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

Options

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

Answer

D

Explanation

`Ti^(2+)=1s^(2),2s^(2)2p^(6), 3s^(2)3p^(6)3d^(2), 4s^(0)`
`V^(3+) = 1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6),3d^(2),4s^(0)`
`Cr^(4+) = 1s^(2), 2s^(2)2p^(6),3s^(2)3p^(6)3d^(2),4s^(0)`
`Mn^(5+) = 1s^(2), 2s^(2)2p^(6),3s^(2)3p^(6)3d^(2),4s^(0)`