Among the following series of transition metal ions, the one where all metal ions have `3d^(2)` electronic configuration is:

Options

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

Answer

B

Explanation

`Ti_(22) = 3d^2 4s^2 , Ti^(2+) = 3d^2`
`V_23 = 3d^3 4s^2 , V^(3+) = 3d^2`
`Cr_(24) = 3d^4 4s^2 , Cr^(4+) = 3d^2`
`Mn_25 = 3d^5 4s^2 , Mn^(5+) = 3d^2`.