An attraction between anions and cations. Stronger than covalent bonds. Commonly called salts. Ionic Bonds
Coulomb’s Law • The strength of an ionic bond can be determined by 2 factors: • The charge of the ions, the higher the numerical charge value the stronger the attraction. • The distance of the ions, the closer they are the stronger the bond.
Lattice Energy • The energy required to break an ionic bond into its gaseous foundational elements:
Covalent Bonding • When two atoms have a small difference in their tendencies to lose or gain electrons, we observe electron sharing and covalent bonding. • This type of bonding most commonly occurs between nonmetal atoms (although a pair of metal atoms can sometimes form a covalent bond). • Each nonmetal atom holds onto its own electrons tightly (high IE) and tends to attract other electrons as well (highly negative EA). • The attraction of each nucleus for the valence electrons of the other draws the atoms together. • A shared electron pair is considered to be localized between the two atoms because it spends most of its time there, linking them in a covalent bond of a particular length and strength.
Bond Energy • Atoms will position themselves so that the system will achieve the lowest possible energy. • The distance where this energy is minimal is the bond length (distance between 2 nuclei, usually in nm or pm). • The energy required to break a bond is called the bond energy (usually in kJ/mol). • The shorter the bond length (the stronger the attraction) the higher the bond energy. • Rank in order of increasing bond energy: • H—F , H—Br, H—Cl
Polarity • The distribution of bonding electrons in a molecule maybe be equal or not depending on the surrounding atoms (and their electronegativities) and lone pairs in a molecule. This leads to a discussion on polarity: • Polar – the unsymmetrical distribution of electrons will cause a negative and positive end creating an overall dipole moment in the molecule. • Nonpolar – the symmetrical distribution of electrons will cause the dipole moments in the molecule to cancel out, and leaving the molecule with no net dipole.
Dipole Moments & Polarity Dipole moments maybe symbolized by either Greek letter delta (lowercase), δ , or arrows crossed at the positive end, or both.
Lewis Structures • Lewis Dot Diagrams represent the number of valence electrons present in an atom. • Lewis structures represent the covalent bonds in a molecule. Lines are used to represent bonds (1 line = single bond, 2 lines = double bond, 3 lines = triple bond) and 2 dots represent lone pairs of electrons
Gilbert Newton Lewis (1875–1946). As early as 1902, nearly a decade before Rutherford proposed the nuclear model of the atom, Lewis’s notebooks show a scheme that involves the filling of outer electron “shells” to explain the way elements combine. His electron-dot symbols and associated structural formulas have become standards for representing bonding. Among his many other contributions is a more general model for the behavior of acids and bases.
Drawing Lewis Structures 1. Count the total number of valence electrons in the molecule. Usually, its easier to divide by 2 and work with pairs of electrons. 2. Create a “skeleton” structure by connecting surrounding (terminal) atoms to the central atom. The central atom is the one that there is one of or the least electronegative (generally the first element in the formula is the central). 3. Place electrons on surrounding atoms so they each may have an octet (remember hydrogen only needs 2). Place the remaining electrons (if any) on the central atom. 4. Once all electrons are used check to see that all atoms have an octet……otherwise known as “the happy check” (copy-write by Mrs. A ). 5. If the central (C, N, O, P, S, Se) does not have an octet must use lone pairs on the surrounding to make multiple bonds (only possible are double and triple).
Examples • Draw a Lewis Structure for: • The hypochlorite ion: • A water molecule: • Ammonia: • Ammonium: • Carbon dioxide: • Methane: • Ethane, C2H6:
Exceptions to the Octet Rule • Electron Deficient • The central atom has less than 8 electrons • Boron Trifluoride • Expanded Octet • The central atom is sharing more than 8 electrons • Xenon tetrafluoride • Odd Electron • A few molecules contain a central atom with an odd number of valence electrons, so they cannot possibly have all their electrons in pairs. Such species, called free radicals, contain a lone (unpaired) electron, which makes them paramagnetic and extremely reactive. • Nitrogen monoxide and nitrogen dioxide
Resonance • Resonance structures express the various plausible locations of a double/triple bond location on a molecule. • Resonance does not imply different kinds of molecules but just reflects plausible Lewis structures.
Examples • Draw a Lewis Structure for: • The nitrite ion: • Sulfur dioxide: • The nitrate ion: • Ozone: • Nitrogen gas: • Benzene*, organic molecule, that forms a hexagonal ring structure.
Formal Charge • Formal charge is used to determine the one structure that exists in nature when more than one arrangement of atoms in the Lewis structure is possible. • i.e. A—A—B or A—B—A • You must determine the formal charge for all central atoms. This is done using the following formula: Cf = X - (Y + Z/2) X = the number of valence electrons on the atom Y = the # if unshared electrons around the atom Z = the # of bonding electrons • The more likely Lewis structure is the one in which: • The formal charges are closest to zero • Any negative formal charge is located on the stronger electronegative atom. • Formal charges must sum to the actual charge on the species: zero for a molecule and the ionic charge for an ion.
Formal Charge in Resonance • Three criteria help us choose the more important resonance structures: • • Smaller formal charges (positive or negative) are preferable to larger ones. • • The same nonzero formal charges on adjacent atoms are not preferred. • • A more negative formal charge should reside on a more electronegative atom.
Example • Decide which is the more relevant Lewis structure for the cyanate ion.
Deadly Free-Radical Activity Free radicals can be extremely dangerous to biological systems because they rupture bonds in cells’ biomolecules. If a free radical reacts with a biomolecule, it typically forms a covalent bond to one of the H atoms and removes it, and the biomolecule is left with an unpaired electron, thereby becoming a new free radical. That species repeats the process and creates other species with lone electrons that proliferate to disrupt chromosomes and cell membranes. We benefit from this activity when we apply the disinfectant hydrogen peroxide to a cut because it forms free radicals that destroy bacterial membranes. (The photo shows H2O2 reacting vigorously with a drop of blood.) On the other hand, recent studies have suggested that several disease states, including certain forms of cancer, may be attributable to free radicals. Also, vitamin E is believed to interrupt free-radical proliferation.
VSEPR • Molecular geometry can be predicted on the basis of VSEPR (valence shell electron pair repulsion theory). The idea that valence electron pairs around an atom repel one another as far apart as possible thus impacting the geometry (shape) of the molecule. • To determine molecular geometry, first determine the type of molecule from the Lewis structure and the rest is really memorization. • You must know the name, 3-D shape, bond angles, the hybridization, polarity, and bond order (to be discussed later this section) for each type of molecule, use the Lewis Structure to determine the type of molecule: • A = central atom • X = surrounding atoms • E = lone pairs of electrons on central atom
Molecules With No Lone Pairs on Central • AX2 - linear - sp • AX3 - triangular planar - sp2 • AX4 - tetrahedron - sp3 • AX5 - triangular bipyramid - sp3d • AX6 - octahedron - sp3d2
Molecules With Lone Pairs on Central • AX2E - bent - sp2 • AX3E - triangular pyramid - sp3 • AX2E2 - bent - sp3 • AX4E - see-saw - sp3d • AX3E2 - T-shaped - sp3d • AX2E3 - linear - sp3d • AX5E - square pyramidal - sp3d2 • AX4E2 - square planar - sp3d2
Molecular Shape & Polarity • If there is no lone pair on the central and all of the surrounding atoms are the same then the molecule will be nonpolar: • linear 1, triangular planar, tetrahedral, triangular bipyramidal, & octahedral. • If there are lone pairs around the central and unsymmetrical distribution of the surrounding atoms the molecule is polar: • bent 1 and bent 2, triangular pyramidal, square pyramidal, t-shaped, & seesaw, • If the polar A—X bonds in a molecule AXmEn are symmetrical around the central atom then the molecule is nonpolar: • square planar and linear 2.
Hybridization • The sharing of electrons leads to the formation of hybrid orbitals:
Bond Order As the bond order increases bond length will decrease thus the bond energy increases. In resonance structures the bond order is calculated by taking the average bond order.
Sigma & Pi Bonds • Sigma bonds – basically all single bonds in a molecule • Pi bonds – one of the electron pairs in a multiple bond is a sigma bond, the rest are pi bonds. • Example 7.11 – page 187
MC #1 • Which of the following molecules has the shortest bond length? (A) N2(B) O2(C) Cl2(D) Br2(E) I2
MC #2 • Which of the following has a zero dipole moment? (A) HCN(B) NH3(C) SO2(D) NO2(E) PF5
MC #3 • Pi bonding occurs in each of the following species EXCEPT (A) CO2(B) C2H4(C) CN¯(D) C6H6(E) CH4
MC #4 • The SbCl5 molecule has trigonal bipyramid structure. Therefore, the hybridization of Sb orbitals should be (A) sp2(B) sp3(C) dsp2(D) dsp3(E) d2sp3
MC #5 • CCl4, CO2, PCl3, PCl5, SF6 Which of the following does not describe any of the molecules above? (A) Linear(B) Octahedral(C) Square planar(D) Tetrahedral(E) Trigonal pyramidal
FRQ #1 • Discuss briefly the relationship between the dipole moment of a molecule and the polar character of the bonds within it. With this as the basis, account for the difference between the dipole moments of CH2F2 and CF4.
FRQ #2 (a) Draw the Lewis electron-dot structures for CO32-, CO2, and CO, including resonance structures where appropriate. (b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer. (c) Predict the molecular shapes for the three species. Explain how you arrived at your predictions.
FRQ #3 CF4 XeF4 ClF3 (a) Draw a Lewis electron-dot structure for each of the molecules above and identify the shape of each. (b) Use the valence shell electron-pair repulsion (VSEPR) model to explain the geometry of each of these molecules.
FRQ #4 • Use simple structure and bonding models to account for each of the following. (a) The bond length between the two carbon atoms is shorter in C2H4 than in C2H6. (b) The H-N-H bond angle is 107.5º, in NH3. (c) The bond lengths in SO3 are all identical and are shorter than a sulfur-oxygen single bond. (d) The I3- ion is linear.
FRQ #5 • Answer the following questions, which pertain to binary compounds. (a) In the box provided below, draw a complete Lewis electron-dot diagram for the IF3 molecule. (b) On the basis of the Lewis electron-dot diagram that you drew in part (a), predict the molecular geometry of the IF3 molecule. (c) In the SO2 molecule, both of the bonds between sulfur and oxygen have the same length. Explain this observation, supporting your explanation by drawing in the box below a Lewis electron-dot diagram (or diagrams) for the SO2 molecule. (d) On the basis of your Lewis electron-dot diagram(s) in part (c), identify the hybridization of the sulfur atom in the SO2 molecule.
FRQ #6 • Answer the following questions about the structures of ions that contain only sulfur and fluorine. (a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation. SF4 + BF3 SF3BF4 (i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4. (ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation. (iii) Identify the geometry of the SF3+ cation that is consistent with the Lewis structure drawn in part (a)(i). (iv) Predict whether the F—S—F bond angle in the SF3+ cation is larger than, equal to, or smaller than 109.50˚. Justify your answer.
FRQ #6 (cont.) (b) The compounds SF4 and CsF react to form an ionic compound according to the following equation. SF4 + CsF CsSF5 (i) Draw a complete Lewis structure for the SF5– anion in CsSF5. (ii) Identify the type of hybridization exhibited by sulfur in the SF5– anion. (iii) Identify the geometry of the SF5– anion that is consistent with the Lewis structure drawn in part (b)(i). (iv) Identify the oxidation number of sulfur in the compound CsSF5.
FRQ #7 Answer the following questions that relate to chemical bonding (a) In the boxes provided, draw the complete Lewis structure (electron-dot diagram) for each of the three molecules represented below. CF4 PF5 SF4 (b) On the basis of the Lewis structures drawn above, answer the following questions about the particular molecule indicated. (i) What is the F-C-F bond angle in CF4? (ii) What is the hybridization of the valence orbitals of P in PF5? (iii) What is the geometric shape formed by the atoms in SF4? (c) Two Lewis structures can be drawn for the OPF3 molecule, as shown below. Structure 1 Structure 2 (i) How many sigma bonds and how many pi bonds are in structure 1? (ii) Which one of the two structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.
Credits • (Silberberg, Martin S.. Chemistry: The Molecular Nature of Matter and Change, 5th Edition. McGraw-Hill, 012009. 9.4.2).