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Molecular geometry – the orientation of atoms in space (how the atoms are arranged in a molecule) VSEPR Theory – Valence Shell Electron Pair Repulsion.

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Presentation on theme: "Molecular geometry – the orientation of atoms in space (how the atoms are arranged in a molecule) VSEPR Theory – Valence Shell Electron Pair Repulsion."— Presentation transcript:

1 molecular geometry – the orientation of atoms in space (how the atoms are arranged in a molecule) VSEPR Theory – Valence Shell Electron Pair Repulsion theory VSEPR is a simple, yet powerful technique to predict the molecular geometry (or shapes) of molecules e - pairs (bonding or nonbonding) repel each other. Thus, they attempt to get as far apart from each other as possible to maximize separation

2 # e - pairs around central element shape geometry name angles 2 pairslinear 180  3 pairs trigonal planar 120  4 pairs tetrahedral 

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4 electron pair geometry must be known before molecular geometry can be predicted To determine molecular geometry (MG) 1.draw the correct Lewis structure 2.determine # of electron pairs around the central element 3.determine how those electron pairs orient around the central element 4.attach terminal atoms to the central element 5.the orientation of the atoms in space determine the molecular geometry

5 determine the molecular geometry of BCl 3 MG = trigonal planar

6 determine the molecular geometry of BCl 2 - (anion) MG = bent bent = 3 atoms that are NOT linear

7 determine the molecular geometry of H 2 O MG = bent

8 determine the molecular geometry of NH 3 MG = pyramidal

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10 determine the molecular geometry of CH 4 MG = tetrahedral

11 tetrahedral pyramidalbent

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14 multiple bonds in VSEPR theory * treat a double or triple bond as if it were a “single bond” from a VSEPR standpoint determine the molecular geometry of CO 2 MG = linear

15 determine the molecular geometry of NO 2 - (anion) MG = bent

16 Cl 2 nonpolar bond – electrons are shared equally in the bond polar bond – electrons are NOT shared equally HCl

17 ++ -- dipole moment – quantitative extent to which polarity is measured Br-Cl has a polar bond

18 Is Br-Cl a polar molecule ? Consider the covalent bond as a rope with each atom “pulling electrons to itself” based on electronegativities of each atom  If the entire molecule moves during the “tug’O’war”….. the molecule is POLAR  If the entire molecule does NOT move during the “tug’O’war”….. the molecule is NONPOLAR Br-Cl is a polar molecule

19 Is CO 2 a polar or nonpolar molecule ? the individual dipoles cancel such that the overall dipole moment = 0 CO 2 is a nonpolar molecule

20 Is H 2 O a polar or nonpolar molecule ? MG = bent H 2 O is a polar molecule

21 Is NF 3 a polar or nonpolar molecule ? MG = pyramidal NF 3 is a polar molecule

22 Is BF 3 a polar or nonpolar molecule ? MG = trigonal planar the individual dipoles cancel such that the overall dipole moment = 0 BF 3 is a nonpolar molecule

23 Is CCl 4 a polar or nonpolar molecule ? the individual dipoles cancel such that the overall dipole moment = 0 CCl 4 is a nonpolar molecule MG = tetrahedral

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25 valence bond theory – describes how atomic (VB theory)orbitals form bonds

26 open valency – unpaired electron in a valence orbital available for bonding valence bond theory – describes how atomic (VB theory)orbitals form bonds

27 FF 2p

28 electron promotion – electron is removed from one orbital and placed in an orbital of higher energy hybridization – simple atomic orbitals on the central atom “mix” to form new “hybrid” orbitals hybrid – something of a mixed origin

29 the two new sp hybrid orbitals are 50% s-character and 50% p-character

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31 Remember, this process is occurring only on the central element of boron, B Determine the hybridization of boron, B in BF 3

32 the 3 new sp 2 hybrid orbitals are 33.3% s-character and 66.6% p-character

33 Remember, this process is occurring only on the central element of carbon, C Determine the hybridization of carbon, C in CH 4

34 the 4 new sp 3 hybrid orbitals are 25% s-character and 75% p-character

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36 octet expansion requires the central atom to have empty d-orbitals Determine the hybridization of P in PF 5 hybridization

37 sigma bond – electron overlap that forms all single bonds   pi bond – electron overlap that forms all double and triple bonds H one  bond and one  bond and two  bonds

38 three  bonds and two  bonds five  bonds and one  bond seven  bonds and one  bond

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