. Oxoacids, can be monoprotic (one ionizable hydrogen) or polyprotic (more than one ionizable hydrogen)
A. A compound that has a specific number of water molecules within its solid structure
B. For example, in its normal state, copper(II) sulfate has five water molecules associated with it
C. Copper(II) sulfate pentahydrate —> CuSO4 x 2H2O
D. When the water molecules are driven off by heating, the resulting compound, Cu(SO)4 is sometimes called anhydrous copper(II) sulfate
E. Anhydrous means the compound no longer has water molecules associated with it
F. Cu(SO)4 is white but Cu(SO)4 x 5H2O is blue
Molecular and Formula Mass
A. The molecular mass is the mass in atomic mass units (amu) of an individual molecule
B. To calculate the molecular mass, multiply the atomic mass for each element in a molecule by the number of atoms of that element and total the masses C. Molecular mass of H2O = 2(atomic mass of H) + atomic mass of 0 D. Although the ionic compound does not have a molecular mass, we can use its empirical formula to calculate its formula mass
E. Because the atomic masses on the periodic table are average atomic masses, the result of such a determination is an average molecular mass, sometimes referred to as the molecular weight.
Percent Composition of Compounds
A. A list of the percent by mass of each element in a compound is known as the compounds percent composition by mass
B. Percent mass of an element = (n x atomic mass of element) / (molecular or formula mass of compound) x 100%
C. We could have also used the empirical formula of hydrogen peroxide (HO) for the calculation
D. In this case, we could have used the empirical formula mass to find the mass in amu of one of the compounds
E. Worked Example 5.13
Molar mass (M )
A. In a substance, the molar mass is the mass in grams of one mole of the substance
B. The molar mass of an element is numerically equal to its atomic mass 1. 1 mol C = 12.01 g
2. 1 C atom = 12.01 amu
C. The molar mass of a compound us the sum of the molar masses of the elements it contains
1. 1 mol H2O = 2 x 1.008 g + 16g = 18.02 g
2. 1 mol NaCl = 22.99 g + 35.45g = 58.44g
The Octet Rule
A. According to the octet rule, atoms will lose, gain, or share electrons in order to achieve a noble gas electron configuration
B. Only valence electrons contribute to bonding
C. Only two valence electrons participate in the formation of the F2 bond
A. A lewis structure is a
representation of covalent
B. Shared electron pairs are
shown either as dashes or as pairs of dots
C. Lone pairs are shown as pairs of dots on individual atoms
D. In a single bond, atoms are held together by one electron pair E. In a double bond, atoms share two pairs of electrons
F. In a triple bond, atoms are held together by three electron pairs G. Bond length is defined as the distance between the nuclei of two covalently bonded atoms
H. Multiple bonds are shorter than single bonds
I. For a given pair of atoms, triple bonds are shorter than double bonds which are shorter than single bonds
J. We quantify bond strength by measuring the quantity of energy required to break it
Electronegativity and Polarity
A. There are two extremes in the spectrum of bonding:
1. Covalent bonds occur between atoms that SHARE electrons
2. Ionic bonds occur between a metal and nonmetal and involve ions B. Bonds that fall between these extremes are polar
C. In polar covalent bonds, electrons are shared but not shared equally (the delta is used to denote partial charges on the atoms)
E. Pure covalent bonds- neutral atoms held together by equally shared electrons
G. Ionic bonds- oppositely charged ions held together by electrostatic attraction H. Electron density maps show the distributions of charge
I. Typically, electrons spend a lot of time in red and very little time in blue
A. Electronegativity is the ability of an atom in a compound to draw electrons to itself
B. There is no sharp distinction between non-polar covalent and polar covalent or between polar covalent and ionic
C. The following rules help distinguish among them:
1. A bond between atoms whose electronegativities differ by less than 0.5 is generally considered purely covalent or nonpolar
2. A bond between atoms who's electronegativities differ by the range of 0.5 to 2.0 is generally considered polar covalent
3. A bond between atoms whose electronegativities differ by 2.0 or more is generally considered ionic
Dipole Movement, Partial Charges, and Percent Ionic Character A. Regions where electrons spend little time is typically blue while regions where electrons spend a lot of time is tropically red
B. An arrow is used to indicate the direction of electron shift
C. A quantitative measure of the polarity of a bond is its dipole movement (mu) D. mu = Q x r (Q is the charge; r is the distance between the charges, and mu is always positive and expressed in debye units denoted by the letter D) E. 1 D = 3.336 × 10-30 coulomb meter.
F. Although the designations “covalent,” “polar covalent,” and “ionic” can be useful, sometimes chemists wish to describe and compare chemical bonds with more precision.
G. Comparing the calculated dipole moment with the measured values gives us a quantitative way to describe the nature of a bond using the term percent ionic character.
Lewis Structures and Formal Charge
A. Formal charge can be used to determine the most plausible Lewis Structure when more than one possibility exists for a compound.
B. To determine associated electrons:
1. All the atom’s nonbonding electrons are associated with the atom 2. Half the atom’s bonding electrons are associated with the atom C. When there is more than one possible structure, the best arrangement is determined by the following guidelines:
1. A Lewis structure in which all formal charges are zero is preferred 2. Small formal charges are preferred to large formal charges
3. Formal charges are associated with electronegativities.
A. A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure B. Resonance structures are a human invention
C. Resonance structures differ only in the positions of their electrons
Exceptions to the Octet Rule: Incomplete Octets
A. The central atom has fewer than eight electrons due to a shortage of electrons 1. Elements in group 3A also tend to form compounds surrounded by fewer than eight electrons
2. Boron, for example, reacts with halogens to form compounds of the general formula BX3 having six electrons around the boron atom B. The central atom has fewer than eight electrons due to an odd number of electrons. (Molecules with an odd number of electrons are sometimes referred to as free radicals)
C. The central atom has more than eight electrons
1. Atoms in and beyond the third period can have more than 8 valence e’s 2. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals than can be used in bonding
D. The bond between B-N has both the electrons contributed by the N atom 1. This type of bond is a coordinate covalent or dative bond
2. This type of bond formation is an example of a Lewis acid-base process BF3 is a Lewis acid: it can accept a pair of electrons
NH3 is a Lewis Base: it donates a pair of electrons
Worked Example 5.2
Arrange MgO, CaO, and SrO in order of increasing lattice energy.
Consider the charges on the ions and the distances between them. Apply Coulomb’s law to determine the relative lattice energies. All three compounds contain O2- and all three cations are +2. Recalling that lattice energy increases as the distance between ions decreases, we need only consider the radii of the cations as all three contain the same anion. From Figure 4.13, the ionic radii are 0.72 Å (Mg2+), 1.00 Å (Ca2+), and 1.18 Å (Sr2+).
Worked example 5.3
Name the following ionic compounds: (a) CaO, (b) Mg3N2, and (c) Fe2S3.
Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion.
(a) CaO is
(b) Mg3N2 is
(c) Fe2S3 is
Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (c), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Worked Example 5.6
Write the empirical formulas for the following molecules: (a) glucose (C6H12O6), a substance known as blood sugar; (b) adenine (C5H5N5), also known as vitamin B4; and (c) nitrous oxide (N2O), a gas that is used as an anesthetic (“laughing gas”) and as an aerosol propellant for whipped cream.
Worked Example 5.7
Name the following binary molecular compounds: (a) NF3 and (b) N2O4.
Strategy Each compound will be named using the systematic nomenclature including, where necessary, appropriate Greek prefixes.
Think About It Make sure that the prefixes match the subscripts in the molecular formulas and that the word oxide is not preceded immediately by an “a” or an “o”.
Worked Example 5.9
Name the following ionic compounds: (a) Fe2(SO4)3, (b) Al(OH)3, and (c) Hg2O. Strategy Begin by identifying the cation and anion in each compound, and then
combine the names for each, eliminating the word ion.
Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (a), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Worked Example 5.10
Name the following species: (a) BrO4-, (b) HCO3-, and (c) H2CO3.
Strategy Each species is either an oxoanion or an oxoacid. Identify the “reference oxidation” (the one with the –ate ending) for each, and apply the rules to determine appropriate names.
Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
Worked Example 5.11
Determine the formula of sulfurous acid.
Strategy The –ous ending in the name of an acid indicates that the acid is derived from an oxoanion ending in –ite. The oxoanion must be sulfite, SO32-, so add enough hydrogen ions to make a neutral formula.
Worked Example 5.12
Calculate the molecular mass or the formula mass, as appropriate, for each of the following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c) barium acetate, Ba(C2H3O2)2.
Strategy Determine the molecular mass (for each molecular compound) or formula mass (for each ionic compound) by summing all the atomic masses.
Worked Example 5.13
Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by the FDA for the treatment of mania and manic-depressive illness, also known as bipolar disorder. Calculate the percent composition by mass of lithium carbonate.
Strategy Use Equation 5.1 to determine the percent by mass contributed by each element in the compound.
Worked Example 5.14
Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and (b) the mass of 0.905 mole of sodium chloride.
Strategy Use molar mass to convert from mass to moles and to convert from moles to mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the molar mass of sodium chloride (NaCl) is 58.44 g/mol.
Think About It Always double-check unit cancellations in problems such as these–errors are common when molar mass is used as a conversion factor. Also make sure that the results make sense. In both cases, a mass smaller than the molar mass corresponds to less than a mole of substance.
Worked Example 5.15
(a) Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water.
(b) Determine the mass of 7.92×1019 carbon dioxide molecules.
Strategy Use molar mass and Avogadro’s number to convert from mass to molecules, and vice versa. Use the molecular formula of water to determine the numbers of H and O atoms. 2 23
(b) 7.92×10 CO molecules × 6.022×10 CO molecules × 1 mol CO
= 5.79×10-3 g CO2
Think About It Again, check the cancellation of units carefully and make sure that the magnitudes of your results are reasonable.
Worked Example 6.1
Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond in CsBr, and (c) the carbon-carbon double bond in C2H4.
Strategy Electronegativity values are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5). Use this information to determine which bonds have identical, similar, and widely different electronegativities.
(a) The difference between the electronegativies of F and Cl is 4.0 – 3.0 = 1.0, making the bond in ClF polar.
(b) In CsBr, the difference is 2.8 – 0.7 = 2.1, making the bond ionic. (c) In C2H4, the two atoms are identical. (Not only are they the same element, but each C atom is bonded to two H atoms.) The carbon-carbon double bond is C2H4 is nonpolar.
Worked Example 6.2
Burns caused by hydrofluoric acid [HF(aq)] are unlike any other acid burns and present unique medical complications. HF solutions typically penetrate the skin and damage internal tissues, including bone, often with minimal surface damage. Less concentrated solutions actually can cause greater injury than more concentrated ones by penetrating more deeply before causing injury, thus delaying the onset of symptoms and preventing timely treatment. Determine the magnitude of the partial positive and partial negative charges in the HF molecule.
Strategy Solve for Q. Convert the resulting charge in coulombs to units of electronic charge. According to Table 6.2, μ = 1.82 D and r = 0.92 Å for HF. The dipole moment must be converted from debye to C·m and the distance between ions must be converted to meters.
Worked Example 6.4
Draw the Lewis structure for carbon disulfide (CS2).
Step 1: C and S have identical electronegativities. We will draw the skeletal structure with the unique atom, C, at the center.
Step 2: The total number of valence electrons is 16: 6 from each S atom and 4 from the C atom [2(6) + 4 = 16].
Step 3: Subtract 4 electrons to account for the bonds in the skeletal structure, leaving us 12 electrons to distribute.
Step 4: Distribute the 12 remaining electrons as 3 lone pairs on each S atom. Step 5: There are no electrons remaining after step 4, so step 5 does not apply.
Step 6: To complete carbon’s octet, use one lone pair from each S atom to make a double bond to the C atom.
Worked Example 6.5
The widespread use of fertilizers has resulted in the contamination of some groundwater with nitrates, which are potentially harmful. Nitrate toxicity is due primarily to its conversion in the body to nitrite (NO2-), which interferes with the ability of hemoglobin to transport oxygen. Determine the formal charges on each atom in the nitrate ion (NO3-).
Strategy Follow the six steps to draw the Lewis structure of (NO3-). For each atom, subtract the associated electrons from the valence electrons.
Setup The N atom has five valence electrons and four associated electrons (one from each single bond and two from the double bond). Each singly bonded O atom has six valence electrons and seven associated electrons (six in three lone pairs and one from the single bond). The doubly bonded O atom has six valence electrons and six associated electrons (four in two lone pairs and two from the double bond.)
The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and 0 (doubly bonded O atom).
Worked Example 6.6
Formaldehyde (CH2O), which can be used to preserve biological specimens, is commonly sold as a 37 percent aqueous solution. Use formal charges to determine which skeletal arrangements of atoms shown here is the best choice for the Lewis structure of CH2O.
Solution Of the two possible arrangements, the structure on the left has an O atom with a positive formal charge, which is inconsistent with oxygen’s high electronegativity. Therefore, the structure on the right, in which both H atoms are attached directly to the C atoms and all atoms have a formal charge of zero, is the better choice for the Lewis structure of CH2O.
H C O H
H C H
Worked Example 6.7
High oil and gasoline prices have renewed interest in alternative methods of producing energy, including the “clean” burning of coal. Part of what makes “dirty” coal dirty is its high sulfur content. Burning dirty coal produces sulfur dioxide (SO2), among other pollutants. Sulfur dioxide is oxidized in the atmosphere to form sulfur trioxide (SO3), which subsequently combines with water to produce sulfuric acid – a major component of acid rain. Draw all possible resonance structures of sulfur trioxide.
Strategy Following the steps for drawing Lewis structures, we determine that a correct Lewis structure for SO3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond.
O S O
O S O
O S O