University:
Amherst College
Course:
CHEM 371 | Inorganic Chemistry
Academic year:

2019
Views:

203

Pages:

8
Author:

makket213

Polarity, Formal Charge, and Resonance Flash Review CHEM 371 Polarity H H δ+ δ- δ+ δ- δ+ δ- H S H O H F Covalent Bond Polar Covalent Bond Increasing Electronegativity Difference Tetrahedral Trigonal Pyramid Bent Methane Ammonia Water Non-Polar Polar Polar Non-Polar Polar Bond dipoles are equivalent and cancel out Bond dipoles are inequivalent and do not cancel out 2 Linear Linear Trigonal Planar Trigonal Planar 3 4 Tetrahedral Tetrahedral Trigonal Bipyramidal Trigonal Bipyramidal Octahedral Octahedral 5 6 Polarity Number of Lone Pairs 0 Non-Polar 2 Polar Linear 1 Trigonal Planar Bent 2 Tetrahedral Trigonal Pyramid Bent 3 Trigonal Bipyramidal Seesaw T-Shaped Linear 4 Octahedral Square Pyramidal Square Planar T-Shaped Linear Steric Number 3 4 5 6 Calculating Formal Charge # of electrons in free atom Formal Charge = # of electrons # of electrons = assigned in molecule from lone pairs – # of electrons assigned in molecule + ½ # of electrons shared in bonds F.C. (O) = 6 – (6 + ½ (2)) F.C. (O) = -1 2 electrons shared in bonds 6 electrons from lone pairs F.C. (S) = 6 – (0 + ½ (8)) F.C. (S) = +2 8 electrons shared in bonds 0 electrons from lone pairs -1 -1 +2 -1 Sum of formal charges agrees with overall charge Large Absolute Formal Charge -1 Calculating Formal Charge # of electrons in free atom Formal Charge = # of electrons # of electrons = assigned in molecule from lone pairs – # of electrons assigned in molecule + ½ # of electrons shared in bonds 4 electrons from lone pairs 2 electrons shared in bonds F.C. (O1) = 6 – (6 + ½ (2)) F.C. (O1) = -1 F.C. (O2) = 6 – (4 + ½ (4)) 1 4 electrons shared in bonds F.C. (O2) = 0 2 4 electrons from lone pairs F.C. (S) = 6 – (0 + ½ (12)) F.C. (S) = 0 12 electrons shared in bonds 0 electrons from lone pairs 0 -1 0 -1 Sum of formal charges agrees with overall charge Reduced Formal Charge 0 Examples of Formal Charge N = 5 – (2 + ½(6)) N=0 N = 5 – (2 + ½(6)) N=0 C = 4 – (2 + ½(6)) N = -1 N = 5 – (2 + ½(6)) N=0 C = 4 – (2 + ½(6)) N = -1 O = 6 – (2 + ½(6)) N = +1 O = 6 – (2 + ½(6)) N = +1 0 N N No Net Dipole C 0 -1 -1 N C 0 Dipole Towards C O 0 N +1 Dipole Towards C O +1 Dipole Towards N Resonance Resonance Hybrids result from two or more Resonance Contributors. Please note, electrons are not shifting back and forth, this is importantly not an interconverting equilibrium. 2 Bonds 1 Bond 1 1/2 Bonds - -½ -½ Resonance Hybrid Resonance Contributors Nitrite Ion - 1 Bond 2 Bonds - 1 1/3 Bonds -2/3 - -2/3 - - Carbonate Ion -2/3 - 2 Bonds 1 1/2 Bonds 1 Bond Benzene

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