Physical Chemistry Third Edition

880 21 The Electronic Structure of Polyatomic Molecules

2 pxand 2pyorbitals, exactly as with the N 2 molecule. On each carbon atom, this leaves a

2 sphybrid orbital with an orbital region that is directed along thezaxis away from the C–C

bond. We place the hydrogen atoms on thezaxis, achieving maximum overlap. From the

2 sphybrid on carbon A and a hydrogen 1sorbital we form a carbon–hydrogenσbonding

orbital that we callσA. From the 2sphybrid on carbon B and a hydrogen 1sorbital we form

a carbon–hydrogenσbonding orbital that we callσB. The electron configuration is now

(1sA)^2 (1sB)^2 (1σ)^2 (πu 2 px)^2 (πu 2 py)^2 (σA)^2 (σB)^2

where we place four electrons in the nonbonding inner-shell carbon 1sorbitals instead of

using theσg 1 sandσ∗u 1 sLCAOMOs. The triple bond consists of a sigma bond made from

two hybridsporbitals and two pi bonds that are made from unhybridizedporbitals.

Exercise 21.7

Describe the bonding in N 2 using LCAOMOs similar to those used in the description of acetylene.

Why is this description superior to the previous description of N 2 in Chapter 20?

The energy-localized orbitals for double and triple bonds do not correspond to a

sigma bond plus pi bonds. The three energy-localized orbitals in the triple bond in acety-

lene (ethyne) have the same energy and have the same shape, with the orbital regions

lying 120◦from each other around the bond axis. These energy-localized orbitals are

sometimes called “banana orbitals” because of their shape.^3 These orbitals are superior

to our description of the triple bond as a sigma and two pi bonds, since they allow

the electrons to be as far from each other as possible. However, our goal is simply to

provide a way to describe chemical bonding without calculation, and our method is

adequate for that purpose.

Fourth- and Fifth-Order Bonds

Elements in the second row of the period table do not form higher-order bonds than

triple bonds, but elements in the higher rows can form higher-order bonds, which can

be represented with linear combinations usingdorbitals as well assandporbitals.

A fifth-order bond has been reported between two chromium(I) ions.^4

PROBLEMS

Section 21.4: Molecules with Multiple Bonds

21.15Write an approximate LCAOMO wave function without
antisymmetrization for the formaldehyde molecule, using
no more than two atomic orbitals in an LCAOMO. Place
the C–O bond on thezaxis and the H atoms in theyz
plane. Use hybrid orbitals where appropriate. Specify
which atomic orbitals make up each LCAOMO and

identify each with an index and the designationσorπ.

Give the shape of the molecule.

21.16Using LCAOMOs made with hybrid orbitals, describe the

bonding and molecular shape of each of the molecules:

a.H 2 CCCH 2

b.NO 2

(^3) Levine,op. cit., p. 527 (note 2).
(^4) T. Nguyen,Science, 310 , 844 (2005).