Chapter 11 Chemical Bonds: The Formation of Compounds from Atoms 1
11.1 Periodic Trends in atomic properties 11.1 Periodic Trends in atomic properties design of periodic table is based on observing properties of the elements periodic trends allow us to use the periodic table to predict properties and reactions of a wide variety of substances 1. metals and nonmetals metals are usually lustrous, malleable and good conductors of heat and electricity nonmetals are nonlustrous, brittle and poor conductors metals tend to lose electrons and form cations nonmetals tend to gain electrons and form anions a metal reacts with a nonmetal, electrons are often transferred from the metal to the nonmetal hydrogen is a unique element 2
2. atomic radius the increase in radius down a group for each step down a group, an additional energy level is added to the atom decrease in atomic radius across a period electrons are added the same energy level each time an electron is added, a proton is added to the nucleus and increase in positive charge that pulls the electrons closer to the nucleus 3
3. ionization energy the energy required to remove an electron from the (gaseous) atom Na + ionization energy Na + + e - first ionization energy for the elements in the first four periods ionization energy in Group A elements decreases from top to bottom ionization energy gradually increases from left 4 to right across a period
11.2 Lewis structures of atoms metals tend to form cations and nonmetals tend to form anions in order to attain a stable valence electron structure which contains 8 electrons this rearrangements are accomplished by losing, gaining or sharing electrons with other atoms Lewis structure using the symbol for element and dots for valence electrons paired dots represent paired electrons unpaired dots represent unpaired electrons ex. unpaired electron :B paired electron symbol for element Lewis structures for the first 20 elements 5
11.3 The ionic bond: transfer of electrons from one atom to another the chemistry of many elements is to attain an outer electron structure like that of noble gases ex. Na atom loses 3s electron to form Na + this process requires energy Cl atom gains an electron to form Cl - this process releases energy 6
consider Na and Cl atoms react with each other Na + and Cl - strongly are attracted to each other by their opposite electrostatic charges Lewis representation:.... Na + Cl: [Na] + [:Cl:] - NaCl is made up of cubic crystals each Na + is surrounded by 6 Cl -, each Cl - is surrounded by 6 Na + 7
relative sizes of Na and Cl atoms with those of their ions Na + is smaller than Na atom Cl - is larger than Cl atom ionic bond the attraction between oppositely charged ions 8
ex. 11.2 how Mg and Cl combine to form MgCl 2 ex. 11.3 formation of NaF from its elements ex. 11.5 formation of MgO from its elements 9
11.4 Predicting formulas of ionic compounds the concept forms the basis for our understanding of chemical bonding: in almost all stable chemical compounds of representative elements, each atom attains a noble gas electron configuration predicting the formulas of ionic compounds ex. Compound formed between Ba and S Ba [Xe]6s 2 S [Ne]3s 2 3p 4 loses 2 electrons to achieve Xe configuration Ba 2+ gains 2 electrons to achieve Ar configuration S 2- BaS ex. 11.8 predict formulas for (a) magnesium sulfide Mg 2+ S 2- MgS (b) potassium phosphide K + P 3- K10 3 P (c) magnesium selenide Mg 2+ Se 2- MgSe
11.5 The covalent bond: sharing electrons electron transfer between atoms does not occur in molecular compounds in which a chemical bond formed by sharing pairs of electrons between atoms 1916 G. N. Lewis introduced covalent bond consists of a pair of electrons shared between two atoms examples of molecular compounds: H 2, Cl 2, H 2 O, HCl, CO 2, sugar ionic compound formation of hydrogen molecule (H 2 ) involves overlapping and pairing of 1s electron orbitals from two H atoms 11
formation of chlorine molecule (Cl 2 ) involves overlapping of unpaired 3p electron orbitals from two Cl atoms other examples............ :F : F: : I : I : :O :: O: :N ::: N: fluorine iodine oxygen nitrogen using a dash replace the pair of dots............ :F F: : I I : :O = O: :N N: ionic bond and covalent bond represent two extremes between two extremes polar covalent bond unequal sharing of electrons between two atoms.. ex. H:Cl: :C:::O: 12
11.6 Electronegativity the bond formed between two different kinds of atoms which exert unequal attraction for the pair of electron one atom assume a partial positive charge, the other a partial negative charge the attractive force that an element has for shared electrons in a molecule or polyatomic ion is known as its electronegativity ex. hydrogen chloride HCl Pauling scale assign electronegativity of F 4.0 increase decrease 13
the highest electronegativity is 4.0 for F the lowest electronegativity is 0.7 for Cs and Fr the higher the electronegativity, the stronger an atom attracts electrons the polarity of a bond is determined by the difference in electronegativity values of the atoms forming the bond nonpolar covalent bond polar covalent bond dipole + - H Cl H Br I Cl O H H if the electronegativity difference between two bonded atoms is greater than 1.7~1.9, the bond will be more ionic than covalent 0 0.9 2.1 14
carbon dioxide CO 2 is nonpolar O = C = O CCl 4 is nonpolar H 2 O is a polar molecule O H H 15
11.7 Lewis structures of compounds the procedure helps to write the Lewis structure ex. 11.9 how man valence electrons in each of the following atoms: Cl H C O N S P I 7 1 4 6 5 6 5 7 16
ex. 11.10 write the Lewis structure for H 2 O step 1 total valence electrons = 2 1 + 6 = 8 step 2 the skeleton structure H O H : O H H step 3 subtract 4 electrons from 8 8 4 = 4 step 4 distribute 4 electrons around O atom H : O : H O : H H ex.11.11 write the Lewis structure for CH 4 step 1 total valence electrons = 4 1 + 4 = 8 step 2 H H H C H H : C : H H H step 3 8 8 = 0 H H C H H 17
ex.11.11 write the Lewis structure for CO 2 step 1 total valence electrons = 4 + 2 6 = 16 step 2 O C O O : C : O step 3 16 4 = 12 step 4................ : O : C : O : : O : C : O : : O : C : O : I II III step 5 : O :: C :: O : : O = C = O: 18
11.8 Complex Lewis structures for some molecules or polyatomic ions, no single Lewis structure consistent with all characteristics and bonding information can be written ex. nitrate ion NO - 3 step 1 total valence electrons = 5 + 3 6 + 1 = 24 step 2 O O O N O O : N : O step 3 24-6 = 18 step 4 : O electron deficient : O : N : O : step 5 - - - : O : O : : O : : O : N : O : : O =N : O : : O : N= O : there are three possible Lewis structures a molecule or ion that has multiple correct Lewis structures shows resonance each of these Lewis structures is called a resonance structure 19
ex. 11.14 write the Lewis structure for CO 3 2- step 1 total valence electrons = 4 + 3 6 + 2 = 24 step 2 O O O C O O : C : O step 3 24-6 = 18 step 4 : O : : O : C : O : C atom is electron deficient step 5 2-2- 2- : O : O : : O : : O : C : O : : O =C : O : : O : C= O : 20
11.9 Compounds containing polyatomic ions sodium carbonate Na 2 CO 3 has both ionic and covalent bonds ionic bonds exist between Na + and CO 3 2- covalent bonds present between C and O atoms an important difference between ionic and covalent bonds can be demonstrated by dissolving Na 2 CO 3 in water: Na 2 CO 3 (s) water 2 Na + (aq) + CO 3 2- (aq) 21
11.10 Molecular shape Lewis structures do not indicate anything regarding the geometric shape of a molecule geometric shapes for several common examples how do we predict the geometric shape of a molecule? 22
11.11 The valence shell electron pair repulsion (VSEPR) model VSEPR model is based on the idea that electron pairs will repel each other electrically and will seek to minimize this repulsion to accomplish this minimization, the electron pairs will arrange around the central atom as far apart as possible 1. linear structure 180 o ex. BeCl 2 Cl Be Cl 2. trigonal planar 120 o ex. BF 3 F F B F 3. tetrahedral 109.5 o ex. CH 4 H H C H H 23
one or more electron pairs may be nonbonding or lone pairs ex. NH 3.. H : N : H.. H ex. H 2 O.. H : O :.. H 24
ex. 11.15 predict the molecular shape for (1) H 2 S.. H : S : H tetrahedral bent.. (2) CCl 4.. tetrahedral : Cl :...... : Cl : C : Cl :...... : Cl :.. (3) AlF 3.. trigonal planar : F :...... : F : Al : F :.... 25