Cl C Cl Cl Cl Note For identical terminal atoms all VSEPR formulas with m 0

Cl c cl cl cl note for identical terminal atoms all

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Cl C Cl Cl Cl Note: For identical terminal atoms, all VSEPR formulas with m = 0 yield nonpolar molecules. Assuming that the bonds are all polar , the formulas with m 0 all yield polar molecules, except for AX 2 E 3 and AX 4 E 2 , which are nonpolar . Example 3.3b Identify which, if any, of the following molecules are polar: (1) water, H 2 O ; (2) methane, CH 4 ; (3) iodine trichloride, ICl 3 . Answer (1) AX 2 E 2 , angular, polar; (2) AX 4 , tetrahedral, nonpolar; (3) AX 3 E 2 , T-shaped, polar . Note: The C–H bond is essentially nonpolar . Molecules containing only C and H that have nonsymmetrical arrangements of atoms are essentially nonpolar . 4
Molecular Shape and Structure Valence-Bond (VB) Theory (Sections 3.4–3.8) K ey Concepts valence-bond theory, localized electron model, σ -bond, π -bond, nodal surface, overlap of atomic orbitals, single bond, double bond, triple bond, promotion, hybridization of atomic orbitals, hybrid atomic orbitals, hybridization schemes and relation to geometry, s p , s p 2 , s p 3 , s p 3 d , s p 3 d 2 , hydrocarbons, alkanes, alkenes, alkynes, properties of double bonds 3.4 Sigma ( σ ) and Pi ( π ) Bonds E xample 3.4a Use the valence-bond theory to describe the bonding in (1) H 2 ; (2) HCl; (3) O 2 . Answer (1) In H 2 , each H atom has one unpaired electron in a 1 s -orbital. Two H atoms with opposite spins are brought together to form H 2 . The two 1 s -orbitals overlap to form one σ - bond (paired electrons). (2) In HCl, the H atom has one unpaired electron in a 1 s -orbital; the Cl atom has one unpaired electron in a 3 p -orbital. The two orbitals overlap end-to-end to form one σ - bond (paired electrons). (3) In O 2 , the O atom (1 s 2 2 s 2 2 p x 2 2 p y 1 2 p z 1 ) has two unpaired electrons. The half-occupied 2 p z -orbitals overlap end-to-end with each other to form one σ - bond . The half- occupied 2 p y -orbitals overlap side-by-side to form one π - bond . Thus, O 2 has a double bond consisting of one σ - bond and one π - bond (an electron pair in each bond). The two electrons in a bonding pair have opposite spins. Note: The z -axis is customarily chosen to be the bond axis. Note: The molecular orbital theory has an improved description of the bonding in the O 2 molecule, which is a biradical (see Sections 2.10 and 3.11). Example 3.4b Use the valence-bond theory to describe the bonding in (1) CO; (2) NO ; (3) Li 2 (gas-phase, high temperature molecule). Answer (1) triple bond: one σ - bond and two π - bonds [isoelectronic with N 2 ] (2) double bond: one σ - bond and one π - bond [isoelectronic with O 2 ] (3) single bond: one σ - bond [two 2 s -orbitals overlapping] 3.5 Hybridization of Orbitals Example 3.5a Use the valence-bond theory and hybridization to describe the bonding in BF 3 . Solution The ground-state electron configuration of B is [He]2 s 2 2 p x 1 . The molecule BF 3 has a trigonal-planar shape with three equal B–F single bonds (using the VSPER technique). To account for this with the VB theory, promote one 2 s -electron on B to a vacant 2 p - orbital {2 s 2 2 p x 1 2 s 1 2 p x 1 2 p y 1 }. Mix or hybridize the resulting half-filled valence orbitals (one 2 s - and two 2 p -orbitals) to get three s p 2 hybrid orbitals, each with one unpaired electron. Now overlap one

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