Chromatography

8 MCQs2 revision cards9-step worked example
Source: NCERT Purification and Characterisation of Organic CompoundsPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 27 Sep 2026

Chromatography, explained for NEET

The Rf trap is a ratio you can write upside down and never notice, because both numbers are distances in centimetres and the arithmetic works either way. Rf is the distance moved by the substance divided by the distance moved by the solvent front — substance on top. The solvent front always travels at least as far as any spot it carries, so Rf can never exceed 1. An answer of 1.4 is not a "fast-moving component"; it is an inverted ratio.

Chromatography separates a mixture by distributing its components between two phases: one stationary, one mobile. NCERT Class 11 Chemistry, Chapter 8, page 282 classifies techniques by how that distribution happens. In adsorption chromatography — column and thin layer — components stick to the surface of a solid adsorbent such as silica gel or alumina, and differential adsorption pulls them apart. In partition chromatography — paper chromatography — components dissolve to different extents between two liquids.

That second classification is where the marks go. Paper chromatography is partition, not adsorption (page 283). The chromatography paper is not the stationary phase; the water trapped in the pores of the paper is. The paper is only the support holding it. The mobile phase is the developing solvent that rises through the sheet and carries the components with it.

For NEET this topic pays in two currencies: a one-step Rf calculation, and a classification question asking which technique belongs to which class. Both are recall-plus-one-step, and both have distractors built from exactly the confusions above.

Watch-out: in a chromatogram the spot's distance is measured from the baseline to the centre of the spot, and the solvent distance is measured from the same baseline to the solvent front — not from the bottom edge of the paper.

Can you answer these Chromatography MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

In paper chromatography, the stationary phase is:

Show answer and why every option is right or wrong

Answer: C. Paper chromatography is a partition technique in which water held in the pores of the paper acts as the stationary phase, with the paper serving only as the support — NCERT Class 11 Chemistry, Chapter 8, page 284.

Why A is wrong: A is wrong because the cellulose paper is only the inert support that holds the stationary liquid; naming the paper itself as the stationary phase is the single most-penalised confusion in this topic.

Why B is wrong: B is wrong because the developing solvent that ascends the sheet is the mobile phase — it is the phase that moves, which is the definition of mobile.

Why D is wrong: D is wrong because silica gel is the adsorbent used in thin layer and column chromatography, not in paper chromatography, and it is not coated on chromatography paper.

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MCQ 2Direct ApplicationPractice

On a developed chromatogram the solvent front has risen 8.0 cm from the baseline, and the centre of a component's spot lies 3.2 cm from the same baseline. The Rf value of that component is:

Show answer and why every option is right or wrong

Answer: B. Rf = distance moved by substance ÷ distance moved by solvent = 3.2 cm ÷ 8.0 cm = 0.40 — the retardation factor defined in NCERT Class 11 Chemistry, Chapter 8, page 283.

Why A is wrong: A is wrong because 8.0 ÷ 3.2 = 2.5 inverts the ratio, putting the solvent distance on top; a value above 1 is impossible since the solvent front always travels at least as far as the spot it carries.

Why C is wrong: C is wrong because 8.0 − 3.2 = 4.8 cm subtracts the distances instead of dividing them; Rf is a ratio and is dimensionless.

Why D is wrong: D is wrong because 0.25 comes from reading the spot as 2.0 cm rather than 3.2 cm; the spot distance is measured to the centre of the spot from the baseline.

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MCQ 3Easy RecallPractice

Which of the following pairs correctly matches a chromatographic technique with its class?

Show answer and why every option is right or wrong

Answer: D. Thin layer chromatography uses a thin layer of adsorbent such as silica gel or alumina on a glass plate, so it is adsorption chromatography along with column chromatography — NCERT Class 11 Chemistry, Chapter 8, page 283.

Why A is wrong: A is wrong because column chromatography separates components by differential adsorption on a solid adsorbent packed in the column, placing it in the adsorption class, not the partition class.

Why B is wrong: B is wrong because paper chromatography works by partition of the components between trapped water and the developing solvent; classing it as adsorption is an explicitly flagged negative-marking trap in this topic.

Why C is wrong: C is wrong because every chromatographic technique has a mobile phase by definition — in paper chromatography it is the developing solvent rising through the sheet.

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MCQ 4Direct ApplicationPractice

A student reports Rf values of 0.62 and 1.30 for the two components of a mixture separated by paper chromatography. The correct conclusion is:

Show answer and why every option is right or wrong

Answer: C. Rf is the substance distance divided by the solvent distance, and the solvent front is always at or ahead of every spot, so Rf lies between 0 and 1 — NCERT Class 11 Chemistry, Chapter 8, page 283.

Why A is wrong: A is wrong because no component can outrun the solvent that transports it up the paper; high solubility in the mobile phase pushes Rf towards 1, never past it.

Why B is wrong: B is wrong because Rf is bounded above by 1 by construction — its numerator is always less than or equal to its denominator.

Why D is wrong: D is wrong because it reverses the bound; 0.62 is an entirely normal Rf, and a value exceeding 1 is the impossible one.

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MCQ 5Easy RecallPractice

In adsorption chromatography, the separation of components depends on:

Show answer and why every option is right or wrong

Answer: B. Adsorption chromatography rests on differential adsorption of the components on the surface of a solid adsorbent such as silica gel or alumina — NCERT Class 11 Chemistry, Chapter 8, page 282.

Why A is wrong: A is wrong because separation by boiling-point difference is the principle of distillation, a different purification technique entirely.

Why C is wrong: C is wrong because density difference between immiscible liquids is what separates layers in differential extraction, not what drives adsorption chromatography.

Why D is wrong: D is wrong because sublimation separates a solid that passes directly to vapour; adsorption chromatography involves no phase change of that kind.

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MCQ 6Direct ApplicationPractice

Two components P and Q are separated on the same chromatogram. P has Rf = 0.75 and Q has Rf = 0.25. Which statement follows?

Show answer and why every option is right or wrong

Answer: A. A larger Rf means the component moved a larger fraction of the solvent distance, which happens when it is held less strongly by the stationary phase and carried more readily by the mobile phase — NCERT Class 11 Chemistry, Chapter 8, page 283.

Why B is wrong: B is wrong because equal affinities would give equal Rf values on the same chromatogram; the differing Rf values are themselves evidence of differing affinities.

Why C is wrong: C is wrong because the lower Rf belongs to the component held back more strongly by the stationary phase; Q has the greater affinity for the stationary phase, not the mobile one.

Why D is wrong: D is wrong because no component outruns the solvent front; an Rf of 0.75 means P travelled three-quarters of the solvent distance, not more than it.

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MCQ 7CalculationPractice

On a chromatogram the solvent front is 1.20 × 10¹ cm from the baseline. Component X has Rf = 0.45 and component Y has Rf = 0.70. The separation between the centres of the two spots is:

Show answer and why every option is right or wrong

Answer: D. Each spot distance is Rf × solvent distance, so X is at 0.45 × 1.20 × 10¹ cm = 5.4 cm and Y is at 0.70 × 1.20 × 10¹ cm = 8.4 cm; the separation is 8.4 − 5.4 = 3.0 cm, using the Rf definition on NCERT Class 11 Chemistry, Chapter 8, page 283.

Why A is wrong: A is wrong because 8.4 cm is the distance of Y from the baseline, not the gap between the two spots; the first step must be completed for both components and then subtracted.

Why B is wrong: B is wrong because 0.70 − 0.45 = 0.25 is a difference of Rf values, which is dimensionless; it must still be multiplied by the solvent distance to give a length in centimetres.

Why C is wrong: C is wrong because 5.4 cm is the distance of X from the baseline, the other single-component result rather than the difference asked for.

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MCQ 8CalculationPractice

A colourless mixture is separated by paper chromatography and the developed paper shows no visible spots. The solvent front lies 1.00 × 10¹ cm from the baseline. After the paper is examined under ultraviolet light, one spot fluoresces with its centre 6.5 cm from the baseline. Which pair of statements is correct?

Show answer and why every option is right or wrong

Answer: A. Rf = 6.5 cm ÷ 1.00 × 10¹ cm = 0.65, and colourless components must be made visible by a detection method such as ultraviolet light before the spot distance can be measured — NCERT Class 11 Chemistry, Chapter 8, pages 283 and 284.

Why B is wrong: B is wrong because 1.00 × 10¹ ÷ 6.5 = 1.54 inverts the ratio; the detection half of the statement is correct but the Rf is the inverted value, and one wrong half makes the option wrong.

Why C is wrong: C is wrong because an Rf below 1 is the normal and expected result, not an error signal; values between 0 and 1 are exactly what the definition produces.

Why D is wrong: D is wrong because Rf is a ratio of two distances and is therefore dimensionless; quoting the raw spot distance as the Rf drops the division entirely.

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Chromatography: quick recall before you leave

How do you solve a Chromatography question? A worked example

  1. 1

    Given

    A mixture of two dyes is separated by paper chromatography. Measured from the baseline on the developed chromatogram:
    • Distance moved by the solvent front = 1.50 × 10¹ cm• Distance moved by the centre of dye A's spot = 9.0 cm• Distance moved by the centre of dye B's spot = 4.5 cm

  2. 2

    Required

    The Rf value of each dye, and which dye is held more strongly by the stationary phase.

  3. 3

    Concept

    Paper chromatography separates by partition: each component distributes itself between the stationary phase (water trapped in the paper) and the mobile phase (the developing solvent). A component held strongly by the stationary phase lags behind; one that favours the mobile phase travels nearer the solvent front. The retardation factor Rf quantifies this position on a scale from 0 to 1.

  4. 4

    Formula

    $$R_f = \frac{\text{Distance moved by the substance}}{\text{Distance moved by the solvent}}$$

  5. 5

    Substitution

    $$R_{f(A)} = \frac{9.0 \text{ cm}}{1.50 \times 10^1 \text{ cm}} \qquad R_{f(B)} = \frac{4.5 \text{ cm}}{1.50 \times 10^1 \text{ cm}}$$

  6. 6

    Calculation

    $$R_{f(A)} = 0.60 \qquad R_{f(B)} = 0.30$$

    The centimetre units cancel, leaving a pure number — Rf carries no unit. Both measured distances are quoted to two significant figures (the solvent distance is written as 1.50 × 10¹ cm precisely so that its two trailing digits are not ambiguous), so each Rf is reported to two significant figures.

  7. 7

    Final answer

    Dye A: Rf = 0.60. Dye B: Rf = 0.30. Dye B has the smaller Rf, so it travelled less far and is held more strongly by the stationary phase — the trapped water.

  8. 8

    Common trap

    Writing the ratio upside down. Both quantities are distances in centimetres, so 1.50 × 10¹ ÷ 9.0 = 1.7 looks like an ordinary answer and the units give no warning. The check that costs one second: if Rf > 1, the ratio is inverted — nothing on the paper can outrun the solvent that carries it. The second trap is measuring the spot from the bottom edge of the paper rather than from the baseline, which inflates both spot distances by the same amount and so corrupts every Rf on the chromatogram.

  9. 9

    Similar NEET-style question

    In a paper chromatogram the solvent front travels 1.20 × 10¹ cm from the baseline and a component's spot centre is found at 9.6 cm from the baseline. A second component on the same paper has Rf = 0.35. How much further from the baseline is the first component than the second?
    *(Answer: first component Rf = 0.80, so it lies at 9.6 cm; second lies at 0.35 × 1.20 × 10¹ = 4.2 cm; separation = 5.4 cm.)*

What to remember before solving Chromatography questions

Key Fact

Chromatography

Chromatography is an important technique extensively used to separate mixtures into their components, purify compounds and also to test the purity of compounds. The name chromatography is based on the Greek word chroma, for colour since the method was first used for the separation of coloured substances found in plants. In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one another. The moving phase is called the mobile phase. Based on the principle involved, chromatography is classified into different categories. Two of these are: (a) Adsorption chromatography, and (b) Partition chromatography.

-- NCERT Class 11 Chemistry, Ch. 8, p. 282
Key Fact

Adsorption chromatography

a) Adsorption Chromatography: Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina. When a mobile phase is allowed to move over a stationary phase (adsorbent), the components of the mixture move by varying distances over the stationary phase. Following are two main types of chromatographic techniques based on the principle of differential adsorption. (a) Column chromatography, and (b) Thin layer chromatography.

-- NCERT Class 11 Chemistry, Ch. 8, p. 283
Key Fact

Column chromatography

Column Chromatography: Column chromatography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (Fig. 8.11). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column (Fig.8.11). Fig.8.11 Column chromatography. Different stages of separation of components of a mixture. The column is labelled with solvent at the top, the mixture of compounds (a+b+c), the adsorbent (stationary phase), and glass wool above the stopcock.

-- NCERT Class 11 Chemistry, Ch. 8, p. 283
Key Fact

Thin layer chromatography

Thin Layer Chromatography: Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer (about 0.2mm thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate or chromaplate. The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of the TLC plate. The glass plate is then placed in a closed jar containing the eluant (Fig. 8.12a). As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of colourless compounds, which are invisible to the eye but fluoresce in ultraviolet light, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds, which adsorb iodine, will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example, amino acids may be detected by spraying the plate with ninhydrin solution (Fig.8.12b).

-- NCERT Class 11 Chemistry, Ch. 8, p. 283
Formula

Retardation factor (Rf value)

The relative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. Rf value (Fig.8.12 b). Rf = Distance moved by the substance from base line (x) / Distance moved by the solvent from base line (y) Fig.8.12 (a) Thin layer chromatography. Chromatogram being developed. Fig.8.12 (b) Developed chromatogram, labelled with solvent front, spot, base line, x and y.

-- NCERT Class 11 Chemistry, Ch. 8, p. 283
Key Fact

Partition chromatography (paper chromatography)

Partition Chromatography: Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents (Fig. 8.13). This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed under thin layer chromatography. Fig.8.13 Paper chromatography. Chromatography paper in two different shapes.

-- NCERT Class 11 Chemistry, Ch. 8, p. 284

Chromatography questions from past NEET papers

1 question from NEET 2020. Answers verified against NTA official keys.

All 7 past-paper questions from Purification and Characterisation of Organic Compounds →

Sources

NCERT refs: Class 11 Chemistry Chapter 8, p.282

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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