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Topic: Comparing LFSE of transition metal complexes  (Read 1525 times)

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Offline Triangel

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Comparing LFSE of transition metal complexes
« on: August 02, 2020, 01:30:45 PM »
Hi all,

I need help checking my answers for my Inorganic chemistry course, and I've run into a few things that I am unsure of. I would very much appriciate if somebody could check my reasoning in this problem.

The task is to compare the LFSE of three pairs of metal complexes, and determine which complex has the highest LFSE. The pairs are:

  • [Cr(OH2)6]2+ vs. [Cr(OH2)6]2+
  • [CrCl6]3- vs. [Cr(NH3)]3+
  • [Ni(OH2)]2+ vs. [Ni(en)3]3+

I have put down the electron configurations of the complexes as follows in the list below, and calculated the LFSE from the formula LFSE = -0,4x + 0,6y, if x is the number of electrons in the t2g-orbitals, and y is the number of electrons in eg-orbitals.

  • [Cr(OH2)6]2+: t2g3eg1, LFSE = -0,6
  • [Cr(OH2)6]2+: t2g3eg0, LFSE = -1,2
  • [CrCl6]3-: t2g3eg0, LFSE = -1,2
  • [Cr(NH3)]3+: t2g3eg0, LFSE = -1,2
  • Ni(OH2)]2+:  t2g6eg2, LFSE = -1,2
  • [Ni(en)3]3+:  t2g6eg2, LFSE = -1,2

As you can see, the last two pairs of complexes then have the same LFSE, which makes me think I have made a mistake somewhere. I am also concerned about the first pair, as a higher oxidation number is supposed to create a higher (more positive?) LFSE with the same ligands, and here the loss of a electron makes it lower.

Any feedback is much appreciated. Cheers!

Offline mjc123

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Re: Comparing LFSE of transition metal complexes
« Reply #1 on: August 03, 2020, 04:17:56 AM »
Quote
[Cr(OH2)6]2+ vs. [Cr(OH2)6]2+
That's a difficult one! I suppose one of them should be 3+, in which case your answer looks right. (I also assume the Ni(en)3 complex should be 2+ not 3+? And you have missed out a 6 in Cr(NH3) and Ni(OH2). Take more care.)

For the last two pairs, LFSE is the same multiple of Δ; you should consider which ligands give the greater magnitude of Δ.

Offline Triangel

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Re: Comparing LFSE of transition metal complexes
« Reply #2 on: August 10, 2020, 04:50:06 PM »
Hi, and sorry for the late reply.

Quote
[Cr(OH2)6]2+ vs. [Cr(OH2)6]2+
That's a difficult one! I suppose one of them should be 3+, in which case your answer looks right. (I also assume the Ni(en)3 complex should be 2+ not 3+? And you have missed out a 6 in Cr(NH3) and Ni(OH2). Take more care.)

Yes, of course, the pairs should read as follows:
  • [Cr(OH2)6]+2 vs. [Cr(OH2)6]+3
  • [CrCl6]3- vs. [Cr(NH3)6]+3
  • [Ni(OH2)6]+2 vs. [Ni(en)3]+3
I think I missed the higher oxidation state in [Ni(en)3]+3 in the electron configuration too, so that should be one less electron in eg.

For the last two pairs, LFSE is the same multiple of Δ; you should consider which ligands give the greater magnitude of Δ.

That helps a lot. So since NH3 is higher in the spectrochemical series than Cl, the LFSE of that complex should be higher, and the same reasoning goes for pair 3, where en > OH2.

Thank you!

Offline mjc123

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Re: Comparing LFSE of transition metal complexes
« Reply #3 on: August 10, 2020, 05:20:13 PM »
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and the same reasoning goes for pair 3, where en > OH2.
Well, the reasoning isn't the same if the en complex really is 3+

Offline Triangel

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Re: Comparing LFSE of transition metal complexes
« Reply #4 on: August 11, 2020, 03:22:19 AM »
...I must have been tired last night. The question has the en complex as +2. Then the logic applies. Lesson learned, double check everything.

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