Question

The correct electronic configuration of $V^{2+}$ is:

• A. 1 2 2 2 3
• B. 1 2 2 3 3 3
• C. 1 2 2 3 3 4 3
• D. None of the above

Answer 1

The correct option is C

1 2 2 3 3 4 3

The order in which the orbitals are filled is as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s

Atomic number of V = 23

Number of electrons in $V^{2+}$ = 21

$\therefore$ $V^{2+}$ = 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$ 3$p^6$ 4$s^0$ 3$d^3$!

Surprised?
See, the thing is: although 4s has a lower energy than 3d, it is the outermost shell. So any electrons to leave would be the ones from this shell first.