[Ar]3d34s2
The electron configuration of vanadium in its 3 oxidation state is Ar 3d2.
The electron configuration diagram for vanadium is 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
The electron configuration of a vanadium atom in its ground state in the V3 oxidation state is Ar 3d2.
Vanadium ( V) has that configuration. Its atomic number is 23.
There are 3 unpaired electrons in a vanadium atom, as vanadium has an electron configuration of [Ar] 3d^3 4s^2.
1s2 2s2 2p6 3s2 3p6 4s2 3d3A+
The orbital diagram for vanadium shows five electrons in the 3d orbital and two electrons in the 4s orbital. This arrangement reflects the electron configuration of vanadium, which is Ar 3d3 4s2.
The magnetic moment of 1.73 μB suggests that the vanadium compound has one unpaired electron, which typically corresponds to a +4 oxidation state. Vanadium in the +4 oxidation state (V^4+) has the electron configuration of [Ar] 3d^1 4s^0, leading to one unpaired electron in the d-orbital. Therefore, the oxidation state of vanadium in this compound is +4.
1s2 2s2 2p6 3s2 3p6 4s2 3d3
[Ar]3d34s2 is the shorthand electron configuration for Vanadium (V).
The orbital configuration of vanadium is 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
The element with the electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d3 is Vanadium with atomic number 23.