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Solutions to Problems 13 end-to-end (o) overlap of the Px orbitals is shown first, giving σ and (bonding and antibonding) molecular orbitals, followed by the two overlaps of the pairs of and P₂ orbitals, giving two sets of and π* molecular orbitals, respectively. Because σ overlap is generally better than TT overlap, the diagrams are shown here with a larger energy gap between the σ and orbitals than between the and π* orbitals-recall from Figure that the difference in energy between atomic and molecular orbitals is related to the strength of the bonding-the change in energy going from the atoms to the molecule. (More sophisticated forms of theoretical analysis reveal that the actual ordering of orbital energies is not quite the same as that shown here, but you don't need to worry about that.) σ* π* π* π* π* + + 2px 2p₂ 2p 2p π π σ For (c), O₂, 4 net bonding electrons O₂⁺, 5 net bonding electrons. So O₂⁺ has the stronger bond. For (d), N₂, 6 net bonding electrons 5 net bonding electrons. So N₂ is better. 35. Use valence shell electron pair repulsion (VSEPR) to predict geometry about any carbon or nitrogen atom. Count the number of other atoms attached to it and add to that the number of lone pair(s) it may contain. Two = linear and sp hybridized; 3 = trigonal planar and sp² hybridized; 4 = tetrahedral and sp³. Not complicated. (a) The four atoms connected to this carbon will be attached by four single bonds and will be arranged in roughly a tetrahedral geometry, which is explained by sp³ hybridization. It won't be exactly tetrahedral because the four atoms aren't identical (two hydrogens, a carbon, and a Br). (b) Don't worry about multiple bonds. The carbon in question is attached to three other atoms; therefore it is approximately trigonal planar with sp² hybridization. (c) This carbon is attached to three other atoms. So, as in (b), trigonal planar, sp² hybridization. (d) The nitrogen atom is attached to three other atoms and has one lone pair, so it is sp³ hybridized. We don't call it tetrahedral, however. When we choose a word to describe the geometry around an atom we usually consider only the atoms to which it's attached, not the lone pairs. So, the nitrogen in CH₃NH₂ is best described as having a pyramidal geometry, like the nitrogen in ammonia, NH₃. (e) This carbon atom sits between two other atoms. Again. VSEPR disregards multiple bonds. The geometry here will be linear, and the carbon sp hybridized. (f) Nitrogen, bonded to three other atoms, is trigonal planar and sp² hybridized. 36. (a) This carbon atom uses sp³ hybrid orbitals for its four sigma (σ) bonds. So does the other carbon atom in the molecule, so the bond between them is formed by overlap between two sp³ hybrid orbitals. The C-H bonds use overlap between the sp³ hybrid orbital on carbon and a hydrogen atomic orbital. The C-Br bond forms from overlap of the carbon sp³ hybrid and an atomic p orbital of bromine. (b) Two of the three sp² orbitals on the indicated carbon atom go to ordinary tetrahedral carbons. Those σ bonds therefore involve sp²-sp³ overlap. The oxygen atom is bonded to one other atom (our carbon) and has two lone pairs. + 2 = 3. It may be considered to be sp² hybridized, in which case the σ bond will be However, as discussed in Exercise 1-17. oxygen is energetically reluctant to undergo orbital hybridization. It is therefore more accurate to consider this σ bond to involve overlap. One of the remaining p orbitals on oxygen and the carbon orbital align in a parallel (side-by-side) manner, giving the second pi (π) bond between them.