physical chemistry - Reaction molecularity and order

Question:

A reaction involving two different reactants can never be a

1. unimolecular reaction


2. bimolecular reaction


3. second order reaction


4. first order reaction

The answer as per my book is 1. unimolecular.
But the nucleophilic substitution S_N1 is unimolecular but it involves two different reactants. Let us take an example, $$\ce{(CH3)3C-Cl + Br- -> (CH3)3C-Br + Cl-}$$ This follows S_N1 mechanism. Yet, it involves two different reactants even though the R.D.S depends only on the formation of carbocation and the rate law expression being $$\text{rate} = k[\ce{(CH3)3C-Cl}]$$ where $k$ is the rate constant. So, is the answer that unimolecular reactions can never involve two different reactants wrong? Or am I missing something trivial?
Also it would be great if other options are discussed.

Answer

My views are slightly different from the ones expressed in other answers.

First of all,

Molecularity isn't defined for a complex reaction.

From Physical Chemistry by Ira. N Levine: (Page 531, Rate Laws And Equilibrium Constants For Elementary Reactions ):

The number of molecules that react in an elementary step is the molecularity of the elementary reaction. Molecularity is defined only for elementary reactions and should not be used to describe overall reactions that consist of more than one elementary step. The elementary reaction $\ce{ A → products}$ is unimolecular. The elementary reactions $\ce{A + B → products}$ and $\ce{ 2A → products}$ are bimolecular. The elementary reactions $\ce{A + B + C → products}$, $\ce{2A + B → products}$, and $\ce{3A → products}$ are trimolecular (or termolecular).

Atkins' mentions: (page 810, Elementary Reactions):

The molecularity of an elementary reaction is the number of molecules coming together to react in an elementary reaction.It is most important to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

Now, from Wikipedia:

This difference can be illustrated on the reaction between nitric oxide and hydrogen:

${\displaystyle {\ce {2NO + 2H2 -> N2 + 2H2O}}}$

The observed rate law is ${\displaystyle v=k{\ce {[NO]^2[H2]}}}$, so that the reaction is third order. Since the order does not equal the sum of reactant stoichiometric coefficients, the reaction must involve more than one step. The proposed two-step mechanism is

$${\displaystyle {\ce {2 NO + H2 -> N2 + H2O2}}}$$

$${\displaystyle {\ce {H2O2 + H2 -> 2H2O}}}$$

On the other hand, the molecularity of this reaction is undefined, because it involves a mechanism of more than one step. However, we can consider the molecularity of the individual elementary reactions that make up this mechanism:

The first step is termolecular because it involves three reactant molecules, while the second step is bimolecular because it involves two reactant molecules.

As far as your example of S_N1 reactions is concerned, Peter Skyes' acknowledges while discussing the kinetics of S_N1:

The molecularity refers to the number of species that are undergoing bond-breaking or making in one step of the reaction, usually, in the rate limiting step.

Thus, when we say S_N1 (S= substituition, N= nucleophilic, 1= unimolecular) we are essentially talking with respect to the RDS in which the carbocation is formed. Note that though most books identify 1 for unimolecular, Clayden's book refers to it as 1st order, which, in my honest opinion, is much more clearer.

The slope of the first graph is simply the first-order rate constant because rate = $k_1[t\text{-}\ce{BuBr]}$. But the slope of the second graph is zero. The rate determining step does not involve $\ce{NaOH}$ so adding more of it does not speed up the reaction. The reaction shows first-order kinetics (the rate is proportional to one concentration only) and the mechanism is called S_N1, that is, Substitution, Nucleophilic, 1st order.

The question is poorly framed. However, since molecularity is defined only for an elementary reaction, maybe after looking at the options, the question setters expect you to assume that they are talking about an elementary reaction. If this is the case, only then option 1 (but along with option 4) can be marked correct. However, because they haven't mentioned anything, the candidate attempting the question is bound to get confused. If they are talking about the RDS step, then your argument is valid. An S_N1 reaction is indeed unimolecular wrt the RDS step (according to all standard books like Solomon Frhyle, Peter Skyes', etc.) which would deem all of the options incorrect.