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Topic: [Co(NH3)5(H2O)]3+ is inert or labile?  (Read 7694 times)

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Offline sahar58

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[Co(NH3)5(H2O)]3+ is inert or labile?
« on: February 03, 2010, 09:26:04 PM »
As I know [Co(NH3)5(H2O)]3+ is labile, but why has half-life more than 1 day?

Offline Dolphinsiu

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Re: [Co(NH3)5(H2O)]3+ is inert or labile?
« Reply #1 on: February 05, 2010, 11:54:05 AM »
For octahedral geometry, three d orbital is at low lying T2g level, leaving two at high lying eg* level. since all ligands are neutral. Formal oxidation number of Co is Co(III), d6 system, Hence it is definitely inert species, Why it is labile?

Offline sahar58

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Re: [Co(NH3)5(H2O)]3+ is inert or labile?
« Reply #2 on: February 05, 2010, 02:56:41 PM »
For octahedral geometry, three d orbital is at low lying T2g level, leaving two at high lying eg* level. since all ligands are neutral. Formal oxidation number of Co is Co(III), d6 system, Hence it is definitely inert species, Why it is labile?

I think because of H2O ligands that can exchange easily.

for labile and inert, I determine no. of electrons in d, and if it was d4-d6, then of high spin or low spin, consider ligands.

Is that wrong?

Offline Dolphinsiu

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Re: [Co(NH3)5(H2O)]3+ is inert or labile?
« Reply #3 on: February 07, 2010, 12:28:44 PM »
Yes, you remind me one important thing. Ligand is our second concern when we determine whether the system is of high spin or low spin.

For high-spin d6 system, it is really an inert species.

In your case, since all ligands are weak field ligand, low CFSE, low 10Dq, hence, it is low-spin d6 system.
The resulting complex becomes labile (undergoes associative or dissociative pathway more readily). It does find some ways to obtain d5 system, coordinatively inert system.

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