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Author Topic: Inert and labile complexes  (Read 9798 times)

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NTUstudent

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Inert and labile complexes
« on: December 07, 2010, 11:25:28 PM »

Can somebody explain about inertness and lability of complexes and please help with this question:

Among the electronic configurations of octahedral complexes of
a)low-spin d^2
b)high-spin d^5
c)low-spin d^5
d)high-spin d^7
e)low-spin d^9 which ones are generally inert?
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AWK

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TheUnfocusedOne

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Re: Inert and labile complexes
« Reply #2 on: December 08, 2010, 05:40:26 AM »

Just remember that the reorganization of orbitals and electron structure costs energy. Nature is lazy. The ones that will experience any type of reorganization will be much less likely to drop a ligand than those who don't experience such changes.

Taube used the relationship of electronic structure and substitution reactivity to prove the existence of outersphere electronic transfers. It won him a nobel prize, so it couldn't of been to bad of an experiment.
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darko

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Re: Inert and labile complexes
« Reply #3 on: December 08, 2010, 06:58:17 AM »

Can somebody explain about inertness and lability of complexes and please help with this question:

Among the electronic configurations of octahedral complexes of
a)low-spin d^2
b)high-spin d^5
c)low-spin d^5
d)high-spin d^7
e)low-spin d^9 which ones are generally inert?

I am little bit confused about low-spin d2, how is posiblle to get pairing of two electrons i three degenerate orbitals (or two)?
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TheUnfocusedOne

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Re: Inert and labile complexes
« Reply #4 on: December 08, 2010, 07:58:49 AM »

What makes you think that they would be paired?

A paired state, though possible, would be very unfavored for this configuration.

The lowest states would be the ones that have the electrons seperate and has the third empty. Low spin doesn't mean that the electrons are paired.  Low spin just means that the energy of repulsion between paired electron is not greater than the energy required to have some of the electrons in the higher energy orbitals. In other words, the molecule wouldn't be any lower in energy with its electrons split between the two sets of orbitals. The molecules are trying to get their energy to the lowest state possible.
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darko

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Re: Inert and labile complexes
« Reply #5 on: December 08, 2010, 10:30:38 AM »

you did not understand me, I think that d2 low spin should refer to situation when two electrons are paired, high spin is when thay are unpaird, in general, for me it make no sence to refer to d2 low spin (althuogh is maybe in some special case possible), it is in general same for d1 ,d2, d3, considering 1.Hundt rule, or I am wrong?

What makes you think that they would be paired?

A paired state, though possible, would be very unfavored for this configuration.

The lowest states would be the ones that have the electrons seperate and has the third empty. Low spin doesn't mean that the electrons are paired.  Low spin just means that the energy of repulsion between paired electron is not greater than the energy required to have some of the electrons in the higher energy orbitals. In other words, the molecule wouldn't be any lower in energy with its electrons split between the two sets of orbitals. The molecules are trying to get their energy to the lowest state possible.


but in case of d2 there are three degenerated (same energy) orbitals t2g (or at leest two) , what should be high-spin d2? It is only one state (I am not so sure for excaptions) possible , i am not sure that is posible to hawe one electron in t2g and one in eg orbitals  (that will be higher energy)  :-\
 if exceptions are posible how you will call dxy1 dxz1 and how you will call dxy2 is the first one high spin (orbitals are of same energy) or low spin? is the second one high or low spin? the same situation is with d9, it is only one configuration possible t2g6 eg3, it is not low spin it is only one possible

I think that high and low spin classification is possible for d4,d5,d6,d7 configurations

 
« Last Edit: December 08, 2010, 10:58:55 AM by darko »
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TheUnfocusedOne

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Re: Inert and labile complexes
« Reply #6 on: December 08, 2010, 11:14:37 AM »


I think that high and low spin classification is possible for d4,d5,d6,d7 configurations


Short answer yes. You can have configurations where the two electrons are populating both the t2g and eg, but im not sure if you would consider it high spin or low spin. 
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NTUstudent

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NTUstudent

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Re: Inert and labile complexes
« Reply #8 on: December 08, 2010, 03:11:42 PM »

Can somebody explain about inertness and lability of complexes and please help with this question:

Among the electronic configurations of octahedral complexes of
a)low-spin d^2
b)high-spin d^5
c)low-spin d^5
d)high-spin d^7
e)low-spin d^9 which ones are generally inert?


i think there was mistake, actually i copied original question,
if you are confused read about it more! solve it! :)

I am little bit confused about low-spin d2, how is posiblle to get pairing of two electrons i three degenerate orbitals (or two)?
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