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Unformatted text preview: should be:
Shortest → longest bond length: CO < CO+ < CO2+
46. There are 14 valence electrons in the MO electron configuration. Also, the valence shell is n
= 3 . Some possibilities from Row 3 having 14 valence electrons are Cl2, SCl−, S22−, and Ar22+. 48. O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2; O2 should have a lower ionization energy than O. The
electron removed from O2 is in a π2p* antibonding molecular orbital, which is higher in
energy than the 2p atomic orbitals from which the electron in atomic oxygen is removed.
Because the electron removed from O2 is higher in energy than the electron removed from O,
it should be easier to remove an electron from O2 than from O. 50. a. See the illustrations in the solution to Exercise 14.39 for the bonding and antibonding
MOs in OH.
b. The antibonding MO will have more hydrogen 1s character since the hydrogen 1s atomic
orbital is closer in energy to the antibonding MO.
c. No, the overall overlap is zero. The px orbital does not have proper symmetry to overlap
with a 1s orbital. The 2px and 2py orbitals are called nonbonding orbitals.
1s HO !# O "* ! ! 2px 2py
!# 2s ! 2pz 2px 2py !# 2s " !# !# CHAPTER 14 COVALENT BONDING: ORBITALS 585 50.
e. Bond order =
= 1. Note: The 2s, 2px, and 2py electrons have no effect on the bond
f. 52. To form OH+, a nonbonding electron is removed from OH. Because the number of
bonding electrons and antibonding electrons is unchanged, the bond order is still equal to
one. O3 and NO2−are isoelectronic, so we only need consider one of them since the same bonding
ideas apply to both. The Lewis structures for O3 are: O
O O O For each of the two resonance forms, the central O atom is sp2 hybridized with one
unhybridized p atomic orbital. The sp2 hybrid orbitals are used to form the two sigma bonds
to the central atom. The localized electron view of the π bond utilizes unhybridized p atomic
orbitals. The π bond resonates between the two positions in the Lewis structures: In the MO picture of the π bond, all thre...
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