Lecturenote 3 General Chemistry Atoms Dr Sadeem M

Lecturenote 3 General Chemistry: Atoms & Dr. Sadeem M. Al-barody

3 Chemical Bonding I: Basic Concepts 3. 1 The Octet Rule Lewis Structures Multiple Bonds 3. 2 Electronegativity and Polarity Electronegativity Dipole Moment, Partial Charges and Percent Ionic Character 3. 3 Drawing Lewis Structures 3. 4 Lewis Structures and Formal Charge 3. 5 Resonance 3. 6 Exceptions to the Octet Rule Incomplete Octets Odd Numbers of Electrons Expanded Octets

3. 1 The Octet Rule According to the octet rule, atoms will lose, gain, or share electrons in order to achieve a noble gas electron configuration. • • F • • • • + • • • • F • Each F counts both shared electrons to “feel” as though it has a Ne configuration. Only valence electrons contribute to bonding. F 1 s 22 p 5 valence electrons

The Octet Rule Only two valence electrons participate in the formation of the F 2 bond. • • or • • F F • • • • F • • Pairs of valence electrons not involved in bonding are called lone pairs. • • • • F • •

Lewis Structures A Lewis structure is a representation of covalent bonding. Shared electron pairs are shown either as dashes or as pairs of dots. Lone pairs are shown as pairs of dots on individual atoms. H with 2 e− • • HOH • • O with 8 e− • • H O H • • Shared electrons shown as dashes (bonds)

Multiple Bonds In a single bond, atoms are held together by one electron pair. In a double bond, atoms share two pairs of electrons. • • 2 shared pairs of electrons result in double bonds • • O=C=O • • H O H • • O with 8 e− O C O • • one shared pair of electrons results in a single bond • • HOH • • O with 8 e− C with 8 e−

Multiple Bonds A triple bond occurs when atoms are held together by three electron pairs. each N has 8 e− • • 3 shared pairs of electrons result in a triple bond N≡N • • • • N

Multiple Bonds Bond length is defined as the distance between the nuclei of two covalently bonded atoms.

Multiple Bonds Multiple bonds are shorter than single bonds. For a given pair of atoms: triple bonds are shorter than double bonds are shorter than single bonds N≡N N=N < < 110 pm 124 pm N−N 147 pm

Multiple Bonds The shorter multiple bonds are also stronger than single bonds. We quantify bond strength by measuring the quantity of energy required to break it. H 2(g) → H(g) + H(g) Bond energy = 436. 4 k. J/mol We will examine this in more detail in Chapter 10.

3. 2 Electronegativity and Polarity There are two extremes in the spectrum of bonding: covalent bonds occur between atoms that share electrons ionic bonds occur between a metal and a nonmetal and involve ions Bonds that fall between these extremes are polar. In polar covalent bonds, electrons are shared but not shared equally. M: X Pure covalent bond Neutral atoms held together by equally shared electrons Mδ+Xδ− M+ X − Polar covalent bond Ionic bond Partially charged atoms Oppositely charged held together by ions held together by unequally shared electrostatic attraction electrons

Electronegativity and Polarity Electron density maps show the distributions of charge. Electrons spend a lot of time in red and very little time in blue. Electrons are shared equally nonpolar covalent Electrons are not shared equally and are more likely to be associated with F Electrons are not shared but rather transferred from Na to F ionic polar covalent

Electronegativity is the ability of an atom in a compound to draw electrons to itself.

Electronegativity and Polarity Electronegativity varies with atomic number.

Electronegativity There is no sharp distinction between nonpolar covalent and polar covalent or between polar covalent and ionic. The following rules help distinguish among them: A bond between atoms whose electronegativites differ by less than 0. 5 is general considered purely covalent or nonpolar. A bond between atoms who’s electronegativies differ by the range of 0. 5 to 2. 0 is generally considered polar covalent. A bond between atoms whose electronegativities differ by 2. 0 or more is generally considered ionic.

Worked Example 3. 1 Classify the following bonds as nonpolar, or ionic: (a) the bond in Cl. F, (b) the bond in Cs. Br, and (c) the carbon-carbon double bond in C 2 H 4. 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. Solution (a)The difference between the electronegativies of F and Cl is 4. 0 – 3. 0 = 1. 0, making the bond in Cl. F polar. (b)In Cs. Br, the difference is 2. 8 – 0. 7 = 2. 1, making the bond ionic. (c)In C 2 H 4, 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 C 2 H 4 is nonpolar. Think About It By convention, the difference in electronegativity is always calculated by subtracting the smaller number from the larger one, so the result is always positive.

Dipole Moment, Partial Charges, and Percent Ionic Character Regions where electrons spend little time Regions where electrons spend a lot of time • • The consequent charge separation can be represented as: Deltas (δ) denote a partial positive or negative charge. • • H−F • • δ+ δ− • • H−F • • An arrow is used to indicate the direction of electron shift in polar covalent molecules.

Dipole Moment, Partial Charges, and Percent Ionic Character A quantitative measure of the polarity of a bond is its dipole moment (μ). μ=Qxr Q is the charge (C). r is the distance between the charges (m). μ is always positive and expressed in debye units (D). 1 D = 3. 336× 10− 30 C∙m

Dipole Moment, Partial Charges, and Percent Ionic Character A quantitative measure of the polarity of a bond is its dipole moment (μ). μ=Qxr

Worked Example 3. 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 3. 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. μ = 1. 82 D 3. 336× 10 -30 C∙m r = 0. 92 Å × = 6. 07× 10 -30 C∙m 1× 10 -10 m = 9. 2× 10 -11 1Å

Worked Example 3. 2 (cont. ) Solution In coulombs: μ = Q= r 6. 07× 10 -30 C∙m = 6. 598× 10 -20 C -11 9. 2× 10 m In units of electronic charge: 6. 598× 10 -20 1 e. C× = 0. 41 e-19 1. 6022× 10 C Therefore, the partial charges in HF are +0. 41 and -0. 41 on H and F, respectively. • • H−F • • -0. 41 • • +0. 41 Think About It Calculated partial charges should always be less than 1. If a “partial” charge were 1 or greater, it would indicate that at least one electron had been transferred from one atom to the other. Remember that polar bonds involve unequal sharing of electrons, not a complete transfer of electrons.

Dipole Moment, Partial Charges, and Percent Ionic Character Although the designations “covalent, ” “polar covalent, ” and “ionic” can be useful, sometimes chemists wish to describe and compare chemical bonds with more precision. 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. μ (observed) percent ionic character = μ (calculated assuming discrete charges) × 100%

Dipole Moment, Partial Charges, and Percent Ionic Character The figure below demonstrates the relationship between percent ionic character and the electronegativity difference in a heteronuclear diatomic molecule.

Worked Example 3. 3 Using data from Table 3. 2, calculate the percent ionic character of the bond in HI. Strategy Calculate the dipole moment in HI assuming that the charges on H and I are +1 and -1, respectively; bond length in HI is 1. 61 Å , then calculate the percent ionic character. The magnitude of the charges must be expressed in coulombs Setup From Table 3. 2, the bond length in HI is 1. 61 Å (1. 61× 10 -10 m) and the measured dipole moment of HI is 0. 44 D. (1 e- = 1. 6022× 10 -19 C); the bond length (r) must be expressed as meters (1 Å = 1× 10 -10 m); and the calculated dipole moment should be expressed as debyes (1 D = 3. 336× 10 -30 C∙m).

Worked Example 3. 3 (cont. ) Solution The dipole we would expect if the magnitude of the charges were 1. 6022× 10 -19 C is μ = Q × r = (1. 6022× 10 -19 C) × (1. 61× 10 -10 m) = 2. 58× 10 -29 C∙m Converting to debyes gives 2. 58× 10 -29 C∙m × 1 D 3. 336× 10 -30 C∙m The percent ionic character of the H–I bond is = 7. 73 D 0. 44 D × 100% = 5. 7%. 7. 73 D Think About It A purely covalent bond (in a homonuclear diatomic molecule such as H 2) would have 0 percent ionic character. In theory, a purely ionic bond would be expected to have 100 percent ionic character, although no such bond is known.

3. 3 Drawing Lewis Structures Follow these steps when drawing Lewis structure for molecules and polyatomic ions. 1)Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms. 1)Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges. 1)For each bond in the skeletal structure, subtract two electrons from the total valence electrons. 1)Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each atom. Complete the octets of the most electronegative atom first. 1)Place any remaining electrons in pairs on the central atom. 1)If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms.

Drawing Lewis Structures Draw the Lewis Structure for Cl. O 3−. Solution: Step 1 Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms. chlorine’s electronegativity = 3. 0 oxygen’s electronegativity = 3. 5

Drawing Lewis Structures Draw the Lewis Structure for Cl. O 3−. Solution: Step 2 Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges. Cl: 1 x 7 = 7 e− O: 3 x 6 = 18 e− Charge: 1 e− Total: 26 valence e−

Drawing Lewis Structures

Drawing Lewis Structures Draw the Lewis Structure for Cl. O 3−. Solution: Step 4 Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each atom. Complete the octets of the most electronegative atom first. 6 electrons used in single bonds 20 electrons must be placed in the structure starting with terminal atoms The structure now has a total of 24 valence electrons of the available 26.

Drawing Lewis Structures Draw the Lewis Structure for Cl. O 3−. Solution: Step 5 Place any remaining electrons on the central atom. The structure now has a total of 24 valence electrons of the available 26. 2 e− remain to be added to the structure. Lewis structures for polyatomic ions are enclosed by square brackets.

Drawing Lewis Structures Draw the Lewis Structure for Cl. O 3−. Solution: Step 6 If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms. All atoms already have an octet in this structure.

Drawing Lewis Structures

Worked Example 3. 4 Draw the Lewis structure for carbon disulfide (CS 2). Setup 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.

Worked Example 3. 4 (cont. ) Draw the Lewis structure for carbon disulfide (CS 2). Setup 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. Solution Think About It Counting the total number of valence electrons should be relatively simple to do, but it is often done hastily and is therefore a potential source of error in this type of problem. Remember that the number of valence electrons for each element is equal to the group number of that element.

3. 4 Lewis Structures and Formal Charge Formal charge can be used to determine the most plausible Lewis Structure when more than one possibility exists for a compound. Formal charge = valence electrons – associated electrons 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.

Lewis Structures and Formal Charge Determine the formal charges on each oxygen atom in the ozone molecule (O 3). • • • • O =O− O • • 4 unshared + 4 shared = 6 e− 2 2 unshared + 6 shared = 5 e− 6 unshared + 2 shared = 7 e− 2 2 Valence e− 6 6 6 e− associated with atom 6 5 7 Difference (formal charge) 0 +1 − 1

Worked Example 3. 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 (NO 2 -), which interferes with the ability of hemoglobin to transport oxygen. Determine the formal charges on each atom in the nitrate ion (NO 3 -). Strategy Follow the six steps to draw the Lewis structure of (NO 3 -). 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. )

Worked Example 3. 5 (cont. ) Solution The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and 0 (doubly bonded O atom). Think About It The sum of formal charges (+1) + (-1) + (0) = -1 is equal to the overall charge on the nitrate ion.

Lewis Structures and Formal Charge 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 should be consistent with electronegativities. • • 0 0 better structure (based on formal charge) +1 • • 0 O≡C−O • • Formal charge • • O=C=O 0 − 1

Lewis Structures and Formal Charge Based on formal charge, identify the best and the worst structures for the isocyanate ion below: Solution: Step 1 Assign formal charges on each atom using the formula Formal charge = valence electrons – associated electrons Ve− Ae− FC 4 6 5 4 6 6 − 2 +1 0 4 5 − 1 5 4 +1 6 7 − 1 4 7 5 4 6 5 − 3 +1 +1

Lewis Structures and Formal Charge Based on formal charge, identify the best and the worst structures for the isocyanate ion below: Solution: Step 2 Determine the best and worst structure FC − 2 +1 0 − 1 +1 − 3 +1 +1 Best structure: Worst structure: Small formal charges Large formal charges Formal charges more consistent with electronegativities Formal charges inconsistent with electronegativities

Worked Example 3. 6 Formaldehyde (CH 2 O), 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 CH 2 O. Strategy The complete Lewis structures for the skeletons shown are:

Worked Example 3. 6 (cont. ) Strategy The complete Lewis structures for the skeletons shown are: In the structure on the left, the formal charges are as follows: Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0 C atom: 4 valence e- – 5 associated e- (two in the lone pair, one from the single bond, and two from the double bond) = -1 O atom: 6 valence e- – 5 associated e- (two from the lone pair, one from the single bond, and two from the double bond) = +1 Formal charges 0 -1 +1 0

Worked Example 3. 6 (cont. ) Strategy The complete Lewis structures for the skeletons shown are: In the structure on the right, the formal charges are as follows: Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0 C atom: 4 valence e- – 4 associated e- (one from each single bond, and two from the double bond) = 0 O atom: 6 valence e- – 6 associated e- (four from the two lone pairs and two from the double bond) = 0 Formal charges all zero

Worked Example 3. 6 (cont. ) 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 CH 2 O. Think About It For a molecule, formal charges of zero are preferred. When there are nonzero formal charges, they should be consistent with the electronegativities of the atoms in the molecules. A positive formal charge on oxygen, for example, is inconsistent with oxygen’s high electronegativity.

3. 5 Resonance A resonance structure is one of two or more Lewis structures for a single molecule that cannot by represented accurately by only one Lewis structure. • • • • O −O= O • • • • • O =O− O Resonance structures are a human invention. Resonance structures differ only in the positions of their electrons.

Worked Example 3. 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 (SO 2), among other pollutants. Sulfur dioxide is oxidized in the atmosphere to form sulfur trioxide (SO 3), 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 SO 3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond. But the double bond can be put in any one of three positions in the molecule.

Worked Example 3. 7 (cont. ) Solution Think About It Always make sure that resonance structures differ only in the position of the electrons, not in the positions of the atoms.

3. 6 Exceptions to the Octet Rule Exceptions to the octet rule fall into three categories: 1) The central atom has fewer than eight electrons due to a shortage of electrons. H Be H only 4 total valence electrons in the system Elements in group 3 A also tend to form compounds surrounded by fewer than eight electrons. Boron, for example, reacts with halogens to form compounds of the general formula BX 3 having six electrons around the boron atom.

Odd Numbers of Electrons 2) The central atom has fewer than eight electrons due to an odd number of electrons. • • • • O=N−O 17 valence electrons in the system Molecules with an odd number of electrons are sometimes referred to as free radicals.

Worked Example 3. 8 Draw the Lewis structure of chlorine dioxide (Cl. O 2). Strategy The skeletal structure is This puts the unique atom, Cl, in the center and puts the more electronegative O atoms in terminal positions. Think About It Cl. O 2 is used primarily to bleach wood pulp in the amanufacture of paper, but it(7 isfrom alsothe used to bleach There are total of 19 valence electrons Cl and 6 from flour, each of the drinking and deodorize industrial two O disinfect atoms). We subtractwater, 4 electrons to account certain for the two bonds in the facilities. has been used to eradicate toxic skeleton, leaving. Recently, us with 15 itelectrons to distribute as follows: the three lone pairs on homes in New Orleans that damaged by electron the each Omold atom, inone lone pair on the Cl atom, andwere the last remaining also on floodwaters of Hurricane Katrina in 2005. the Cl devastating atom. Solution

Expanded Octets 3) The central atom has more than eight electrons. Sulfur has 6 bonds corresponding to 12 electrons Atoms in and beyond the third period can have more than eight valence electrons. In addition to the 3 s and 3 p orbitals, elements in the third period also have 3 d orbitals that can be used in bonding.

Worked Example 3. 9 Draw the Lewis structure of boron triiodide (BI 3). Strategy The skeletal structure is There a total of 24 valence electrons (3 from the B and 7 from each of the three I atoms). We subtract 6 electrons to account for the three bonds in the skeleton, leaving 18 electrons to distribute in lone pairs on each I atom. Solution

Worked Example 3. 9 (cont. ) Think About It Boron is one of the elements that does not always follow the octet rule. Like BF 3, however, BI 3 can be drawn with a double bond in order to satisfy the octet of boron. This gives rise to a total of four resonance structures.

Worked Example 3. 10 Draw the Lewis structure of arsenic pentafluoride (As. F 5). Strategy The skeletal structure already has more than an octet around the As atom. Think About It Always make sure that the number of There are 40 totalrepresented valence electrons [5 from (Group 5 A) and 7 from each of electrons in your final As Lewis structure matches the fivethe F atoms (Group 7 A)]. We subtract 10 electrons account fortothe five total number of valence electrons you aretosupposed bonds have. in the skeleton, leaving 30 to be distributed. Next, we place three lone pairs on each F atom, thereby completing all their octets and using up all the electrons. Solution

Worked Example 3. 11 Draw two resonance structures for sulfurous acid (H 2 SO 3): one that obeys the octet rule for the central atom, and one that minimizes the formal charges. Determine the formal charges on each atom in both structures. Strategy Begin by drawing the skeletal structure and counting the total number of valence electrons. Use the steps outlined in Section 6. 4 to draw the first structure and reposition one or more lone pairs to adjust the formal charges for the second structure. Note that each hydrogen in an oxoacid is attached to an oxygen atom, and not directly to the central atom. The total number of valence electrons is 26 (6 from S, 6 from each O, and 1 from each H).

Worked Example 3. 11 (cont. ) Solution Following the steps we get the first structure: -1 +1 From the top O atom, which has three lone pairs, we reposition one lone pair to create a double bond between O and S to get the second structure. Think About It In some species, such as the sulfate ion, it is possible to incorporate too many double bonds. Structures with three and four double bonds to sulfur would give formal charges on S and O that are inconsistent with the electronegativities of these elements. In general, you are trying to minimize formal charges by charge of Incorporating the if double bond results in every atom having a formal zero. expanding the central atom’s octet, only add enough double bonds to make the formal charge on the central atom zero.

Worked Example 3. 12 Draw the Lewis structure of xenon tetrafluoride (Xe. F 4). Strategy Follow the steps for drawing Lewis structures. The skeletal structure is There are 36 total valence electrons (8 from Xe and 7 from each of the four F atoms). We subtract 8 electrons to account for the bonds in the skeleton, leaving 28 to distribute. We first complete the octets of all four F atoms. When this is About It Atoms thelone second done, 4 Think electrons remain, so we beyond place two pairsperiod on thecan Xe atom. accommodate more than an octet of electrons, whether those Solution electrons are used in bonds or reside on the atoms as lone pairs.

3 Chapter Summary: Key Points Lewis Structures Octet Rule Bond Order Electronegativity and Polarity Dipole Moment Percent Ionic Character Drawing Lewis Structures Formal Charge Resonance Structures Exceptions to the Octet Rule Incomplete Octets Odd Numbers of Electrons Expanded Octets

Group Quiz • Draw Lewis Structures for the following substances: • CO 2 • CO • HCN • SO 32 - 61

Group Quiz • Draw all possible Lewis Structures for: • • • POCl 3 BF 3 NO 2 PF 5 SO 3