Question

1) What is the electron configuration of
`\mathrm{V}^(3 *)` ?


`[\mathrm{Ar}] 4 s^2 3 d^3`

`[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^0`

`[\mathrm{Ar}\space ] 45^0 3 d^2`

`[\mathrm{Ar}] 4 s^2 3 d^5`

Answer 1

V³⁺: [Ar]4s⁰3d²

Step-by-step explanation:

For V³⁺, 3 electrons are transferred, first from the valence level followed by lower energy sublevels. That is:

V⁰ ground state is [Ar]4s²3d³ => [Ar]4s⁰3d² + 3e⁻. The higher principle quantum number is n = 4 valence shell or the 4s² orbital and the next lower energy level is the 3d³ series, or 3d¹d¹d¹d⁰d⁰ for the paramagnetic electrons. Ionization, again begins at the 4s² level giving 4s⁰ followed by the 3d³ level giving 3d¹d¹d⁰d⁰d⁰. The resulting configuration is therefore

V³⁺:[Ar]4s⁰3d¹d¹d⁰d⁰d⁰.