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As far as your example of $$S_N1$$SN1 reactions is concerned, Peter Skyes' acknowledges while discussing the kinetics of $$S_N1$$ SN1:

Thus, when we say $$S_N1$$SN1 (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 $$1$$1 for unimolecular, Clayden's book refers to it as $$1^{st}$$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 tethe slope of the second grathgraph 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 $$\text{S}_{\text{N}}1$$SN1, 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 $$4$$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 $$\text{RDS}$$RDS step, then your argument is valid. An $$S_N1$$SN1 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.

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

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 $$1^{st}$$ 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 te slope of the second grath 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 $$\text{S}_{\text{N}}1$$, 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 $$\text{RDS}$$ 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.

As far as your example of SN1 reactions is concerned, Peter Skyes' acknowledges while discussing the kinetics of SN1:

Thus, when we say SN1 (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 SN1, 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 SN1 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.

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The slope of the first graph is simply the first-order rate constant because rate = $$k_1[t\text{-}\ce{BuBr]}$$. But te slope of the second grath 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 $$\text{S}_{\text{N}}1$$, that is, Substitution, Nucleophilic, 1st order.

The slope of the first graph is simply the first-order rate constant because rate = $$k_1[t\text{-}\ce{BuBr]}$$. But te slope of the second grath 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 $$\text{S}_{\text{N}}1$$, that is, Substitution, Nucleophilic, 1st order.

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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 $$1^{st}$$ order, which, in my honest opinion, is much more clearer.

# Conclusion:

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 $$\text{RDS}$$ 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.

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 $$1^{st}$$ order, which, in my honest opinion, is much more clearer.

# Conclusion:

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 $$\text{RDS}$$ 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.

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 $$1^{st}$$ order, which, in my honest opinion, is much more clearer.

# Conclusion:

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 $$\text{RDS}$$ 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.

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