Which of the following is correct condensed electronic configuration for ground state V3+ion?
Select one:
a. [Ar], 4s0, 3d2
b. [Ar], 4s2, 3d0
c. [Kr], 4s0, 3d2
d. [Ne], 4s0, 3d2
Vanadium (V) has atomic number =23.
Electronic configuration of vanadium (V) = [Ar] 3d3 4s2
When vanadium is in +3 oxidation state then 3 electrons are removed from its electronic configuration & first from the outermost shell.
That's why, Electronic configuration of V3+ = [Ar] 3d2 4s0.
Hence, correct option is (a).
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