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Lecture

CHEM - L5.pdf

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Department
Chemistry
Course
CHEM 1E03
Professor
Randall S Dumont
Semester
Fall

Description
CHEMICAL BONDING We have seen how it is most favorable for atoms to have a full outer shell (valence shell) of electrons. This is the driving force for atomic reactivity. Atoms seek to attain this electron configuration - either by losing, gaining or sharing outer shell electrons. The number of outer shell electrons accounts for the observed valence of atoms. Atomic Lewis Symbols: Outer Shell Observed Atomic Electron Lewis Configuration valence Symbol 2s 2p 1 Li⋅ excited state 2 ⋅Be ⋅ configurations allow ⋅ greater valence & better 3 ⋅B⋅ achieve (on bonding) a ⋅ full (or almost full valence 4 ⋅⋅ ⋅ shell) . 3 :. ⋅ . 2 :..⋅ .. 1 ..⋅ .. 0 :.. : 1 Ionic Bonding Atoms to right of the periodic table (metals) have few valence electrons, whereas atoms to the left (nonmetals) have nearly filled valence shells. It is favorable for the metal to lose its valence electrons to the non-metal to form ions with filled outer shells - octets ♣ These ions arrange themselves alternately in a crystal lattice to maximize the electrostatic attraction between the ions of opposite charge, while minimizing the repulsion between ions of the same charge. ♣ 2 The cost of forming ions from atoms is the ionization energy of the metal atom (or, eg., the sum of the 1 & 2 ionization energies in the case of forming 2+ ions) minus the electron affinity of the nonmetal atom (or sum of successive electron affinities for multiply negatively charged anions). Low ionization energy of metal + high electron affinity of non-metal makes ionic bond formation favorable. There are other contributing terms (i.e. contributing to ∆H° off ionic solid) ... • we must break bonds of elemental materials to form ionic compound from elements • having produced gaseous ions, there is a "lattice energy", ∆H° latticeassociated with the electrostatic forces between ions - attractive forces emphasized over repulsive ♣ 3 Lattice Energy The lattice energy... • increases with increasing charge on ions • decreases with increasing distance between ions (sum of ionic radii) + - E = k q q /r Of course, the lattice energy is more complex involving a sum over all pairs of ions in the crystal. But, the electrostatic attraction between neighboring ions of opposite charge gives the right trends. eg. ♣ MgCl 22 2J/mol Na 2O 2570 kJ/mol MgO 3890 kJ/mol 4 Covalent Bonding Bonding between non-metal atoms requires both atoms attempting to complete their octets. This is possible if the atoms share some of their valence electrons - the covalent bond ... ♣ eg. .. .. :..⋅ + ..: :F.:F: .. .. .. .. :.. ..: electrons of covalent bond pair of electrons 5 ♣ egs. of covalent bonds ... Ammonia: NH N H 3 2 2 3 1 1s 2s 2p 1s # of valence 5 1 electrons - also group # .H H :N⋅ .H :N − H three single bonds . .H H one lone pair Carbon dioxide: CO 2 2 2 2 1O 2 4 1s 2s 2p 1s 2s 2p . . .. . . .. . .. . . . ... ..=C=O .. two double bonds four lone pairs one triple bond Nitrogen: :N≡N : two lone pairs 6 Polyatomic Ions eg. Ammonium: ♣ The bonding in NH Cl is ionic between NH and Cl , and − 4+ 4 covalent within NH .4 + The positive charge in NH 4 can be associated with the N atom - because N is sharing more than three (its usual valence) of its valence electrons. 7 Formal Charge = − − core charge − (# of unshared e s) − (# of shared e s) 2 nuclear charge − (# core electrons)unt the non-bonding i.e. Group # electrons count the non-bonding - lone pairs (2 per paielectrons … & unpaired (case of - two per bond free radicals) - double bonds = 4 electrons, etc. … & divide by 2 egs. N in NH 3 H :N− H three single bonds one lone pair H Formal Charge on N = 5 − 2 − (3 2)/2 = 0 + N in NH 4 H H− N− H four single bonds no lone pairs H Formal Charge on N = 5 − (4 ⋅2)/2 = +1 8 Sum of formal charges (over all atoms) on neutral molecule = 0 Sum of formal charges on (polyatomic) ion = charge on ion Formal charge is a guide to electronic structure (via Lewis structures), not a true charge on the atom Correct Lewis structures (almost) never have adjacent atoms with the same sign formal charge Generally … try to reduce formal charges as much as possible when writing Lewis structures 9 Electronegativity • Another tool to predict molecular structure (via Lewis structures) • Measures strength of attraction of an atom – in a molecule – for the electrons in its bonds ♣ 10 Difference of electronegativities of bonded atoms determines the type of bonding ♣ egs. Cl−Cl pure covalent H−Cl polar covalent +∆E− = 3.0 – 2.1 = 0.9 Li Cl ionic ∆EN = 3.0 – 1.0 = 2.0 ♣ 11 Lewis Structures 1. Draw structure with only single bonds • H is always terminal – i.e. connected to only one other atom • the central atom is generally the least electronegative 2. Find the total # of valence electrons • i.e. add valence electrons of all the atoms … • … plus 1 for each net negative charge – polyatomic anions • … minus 1 for each net positive charge – polyatomic cations 3. Subtract # of electrons already used to form single bonds in current structure (2 electrons per bond) • use remaining electrons (in pairs) to form octets about each atom (except H) – start with the most electronegative atom • if any electrons remain, add them to the central atom 4. If any atom (other than those in group II or III) still has an incomplete octet, convert non-bonding pairs (i.e. lone pairs) to bonding pairs, creating double and/or triple bonds 5. Calculate formal charges • If central atom is from period 3 or higher, then the octet rule can be broken • Form additional multiple bonds to central atom in order to eliminate as many formal charges as possible 12 H2 4 − Step 1: O H H O P O Note - H s are terminal Step 2: O & P is the central atom # of valence electrons - lesthan Otronegative = 5 + 4 6 + 21 + 1 = 32 P has 52 H s - each with 1is a net −1 valence e salence electronrge on the group 5 H is group 1 anion valence electrons 6 O is group 6 Step 3: 6 2 = 12 electrons used to make the 6 single bonds 32 − 12 = 20 electrons remain O is the most ..ectronegative element −− H→−− O.. & ... O. . −O− O.. .. .. H H O.. P O.. we use 2• + 2 6 = 20 electrons, .. no electrons remain O.. 13 Step 5: Calculate formal charges ▯ H’s … FC = 1- 2/2 = 0 ▯ −− H O’s … FC −6− 4 4/2 = 0 ▯ −O (i.e. termina… FC = 6 −6 2/2 = 1 ▯ P … FC = 5 4•2/2 = +1 ....
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