An explanation that makes reference to the following
- +5 (1)
- vanadium has 3 (unpaired) electrons in the (3)d subshell and 2 electrons in the 4s (orbital / subshell)
(1)
- the total number of 3d and 4s electrons gives the highest (stable) oxidation state
(1)
Allow 5+ / 5 /+V / V / V^5+
Accept electronic configuration of V is [Ar] 3 d^3 4 s^2 /[Ar] 4 s^2 3 d^3 / 'electrons in boxes' Ignore errors in [Ar] configuration Do not award incorrect electronic configurations of 3d and 4s
Allow electronic configuration of V^5+ shown if electronic configuration of V given for M2
Allow vanadium has 5 electrons in its valence shell Allow vanadium has 5 outer electrons
Allow loss of 5 electrons from electronic configuration [Ar] 3 d^5
Do not award 'orbit' for orbital / shell / subshell
No TE on other incorrect electronic configurations Ignore reference to stability of V^5+ Ignore just ‘loses 5 electrons’