In the photoelectron spectrum of carbon disulfide, CS₂, ionization from the highest occupied, 1 $\pi$_g, molecular orbital results in little vibrational structure. In contrast, ionization from the next lowest occupied 1 $\pi$_u molecular orbital results in a long vibrational progression in the $\pi$₁ symmetric stretching mode of the CS₂⁺ ion. Which of the following descriptions of the bonding in neutral CS₂ best explains these observations?

Question

Quizplus · Accepted Answer

Ionization from a bonding molecular orbital (MO) typically results in a vibrational progression due to the change in bond order and bond length upon ionization. The observation of a long vibrational progression upon ionization from the 1πu MO suggests it is bonding in character, as its removal increases the bond length and decreases the bond order, leading to significant vibrational effects. The lack of vibrational structure upon ionization from the 1πg MO suggests it is non-bonding, as its removal does not significantly affect the bond order or length.

In the Photoelectron Spectrum of Carbon Disulfide, CS2, Ionization from the Highest

In the Photoelectron Spectrum of Carbon Disulfide, CS₂, Ionization from the Highest