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Unformatted text preview: 9 sp2 hybridized C and N atoms (9 from C’s and 0 from N’s)
7 sp3 hybridized C and O atoms (5 from C’s and 2 from O’s)
39 σ bonds and 6 π bonds (this includes the 3 π bonds in the benzene ring that are
delocalized) CHAP. 14
594 COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS O=N‒Cl: The bond order of the NO bond in NOCl is 2 (a double bond). NO: 74. From molecular orbital theory, the bond order of this NO bond is 2.5 (see Figure
14.43 of the text). Both reactions apparently only involve the breaking of the N‒Cl bond. However, in the
reaction ONCl → NO + Cl, some energy is released in forming the stronger NO bond,
lowering the value of ΔH. Therefore, the apparent N‒Cl bond energy is artificially low for
this reaction. The first reaction only involves the breaking of the N‒Cl bond. Challenge Problems
76. The ground state MO electron configuration for He2 is (σ1s)2(σ1s*)2 giving a bond order of 0.
Therefore, He2 molecules are not predicted to be stable (and are not stable) in the lowestenergy ground state. However, in a high-energy environment, electron(s) from the antibonding orbitals in He2 can be promoted into higher-energy bonding orbitals, thus giving a
nonzero bond order and a “reason” to form. For example, a possible excited-state MO
electron configuration for He2 would be (σ1s)2(σ1s*)1(σ2s)1, giving a bond order of (3 – 1)/2 =
1. Thus excited He2 molecules can form, but they spontaneously break apart as the
electron(s) fall back to the ground state, where the bond order equals zero. 78. The electron configurations are:
N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2 N22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 Note: The ordering of the σ2p and π2p orbitals
is not important to this question. N2−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1
The species with the smallest ionization energy has the electron that is easiest to remove.
From the MO electron configurations, O2, N22−, N2−, and O2+ all contain electrons in the same
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