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Chapter 10 Molecular Structure and Bonding Theories. VSEPR. Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures. VSEPR Rule 1 : A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs.
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VSEPR Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures. • VSEPR Rule 1: A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs. • Minimum repulsion results when the electron pairs are as far apart as possible.
Steric Number • Steric number = (number of lone pairs on central atom) + (number of atoms bonded to central atom) • The steric number is determined from the Lewis structure. • Steric number determines the bonded-atom lone-pair arrangement, the shape that maximizes the distances between the valence-shell electron pairs.
Steric Number = 2 • In the Lewis structure of BeCl2, beryllium has two bonded atoms and no lone pairs, steric number = 2. • A linear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible.
Molecules with Multiple Bonds • The Lewis structure of HCN (H-Cº N:) shows that the carbon atom is bonded to two atoms and has no lone pairs, steric number = 2. • The bonded-atom lone-pair arrangement is linear. • The number of bonded atoms, not the number of bonds, determines the steric number.
Steric Number = 3 • The Lewis structure of BF3 shows the boron atom has a steric number = 3; the bonded-atom lone-pair arrangement is trigonal planar.
Steric Number = 4 • The Lewis structure of CH4 shows the carbon atom has a steric number = 4; the bonded-atom lone-pair arrangement is tetrahedral.
Steric Number = 5 • The phosphorus atom in PF5 has a steric number = 5; the bonded-atom lone-pair arrangement is trigonal bipyramidal.
Steric Number = 6 • The sulfur atom in SF6 has a steric number = 6; the bonded-atom lone-pair arrangement is octahedral.
O H H Central Atoms with Lone Pairs • The Lewis structure of H2O is • Steric number = 4, 2 bonded atoms and 2 lone pairs. • The bonded-atom lone-pair arrangement is tetrahedral.
Molecular Shape of H2O • Molecular shapeis the arrangement of the atoms in a species. • The bonded-atom lone-pair arrangement of H2O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom).
Molecular Shape of NH3 • What is the electron pair geometry and molecular shape of NH3?
Electron Pair Repulsions • The measured bond angle in H2O (104.5o) is smaller than the predicted angle (109.5o) • Explanation: repulsions vary lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair
Location of Lone Pair in SF4 • The favored structure for a trigonal bipyramid minimizes 90o lone pair interactions – the one on the right. Two structures are possible:
Lone Pairs in Trigonal Bipyramids • Lone pairs always occupy the equatorial positions in a trigonal bipyramid so that lone pair-lone pair repulsions are oriented at 120o.
Location of Lone Pairs in XeF4 • The structure on right has no 90o lone pair-lone pair interactions and is favored.
Test Your Skill • What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF3?
Multiple Central Atoms • The geometry of each central atom is determined separately. • The CH3 carbon in CH3CN has tetrahedral geometry and the other carbon has linear geometry.
Shapes of Molecules H H N S H • What are the bonded-atom lone-pairarrangements and the shapes about each central atom in NH2SH? • Draw the Lewis structure. • The bonded-atom lone-pairarrangements of both are tetrahedral, the nitrogen shape is trigonal pyramidal and sulfur is “V” shaped.
Overall Shape of C2H4 • Ethylene, C2H4 , could be planar (left) or nonplanar (right). The VSEPR model does not predict which is preferred.
Polarity of Molecules • The bond dipoles in CO2 cancel because the linear shape orients the equal magnitude bond dipoles in exactly opposite directions.
Polarity of Molecules • The bond dipoles do not cancel in COSe; they are oriented in the same direction and are of unequal length. They do not cancel in OF2 because the V-shape of the molecule does not orient them in opposite directions.
Polarity of Molecules • The bond dipoles in BCl3 and CCl4 cancel because of the regular shape and equal magnitude.
Polarity of Molecules • The bond dipoles in BCl2F and CHCl3 do not cancel because they are not of the same magnitude.
Test Your Skill • Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?
Valence Bond Theory • Valence bond theory describes bonds as being formed by overlap of partially filled valence orbitals.
Test Your Skill • Identify the orbitals that form the bond in HCl.
Bonding in NH3 • The observed bond angles of 107.5o in NH3 are not consistent with the angles of90o expected if the bonds formed from N 2p orbitals.
Hybrid Orbitals • Hybrid orbitalsare orbitals obtained by mixing two or more atomic orbitals on the same central atom. • Appropriate hybrid orbitals formed by mixing one s and xp atomicorbitals make bonds at either 180o (x = 1), 120o (x = 2), or 109.5o (x = 3).
Shape of Hybrid Orbitals • For clarity, hybrid orbitals are pictured as elongated with the small lobe omitted.
Bonding in BeCl2 • The bonds in BeCl2 arise from the overlap of two sp hybrid orbitals on the beryllium atom with the 3p orbitals on the two chlorine atoms.
Bonding in BF3 • The bonds in BF3 arise from the overlap of three sp2 hybrid orbitals on the boron atom with 2p orbitals on the three fluorine atoms.
Bonding in CH4 • The bonds in CH4 arise from the overlap of four sp3 hybrid orbitals on the carbon atom with 1s orbitals on the four hydrogen atoms.
Lone Pairs and Hybrid Orbitals • Hybrid orbitals can hold lone pairs as well as make bonds.
Hybridization with d Orbitals • Hybrid orbitals of central atoms with steric numbers of 5 or 6 involve d orbitals.
Test Your Skill • Identify the hybrid orbitals on the central atoms in SiH4 and HCN.
Types of Bonds: Sigma Bonds • Sigma bonds (s): the shared pair of electrons is symmetric about the line joining the two nuclei of the bonded atoms.
Bonding in C2H4 • The C-C sigma bond in C2H4 arises from overlap of sp2 hybrid orbitals and the four C-H sigma bonds from overlap sp2 hybrid orbitals on C with 1s orbitals on H. • The second C-C bond forms from sideways overlap of p orbitals.
Types of Bonds: Pi Bonds • Pi bonds (p) places electron density above and below the line joining the bonded atoms – they form by sideways overlap of p orbitals.
Bonding in C2H4 • The double bond in C2H4 is one sigma bond and one pi bond – each bond is of similar strength.
Proof of Pi Bonds: Shape of C2H4 • C2H4 isplanar(A) because pi overlap is at a maximum. Rotation of one end by 90o (B) reduces pi overlap to zero.
Triple Bonds • The triple bond in C2H2 is one sigma bond and two pi bonds between the sp hybridized carbon atoms.
Sigma Bonds in Benzene • Each carbon atom in benzene, C6H6, forms three sigma bonds with sp2 hybrid orbitals.