Bonding in Polyatomic Polyatomic Molecules Molecules Bonding in - PowerPoint PPT Presentation

Bonding in Polyatomic Polyatomic Molecules Molecules Bonding in Basically two ways to approach polyatomics polyatomics. . Basically two ways to approach - are not confined to a First is to use delocalized delocalized M.O.’s M.O.’s where where e are not confined to a First is to use e - single bond (region between 2 atoms) but can wander over 3 or single bond (region between 2 atoms) but can wander over 3 or more atoms. We will use this approach later for C bonding. more atoms. We will use this approach later for C bonding. Second is to use hybridization of atomic hybridization of atomic orbitals orbitals and then use these and then use these Second is to use to form localized (usually) bonds. to form localized (usually) bonds. SAME nucleus to form on the SAME Hybridization combines orbitals orbitals on the nucleus to form Hybridization combines new orbitals orbitals called hybrids. Hybrids have characteristics of both called hybrids. Hybrids have characteristics of both new the atomic orbitals orbitals from which they are formed. from which they are formed. the atomic Example: sp sp hybrid is formed by combining (adding) a 2s and a 2p hybrid is formed by combining (adding) a 2s and a 2p Example: wave function or orbital on a single atom. wave function or orbital on a single atom.

Localized BeH 2 orbitals: : Localized BeH 2 orbitals Begin by hybridizing hybridizing Be 2s, 2p Be 2s, 2p orbitals orbitals to give two to give two sp sp hybrid hybrid Begin by orbitals -- -- each pointing in opposite direction: each pointing in opposite direction: orbitals + = 2s+2p - = 2s ≡ ≡ sp and sp ≡ sp ≡ = 2s+2p ≡ ≡ ≡ ≡ a and 2p ≡ ≡ ≡ ≡ sp + ≡ ≡ sp a sp - = 2s- -2p ≡ ≡ sp b sp b ≡ sp ≡ + ≡ ≡ ≡ ≡ sp + ≡ ≡ sp a sp a Add sp a Add sp a + 1s + 1s } . . + + . . to get localized to get localized – – + + + + Be- -H bond H bond Be H H H H Be Be Add sp Add sp b b + 1s + 1s . + + . . . } + – – + + + to get localized to get localized Be- -H bond H bond Be - ≡ ≡ sp ≡ ≡ ≡ ≡ sp - ≡ ≡ sp b sp b

sp hybrids sp hybrids on Be give localized bonds and a linear structure on Be give localized bonds and a linear structure - ≡ + ≡ ≡ ≡ sp ≡ ≡ ≡ ≡ sp ≡ ≡ H1s + (sp sp - ≡ ≡ ≡ sp b ) and H1s + (sp sp + ≡ ≡ ≡ sp a ) H1s + ( b ) and H1s + ( a ) . . . . + . . . . + + + – – – – H Be Be H H Be Be H

BH 3 BH 3 Fragment: Fragment: 2 hybrid Can explain by sp Can explain by sp 2 hybrid which provides which provides 3 3 localized bonds. localized bonds. Note: (B 1s 2 Note: (B 1s 2 2s 2s 2 2 2p 2p - - 3 valence 3 valence e - ) ) e - 120˚ 120˚ sp 2 sp 2 hybrid hybrid Third sp 2 Third sp 2 B sp B sp 2 2 : Combine : Combine points along points along 2s + 2p x + 2p y 2s + 2p x + 2p y this direction this direction 2 points Second sp 2 Second sp points + + along this direction along this direction • - - Boron Nucleus Boron Nucleus

Three of these in a plane Three of these in a plane pointing at 120˚ with respect pointing at 120˚ with respect to each other to each other 3 sp 3 sp 2 2 120˚ 120˚ hybrids hybrids

Overlap with H 1s to give 3 B- Overlap with H 1s to give 3 B -H bonds H bonds in a plane in a plane pointing pointing at 120º 120º with respect to each other: BH with respect to each other: BH 3 at 3 120˚ Predicts planar Predicts planar trigonal structure structure trigonal +H +H + + with equal with equal H H 120º H- -B B- -H angles. H angles. 120º H 2 hybrids + + + + (sp sp 2 hybrids) ) ( (1/3 s, 2/3 p) (1/3 s, 2/3 p) B B Note B has 3 Note B has 3 valence electrons valence electrons + + 1s 2 1s 2 2s 2s 2 2 2p 2p Put 1 valence e - from +H + H B and 1 e - from H in each of 3 bonds.

CH 4 CH 4 Structure. If all C Structure. If all C- -H bonds same, need H bonds same, need hybridization hybridization that that 3 does the job. provides 4 equal localized bonds. sp sp 3 does the job. provides 4 equal localized bonds. 3 gives ≡ sp ≡ ≡ ≡ z ≡ ≡ ≡ ≡ 2s + 2p x + 2p y + 2p z sp 3 gives 4 hybrid 4 hybrid orbitals orbitals which point to which point to 2s + 2p x + 2p y + 2p 3 [1/4 s, the corners of a tetrahedron. Angle between is 109 109º º28 28’ ’ sp sp 3 [1/4 s, the corners of a tetrahedron. Angle between is 3/4 p]. Tetrahedral hybrids. 3/4 p]. Tetrahedral hybrids. 2 → → → 4 valence → → Since C has Z = 6, 1s 2 2s 2 2p 2 → → → 4 valence e Since C has Z = 6, 1s 2 2s 2 2p e - - → → 4 valence → → 4 H atoms have 4 x 1s → → → → 4 valence e 4 H atoms have 4 x 1s - e - - go into 4 bonds constructed of H 1s and each sp 3 gives 8 valence e - go into 4 bonds constructed of H 1s and each sp 3 gives 8 valence e 3 + H (1s). 4 equivalent localized electron pair bonds sp 3 + H (1s). 4 equivalent localized electron pair bonds sp

Experimentally, find CH 4 has 4 bonds 4 bonds of equal length (1.043Å, of equal length (1.043Å, Experimentally, find CH 4 has arranged at corners of tetrahedron (109º28’ apart). arranged at corners of tetrahedron (109º28’ apart). Isolated Carbon atom Isolated Carbon atom C atom in CH 4 C atom in CH 4 ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ 2p 2p 2s 2 2s 2 E E sp 3 sp 3 sp 3 sp 3 sp 3 sp 3 sp 3 sp 3 1s 2 1s 2 1s 2 1s 2        

Geometry of carbon sp 3 /H 1s Bonds in methane (CH 4 ): Geometry of carbon sp 3 /H 1s Bonds in methane (CH 4 ): 3 hybridization on C leads to 4 bonds. CH sp 3 hybridization on C leads to 4 bonds. CH 4 is a sp 4 is a good example. good example. 4 C sp 3 /H 1s All H- -C C- -H H All H H H bonds angles are angles are ′ ′ ′ ′ 109º 28 ′ ′ ′ ′ 109º 28 C C H H H H H H

Summary of Hybridization Results Summary of Hybridization Results Example Example Groups Attached Groups Attached Hybrid Hybrid Geometry Geometry to Center Atom to Center Atom BeH 2 BeH 2 2 sp sp linear H- linear H -Be Be- -H H 2 BH 3 3 sp 2 2 trig. plane (120º) BH 3 sp trig. plane (120º) 3 CH 4 4 sp 3 tetrahedral CH 4 sp 3 tetrahedral 4 (109º28’ (109º28’ H- -C C- -H angles) H angles) H

Localized Bonds and Lone Pair Electrons Localized Bonds and Lone Pair Electrons Lone pair electrons are essentially valence electrons which Lone pair electrons are essentially valence electrons which do not become involved in bonding. These are handled These are handled do not become involved in bonding. beautifully by hybrid orbital picture. beautifully by hybrid orbital picture. NH 3 3H, 1s (no choice) N: 1s 2 2s 2 2p 3 Nitrogen Nitrogen Could make 3 Could make 3 bonds bonds 1s nucleus nucleus From 3, 2p orbitals orbitals From 3, 2p and 3, 1s H orbital: and 3, 1s H orbital: 2p z 2p y 1s This predicts 2p x 90º geometry H atom 1s 2 e - remaining 1s orbital in 2s 2 of N.

Geometry of NH 3 Geometry of NH 3 found to be found to be Trigonal Trigonal Pyramidal Pyramidal N H H H 107˚ 3 H project down. 107º angle is suspiciously close to 109º28” 107º angle is suspiciously close to 109º28” 3 H project down. 3 hybridization! predicted by sp 3 hybridization! predicted by sp 3 combinations. Try hybridizing sp sp orbitals orbitals on N to form on N to form 4 sp 4 sp 3 combinations. Try hybridizing - ! Remember, must account for 5 5 N valence N valence e ! Remember, must account for e -

3 orbitals 4N sp 3 orbitals combine with 3 1s H combine with 3 1s H orbitals orbitals to give 3 sp to give 3 sp 3 , 1s M.O.’s 4N sp 3 , 1s M.O.’s 3 hybrid left leaving one sp 3 hybrid left leaving one sp - from Put 1 e from N N into each M.O. 3 into each M.O. 3 e - Put 1 e - e - - from each H into M.O.’s 3 Put 1 e from each H into M.O.’s 3 e - Put 1 e - e - - from 3 orbital. Put 2 e from N N into free sp into free sp 3 orbital. Put 2 e - - in N, 2 go into one sp 3 orbital, 3 go into other 3 sp Of 5 valence e in N, 2 go into one sp 3 orbital, 3 go into other 3 sp 3 ,s. Of 5 valence e - 3 ,s. (combined with H (1s)) (combined with H (1s)) One of the driving forces for the tetrahedral configuration is that hat One of the driving forces for the tetrahedral configuration is t it puts bonding and lone pair electron groups as far away from it puts bonding and lone pair electron groups as far away from each other as possible. each other as possible. repulsions . This minimizes repulsions . This minimizes

Geometry of NH 3 : Trigonal Pyramidal • • • • Lone Pair Lone Pair N H H H Essentially a tetrahedral arrangement of bonds with lone pair taking the 4th position of the tetrahedron.

Approximately Approximately Tetrahedral Tetrahedral N N ~ 109° 28’ ~ 109° 28’ Electron pair repulsion effect is largest for small central atoms Electron pair repulsion effect is largest for small central atom s like B, N, O. like B, N, O. As go to larger central atoms (e.g. S or metals) frequently As go to larger central atoms (e.g. S or metals) frequently find this effect not so large and start to get things closer find this effect not so large and start to get things closer to pure p orbital bonds (90º structures) to pure p orbital bonds (90º structures) For example, H 2 S has H- -S S- -H angle of 92º. H angle of 92º. For example, H 2 S has H