(a). Why `Mn^(2+)` compounds are more stable than `Fe^(2+)` towards oxidation to their `+3` state?
(b). Calculate the magnetic moment of `V^(3+)` ion.
(c). `[Ti(H_2O)_6_^(3+)` gives violet coloured aqueous solution but
`[Mg(H_2O)_6]^(2+)` solution is colourless.
(d). `[Ti(H_2O)_6]^(3+)` is coloured whereas `[Sc(H_2O)_6]^(3+)` is colourless. Why?

Explanation

(a). Electronic configuration of `Mn^(2+)` is `3d^5` which is half filled and hence stable. Therefore `IF_3` is very high i.e., 3rd electron cannot be lost easily. In case of `Fe^(2+)`, electronic configuration is `3D^6`.
(b). `V(Z=23)implies3d^34s^2,V^(3+)=3d^2(n=2)`
`mu=sqrt(n(n+2))BM=sqrt(2(2+2))=sqrt(8)=2.73BM`.
(c). In `[Ti(H_2O)_6]^(3+)`, the `d^1` electron occupies `t_(2g)` orbital in octahedral field. On irradiation with light, the `t_(2g)` electron is promoted to `e_g` orbital and the resulting absorption band goves violet colour.
In case of `[Mg(H_2O)_6]^(2+)` the electronic configuration of `Mg^(2+)` is `1s^2`, `2p^6` which does not permit any electronic transition `(2pto3s)` as the energy gap is very large or there does not occur d-d transition of electron. Hence, it gives colourless solution.
(d). A in `[Ti(H_2O)_6]^(3+)`, the titanium ion is present in `Ti^(3+` form with `3d^1`, `4s^0` configuration. The single upaired electron of `3d^1` orbital makes the compound to show colour. In `[Sc(H_2O)_6]^(3+)` scandium is present as `Sc^(3+)` state with `3d^0`,`4s^0` configuration. Since no electron is present in the 3d and 4 s orbitals, it remains colourless.