Assertion (A): Order and molecularity of a reaction are always same. Reason (R): Complex reactions involve a s

Question

Q. Assertion (A): Order and molecularity of a reaction are always same. Reason (R): Complex reactions involve a sequence of elementary reactions and the slowest step is rate determining. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

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Concept understanding — Order and Molecularity

Order and Molecularity: The Two Faces of a Reaction

Imagine you are watching a chemical reaction happen. Two molecules of A collide, rearrange, and become B. That collision — the actual event where bonds break and form — is an elementary step. The number of molecules that must come together in that single step is its molecularity.

Now imagine a different reaction. You mix A and B, but the product appears at a rate that depends on the square of A and not at all on B. That dependence — how the speed of the overall reaction changes when you change concentrations — is its order.

These two ideas answer different questions. Molecularity asks: How many particles actually meet in one step? Order asks: What is the mathematical relationship between concentration and rate for the overall reaction?

Molecularity — The Microscope View

Molecularity is a property of an elementary reaction only. It is the number of reactant particles (atoms, ions, molecules) that collide in that step.

Unimolecular (1): A single molecule decomposes or rearranges. Example: $N_2O_5 \rightarrow NO_2 + NO_3$
Bimolecular (2): Two particles collide. Example: $NO_2 + CO \rightarrow NO + CO_2$
Termolecular (3): Three particles collide simultaneously — very rare because three-way collisions are improbable. Example: $2NO + O_2 \rightarrow 2NO_2$

Watch out

Molecularity is never zero, never fractional, and never greater than 3. You cannot have a molecularity of 1.5 or 4 — collisions of four particles at once are astronomically unlikely.

Order — The Macroscope View

Order is an experimental quantity. You run the reaction, measure the initial rate at different concentrations, and deduce the rate law:

\text{Rate} = k[A]^m[B]^n

The order with respect to A is $m$ , the order with respect to B is $n$ , and the overall order is $m + n$ .

Order can be:

Integer (0, 1, 2, 3)
Fractional (e.g., 0.5 for a reaction like $H_2 + Br_2 \rightarrow 2HBr$ )
Negative (if increasing a reactant slows the reaction)
Zero (if concentration does not affect rate)

Important

Order is determined by experiment, not by the balanced chemical equation. A reaction written as $2A + B \rightarrow C$ does not automatically have order 3. The actual rate law could be $\text{Rate} = k[A]^2$ (order 2) or $\text{Rate} = k[A]$ (order 1) — only the lab can tell.

The Critical Difference

Property	Molecularity	Order
Applies to	Elementary steps only	Overall reaction (or any step)
Determined by	Stoichiometry of the step	Experiment
Can be fractional?	No	Yes
Can be zero?	No	Yes
Can be negative?	No	Yes
Maximum value	3 (rarely)	Any number

Why the Confusion?

Many students think: "If the balanced equation says 2A + B, then molecularity is 3 and order is 3." This is wrong for two reasons.

First, most reactions are multi-step. The balanced equation shows only the net change, not the actual collision events. A reaction like $2NO + O_2 \rightarrow 2NO_2$ appears termolecular, but it actually happens in two bimolecular steps:

NO + NO \rightarrow N_2O_2 \quad (\text{fast})

N_2O_2 + O_2 \rightarrow 2NO_2 \quad (\text{slow})

The molecularity of each step is 2 (bimolecular). The overall order, determined by the slow step, is $2 + 1 = 3$ — but that is a coincidence, not a rule. …

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Order and molecularity are not always the same — they coincide only for elementary reactions. Complex reactions follow the rate-determining step, making the assertion false while the reason is true. The answer is (D).

The heart of this question lies in understanding two fundamental kinetic parameters: order and molecularity, and recognizing when they align and when they diverge.

Molecularity is a theoretical concept that counts the number of reactant molecules (or atoms, ions) that must collide simultaneously in a single elementary step. It's always a small whole number (1, 2, or rarely 3) because the probability of more than three particles colliding at once is vanishingly small. Molecularity is defined only for elementary reactions.

Order is an experimental quantity — the sum of the powers of concentration terms in the rate law. It can be fractional, zero, or even negative, and is determined from actual kinetic data.

For an elementary reaction, the rate law can be written directly from the stoichiometry, so order equals molecularity. For example, the elementary step $\ce{2NO + O2 -> 2NO2}$ has molecularity 3 and is third-order overall.

But here's where it gets interesting: most reactions we encounter are complex (multi-step). The overall rate law is governed by the slowest step — the rate-determining step (RDS). The order of the overall reaction reflects the RDS and any pre-equilibrium steps, not the overall stoichiometry.

Let's examine the assertion and reason:

Assertion (A): "Order and molecularity of a reaction are always same."

This is false. Consider the decomposition of hydrogen peroxide catalyzed by iodide:

\ce{2H2O2 -> 2H2O + O2}

The mechanism involves:

Step 1 (slow): $\ce{H2O2 + I^- -> H2O + IO^-}$ (molecularity = 2)
Step 2 (fast): $\ce{H2O2 + IO^- -> H2O + O2 + I^-}$

The rate law is $\text{rate} = k[\ce{H2O2}][\ce{I^-}]$ , giving order = 2. The RDS has molecularity 2, so they match here. But the overall reaction has stoichiometric coefficient 2 for $\ce{H2O2}$ , yet molecularity isn't defined for the overall process — only for elementary steps.

A clearer counter-example: the reaction $\ce{2NO2 + F2 -> 2NO2F}$ proceeds via:

$\ce{NO2 + F2 -> NO2F + F}$ (slow)
$\ce{NO2 + F -> NO2F}$ (fast)

The rate law is $\text{rate} = k[\ce{NO2}][\ce{F2}]$ (order = 2), but the overall stoichiometry suggests molecularity would be 3 if it were elementary — which it isn't. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 65 on this concept.

CBSE 2026Set 56/2/11 markMCQ
$Q.Identify the correct statement : (A) Molecularity of a reaction is an experimental quantity. (B) For complex reactions molecularity has no meaning. (C) Molecularity of a reaction can be zero or even a fraction. (D) Molecularity more than three is very common in chemical reactions.$
›Reveal solutionSolution

Molecularity is a theoretical concept tied to the elementary step, not an experimental quantity; it is always a positive integer (1, 2, or 3) and has no meaning for complex reactions. The correct statement is (B).

The Concept: What Molecularity Really Means

Molecularity is one of those terms that sounds like it should be complicated, but it’s actually very simple — provided you remember one crucial thing: molecularity only applies to an elementary step, never to the overall reaction.

An elementary step is a single collision or a single molecular event. If two molecules collide and react in one step, that step is bimolecular (molecularity = 2). If one molecule breaks apart on its own, it’s unimolecular (molecularity = 1). That’s it. There is no such thing as a termolecular step involving four or five molecules simultaneously — the probability of that many particles colliding at the exact same instant is astronomically small. So molecularity is always a small positive integer: 1, 2, or rarely 3.

Now contrast this with order of reaction, which is an experimental quantity. Order can be zero, fractional, or even negative — it comes from the rate law you determine in the lab. Molecularity is a theoretical label for a step, not something you measure.

Let’s examine each option carefully.

Step-by-Step Analysis

1. Option (A): “Molecularity of a reaction is an experimental quantity.”

This is false. You cannot measure molecularity by doing an experiment. You can measure the rate and find the order, but molecularity is deduced from the mechanism — it’s a theoretical count of how many molecules participate in an elementary step. For example, if you propose that $\mathrm{NO_2 + CO \rightarrow NO + CO_2}$ happens in one step, you assign it a molecularity of 2; you don’t “find” it experimentally.

2. Option (B): “For complex reactions molecularity has no meaning.”

This is true. A complex reaction proceeds through multiple elementary steps. The overall reaction has no single molecularity — each step has its own. Asking “what is the molecularity of the overall reaction?” is like asking “what is the age of a family?” — it doesn’t apply. Molecularity is defined only for an elementary step. …
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CBSE 2026Set 56/2/11 markMCQ
$2HI \rightarrow H_2 + I_2$
›Reveal solutionSolution
The assertion is true (the reaction is bimolecular), and the reason correctly explains why — two HI molecules must collide simultaneously. The correct choice is (A).

Concept and Intuition

Molecularity of a reaction is a fundamental concept in chemical kinetics. It refers to the number of reactant particles (atoms, ions, or molecules) that must collide simultaneously in the rate-determining step (the slowest elementary step) for the reaction to occur. It is always a whole number (1, 2, or rarely 3) and is defined only for elementary reactions — not for overall balanced equations that may represent a sequence of steps.

The given reaction is:
$2HI \rightarrow H_2 + I_2$
This is an elementary reaction — it occurs in a single step. Two molecules of HI must come together and collide with sufficient energy and proper orientation to break the H–I bonds and form H–H and I–I bonds. Since two reactant molecules are involved in this single collision event, the molecularity is 2 (bimolecular).

The reason statement says exactly this: "Two molecules of the reactants are involved in simultaneous collision between them." That is the very definition of molecularity = 2 for this reaction.

Watch out

A common mistake is to confuse molecularity with order of reaction. Molecularity is a theoretical concept based on the reaction mechanism (always an integer), while order is experimental (can be fractional, zero, or negative). For this elementary reaction, the order also happens to be 2, but that is not what the assertion is about.

Step-by-step reasoning

Identify the nature of the reaction.

The reaction $2HI \rightarrow H_2 + I_2$ is known to be an elementary reaction — it proceeds in a single step without any intermediate species. This is a standard fact from chemical kinetics.

Define molecularity.

Molecularity is the number of reactant particles that come together in the rate-determining step of an elementary reaction. For a single-step reaction, it equals the number of molecules on the reactant side of the balanced elementary equation.

Count the reactant molecules in the collision. …
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CBSE 2026Set V11 markMCQ
$Q. Which of the following statement/s is/are incorrect about order and molecularity of a chemical reaction? I. The probability that more than three molecules can collide and react simultaneously is very small. II. There can be only three values for the order of the reaction, that is zero, first and second. III. The order depends on rate equation while molecularity does not. IV. There can be infinite number of values for order.$
(a) I and IV
(b) III only
(c) II and IV
(d) II only
›Reveal solutionSolution
Only statement II is wrong — the order of a reaction is not restricted to just zero, first and second.

Evaluate each statement:

I. Correct. The chance of more than three molecules colliding and reacting simultaneously (a termolecular-plus event) is indeed very small; this is why high molecularities are not observed.

II. Incorrect. The order of a reaction is not limited to only three values (0, 1, 2). Order is determined experimentally and can be 0, 1, 2, 3, a fraction (e.g. 1/2, 3/2) or even negative. This statement is false. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Rate = K [X]^3/2 [Y]^-1. The overall order of reaction for the above rate expression is$
(a) 1/2
(b) 1
(c) 3/2
(d) 5/2
›Reveal solutionSolution

The overall order of a reaction is the sum of the powers (exponents) of the concentration terms in the experimentally determined rate law.

…

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CBSE 2026Set ANNUAL1 mark
$Q. If the value of rate constant for any reaction is found to be 1.75 x 10^-5 L mol^-1 s^-1, then order of reaction will be ______ (fill in the blank).$
›Reveal solutionSolution
The unit of the rate constant k reveals the overall order n of a reaction, since k has units of (concentration)^(1-n) time^-1.

Zero order: mol L⁻1 s⁻1

First order: s⁻1

Second order: L mol⁻1 s⁻1 (mol⁻1 L s⁻1) …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Calculate the overall order of a reaction which has the rate expression: Rate = K[A]^1/2 [B]^3/2$
(a) Second order
(b) First order
(c) Zero order
(d) Third order
›Reveal solutionSolution

The overall order of a reaction is the sum of the powers (exponents) of the concentration terms in the experimentally determined rate law.

Given rate law: Rate = K[A]^(1/2)[B]^(3/2)

Order with respect to A = 1/2 …

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CBSE 2026Set ANNUAL1 mark
$Q.Fill in the blank: For a chemical reaction ............... can never be a fraction.$
›Reveal solutionSolution
Molecularity of an elementary reaction (the number of reacting species colliding simultaneously) is always a positive whole number — it can never be zero, negative, or a fraction. Order, by contrast, CAN be zero, negative, or fractional.

It is important not to confuse order and molecularity:

Order of a reaction is an experimentally determined quantity (sum of the powers of concentration terms in the rate law). It can be zero, a whole number, or even a fraction (e.g., order 1.5 in some complex reactions), and is defined for the overall reaction (or steps of a mechanism). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. For a reaction A+B+C\to Product, the rate law is expressed as, r=k[A]^(1/2)[B]^(1/3)[C]^(1/4), then the total order of reaction is-$
(a)
(i) 13/11
(b)
(ii) 13/14
(c)
(iii) 12/13
(d)
(iv) 13/12
›Reveal solutionSolution

Overall order $= \tfrac12 + \tfrac13 + \tfrac14 = \tfrac{13}{12}$ . Correct option: (iv).

Concept. The order of a reaction is the sum of the powers (exponents) to which the concentration terms are raised in the experimentally determined rate law.

Steps.

$\text{Order} = \frac12 + \frac13 + \frac14$

Take LCM $= 12$ : …

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CBSE 2026Set ANNUAL1 mark
$Q. Write answer in one word/sentence: Write unit of rate constant for second order reaction.$
›Reveal solutionSolution

The rate constant of a second-order reaction has units mol^-1 L s^-1.

For an n-th order reaction the unit of k is (concentration)^(1-n) time^-1, i.e. (mol L^-1)^(1-n) s^-1.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Why are reactions with three or higher molecularity very rare?$
›Reveal solutionSolution
A trimolecular (or higher) step needs three or more particles to collide at the same instant with proper orientation and enough energy — a very low-probability event — so such reactions are rare.

Molecularity is the number of reacting species (atoms, ions or molecules) that come together in a single elementary reaction step.

A bimolecular collision (two particles) is common.

A termolecular step requires three particles to be at the same point in space at the same instant, all with the correct orientation and sufficient energy. The probability of such a simultaneous three-body collision is very small compared with a two-body collision. …
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CBSE 2026Set ANNUAL1 mark
$Q. A reaction is found to be of ½ order. What is the unit of its rate constant?$
›Reveal solutionSolution

Using rate $= k[A]^{1/2}$ with rate in $mol\,L^{-1}s^{-1}$ and concentration in $mol\,L^{-1}$ , the unit of $k$ is $mol^{1/2}\,L^{-1/2}\,s^{-1}$ .

The general rule for an $n^{th}$ -order reaction is that the unit of the rate constant is

$k \;=\; mol^{(1-n)}\,L^{(n-1)}\,s^{-1}$

Derivation for $n = \tfrac12$ : the rate law is

$\text{rate} = k\,[A]^{1/2}$

so …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In reaction A\toB, the rate of reaction is doubled on increasing concentration of reactant four times. The order of reaction is-$
(a) 1/2
(b) 2
(c) 4
(d) Zero
›Reveal solutionSolution

Order found from rate = k[A]^n using the given fold-changes.

Rate law: rate $= k[A]^n$ . When $[A]$ is increased 4 times, rate becomes 2 times.

$\dfrac{2\,\text{rate}}{\text{rate}} = \dfrac{k(4[A])^n}{k[A]^n} = 4^n$

…

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