Answer :
Explanation:
Atomic number of Vanadium is 23 and its electronic distribution is 2, 8, 8, 5. And, we know that when an atom has a positive charge then it means electrons have been lost by the atom.
As the given Vanadium atom has a charge of +3 which means that 3 electrons have been lost by Vanadium. Hence, electronic configuration of Vanadium in both neutral and ionic form are as follows.
Neutral state: [tex][Ar]4s^{2} 3d^{3}[/tex]
Ionic state ([tex]V^{3+}[/tex]): [tex][Ar]4s^{0} 3d^{2}[/tex]
When an atom is neutral then its number of protons is equal to the number of electrons.
Hence, we can conclude that in [tex]V^{3+}[/tex] number of protons is 23 and number of electrons is 20.