Valence Shell lectron Pair Repulsion Theory (VSPR) Used to predict the 3-D shape of molecules and ions. Postulates: 1. Shape of molecule/ion is determined by the total number of regions of electron density about the central atom. Note: Lone pairs and bonding e- pairs each count as regions of e- density. 2. Regions of e- density repel each other and adopt an arrangement so as to be as far apart as possible. 3. There are five basic arrangements for the regions of e- density. Five Basic rrangements of Regions of lectron Density # Regions of e- Density 3-D rrangement 180 Possible Molecular Shapes 2 linear * linear 3 trigonal planar 120 * trigonal planar * bent 4 tetrahedral 109.5 * tetrahedral * trigonal pyramidal * bent 5 trigonal bipyramidal axial position 120 90 equatorial positions * trigonal bipyramidal * distorted tetrahedral (or seesaw shaped) * T-shaped * linear 6 octahedral 90 * octahedral * square pyramidal * square planar 1
Predicting Molecular Shape Molecular shape is the 3-D arrangement of atoms about the central atom. Lewis structure alone does not give the 3-D molecular shape of molecule/ion. VSPR theory must be used to predict the molecular shape. To predict the Molecular Shape: 1. Draw a correct Lewis structure. 2. Count up the total # regions of e- density on central atom. * Lone pairs and bonding e- pairs count as regions of e- density. * Double/triple bonds count as one region of e- density. 3. rrange the regions of e- density in one of five basic arrangements. 4. The molecular shape is the resulting arrangement of atoms in the molecule or ion. xamples: Predict the 3-D arrangement of the regions of e- density, molecular shape, and bond angles for each of the following. CO 2 BeCl 2 BF 3 PbCl 2 CH 4 H 3 O + H 2 O NH 2 - PCl 5 IF 4 + ClF 3 IF 2 - SF 6 TeF 5 - ICl 4 - XeF 4 CH 2 O H 2 2
Questions?. In what positions are one, two, and three lone pairs of electrons placed in trigonal bipyramidal arrangements? Why? B. In what positions are two lone pairs of electrons placed in octahedral arrangements? Why? C. The molecular shapes of CH 4, NH 3, and H 2 O are derived from a tetrahedral arrangement of four regions of e- density. xplain why the trend in bond angles is as follows: 109.5 (in CH 4 ), 107 (in NH 3 ), and 104.5 (in H 2 O). Polarity of Molecules Nonpolar Molecule: * centers of positive and negative charge coincide * vector sum of bond dipoles is zero * highly symmetric molecular shape xample: Consider BeCl 2 (electronegativities: Be=1.5, Cl=2.9) Polar Molecule: * centers of positive and negative charge do NOT coincide * vector sum of bond dipoles is nonzero * unsymmetric (asymmetric) molecular shape xample: Consider BeFCl (electronegativities: Be=1.5, Cl=2.9, F=4.1) 3
Predicting Molecular Polarity Steps: 1. Draw a correct Lewis structure. 2. Use VSPR theory to predict the molecular shape. 3. Predict polarity by looking at molecular symmetry or vector sum of bond dipoles. * nonpolar: highly symmetric or vector sum of bond dipoles=0 * polar: asymmetric or vector sum of bond dipoles 0 xamples: Decide whether each of the following molecules is polar or nonpolar. CH 4 CH 3 Cl CO 2 NH 3 SF 6 H 2 O SF 4 XeO 2 XeF 4 Valence Bond Theory VB Theory gives information on the nature/makeup of the chemical bond. Postulates: 1. Covalent bonds are formed by overlap of singly occupied (1 e-) valence atomic orbitals on adjacent atoms. 2. lectron spins are paired and the pair of electrons is shared in the overlap region. 3. The larger the orbital overlap, the stronger the covalent bond. xamples:. Use VB theory to describe the nature of the covalent bonds in H 2, F 2, and HF. B. If VB theory is used to describe the bonding in H 2 O are the correct bond angles found? The experimental bond angle in water is 104.5. 4
Hybridization Must be invoked to describe the nature/makeup of the covalent bond in molecules or ions with more than two atoms. Hybrid Orbitals: orbitals centered on the same atom interact (mathematically combine) during bond formation to form new orbitals called hybrid orbitals. xample:. For H 2 O, what type of hybrid orbitals does the oxygen atom use for bonding? B. List the five types of hybrid orbitals that can be used to form covalent bonds. Give information on.. atomic orbitals used to form each hybrid # of equivalent hybrids formed arrangement of hybrids angle between adjacent hybrids sp: tomic Orbitals # hybrids rrangement ngle sp 2 : sp 3 : sp 3 d: sp 3 d 2 : 5
Predicting Hybridization To predict hybridization of central atom.. 1. Draw a correct Lewis structure. 2. Count up total # regions of e- density. 3. # hybrids needed is equal to # regions of e- density. 4. Choose hybridization that gives only this # of hybrids. xamples: Predict what type of hybridization (sp, sp 2, etc) is used by the central atom during bond formation in each of the following. BCl 3 XeF 4 H 2 H 4 SF 4 PCl 3 Bond Properties Properties that depend on the # e- pairs shared between two bonded atoms.. Bond Order (BO): gives the number of e- pairs shared between two bonded atoms. B. Bond Length (BL): distance between nuclei of two bonded atoms. Depends on. BL = R 1 + R 2 atomic radii bond order xamples:. rrange the following molecules in order of decreasing bond length: HBr, HCl, HF, and HI. B. rrange the following bonds in order of decreasing length: C-C, C-O, C-F, C-N. C. rrange the following molecules in order of decreasing carbon-carbon bond length: H 6 H 2 H 4 6
C. Bond nergy (B): energy required to separate the bonded atoms to give neutral species ( needed to break the bond). lso called bond dissociation energy. x. H 2 2 H B = + 435 kj/mol Depends on.. 1. bond order x. B BO H 6 348 kj/mol 1 (carbon-carbon single bond) H 4 612 kj/mol 2 (carbon-carbon double bond) H 2 960 kj/mol 3 (carbon-carbon triple bond) Questions? Why doesn t it take twice as much energy to break a double bond as a single bond? and three times as much to break the triple bond as the single bond? Calculation of nthalpy Change from B Bond breaking is NDOTHRMIC. Bond making is XOTHRMIC. In chemical reactions..bonds are broken in reactants ( equal to B absorbed) and bonds are formed in products ( equal to B released). xample: Calculate the enthalpy change for the reaction, 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Given: H-H B = 436 kj/mol H-O B = 463 kj/mol O=O B = 502 kj/mol O-O B = 138 kj/mol 7
xamples: Calculate the enthalpy changes for the reactions in and B given the following bond energies. Bond B (kj/mol) Bond B (kj/mol) H-H 436 CΞC 960 H-O 463 C=O 743 C-H 412 O=O 502 C-C 348 O-O 138 C=C 612. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) B. 2 H 2 (g) + H 2 (g) CH 3 CH 3 (g) 8