Question

An electron in the$\mathbf{\pi }$orbital of ethylene $\left(C_2{H}_4\right)$is excited by a photon to the ${\mathbf{\pi }}^{\mathbf{*}}$orbital. Do you expect the equilibrium bond length in the excited ethene molecule to be greater or less than that in ground-state ethene? Will the vibrational frequency in the excited state be higher or lower than in the ground state? Explain your reasoning.

Short answer

The bond length in excited ethylene molecule will be longer and vibrational frequency will be smaller.

Step-by-step solution

Bond Length and vibrational frequency

The frequency of molecule by which it vibrates is known as vibrational frequency. Thelength between centre of nucleus of two atoms in a molecule is known as bond length.

With increase in bond length, vibrational frequency of molecule decreases because distance between electron and nuclei increases, hence a weaker bond is formed resulting in lower vibrational frequency.

Equilibrium Bond Length

The electron in ethylene is excited from$\mathrm{\pi }$ to ${\mathrm{\pi }}^{*}$orbital. In a$\mathrm{\pi }\to {\mathrm{\pi }}^{*}$ transition, an electron is moved from a bonding to anti-bonding orbital. The net bond order is changed from 2 to 1.

The C-C bond in ethylene, in the excited state should be longer than it was in the ground state because distance increases between electron in nucleus in excited state. Thus, the bond length will be increased. Longer bond length corresponds to weaker bonds, so nuclei will not be as tightly bound.

Vibrational Frequency in Excited State

As the bond is weaker it is not as ‘stiff’ as the stronger bond. As a result, the force constant will be reduced hence, reducing the vibrational frequency. Therefore, the vibrational frequency is smaller in the excited state.

Representation of ground and excited state