   Chapter 13, Problem 13.72E

Chapter
Section
Textbook Problem

# Referring to exercise 13.71: How would the symmetry-adapted linear combinations for the molecular orbitals of H 2 S differ if the core atomic orbitals of S were included?

Interpretation Introduction

Interpretation:

The effect of the involvement of orbitals for sulfur on the symmetry-adapted linear combination of molecular orbitals for hydrogen sulfide is to be stated.

Concept introduction:

Symmetry-adapted linear combination is referred to as the wavefunction of the symmetry equivalent orbitals. The symmetry-adapted linear combination is used to determine the binding schemes and symmetries of the molecule. The SALC’s are determined with the help of character tables and great orthogonality theorem.

Explanation

The structure of H2S is shown below.

Figure 1

The geometry of hydrogen selenide is bent due to presence of two lone pair of electrons. The 180° out of plane rotation of the given molecule produces same structure. Hence, it contains C2 symmetry operation. It contains vertical plane of symmetry, σv and σv' in two different planes. Therefore, the point group of the given molecule is C2v.

The atomic orbitals of H which participates in bonding are 1sH1 and 1sH2. The atomic orbitals of oxygen atom which involves in the bond formation are 1sS, 2sS, 2px,S, 2py,S, 2pz,S, 3sS, 3px,S, 3py,S and 3py,S. The symmetry operation E has no effect on the orbitals. The symmetry has no effect on the hydrogen 1s orbitals but can reverse 2py,S and 3py,S orbitals. The symmetry labels are multiplied with the character for the symmetry operation of that irreducible representation. The table obtained for A1 symmetry is shown below.

1sH11sH21sS2sS2px,S2py,S2pz,S3sS3px,S3py,S3pz,SE1sH11sH21sS2sS2px,S2py,S2pz,S3sS3px,S3py,S3pz,SC21sH21sH11sS2sS12px,S12py,S2pz,S3sS13px,S13py,S3pz,Sσv1sH11sH21sS2sS2px,S12py,S2pz,S3sS3px,S13py,S3pz,Sσv'1sH21sH11sS2sS12px,S2py,S2pz,S3sS13px,S3py,S3pz,S

The summation of each column in the table gives the eleven linear combinations as shown below.

ΨA1=14(1sH1+1sH2+1sH1+1sH2)=12(1sH1+1sH2)ΨA1=14(1sH2+1sH1+1sH2+1sH1)=12(1sH1+1sH2)ΨA1=14(1sS+1sS+1sS+1sS)=1sSΨA1=14(2sS+2sS+2sS+2sS)=2sS

ΨA1=14(2px,S2px,S+2px,S2px,S)=0ΨA1=14(2py,S2py,S2py,S+2py,S)=0ΨA1=14(2pz,S+2pz,S+2pz,S+2pz,S)=2pz,SΨA1=14(3sS+3sS+3sS+3sS)=3sS

ΨA1=14(3px,S3px,S+3px,S3px,S)=0ΨA1=14(3py,S3py,S3py,S+3py,S)=0ΨA1=14(3pz,S+3pz,S+3pz,S+3pz,S)=3pz,S

Similarly the table obtained for A2 symmetry is shown below.

1sH11sH21sS2sS2px,S2py,S2pz,S3sS3px,S3py,S3pz,SE1sH11sH21sS2sS2px,S2py,S2pz,S3sS3px,S3py,S3pz,SC21sH21sH11sS2sS12px,S12py,S2pz,S3sS13px,S13py,S3pz,Sσv1sH11sH21sS2sS2px,S12py,S2pz,S3sS3px,S13py,S3pz,Sσv'1sH21sH11sS2sS12px,S2py,S2pz,S3sS13px,S3py,S3pz,S

The summation of each column in the table gives the eleven linear combinations as shown below.

ΨA2=14(1sH1+1sH21sH11sH2)=0ΨA2=14(1sH2+1sH11sH21sH1)=0ΨA2=14(1sS+1sS1sS1sS)=0ΨA2=14(2sS+2sS2sS2sS)=0

ΨA2=14(2px,S2px,S2px,S+2px,S)=0ΨA2=14(2py,S2py,S+2py,S2py,S)=0ΨA2=14(2pz,S+2pz,S2pz,S2pz,S)=0ΨA2=14(3sS+3sS3sS3sS)=0

ΨA2=14(3px,S3px,S3px,S+3px,S)=0ΨA2=14(3py,S3py,S+3py,S3py,S)=0ΨA2=14(3pz,S+3pz,S3pz,S3pz,S)=0

Similarly the table obtained for B1 symmetry is shown below

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