[sahar58]

As I know [Co(NH3)5(H2O)]3+ is labile, but why has half-life more than 1 day?

[Dolphinsiu]

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?

[sahar58]

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?

[Dolphinsiu]

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.