Presentation on theme: "Bonding: General Concept Types of Chemical Bonds The Formation of Ions and Their Electron Configurations Ionic Size and Charges, and the Relative Strength."— Presentation transcript:
Bonding: General Concept Types of Chemical Bonds The Formation of Ions and Their Electron Configurations Ionic Size and Charges, and the Relative Strength of Ionic Bonds Energy of Formation of Ionic Compounds – The Born-Haber Cycle Covalent Bond: Electronegativity, and Bond Polarity Lewis Structures and the Octet Rule Exceptions to the Octet Rule Resonance Lewis Structures Bond Energies The Calculation of Enthalpy of Reaction from Bond Energy The VSEPR Model and Molecular Shapes
Ionic bonds electrostatic attractions between cations and anions; Bonds formed between metals and nonmetals Reactions that produce ionic bonds involve the transfer of one, two, or three electrons from a metal atom to nonmetal atom
Covalent Bonds One, two or three pairs of electrons shared between two atoms Bonds between two nonmetals or between semimetal and nonmetal atoms Bonds are formed when two atoms share electron pairs
The Formation of Ions Ions are formed when metals react with nonmetals, in which the metal atoms donate their valence electrons to the nonmetals; Atoms of the representative metals lose their valence electrons to become cations that have the electron configurations of noble gases; The nonmetal atoms gain a number of electrons to become anions that also have the electron configuration of the noble gases;
Formation of Cations From the alkali metals (1A): –M M + + e - From the alkaline Earth metals (2A): –M M 2 + + 2e - From Group 3A metals: M M 3+ + 3e - ;
Formation of Anions From the halogen family (VIIA): –X + e - X - From the oxygen family (VIA): –X + 2e - X 2- From N and P (in Group VA): –X + 3e - X 3-
Common Ions of the Representative Elements Ions with the electron configuration of He = 1s 2 – Li + and H - Ions with the electron configuration of Ne = 1s 2 2s 2 2p 6 –Na +, Mg 2+, Al 3+, F -, O 2 -, and N 3 - Ions with the electron configuration of Ar = 1s 2 2s 2 2p 6 3s 2 3p 6 –K +, Ca 2+, Sc 3+, Cl -, S 2 -, and P 3 - Ions with the electron configuration of Kr = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 –Rb +, Sr 2+, Y 3+, Br -, and Se 2 - ; Ions with the electron configuration of Xe = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 –Cs +, Ba 2+, La 3+, I -, and Te 2 - ;
Cations From the Transition Metals Transition metal atom loses variable number of electrons Cations derived from transition metals have variable charges Cations of transition metals do not acquire the electron configurations of noble gases Examples: –Cr Cr 2+ + 2e - ;Cr 2+ : [Ar] 3d 4 –Cr Cr 3+ + 3e - ;Cr 3+ : [Ar] 3d 3 –Fe Fe 2+ + 2e - ;Fe 2+ : [Ar] 3d 6 –Fe Fe 3+ + 3e - ;Fe 3+ : [Ar] 3d 5 (Note that these cations do not have the 4s electrons in their electron configurations.)
Charge Density and The Strength of Ionic Bonds Relative sizes of isoelectronic ions: Al 3+ < Mg 2+ < Na + < Ne < F - < O 2- < N 3- ; Sc 3+ < Ca 2+ < K + < Ar < Cl - < S 2- < P 3- ; Trend of ionic radii within a group: Li + < Na + < K + < Rb + < Cs + ; F - < Cl - < Br - < I - ; Ionic Bond Strength: Strength of Ionic bonds is related to charge density of the ions; Greater charge and smaller ions lead to stronger ionic bond;
Ionic Bond Strength and Lattice Energy The strength of ionic bonds is associated with the magnitude of the lattice energy Lattice energy - energy released when gaseous ions combine to form a mole of solid ionic compound: –M + (g) + X - (g) MX (s) ; U L = lattice energy Example: Na + (g) + Cl - (g) NaCl (s) ; U L = -787 kJ/mol Li + (g) + F - (g) LiF (s) ; U L = -1047 kJ/mol Lattice energy = k(q 1 q 2 /r); –where q 1 and q 2 are charge magnitude on ions, r is the internuclear distance, and k is the proportionality constant. –Lattice energy increases with charge magnitude but decreases with ionic size
Lattice Energies of Some Ionic Compounds Lattice Energy, U L (kJ/mol) –The energy required to separate a mole of ionic solids into the gaseous/vapor ions; –MX (s) M + (g) + X - (g) M n+ /X n- F - Cl - Br - I - O 2 - Li + 1047 853 807 7572942 Na + 923 787 747 7042608 K + 821 715 682 6492311 Mg 2+ 2957 2526 2440 23273919 Ca 2+ 2628 2247 2089 20593570 ______________________________________________________________________
The Born-Haber Cycle for NaCl Na + (g) + Cl (g) _______________ -349 kJ +496 kJ _______ Na + (g) + Cl - (g) Na (g) + Cl (g) ___________ +121 kJ Na (g) + ½Cl 2 (g) ________ ? kJ +108 kJ Na (s) + ½Cl 2 (g) ________ -411 kJ NaCl (s) _________________
Chemical Processes in the Formation of NaCl Na (s) Na (g) ; H s = +108 kJ ½Cl 2 (g) Cl (g) ; ½BE = +121 kJ Na (g) Na + (g) + e - ;IE = +496 kJ Cl (g) + e - Cl - (g) ;EA = -349 kJ Na + ( g) + Cl - (g) NaCl (s) ; U L = ? kJ Na (s) + ½Cl 2 (g) NaCl (s) ; H f = -411 kJ –U L = H f – ( H s + ½BE + IE + EA) H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; U L = Lattice energy; H f = Enthalpy of formation)
The Born-Haber Cycle for LiF Li + (g) + F (g) _______________ -328 kJ +520 kJ _______Li + (g) + F - (g) Li (g) + F (g) ___________ +77 kJ Li (g) + ½F 2 (g) ________ ? kJ +161 kJ Li (s) + ½F 2 (g) ________ -617 kJ LiF (s) _________________
Chemical Processes in the Formation of LiF Li (s) Li (g) ; H s = +161 kJ ½F 2 (g) F (g) ; ½BE = +77 kJ Li (g) Li + (g) + e - ;IE = +520 kJ F (g) + e - F - (g) ;EA = -328 kJ Li + ( g) + F - (g) LiF (s) ; U L = ? Li (s) + ½F 2 (g) LiF (s) ; H f = -617 kJ –U L = H f – ( H s + ½BE + IE + EA) H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; U L = Lattice energy; H f = Enthalpy of formation)
The Born-Haber Cycle for MgO Mg 2+ (g) + O 2- (g) _____________ +737 kJ Mg 2+ (g) + O (g) ________ +2180 kJ Mg (g) + O (g) _________ +247 kJ Mg (g) + ½O 2 (g) ________ ? kJ +150 kJ Mg (s) + ½O 2 (g) ________ -602 kJ MgO (s) _________________
Chemical Processes in the Formation of MgO Mg (s) Mg (g) ; H s = +150 kJ ½O 2 (g) O (g) ; ½BE = +247 kJ Mg (g) Mg 2+ (g) + 2 e - ;IE = +2180 kJ O (g) + 2 e - O 2- (g) ;EA = +737 kJ Mg 2+ (g) + O 2- (g) MgO (s) ; U L = ? kJ Mg (s) + ½O 2 (g) MgO (s) ; H f = -602 kJ –U L = H f – ( H s + ½BE + IE + EA) H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; U L = Lattice energy; H f = Enthalpy of formation)
Covalent Bonds Covalent bonds –A bond between two nonmetal atoms or between a semi- metal and a nonmetal atoms; –Bonded atoms may share one, two, or three pairs of electrons. Nonpolar covalent bonds are formed between identical atoms or if bonded atoms have the same electronegativity. Polar covalent bonds are formed when bonded atoms have different electronegativity; Polar covalent bonds are covalent bonds with partial ionic character
Potential Energy Diagram for Covalent Bond Formation
Potential energy of H-atoms during the formation of H 2 molecule
Lewis Model for the Formation of Covalent Bonds and Covalent Molecules
General trends: Electronegativity increases from left to right along a period For the representative elements (s and p block) the electronegativity decreases as you go down a group The transition metal group is not as predictable as far as electronegativity.
Electronegativity Electronegativity is the relative ability of a bonded atom to attract shared electrons closer to itself. –Electronegativity increases going across a period and decreases going down a group. –Most electronegative elements – at top right corner of PT –Least electronegative elements – at bottom left corner of PT –Fluorine (F) is most electronegative with EN value of 4.0 –Francium (Fr) is least electronegative with a value of 0.7 The polarity of a covalent bond depends on the electronegativity difference ( EN) of the two bonded atoms.
Electronegativity and Bond Polarity CompoundF2F2 HFLiF Electronegativity Difference 4.0 - 4.0 = 04.0 - 2.1 = 1.94.0 - 1.0 = 3.0 Type of Bond Nonpolar covalent Polar covalent Ionic (non- covalent) In F 2 the electrons are shared equally between the atoms, the bond is nonpolar covalent In HF the fluorine atom has greater electronegativity than the hydrogen atom. The sharing of electrons in HF is unequal: the fluorine atom attracts electron density away from the hydrogen (the bond is thus a polar covalent bond)
Electronegativity and bond polarity The H-F bond can thus be represented as: The ' + ' and ' - ' symbols indicate partial positive and negative charges. The arrow indicates the "pull" of electrons off the hydrogen and towards the more electronegative atom. In lithium fluoride the much greater relative electronegativity of the fluorine atom completely strips the electron from the lithium and the result is an ionic bond (no sharing of the electron)
Predicting Bond Type From Electronegativity A general rule of thumb for predicting the type of bond based upon electronegativity differences EN) If EN between the two atoms is 0-0.5, the bond is non-polar covalent; If EN between the two atoms is greater than 0.5, but less than 1.5, the bond is polar covalent If EN between the two atoms is 1.5, or greater, the bond is ionic
Bond Length... The bond length is defined as the distance between the nuclei of the two atoms involved in the bond. In general, the larger the atoms involved in a bond, the longer the bond length, and the more bonds between two atoms, the shorter the bond length.
Bond Energy Bond energy is the energy required to break the bond(s) between two atoms. In general, the shorter the bond, the higher the bond energy.
Lewis Symbols for Atoms and The Formation of Covalent Molecules
How to draw a Lewis structure of a simple molecular compound or polyatomic ion? A Lewis structure can be drawn for a molecule or ion by following three steps: 1.Calculate the number of valence electrons (including charges, if any) 2.Write skeleton structure –Lowest EN (electronegative) atom, largest atom, and/or atom forming most bonds is usually the central atom –Hydrogen and Fluorine cannot be the central atom. –Connect all atoms with a single bond – using a line or two dots. –In oxoacids, such as sulfuric acid (H 2 SO 4 ), the oxygen bonds to central atom and the H to oxygen. –Compounds are usually compact and symmetrical structures 3.Count how many electrons have been used –Distribute remaining electrons to terminal atoms filling them until each has eight electrons (octet rule), unless it is hydrogen atom. 4.Central atom octet is filled last –Any remaining electrons become lone pairs on central atom. –If central atoms do not have an octet, move from terminal atoms one pair at a time to form double and triple bonds.
Covalent Bonds and Lewis Structures Some Molecules
Assigning Formal Charges in Lewis Dot Structures? Formal Charge –To determine the formal charge of an atom from a Lewis dot structure we need to assign each electron to an atom in the structure. To do this we use the following rules: 1.All nonbonding electrons (unshared electrons) are assigned to the atom on which they are found. 2.Each atom in a bond is assigned ½ of the total number of electrons in the bond (i.e. for a single bond each atom is assigned 1 electron, for a double bond each atom is assigned 2 electrons, etc.) 3.For each atom the number of electrons assigned in the above steps is subtracted from the number of valence electrons in the atom.
Choosing the correct or best Lewis structures based on formal charges If two or more Lewis dot structures can be drawn which satisfy the octet rule, the most stable one will be the structure where: 1.The formal charges are as small as possible. 2.Any negative charges are located on the more electronegative atoms.
Bond Breaking and Bond Formation in the Reaction to form H 2 O
Using Bond Energies to Calculate Enthalpy of Reactions in Gaseous State Chemical reactions in the gaseous state only involve: –the breaking of covalent bonds in reactants and –the formation of covalent bonds in products. Bond breaking requires energy input - an endothermic process), while bond formation releases energy – an exothermic process; H reaction = (Energy of bond breaking) + (Energy of bond formation)
Calculation of Reaction Enthalpy Using Bond Energies Use bond energies to estimate the H for the reaction: CH 3 OH (g) + 2 O 2 (g) CO 2 (g) + 2H 2 O (g) ; (energy of bond breaking) (in kJ) = 3 x BE(C─H) + BE(C─O) + BE(O─H) + 2 x BE(O═O) = (3 x 413) + 358 + 467 + (2 x 495) = 3054 kJ (energy of bond breaking) (in kJ) = 2 x -BE(C═O)* + 4 x -BE(O─H) = (2 x -799) + (4 x -495) = -3578 kJ H reaction = 3054 + (-3578) = -524 kJ
Molecular Shapes of BeI 2, HCl, IF 2 -, ClF 3, and NO 3 -