Molecular shape: VSEPR: Valence Shel Electron Pair Repulsion

Molecular shape: VSEPR: Valence Shel Electron Pair Repulsion

Molecular shape: VSEPR: Valence Shel Electron Pair Repulsion model. This model is based on an arrangement that minimizes the repulsion of shared & unshared pairs of e- around the central atom. This repulsion forces results in atoms existing at fixed angles to each other. Bond angle: Is the angle formed by any two terminal atoms & the central atom. Facts: - Shared e- pairs repel one another. - Lone pairs occupy a slightly larger orbital than shared e-. Examples: I Molecules than contain no lone pairs of e- in the central atom: 1) BeCl2 8-2 = 6 lone pairs (3 pairs in each Cl) Cl Be Cl The bonding atoms will seek the

maximum separation Bond angle = 180o Geometry = Linear Molecules with: 2 pairs of bonding eand 0 lone pairs BeCl2 Geometry = Linear Bond angles = 180o It has 2 identical covalent bonds. Be --- 1s2 2s2 (Ground state configuration) Be --- 1s2 2s1 2p1 (Excited atom configuration) Beryllium atom undergoes hybridization: 1 atomic orbital s 1 atomic orbitals p 2 hybrid orbitals (sp)

(Linear) Be --- 1s2 2(sp)1 2(sp)1 Hybridization: Hybrid orbitals: Result from combining two or more atomic orbitals. # of hybrid orbitals = Total number of atomic orbitals combined. Examples: 2) AlCl3 I Molecules than contain no lone pairs of e- in the central atom: 12-3 = 9 lone pairs (3 pairs in each Cl) Cl Al Cl Cl The bonding atoms will seek the maximum separation Cl

Al Cl Bond angle = 120o Cl Geometry = Trigonal planar Molecules with: 3 pairs of bonding e- and 0 lone pairs AlCl3 Geometry = Trigonal planar Bond angles = 120o It has 3 identical covalent bonds. Al --- 1s2 2s2 2p6 3s2 3p1 (Ground state configuration) Al --- 1s2 2s2 2p6 3s1 3p1 3p1 (Excited atom configuration) Aluminum atom undergoes hybridization: 1 atomic orbital s

2 atomic orbitals p 3 hybrid orbitals (sp2) (Trigonal planar) Al --- 1s2 2s2 2p6 3(sp2)1 3(sp2)1 3(sp2)1 Examples: I Molecules than contain no lone pairs of e- in the central atom: 3) CH4 4-4 = 0 lone pairs H H C H H C H The bonding atoms will seek the maximum

separation H H 109.5o H Bond angle = 109.5o Geometry = Tetrahedral Molecules with: 4 pairs of bonding e- and 0 lone pairs C --- 1s2 2s2 2p2 (Ground state configuration) C --- 1s2 2s1 2p1 2p1 2p1 (Excited atom configuration) It can form 4 covalent bonds; but not identical. Carbon atom undergoes hybridization: 1 atomic orbital s 3 atomic orbitals p 4 hybrid orbitals (sp3)

(Tetrahedral) C --- 1s2 2(sp3)1 2(sp3)1 2(sp3)1 2(sp3)1 Bond angle = 109.5o Examples: II Molecules than contain one lone pairs of e- in the central atom: 1) PH3 4-3 = 1 lone pairs (in P atom) H P H H The bonding atoms will seek the maximum separation The lone pair occupies larger orbital, so it pushes the bonds down. P

H H H Bond angle = 107.3o Geometry = Trigonal pyramidal Molecules with: 3 pairs of bonding e- and one lone pairs PH3 Geometry = Trigonal pyramidal Bond angles = 107.3o P --- 1s2 2s2 2p6 3s2 3p13p1 3p1 (Ground state configuration) Phosphorus atom undergoes hybridization: 1 atomic orbital s 3 atomic orbitals p 4 hybrid orbitals (sp3)

P --- 1s2 2s2 2p6 3(sp3)2 3(sp3)1 3(sp3)1 3(sp3)1 (Trigonal pyramidal) Examples: III Molecules than contain two lone pairs of e- in the central atom: 1) H2O 4-2 = 2 lone pairs (in O atom) H O H The bonding atoms will seek the maximum separation The two lone pair occupy larger orbitals, so it pushes the bonds down. O H H

Bond angle = 104.5o Geometry = Bent Molecules with: 2 pairs of bonding e- and two lone pairs H2O Geometry = Bent Bond angles = 104.5o O --- 1s2 2s2 2p2 2p1 2p1(Ground state configuration) Oxygen atom undergoes hybridization: 1 atomic orbital s 3 atomic orbitals p 4 hybrid orbitals (sp3) O --- 1s2 2(sp3)2 2(sp3)2 2(sp3)1 2(sp3)1 (Bent) Examples: 1) NbBr5

IV Molecules than contain no lone pairs of e- in the central atom, which are exceptions of the Octet Rule: 20-5 = 15 lone pairs (three in each Br atom) Br Br Br Nb Br Br The bonding atoms will seek the maximum separation Bond angle = 90o /120o Geometry = Trigonal bipyramidal Molecules with:

5 pairs of bonding e- and 0 lone pairs NbBr5 Geometry = Trigonal bipyramidal Bond angles = 90o/120o Nb --- [Kr] 5d3 4s2 (Ground state configuration) Nb --- [Kr] 5d1 4s1 4p1 4p1 4p1 (Excited atom configuration) Phosphorus atom undergoes hybridization: 1 atomic orbital s 3 atomic orbitals p 1 atomic orbital d 5 hybrid orbitals (sp3d) (Trigonal bipyramidal) Nb --- [Kr] 5(sp3d)1 5(sp3d)1 5(sp3d)1 5(sp3d)1 5(sp3d)1 IV Molecules than contain no lone pairs of e- in the central atom, which are exceptions of the Octet Rule:

Examples: 2) SF6 F F F 24-6 = 18 lone pairs (three in each F atom) S F F F The bonding atoms will seek the maximum separation Bond angle = 90o Geometry = Octahedral Molecules with: 6 pairs of bonding e- and 0 lone pairs

SF6 Geometry = Octahedral Bond angles = 90o S --- 1s2 2s2 2p6 3s2 3p4 (Ground state configuration) S --- 1s2 2s2 2p6 3s1 3p1 3p1 3p1 3d1 3d1 (Excited atom configuration) Sulfur atom undergoes hybridization: 1 atomic orbital s 3 atomic orbitals p 2 atomic orbitals d 6 hybrid orbitals (sp3d2) (Octahedral) S 1s2 2s2 2p6 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 SUMMARY: Example: # of shared pairs (single

bonds) Lone pairs Geometry Bond Angle Hybridization BeCl2 2 0 Linear 180o sp AlCl3 3

0 Trigonal planar 120o sp2 CH4 4 0 Tetrahedral 109.5o sp3 PH3 3

1 Trigonal pyramidal H2O 2 2 NbBr5 5 0 SF6 6 0 107.3o sp3

Bent 104.5o sp3 Trigonal bipyramidal 90o/120o sp3d 90o sp3d2 Octahedral Example: Determine the molecule geometry, bond angle & type of hybridization of PH 3 1) PH3 valence e- = 5 + 3(1) = 8 ebonding pairs =

H P H = 4 pairs 4-3 = 1 lone pairs (in P atom) H 3 single covalent bonds. 1 lone pairs of e- in the central atom. Geometry = Trigonal pyramidal Bond angle = 107.3o Hybridization = sp3 P H H H Classification of covalent bonds:

All single covalent bonds are very strong bonds, named sigma (). Any double covalent bond contains one sigma () & one pi () bonds. Any triple covalent bond contains one sigma () & two pi () bonds. Sigma () covalent bonds are stronger than pi () covalent bonds. Exercise: Determine the molecular geometry, bond angle, & type of hybridization for the following: 1) BF3 2) SCl2 3) CF4 4) NH4)+1 5) BeF2

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