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I am a bit confused about how to solve this problem. I can interpret coordination complexes like $\ce{[CoCl6]^4-}$ but I just came across a coordination complex that I am unsure of how to solve.

It is $\ce{[Mn(en)3]^3+}$ where en stands for ethylenediamine. The problem first asks that I draw the structure (not sure how to do this as ethylenediamine is a large molecule). I would also like to know whether this complex would be tetrahedral, octahedral, trigonal planar, etc. I'm just kind of confused as to how to solve this problem when its a large molecule as the ligand rather than a molecule like $\ce{Cl6}$.

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    $\begingroup$ Point of detail: The chloride ligands are not one monolithic counter-ligand. Each ion complexes ~independently with the central metal. $\endgroup$
    – hBy2Py
    Commented Dec 1, 2016 at 1:38
  • $\begingroup$ I was about to contest hexachloridocobaltate(II) when I realised it was cobalt(III) that preferred tetracoordination of chlorido ligands due to its small size. $\endgroup$
    – Jan
    Commented Dec 1, 2016 at 1:42

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In principle, there is no difference between having monodentate ligands such as chlorido ligand or more complex ligands such as ethylenediamine.

The first step is always to draw out the structure of the ligand by itself. Then, identify how many coordinating entities it contains. Ethylenediamine is $\ce{H2N-CH2-CH2-NH2}$ and both nitrogens can form coordinate bonds.

In the next step, identify how many coordinating entities you have in total and thus what the maximum coordination number is. As mentioned, each $\ce{en}$ has two potentially coordinating nitrogens for a total of six dative bonds. Thus, a hexacoordinated octahedron is the maximum possible.

Continue by looking at the metal. What do you know about its properties in complexes? Does it prefer any type of complex? In this case, manganese(III) is absolutely fine with having an octahedral geometry, so that’s taken care of.

Once you’ve done all this, you will notice that the metal enjoys octahedral coordination and there are enough Lewis bases for octahedral coordination to occur — the complex will likely be octahedral. If the complex had been $\ce{[Ni(en)2]^2+}$, we would have arrived at ‘square planar’ by the same methodology.

Next, draw the complex. Start with the metal and the octahedral geometry and put a nitrogen at the end of each line. Then, try to connect the nitrogens with $\ce{CH2-CH2}$ fragments. There should be two possible, enantiomeric outcomes.

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  • $\begingroup$ Hey thanks a lot! One thing that I am curious about is "What do you know about its properties in complexes? Does it prefer any type of complex?" are there some atoms that prefer octahedral/tetrahedral/square planar? Is there a way to tell which prefers which, or is it mostly memorization? $\endgroup$
    – Idiot
    Commented Dec 1, 2016 at 3:35
  • $\begingroup$ @Andrew The most obvious case is nickel(II)’s preference for square planar complexes. And then maybe some small ions that don’t prefer octahedral but are fine with tetrahedral. That’s just an exception phrase to catch those before you start using octahedral for everything ;) $\endgroup$
    – Jan
    Commented Dec 1, 2016 at 18:19
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Basically $\ce{en}$ is a strong field ligand and is a neutral one so the $\ce{Mn}$ in $+3$ state would have $3\mathrm{d^4}$ configuration and $\ce{en}$ would pair them up in the first two orbitals so there would be an inner orbital complex formation with octahedral geometry having $\mathrm{d^2sp^3}$ hybridization.

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    $\begingroup$ Please note that it is practically always a very bad idea to extend hybridisation to transition metal complexes. See my other answers on that topic. $\endgroup$
    – Jan
    Commented Dec 1, 2016 at 18:18

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