UV Visible Spectroscopy Questions and Answers

Q: The absorbance A, of a 5 × 10\(^{-4}\) M solution of the amino acid tyrosine, at a wavelength of 280 nm, is 0.75. The path length of cuvette is 1 cm. What is the molar absorption coefficient (ε) ?

The molar absorption coefficient is 1500 M-1 cm-1.The Beer-Lambert law gives absorbance as A = εcl, so the coefficient is the absorbance divided by concentration and path length:\[ \begin{aligned} \varepsilon &= \frac{A}{c\,l} = \frac{0.75}{(5 \times 10^{-4})(1)} \\ &= 1500 \text{ M}^{-1}\text{cm}^{-1} \end{aligned} \]150 is a tenfold slip in the power of ten. 3000 and 4500 are two and three times the correct value. The units confirm the working: absorbance has none, and c times l is in M cm, so the coefficient is in M-1 cm-1.

Q: Which region of the spectrum does this technique use?

The ultraviolet and visible range runs from about 190 nanometres, where air and solvents begin to absorb, to about 800 where the visible ends. The micrometre range is the infrared, radio frequencies belong to magnetic resonance, and X-rays probe core electrons. Each spectral region probes a different kind of molecular change, in order of the energy each requires, and knowing which region a technique occupies places it in that sequence. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Q: What kind of molecular change does an ultraviolet photon cause?

Ultraviolet and visible photons carry a few electron volts, enough to promote an electron between molecular orbitals. Vibration needs the tenths of an electron volt an infrared photon carries, rotation far less, and nuclear spin transitions a millionth as much. That the energy required orders the techniques from X-ray down to radio frequency is the single idea that organises the whole of spectroscopy.

Q: Which transition requires the most energy?

A sigma bonding electron is the most tightly held, so promoting it to the corresponding antibonding orbital needs the largest energy and the shortest wavelength - below about 150 nanometres, which is why saturated hydrocarbons are transparent in the usual range and make good solvents. The others require progressively less. Ordering the four transitions by energy explains at once which compounds absorb where a spectrophotometer can see them. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Q: Which promotion happens at the longest wavelength of the four?

Promoting a non-bonding electron into a pi antibonding orbital costs the least, giving bands at the long-wavelength end, often above 280 nanometres. It is also a weak transition, with molar absorptivities of tens rather than thousands, because the orbitals overlap poorly. A band that is both long in wavelength and weak in intensity is the signature of this transition, and the two properties together identify it more reliably than either alone. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: Why is the n to pi star transition weak?

The non-bonding electrons sit in an orbital roughly perpendicular to the pi system, so the overlap that determines the transition's probability is small. Electron count, spin and gap size are not the reason - the transition is symmetry-forbidden rather than spin-forbidden. That intensity reflects orbital overlap while wavelength reflects the energy gap means the two axes of a band carry independent information about the transition producing it. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Q: Which transition dominates the spectrum of a conjugated organic compound?

In a conjugated system the pi to pi star transition is both accessible in wavelength and strongly allowed, with absorptivities in the thousands or tens of thousands. Sigma transitions lie too far into the vacuum ultraviolet, and n to pi star bands are weak. Because it is both visible to the instrument and intense, this transition carries nearly all the analytical value of the technique for organic compounds. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Q: What does a molar absorptivity of ten thousand suggest about a transition?

A large absorptivity means a high transition probability, so the transition is allowed by the selection rules and the orbitals overlap well. It says nothing about wavelength, concentration or bandwidth - absorptivity is a property of the substance, not of how much is present. Reading intensity as evidence about the transition rather than about the sample is the distinction between qualitative and quantitative use of the same number. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: What is a chromophore?

A chromophore is the structural feature responsible for the band - a carbonyl, a nitro group, an aromatic ring, a carbon-carbon double bond. Solvent, wavelength and reference are other things entirely. Because a given chromophore absorbs at a broadly similar wavelength wherever it appears, its presence can be inferred from a spectrum, which is what gives the technique qualitative value alongside the quantitative. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Q: What does an auxochrome do?

An auxochrome such as hydroxyl, amino or halogen has no band of its own in the accessible range but shifts and usually intensifies the chromophore's when attached. It does not absorb alone or suppress absorption, and any solubility effect is incidental. A group that modifies a property it does not itself possess earns a name of its own, and the two terms together account for most of what sets an organic compound's colour. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

The absorbance A, of a 5 × 10\(^{-4}\) M solution of the amino acid tyrosine, at a wavelength of 280 nm, is 0.75. The path length of cuvette is 1 cm. What is the molar absorption coefficient (ε) ?

A.150 M\(^{-1}\) cm\(^{-1}\)
B.1500 M\(^{-1}\) cm\(^{-1}\)
C.3000 M\(^{-1}\) cm\(^{-1}\)
D.4500 M\(^{-1}\) cm\(^{-1}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The molar absorption coefficient is 1500 M^-1 cm^-1.

The Beer-Lambert law gives absorbance as A = εcl, so the coefficient is the absorbance divided by concentration and path length:

\[ \begin{aligned} \varepsilon &= \frac{A}{c\,l} = \frac{0.75}{(5 \times 10^{-4})(1)} \\ &= 1500 \text{ M}^{-1}\text{cm}^{-1} \end{aligned} \]

150 is a tenfold slip in the power of ten. 3000 and 4500 are two and three times the correct value. The units confirm the working: absorbance has none, and c times l is in M cm, so the coefficient is in M^-1 cm^-1.

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Which region of the spectrum does this technique use?

A.Roughly 190 to 800 nanometres
B.Roughly 2.5 to 25 micrometres
C.Radio frequencies in a magnetic field
D.X-ray wavelengths below one nanometre

Answer: A

The ultraviolet and visible range runs from about 190 nanometres, where air and solvents begin to absorb, to about 800 where the visible ends. The micrometre range is the infrared, radio frequencies belong to magnetic resonance, and X-rays probe core electrons. Each spectral region probes a different kind of molecular change, in order of the energy each requires, and knowing which region a technique occupies places it in that sequence. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Discuss in Forum

What kind of molecular change does an ultraviolet photon cause?

A.A nuclear spin flips in a magnetic field
B.An electron moves to a higher orbital
C.A bond stretches or bends more vigorously
D.The whole molecule rotates faster

Answer: B

Ultraviolet and visible photons carry a few electron volts, enough to promote an electron between molecular orbitals. Vibration needs the tenths of an electron volt an infrared photon carries, rotation far less, and nuclear spin transitions a millionth as much. That the energy required orders the techniques from X-ray down to radio frequency is the single idea that organises the whole of spectroscopy.

Discuss in Forum

Which transition requires the most energy?

A.n to pi star
B.n to sigma star
C.Sigma to sigma star
D.Pi to pi star

Answer: C

A sigma bonding electron is the most tightly held, so promoting it to the corresponding antibonding orbital needs the largest energy and the shortest wavelength - below about 150 nanometres, which is why saturated hydrocarbons are transparent in the usual range and make good solvents. The others require progressively less. Ordering the four transitions by energy explains at once which compounds absorb where a spectrophotometer can see them. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Discuss in Forum

Which promotion happens at the longest wavelength of the four?

A.Sigma to sigma star
B.Pi to pi star
C.n to sigma star
D.n to pi star

Answer: D

Promoting a non-bonding electron into a pi antibonding orbital costs the least, giving bands at the long-wavelength end, often above 280 nanometres. It is also a weak transition, with molar absorptivities of tens rather than thousands, because the orbitals overlap poorly. A band that is both long in wavelength and weak in intensity is the signature of this transition, and the two properties together identify it more reliably than either alone. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

Why is the n to pi star transition weak?

A.The two orbitals overlap poorly in space
B.Non-bonding electrons are few in number
C.The transition is forbidden by spin selection
D.The energy gap is too small to absorb strongly

Answer: A

The non-bonding electrons sit in an orbital roughly perpendicular to the pi system, so the overlap that determines the transition's probability is small. Electron count, spin and gap size are not the reason - the transition is symmetry-forbidden rather than spin-forbidden. That intensity reflects orbital overlap while wavelength reflects the energy gap means the two axes of a band carry independent information about the transition producing it. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Discuss in Forum

Which transition dominates the spectrum of a conjugated organic compound?

A.n to pi star
B.Pi to pi star
C.Sigma to sigma star
D.n to sigma star

Answer: B

In a conjugated system the pi to pi star transition is both accessible in wavelength and strongly allowed, with absorptivities in the thousands or tens of thousands. Sigma transitions lie too far into the vacuum ultraviolet, and n to pi star bands are weak. Because it is both visible to the instrument and intense, this transition carries nearly all the analytical value of the technique for organic compounds. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Discuss in Forum

What does a molar absorptivity of ten thousand suggest about a transition?

A.The compound is present at high concentration
B.The band is unusually narrow
C.It is strongly allowed
D.It lies in the visible rather than the ultraviolet

Answer: C

A large absorptivity means a high transition probability, so the transition is allowed by the selection rules and the orbitals overlap well. It says nothing about wavelength, concentration or bandwidth - absorptivity is a property of the substance, not of how much is present. Reading intensity as evidence about the transition rather than about the sample is the distinction between qualitative and quantitative use of the same number. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

What is a chromophore?

A.The solvent in which the sample is dissolved
B.The wavelength of the absorption maximum
C.The reference used to zero the instrument
D.The group giving rise to the absorption

Answer: D

A chromophore is the structural feature responsible for the band - a carbonyl, a nitro group, an aromatic ring, a carbon-carbon double bond. Solvent, wavelength and reference are other things entirely. Because a given chromophore absorbs at a broadly similar wavelength wherever it appears, its presence can be inferred from a spectrum, which is what gives the technique qualitative value alongside the quantitative. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Discuss in Forum

10.

What does an auxochrome do?

A.Shifts a chromophore's band without absorbing alone
B.Produces a band of its own in the visible region
C.Prevents a chromophore from absorbing at all
D.Increases the compound's solubility in the solvent

Answer: A

An auxochrome such as hydroxyl, amino or halogen has no band of its own in the accessible range but shifts and usually intensifies the chromophore's when attached. It does not absorb alone or suppress absorption, and any solubility effect is incidental. A group that modifies a property it does not itself possess earns a name of its own, and the two terms together account for most of what sets an organic compound's colour. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Discuss in Forum

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