200 likes | 215 Views
Ch. 8 Chemical Bonding. Chemical bonds hold atoms together. There are 3 types of chemical bonds: Ionic bonds (electrostatic forces that hold ions together…) Example: Na + Cl - , K + Br - Covalent bonds (result from sharing electrons between atoms…) Example: H 2 , NH 3
E N D
Ch. 8 Chemical Bonding Chemical bonds hold atoms together. There are 3 types of chemical bonds: Ionicbonds(electrostatic forces that hold ions together…) Example: Na+Cl-, K+Br- Covalentbonds(result from sharing electrons between atoms…) Example: H2, NH3 Metallicbonds (refers to metal nuclei floating in a sea of electrons…) Example: copper, gold
Lewis Symbols: Electron dots Valence Electrons: outer-most electrons determined by the “Group A #” from the periodic table Exceptions: d or f-block = 2 valence electrons & Helium =2 e-
Octet Rule Ionic Bonding Atoms often gain or lose or share electrons to fill their valence shell with 8 electrons to achieve a noble gas configuration. Exceptions: * Hydrogen needs only 2 e- to be filled. * Some nonmetals can have more or less than 8. • Ionicbonds—transfer of electrons • Form between an element of low ionizationenergy (not much energy required to pull off an electron) and an element of high electron affinity (lots of energy is released when an electron is added to its outer shell). • Usually form between a metal and a nonmetal.
Ionic Bonding Energies Consider the reaction between sodium and chlorine: Na(s) + ½Cl2(g) NaCl(s)∆Hºf = –410.9 kJ/mol - The reaction is violently exothermic. - We infer that the NaCl is more stable than its constituent elements. Here’s another way to look at the energy of ionic bond formation: Sodium loses 1 electron…Na Na+ + 1 e- Requires 5.1 eV of energy Chlorine gains l electron…Cl + 1 e- Cl- Releases 3.6 eV of energy NaCl forms… Na+ + Cl- [Na+][ Cl-] Releases 5.2 eV of energy [1 eV (electron volt) = 1.602 x 10-19 J] The energy released is greater than the energy required, therefore the ionic bond forms… (∆Hf = - 3.7 eV)
Lattice Energy Lattice energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. Lattice energy increases as distance between the ions decreases. Lattice energy increases as charges on the ions increase.
Ions Metals lose electrons to form smaller (+) cations. Nonmetals gain e- to form larger (-) anions. The # of e- gained or lost depends on how many they need to gain or lose to get to a noble gas configuration. Only then will they become stable. Groups of atoms can have charges too. They are called polyatomic ions. The atoms share electrons (covalent bonds) but the group still has an overall charge. Examples: [NH4+] , [CO3-2]
Covalent Bonds - A line can also be used to represent 2 shared e-’s (or one covalent bond.) Atoms share electrons to fill their valence shell. Usually form between 2 nonmetals Lewis Structures: represent covalent bonds as 2 dots between the atoms
Multiple Covalent Bonds Single bond = 2 electrons shared …(1 pair) Double bond= 4 electrons shared…(2 pairs) Triple bond= 6 electrons shared…(3 pairs) Bond Lengths & Bond Strengths Single bonds are the longest and weakest covalent bonds. Triple are the shortest and strongest covalent bonds. Ionic bonds are much stronger than covalent bonds.
Bond Polarity Bond polarity helps to describe the sharing of the electrons between atoms. There are 3 possibilities… Nonpolar covalent: equal sharing of the e- pair Polar covalent: unequal sharing of the e- pair Ionic: transfer of valence e- from the metal to the nonmetal A molecule that has one side slightly positive and one side slightly negative is said to be a “dipole.” The positive end (or pole) in a polar bond is represented + and the negative pole -. Arrow can also show dipoles. + -
Bond Polarity & Electronegativity (How can you tell what type of bond will form?) Electronegativity: describes an atom’s attraction to the e- pair in a bond…(It’s a number from 0 to 4.0) The difference between electronegativities indicate whether a bond will be nonpolar, polar or ionic. There is no sharp distinction between bonding types. In General: Nonpolar = 0-0.4 Polar= 0.5-2.0 Ionic= Above 2.0 Lattice Energy and Polarity Lattice energy increases as the electronegativity between the atoms in an ionic compound increases
Dipole Moments • Consider HF: • The difference in electronegativity leads to a polar bond. • There is more electron density on F than on H. • Since there are two different “ends” of the molecule, we call HF a dipole. • Dipole moment(debyes, D), m, is the magnitude of the dipole: • µ=Qr • where Q is the magnitude of the charges. Chapter 8
Rule 1) How many electrons are possible around an atom? • For hydrogen, only 2 electrons are possible, therefore only one bond! • Second row elements usually try to get 8 e-. Notable Exception: Boron needs only 6. • Third & Fourth Row usually have 8 e- but can expand to get 10 or more. • Rule 2) Drawing the Lewis Structure: • First arrange the atoms around the central atom (usually the highest electronegative one, but never hydrogen!) • Count the total # of valence electrons in the molecule. If it is an ion, add 1 for each (-) charge or subtract 1 for each (+) charge. • Distribute the electrons keeping Rule #1 in mind. If you have too many electrons, look for double or triple bonds, or place the extras around the “3rd or 4th row element.” Rules For Drawing Lewis Structures
Rule 3) For Ionic Compounds: • Indicate the charge on the ions • Do not share the electrons, transfer them! • Rule 4) For odd-numbered valence electrons: • If you must cheat an element out of 8 e- and only give it 7, then cheat the least electronegative element. • Rule 5) Resonance Structures: • If there is more than one way to draw the Lewis structure, show them all. • Rule 6) Nature doesn’t know anything about Rules1-5. • These are just models for us to use in order to represent bonding. Rules For Drawing Lewis Structures
Drawing Lewis Structures • Sum valence electron from all atoms. • Write symbols for the atoms and show which atoms are connected to which. • Complete the octets of the atoms bonded to the central atom. • Place leftover electrons on the central atom, even if it violates the octet rule. • If there are not enough electrons to give the central atom an octet, try multiple bonds. Chapter 8
Resonance Structures: • The ability to draw more than one “correct” Lewis structure. Resonance Benzene (C6H6) • The true structure for the molecule is somewhere “in between” the resonance structures.
Formal Charge: The formal charge of an atom is the charge that an atom (in a molecule) would have if all of the atoms had the same electronegativity. • To calculate formal charge: • (valence e- - # of bonds - lone pair e-) • Practice: Determine the formal charge on C and N. Formal Charge C= 4 – 3 – 2 = -1 N= 5 – 3 – 2 = 0
The most stable Lewis structure has the smallest formal charge on each atom and the most negative formal charge on the most electronegative atoms. • • It is important to keep in mind that formal charges do NOT represent REAL charges on atoms! Formal Charge
The energy required to break a covalent bond is called the bond dissociation enthalpy, D. • That is, for the Cl2 molecule, D(Cl-Cl) is given by H for the reaction: • Cl2(g) 2Cl(g)H = 242 kJ • When more than one bond is broken… • CH4(g) C(g) + 4H(g)H = 1660 kJ • …the bond enthalpy is a fraction of H for the “atomization reaction”: • D(C-H) = ¼H = ¼(1660 kJ) = 415 kJ Strengths of Covalent Bonds
The bond enthalpy for a given set of atoms depends on the rest of the molecule of which it is a part. An average bond enthalpy is therefore shown. The energy released when a bond forms is just the (-) of the value shown in the table! Strengths of Covalent Bonds
∆H(rxn) = D(bonds broken) - D(bonds formed) • Practice Problem • CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)H(rxn) = ? • In this reaction one C-H bond and one Cl-Cl bond is broken while one C-Cl bond and one H-Cl bond is formed. • So… ∆H(rxn) = [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)] • = 413kJ + 242 kJ – 328 kJ – 431 kJ = –104 kJ • The overall reaction is exothermic which means that the bonds formed are stronger than the bonds broken. • The above result is consistent with Hess’s law. Using “D” to determine ∆H(rxn)