Question

The electronic configuration of vanadium is:
A. $[Ar]3d^{4}4s^1$
B. $[Ar]4d^{3}5s^2$
C. $[Ar]3d^{3}4s^2$
D. $[Ar]3d^{5}4s^0$

Answer 1

Hint: Recall the atomic number of the elements vanadium and add the electrons to the shells according to the Aufbau principle. Take note of the block in which it is present.

Complete step by step answer:
Let us look at the periodic table and locate vanadium.

Here, we can see that the atomic number of vanadium is 23 and it is present in the d-block of elements. This implies that the last electron to be filled in the orbitals of vanadium will be in the d-orbital. First, we will see what the configuration of vanadium is, upto argon, or 18 electrons. The configuration of argon is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$, so out of the 23 electrons, 18 electrons have already been placed in orbitals in a stable manner. Only 5 electrons are left to be placed now.
First, let us see the orbitals placed in the order of increasing energy. The order is:
$$1s<2s<2p<3s<3p<4s<3d<4p$$

From this, we can say that after the $3p$ orbital has been filled, the $4s$ orbital has lower energy and will be filled earlier than the $3d$ orbital. So, we will put the 5 electrons as $4s^{2}3d^3$. The total electronic configuration according to this will be $[Ar]4s^{2}3d^3$.
So, the correct answer is “Option C”.

Note: Please don’t get confused if the orbitals are not given in the order specified by the Aufbau principle. Always draw the diagram for the relative energy of the orbitals before attempting these questions.