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.

The bonding atoms will seek the maximum separation 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 Molecules with: 2 pairs of bonding e- and “0” lone pairs Bond angle = 180o Geometry = Linear

Be --- 1s2 2s2 (Ground state configuration) Geometry = Linear It has 2 identical covalent bonds. BeCl2 Bond angles = 180o 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.

The bonding atoms will seek the maximum separation Examples: I – Molecules than contain no lone pairs of e- in the central atom: 2) AlCl3 12-3 = 9 lone pairs (3 pairs in each Cl) Cl – Al –Cl Cl Cl Al Cl Cl Bond angle = 120o The bonding atoms will seek the maximum separation Geometry = Trigonal planar Molecules with: 3 pairs of bonding e- and “0” lone pairs

Al --- 1s2 2s2 2p6 3s2 3p1 (Ground state configuration) Geometry = Trigonal planar It has 3 identical covalent bonds. AlCl3 Bond angles = 120o 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

The bonding atoms will seek the maximum separation 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 H H 109.5o Bond angle = 109.5o The bonding atoms will seek the maximum separation 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

The bonding atoms will seek the maximum separation 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 P H H H The lone pair occupies larger orbital, so it pushes the bonds down. The bonding atoms will seek the maximum separation Bond angle = 107.3o Geometry = Trigonal pyramidal Molecules with: 3 pairs of bonding e- and one lone pairs

P --- 1s2 2s2 2p6 3s2 3p13p1 3p1 (Ground state configuration) Geometry = Trigonal pyramidal PH3 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) (Trigonal pyramidal) P --- 1s2 2s2 2p6 3(sp3)2 3(sp3)1 3(sp3)1 3(sp3)1

The bonding atoms will seek the maximum separation 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 O H H The two lone pair occupy larger orbitals, so it pushes the bonds down. The bonding atoms will seek the maximum separation Bond angle = 104.5o Geometry = Bent Molecules with: 2 pairs of bonding e- and two lone pairs

O --- 1s2 2s2 2p2 2p1 2p1(Ground state configuration) Geometry = Bent H2O 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) (Bent) O --- 1s2 2(sp3)2 2(sp3)2 2(sp3)1 2(sp3)1

The bonding atoms will seek the maximum separation Examples: IV – Molecules than contain no lone pairs of e- in the central atom, which are exceptions of the Octet Rule: 1) NbBr5 20-5 = 15 lone pairs (three in each Br atom) Br Br Br Nb Br Br Bond angle = 90o /120o The bonding atoms will seek the maximum separation Geometry = Trigonal bipyramidal Molecules with: 5 pairs of bonding e- and “0” lone pairs

Nb --- [Kr] 5d3 4s2 (Ground state configuration) Geometry = Trigonal bipyramidal NbBr5 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

The bonding atoms will seek the maximum separation Examples: IV – Molecules than contain no lone pairs of e- in the central atom, which are exceptions of the Octet Rule: 2) SF6 F 24-6 = 18 lone pairs (three in each F atom) F F 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

S --- 1s2 2s2 2p6 3s2 3p4 (Ground state configuration) Geometry = Octahedral SF6 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: BeCl2 2 Linear 180o sp 3 AlCl3 Trigonal planar 120o # of shared pairs (single bonds) Lone pairs Geometry Bond Angle Hybridization Example: BeCl2 2 Linear 180o sp 3 AlCl3 Trigonal planar 120o sp2 109.5o sp3 CH4 4 Tetrahedral Trigonal pyramidal PH3 3 1 107.3o sp3 Bent 104.5o sp3 H2O 2 2 Trigonal bipyramidal 90o/120o sp3d NbBr5 5 6 SF6 Octahedral 90o sp3d2

valence e- = 8 e- bonding pairs = = 4 pairs 4-3 = 1 lone pairs Example: Determine the molecule geometry, bond angle & type of hybridization of PH3 valence e- = 5 + 3(1) = 8 e- 1) PH3 bonding pairs = = 4 pairs 𝟖 𝟐 H – P –H H 4-3 = 1 lone pairs (in P atom) P H H H 3 single covalent bonds. 1 lone pairs of e- in the central atom. Geometry = Trigonal pyramidal Bond angle = 107.3o Hybridization = sp3

Classification of covalent bonds: All single covalent bonds are very strong bonds, named sigma (σ). Any double covalent bond contains one sigma (σ) & one pi (π) bonds. Any triple covalent bond contains one sigma (σ) & two pi (π) bonds. Sigma (σ) covalent bonds are stronger than pi (π) 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