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

(A)Electron configuration of V is [Ar] `3d^3 4s^2` and thus maximum 5 electrons participate in bonding.
(B)Electron configuration of Cr is [Ar] `3d^5 4s^1` and thus maximum 6 electrons participate in bonding.
( C)Electron configuration 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.
(D)Electron configuration of Sc is [Ar] `3d^1 4s^2` and thus maximum 3 electrons participate in bonding.