The maximum oxidation state shown by `V(Z=23),Cr(Z=24),Co(Z=27),Sc(Z=21)`, are respectively

Options

A. `+5,+6,+3,+3`
B. `+3,+4,+5,+2`
C. `+5,+3,+2,+1`
D. `+4` in each case.

Answer

A

Explanation

1. Electron configuration of V is [Ar] `3d^(3)4s^(2)` and thus maximum 5 electrons participate in bonding.
2. electro configuraiton of Cr is [Ar] `3d^(5)4s^(1)` and thus maximum 6 electrons participate in bonding
3. Electron configuraiton of Co is [Ar] `3d^(7)4s^(2)`. In octahedral splitting in presence of ligands, half filled `t_(2g)^(6)` has higher CFSE and thus +3 oxidation state is most stable.
4. Electron configuraiton of Sc is [Ar] `3d^(1)4s^(2)` and thus maximum 3 electrons participate in bonding.