Questions: Electronic Transitions and Excited State Behavior

Question 1

A molecule absorbs UV light at 300 nm but emits light at 600 nm — well beyond the typical Stokes shift from vibrational relaxation. Which process most likely explains this large emission red-shift?

• A. Fluorescence directly from S₁, with an unusually large vibrational relaxation
• B. Phosphorescence from the triplet state T₁, which lies lower in energy than S₁
• C. Internal conversion, which re-emits the excess energy as longer-wavelength light
• D. A second absorption event that promotes the molecule to a higher excited state

Answer: (B)

Fluorescence involves emission from S₁ back to S₀ and always occurs at longer wavelengths than absorption (Stokes shift) due to vibrational relaxation within S₁ — but the shift is typically tens of nanometers, not hundreds. An emission at twice the absorption wavelength indicates phosphorescence from T₁: after intersystem crossing from S₁ to T₁ (which is lower in energy than S₁), emission from T₁ → S₀ occurs at significantly longer wavelengths. The T₁ → S₀ transition is also spin-forbidden, giving phosphorescence a characteristically long lifetime (microseconds to seconds) compared to fluorescence (nanoseconds).

Question 2

After absorbing a photon, formaldehyde (H₂C=O) adopts a geometry in which the molecule is no longer planar — the carbon and oxygen bend out of plane, and the C=O bond lengthens. What is the best explanation for this geometric change?

• A. The absorbed photon heats the molecule, causing random thermal distortion
• B. The excited electronic state has a different electron density distribution, shifting the equilibrium geometry to a new energy minimum
• C. The absorption breaks the π bond, converting it to a single bond that can rotate freely
• D. The Franck-Condon principle requires the geometry to change during electronic transitions

Answer: (B)

In the n→π* transition of formaldehyde, an electron is promoted from a non-bonding lone pair on oxygen into the C=O π* (antibonding) orbital. The new electron configuration weakens the C=O bond (more electron density in an antibonding orbital) and changes the overall electron distribution. The excited state is effectively a different electronic structure with a different potential energy surface, and the geometry that minimizes energy on that new surface is non-planar with a longer C=O bond. This is the key insight: excited states are not just 'hotter' ground states — they are chemically distinct species.

Question 3

The triplet excited state (T₁) of an organic molecule is typically longer-lived than its first singlet excited state (S₁) because the T₁ → S₀ radiative transition is spin-forbidden.

• T. True
• F. False

Answer: True

Selection rules require that electronic transitions conserve electron spin. S₁ is a singlet state (all electrons paired, net spin = 0) and S₀ is also a singlet — so S₁ → S₀ fluorescence is spin-allowed and fast (nanosecond timescale). T₁ is a triplet state (one electron spin-flipped, net spin = 1) and S₀ is a singlet — so T₁ → S₀ phosphorescence is spin-forbidden, proceeding much more slowly (microseconds to seconds). This long lifetime is what makes triplet states so useful in photochemistry: they persist long enough to undergo bimolecular reactions.

Question 4

Fluorescence occurs from the triplet excited state (T₁), while phosphorescence occurs from the singlet excited state (S₁).

• T. True
• F. False

Answer: False

This reverses the assignment. Fluorescence is emission from the first excited singlet state S₁ to the ground singlet state S₀ — it is spin-allowed and fast. Phosphorescence is emission from the first excited triplet state T₁ to the ground singlet state S₀ — it is spin-forbidden and slow. A useful mnemonic: Fluorescence is Fast (nanoseconds), Phosphorescence is Prolonged (microseconds to seconds). The glow-in-the-dark effect of phosphorescent materials demonstrates the long T₁ lifetime directly.

Question 5

Why does an electronically excited molecule behave chemically differently from its ground state, even though both have the same molecular formula and the same atoms?

Model answer: Electronic configuration determines chemical reactivity. In the excited state, an electron has been promoted to a different molecular orbital — typically from a bonding or non-bonding orbital into an antibonding one. This changes electron density distribution across the molecule: bonds may weaken or lengthen, the dipole moment may shift direction, and new reaction pathways become accessible because the electron distribution no longer resembles the ground state. Essentially, the excited state is a different electronic isomer on a different potential energy surface, with different bond strengths, different equilibrium geometry, and different frontier orbitals available for chemical reactions.

This is why photochemistry produces products that cannot be made thermally: reactions proceed on excited-state potential energy surfaces where different transition states and products are accessible. Photosynthesis, vision (retinal isomerization), and DNA damage from UV light all involve chemistry that only becomes possible in the excited state.