Statistical model of ligand substitution












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
    3 hours 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
    3 hours 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 3 hours ago









andselisk

19.2k662125




19.2k662125










asked 3 hours 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
    3 hours 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
    3 hours 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
3 hours 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
3 hours ago










1 Answer
1






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oldest

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3












$begingroup$

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.






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    3












    $begingroup$

    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









    $endgroup$


















      3












      $begingroup$

      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









      $endgroup$
















        3












        3








        3





        $begingroup$

        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












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        answered 2 hours ago









        orthocresolorthocresol

        40.3k7117247




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