Pharmaceutical Chemistry Lecture 2 PDF

PL1001 Pharmaceutical Chemistry ORBITAL HYBRIDISATION Lecture 2 Dr Alberto Di Salvo d orbitals hybrids sp3d hybridisation the octet rule works well when p orbitals are involved, but doesn’t with d orbitals let’s have a look at Phosphorus and try to understand how the orbit...

PL1001 Pharmaceutical Chemistry ORBITAL HYBRIDISATION Lecture 2 Dr Alberto Di Salvo d orbitals hybrids sp3d hybridisation the octet rule works well when p orbitals are involved, but doesn’t with d orbitals let’s have a look at Phosphorus and try to understand how the orbital hybridisation in PCl5 can be explained Adapted from www.ptable.com sp3d hybridisation – PCl5 P has 15 electrons [1s2 2s2 2p6 3s2 3p3], 5 in the outer shell [3s2 3p3] in PCl5, P forms 5 equivalent bonds with 5 Cl atoms 3d E 3p 3s Ground state 3d E 3p E sp3d 3s Electron promotion Hybridisation sp3d hybridisation – PCl5 The five sp3d hybrid orbitals are directed toward the corners of a trigonal bipyramid. Three orbitals are in the same plane at 120o to each other (equatorial P-Cl bonds). The remaining two are perpendicular (90 o) to this plane (axial P-Cl bonds). sp3d2 hybridisation – SF6 S has 16 electrons [1s2 2s2 2p6 3s2 3p4], 6 in the outer shell [3s2 3p4] sp3d2 hybridisation – SF6 3d E 3p 3s Ground state 3d E sp d 3 2 E 3p 3s Electron promotion Hybridisation sp3d2 hybridisation – SF6 The six sp3d2 orbitals are directed to the corners of an octahedron. The shape of SF6 is octahedral and a bond angle of 90o exists between hybrid orbitals. Orbital hybridisation of a Lewis Acid – Boron trifluoride B has 5 electrons [1s2 2s2 2p1], 3 in the outer shell [2s2 2p1] every F atom will have satisfied the octet rule regarding its outer shell B will be 2 electrons short (sextet) of the Neon configuration (hence electrophilic) 3 equivalent bonds require a trigonal planar bond geometry, with 120 o bond angles. Orbital hybridisation of a Lewis Acid – Boron trifluoride p orbitals in white sp2 orbitals in gray here the shape is a less accurate approximation than in the previous slide Orbital hybrids summary N hybrid Molecular orbitals geometry Hybrid Bond angles required 2 Linear sp 180o 3 Trigonal planar sp2 120o 4 Tetrahedral sp3 109.5o Trigonal 5 bipyramidal sp3d 90o and 120o 6 octahedral sp3d2 90o Worked example – chloroform (CHCl3) Predict the hybridisation, shape and bond angles of chloroform H x Cl x C x Cl x Cl Lewis structure the central atom (C) forms 4 equivalent bonds and obeys the octet rule therefore, C is sp3 hybridised this should remind you of C in CH4 tetrahedral geometry and bond angles of approximately 109o Worked examples Predict the hybridisation, shape and bond angles of carbon atoms indicated by arrows in the structures below. I II O III IV Remember that the structures above are equivalent to the ones below! I II H2 H C C H C O C C III IV Worked examples I H2 C I the above carbon atom forms 2 bonds with another 2 carbon atoms and 2 bonds with 2 protons (H). Hence 4 equivalent bonds, sp3 hybridised. H II C H C II the above structure has a C=C double bond (remember ethene?). The carbon atom on the left hand side forms 3 equivalent bonds plus a double bond, hence sp2 hybridised. How about the carbon atom on the right hand side? Worked examples C III III the above structure has a C≡C triple bond (remember ethyne?). The carbon in question forms 1 bond with the terminal carbon and 1 with the adjacent carbon on it’s right (equivalent), plus 2 more bonds with the terminal carbon, hence sp hybridised. O C IV IV the above structure has a C=O double bond (remember ethene?). The carbon atom in question forms 2 bond with adjacent carbons and one with the oxygen (equivalent), plus 1 more bond with the oxygen, hence sp2 hybridised. Worked examples – how many σ and π bonds? I H2 C I 4σ bonds (overlap of sp3 with sp3 and sp3 with s orbitals). H II C H C II left 3σ (overlap of sp2 with sp2, sp2 with sp3 and sp3 with s orbitals) and 1 π (overlap of pz with pz orbitals); right 4σ bonds Worked examples – how many σ and π bonds? C III III 2σ (overlap of sp with sp orbitals) and 2π (overlap of py with py and pz with pz orbitals) O C IV IV 3σ (overlap of sp2 with sp2 orbitals) and 1π (overlap of pz with pz orbitals) Worked examples 1. Choose the molecule that is incorrectly matched with the electronic geometry about the central atom. (a) CF4 - tetrahedral (b) BeBr2 - linear (c) H2O - tetrahedral (d) NH3 - tetrahedral (e) PF3 - octahedral  Worked examples 2. Which of the following pairs of molecules and their molecular geometries is WRONG? (a) NF3 - trigonal planar (b) H2O - bent (c) BF3 - trigonal planar (d) AsF5 - trigonal bipyramidal (e) SeF6 - octahedral 3. Which molecule has a linear arrangement of all component atoms? (a) CH4 (b) H2O (c) CO2 (d) NH3 (e) BF3 Worked examples 4. What kind of hybrid orbitals are utilised by the carbon atom in CF4 molecules? (a) sp (b) sp2 (c) sp3 (d) sp3d (e) sp3d2 5. A neutral molecule having the general formula AB3 has two lone pairs of electrons on A. What is the hybridisation of A? (a) sp (b) sp2 (c) sp3 (d) sp3d (e) sp3d2