UV-Vis spectroscopy — Beer-Lambert law, conjugation and λ_max: Question bank

Everything here rests on three ideas you already built in the parent note:

Absorption is a jump. A photon is only eaten if its energy $\frac{h c}{λ}$ equals the orbital gap $Δ E$ (see Planck relation E = hν).
Longer conjugation = bigger "box" = smaller gap = longer $λ_{ma x}$ (the Particle in a box picture).
Beer-Lambert $A = ε c l$ , with $A = - lo g_{10} T$ .

Before the traps, three quick pictures pin down the words this page keeps reusing.

Figure — UV-Vis spectroscopy — Beer-Lambert law, conjugation and λ_max

Definition "Allowed" vs "forbidden" transitions — what

$ε$ really counts Not every jump is equally likely. Quantum selection rules (based on the symmetry and spin of the two orbitals) decide whether a transition is easy ("allowed") or nearly blocked ("forbidden"). The molar absorptivity $ε$ is a direct measure of that likelihood:

Allowed (symmetry- and spin-permitted, e.g. most $π \to π^{*}$ ): $ε \sim 1 0^{3}$ to $1 0^{5}$ → tall, strong peaks.
Forbidden (symmetry breaks the rule, e.g. many $n \to π^{*}$ carbonyl bands): $ε \sim 10$ to $1 0^{2}$ → short, weak peaks that only appear because real molecules wobble and relax the rule slightly.

So $ε$ does not tell you where the peak is (that's $λ_{ma x}$ ) — it tells you how tall it is. Look at Figure 2: two peaks at the same wavelength, the tall red one allowed, the short black one forbidden.

True or false — justify

Every item below is a statement. Decide true/false and give the reason — a bare verdict scores nothing.

A molecule that absorbs at longer wavelength has a larger HOMO–LUMO gap.

False.

λ = \frac{h c}{Δ E}

, so longer

λ

means the photon energy is smaller, which means the gap

Δ E

(HOMO→LUMO) is smaller, not larger.

A colourless organic solution definitely has no electronic transitions.

False. It just has no transitions in the visible 400–700 nm window; its

π \to π^{*}

absorption may sit in the UV (e.g. below 400 nm), which our eyes can't see.

Doubling the concentration doubles the absorbance (in the valid range).

True.

A = ε c l

is linear in

c

, so at fixed

ε

and

l

, doubling

c

doubles

A

— provided you stay in the low-concentration linear region.

Doubling the concentration doubles the transmittance.

False. Doubling

c

doubles absorbance to

A_{n e w} = 2 A

; then

T_{n e w} = 1 0^{- A_{n e w}} = 1 0^{- 2 A} = (1 0^{- A})^{2} = T^{2}

. So

T

is squared, not doubled — a much steeper drop (e.g.

T = 0.5 \to 0.25

A transition with

ε = 20, 000

is "allowed" and one with

ε = 20

is "forbidden".

True (roughly). Large

ε

(

\sim 1 0^{3}

–

1 0^{5}

) marks a symmetry-allowed transition (like

π \to π^{*}

); tiny

ε

(

\sim 10

–

1 0^{2}

) marks a weak, symmetry-forbidden one (like many

n \to π^{*}

carbonyl bands) — exactly the selection-rule split in Figure 2.

If a sample transmits 100% of the light, its absorbance is 100%.

False.

T = 1

gives

A = - lo g_{10} (1) = 0

. Full transmittance means zero absorbance — the two are inverse, not equal.

Two solutions with the same

λ_{ma x}

must be the same compound.

False.

λ_{ma x}

constrains the chromophore/conjugation length but isn't unique; many different molecules share a similar-length π-system and absorb near the same wavelength.

A benzene ring is a chromophore even though it has no isolated C=C double bond you'd point to.

True. Its delocalised aromatic π-system undergoes

π \to π^{*}

transitions, so it absorbs (around 255 nm) — that's exactly what makes it a chromophore.

Adding an –OH group to benzene always leaves

λ_{ma x}

unchanged because OH has no double bond.

False. The oxygen lone pair conjugates into the ring (an auxochrome), extending the π-system and red-shifting

λ_{ma x}

(phenol absorbs longer than benzene).

The molar absorptivity

ε

depends on how concentrated you make the solution.

False.

ε

is a fixed property of the molecule (and solvent/wavelength); concentration is the separate variable

c

. Confusing them defeats the whole point of Beer-Lambert.

Spot the error

Each line contains flawed reasoning. State what's wrong and why.

"This solution looks red, so it must be absorbing red light strongly."

Wrong: a solution shows the complement of what it absorbs (Figure 4). Looking red means it absorbs green/blue-green light and lets red pass — see Complementary colours.

A = 0.30

means 30% of the light was absorbed."

Wrong:

A

is a logarithm, not a percentage.

A = 0.30

gives

T = 1 0^{- 0.30} \approx 0.50

, so ~50% is transmitted and ~50% absorbed — absorbance is not a fraction.

"β-carotene is orange because it emits orange light after absorbing UV."

Wrong mechanism: its colour is from selective absorption of blue light, so the transmitted mix looks orange. This is subtractive, not emission/fluorescence.

"Since conjugation lowers

Δ E

, an infinitely long polyene would absorb infinitely long wavelengths."

Wrong: real chains eventually deviate — the box isn't perfectly free-electron, and

λ_{ma x}

converges toward a limit rather than growing without bound.

n \to π^{*}

has a smaller gap than

π \to π^{*}

, so it always gives a stronger, taller peak."

Wrong: smaller gap sets the wavelength, not the intensity.

n \to π^{*}

is usually symmetry-forbidden, so it has a low

ε

— a weak peak despite the longer wavelength (Figure 2).

"We should aim for

A

near 2 or 3 for the most accurate measurement — more absorbance, more signal."

Wrong: at high

A

almost no light reaches the detector, so noise dominates and Beer-Lambert curves. The reliable window is roughly

A = 0.2

1.0

"Twisting a biphenyl so its two rings are perpendicular will red-shift

λ_{ma x}

because it's a bigger molecule."

Wrong: size doesn't matter, overlap does. A 90° twist breaks the p-orbital overlap between rings, killing conjugation and causing a blue (hypsochromic) shift.

Why questions

Answer with a causal chain, not a keyword.

Why does UV-Vis focus on

π \to π^{*}

rather than

σ \to σ^{*}

transitions?

σ \to σ^{*}

gaps are so huge they absorb in the vacuum UV (<150 nm), which normal instruments can't reach;

π \to π^{*}

gaps are moderate and land in the measurable 200–800 nm window.

Why does a longer conjugated chain absorb lower-energy photons?

In the Particle in a box picture (Figure 3), a longer chain is a longer box

L

; energy levels scale as

1/ L^{2}

, so they squeeze together, shrinking

Δ E

and needing lower-energy (longer-

λ

) light.

Why do we take a logarithm to define absorbance instead of just using

I_{0} / I

Each thin slice removes a fixed fraction of light, so attenuation is multiplicative (exponential in path). Taking

lo g_{10}

converts that exponential into a quantity linear in

c

and

l

— easy calibration curves (see Quantitative analysis & calibration curves).

Why does the value of

ε

tell you something about the type of transition?

ε

measures how likely the transition is (its "allowedness" under selection rules); high

ε

(

1 0^{3}

–

1 0^{5}

) flags a strongly allowed

π \to π^{*}

, while

ε \sim 10

–

1 0^{2}

flags a weak, symmetry-forbidden

n \to π^{*}

Why does an auxochrome like –NH₂ shift

λ_{ma x}

even though it has no C=C of its own?

Its lone pair delocalises into the adjacent π-system, effectively lengthening the conjugation and lowering

Δ E

— a red shift, without contributing a formal double bond (see Conjugation and resonance).

Why can two conjugation-equivalent dyes still differ in how deeply coloured they look at equal concentration?

Depth of colour depends on

ε

(via

A = ε c l

), not just

λ_{ma x}

; a dye with a higher molar absorptivity absorbs more strongly and looks more intensely coloured at the same

c

Why does the HOMO–LUMO language line up with the box model at all?

Electrons fill box levels in pairs, so the highest filled level is the HOMO and the first empty one the LUMO; the smallest jump is HOMO→LUMO, exactly the gap

Δ E

the photon must match (see HOMO and LUMO orbitals).

Edge cases

Boundary and degenerate scenarios — the ones exams love.

What is

A

when a completely clear (non-absorbing) solvent fills the cuvette?

A = 0

, because

I = I_{0}

gives

T = 1

and

A = - lo g_{10} (1) = 0

. This is why you "blank" the instrument on pure solvent.

What happens to

A

T \to 0

(essentially no light gets through)?

A = - lo g_{10} T \to + \infty

; absorbance diverges. In practice detector noise and stray light break the law long before this, so such readings are meaningless.

For an isolated single double bond (like ethene, one C=C), where does it absorb and can a normal UV-Vis instrument see it?

Around 170 nm — in the vacuum UV, below the ~190 nm cutoff of standard instruments, so it's usually not observed. You need conjugation to pull the band into range.

If you halve the path length but double the concentration, what happens to

A

It stays the same.

A = ε c l

and the product

c \cdot l

is unchanged (

2 c \times \frac{l}{2} = c l

), so absorbance is identical.

At extremely high concentration, does Beer-Lambert over- or under-estimate the true absorbance?

It over-predicts

A

(the plot of

A

c

bends downward / below the line): solute–solute interactions and refractive-index changes reduce actual absorption, so measured

A

falls short of the linear prediction.

What is the colour of a solution whose

λ_{ma x}

lies at 350 nm (pure UV) with no visible absorption?

Colourless / clear. Nothing in 400–700 nm is removed, so all visible wavelengths pass equally and the eye sees no colour.

If a molecule has an odd number of π-electrons, does the simple "HOMO = level

N /2

" filling rule still apply cleanly?

No — with odd

N

the top level is half-filled (a radical), so the pairs-fill assumption breaks; the clean gap formula below is only valid for even

N

Recall One-line survival kit

Colour seen ::: complement of colour absorbed
$A$ and $T$ ::: inverse, via $A = - lo g_{10} T$
Longer conjugation ::: smaller $Δ E$ , larger $λ_{ma x}$ (red shift)
$ε$ ::: fixed molecular fingerprint, sets peak height, not position
Valid $A$ range ::: about 0.2–1.0