1.

Which one of the following ions is the most stable in aqueous solutions ? (At. No. Ti=22, V=23, Cr= 24 , Mn = 25)

Answer»

`Mn^(2+)`
` Cr^(3+)`
`V^(3+)`
`Ti^(3+)`

Solution :Out of the given species, `Cr^(3+)` has the highest negative REDUCTION potential.Hence, it cannot be reducedto `Cr^(2+)` and, THEREFORE, is the most stable ion in aqueous solution.
ALTERNATIVELY,
`Mn^(3+) = [ AAr] 3d^(4) , Cr^(3+)= [AR]3d^(3)`
`V^(3+) =[Ar]3d^(2)Ti ^(3+) = [ Ar] 3d^(1)`
In `Cr^(3+)`, all three d electrons enter intolower energy `t_(2g)` orbitals. The lowering of energy is MAXIMUM and hence stability is maximum.


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