Statistical model of ligand substitution Announcing the arrival of Valued Associate #679:...

How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green

Aligning matrix of nodes with grid

How is simplicity better than precision and clarity in prose?

I'm thinking of a number

What is the largest species of polychaete?

Keep going mode for require-package

Is it possible to ask for a hotel room without minibar/extra services?

Why use gamma over alpha radiation?

Need a suitable toxic chemical for a murder plot in my novel

Active filter with series inductor and resistor - do these exist?

How to retrograde a note sequence in Finale?

Direct Experience of Meditation

Failing to enforce immigration laws?

Can I throw a sword that doesn't have the Thrown property at someone?

Cold is to Refrigerator as warm is to?

A constraint that implies convexity

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Limit for e and 1/e

Strange behaviour of Check

Make it rain characters

Can't figure this one out.. What is the missing box?

Estimated State payment too big --> money back; + 2018 Tax Reform

Losing the Initialization Vector in Cipher Block Chaining

Why is there no army of Iron-Mans in the MCU?



Statistical model of ligand substitution



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a reason for the mathematical form of the equilibrium constant?Why is ligand substitution only partial with copper(II) ions and ammonia?Ligand Binding Paradox?How do we decide which ligand is monodentate, bidentate, etc.?In the reversible reactions of acyl substitution, how do backward reactions happen by being against the forward reaction drives?Why is thiourea a monodentate ligand?Is glycine strong or weak field ligand? If yes, how?What is an uninegative ligandWhy does ligand substitution occur when OH- is a better ligand than NH3?Is there a difference between a chelate ligand and a polydentate ligand?












5












$begingroup$


Recently, I was told that in case of a particular step of a generic ligand substitution reaction:



$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$



The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to



$$frac{N - n}{n + 1}$$



by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
    $endgroup$
    – Tyberius
    59 mins ago
















5












$begingroup$


Recently, I was told that in case of a particular step of a generic ligand substitution reaction:



$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$



The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to



$$frac{N - n}{n + 1}$$



by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
    $endgroup$
    – Tyberius
    59 mins ago














5












5








5





$begingroup$


Recently, I was told that in case of a particular step of a generic ligand substitution reaction:



$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$



The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to



$$frac{N - n}{n + 1}$$



by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?










share|improve this question











$endgroup$




Recently, I was told that in case of a particular step of a generic ligand substitution reaction:



$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$



The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to



$$frac{N - n}{n + 1}$$



by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?







equilibrium coordination-compounds






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









andselisk

19.2k662125




19.2k662125










asked 1 hour ago









Shoubhik Raj MaitiShoubhik Raj Maiti

1,398732




1,398732








  • 1




    $begingroup$
    I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
    $endgroup$
    – Tyberius
    59 mins ago














  • 1




    $begingroup$
    I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
    $endgroup$
    – Tyberius
    59 mins ago








1




1




$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
59 mins ago




$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
59 mins ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,



$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$



and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for



$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$



is given by



$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$



The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.






share|improve this answer









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "431"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112749%2fstatistical-model-of-ligand-substitution%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,



    $$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$



    and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for



    $$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$



    is given by



    $$begin{align}
    K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
    &= expleft(frac{Delta_mathrm r S}{R}right) \
    &= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
    &= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
    &= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
    &= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
    &= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
    &= frac{N-n}{n+1}
    end{align}$$



    The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,



      $$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$



      and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for



      $$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$



      is given by



      $$begin{align}
      K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
      &= expleft(frac{Delta_mathrm r S}{R}right) \
      &= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
      &= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
      &= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
      &= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
      &= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
      &= frac{N-n}{n+1}
      end{align}$$



      The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,



        $$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$



        and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for



        $$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$



        is given by



        $$begin{align}
        K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
        &= expleft(frac{Delta_mathrm r S}{R}right) \
        &= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
        &= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
        &= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
        &= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
        &= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
        &= frac{N-n}{n+1}
        end{align}$$



        The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.






        share|improve this answer









        $endgroup$



        I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,



        $$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$



        and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for



        $$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$



        is given by



        $$begin{align}
        K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
        &= expleft(frac{Delta_mathrm r S}{R}right) \
        &= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
        &= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
        &= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
        &= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
        &= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
        &= frac{N-n}{n+1}
        end{align}$$



        The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 24 mins ago









        orthocresolorthocresol

        40.3k7117247




        40.3k7117247






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Chemistry Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112749%2fstatistical-model-of-ligand-substitution%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            迭戈·戈丁...

            A phrase ”follow into" in a context The 2019 Stack Overflow Developer Survey Results Are...

            1960s short story making fun of James Bond-style spy fiction The 2019 Stack Overflow Developer...