Among the following series of transition metal ions, the one where all metal ion have the sae `3d` electronic configuration is

Options

A. `Ti^(2+), V^(3+), Cr^(4+), Mn^(5+)`
B. `Ti^(3+), V^(2+), Cr^(3+), Mn^(4+)`
C. `Ti^(+),V^(4+), Cr^(6+), Mn^(7+)`
D. `Ti^(4+), V^(3+), Cr^(2+), Mn^(3+)`

Answer

A

Explanation

Using the atomic numbers, we find the configuration of atoms:
`Ti(Z = 22),[Ar]3d^(2)4s^(2)`
`V(Z = 23),[Ar]3d^(3)4s^(2)`
`Cr(Z = 24),[Ar]3d^(5)4s^(1)`
`Mn(Z = 25),[Ar]3d^(5)4s^(2)`
Therefore, `Ti^(2+), V^(3+), Cr^(4+)`, and `Mn^(5+)` posses `[Ar]3d^(2)` electron configuration.