Question

The correct ground state configuration of chromium atom is:
(A) $[Ar]3{d^5}4{s^2}$
(B) $[Ar]3{d^4}4{s^2}$
(C) $[Ar]3{d^6}4{s^0}$
(D) $[Ar]3{d^5}4{s^1}$

Answer 1

Hint : First identify to which block does chromium belong among the four blocks of periodic table which are s,p,d and f. Then write the general electronic configuration of all the blocks and write the ground state configuration of chromium according to its atomic number.

Complete step by step solution
First let us see to which block does chromium belong to. The atomic number of chromium is $24$ . According to the atomic number it belongs to the d-block of the periodic table.
d-block of the periodic table ranges from group 3 group 12. Elements of d-block are also called transition metals. They are called d-block elements because the last valence electron goes in d or s-orbital and their general configuration is $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$ . Now let us look at the electronic configuration of the element just before chromium to predict the configuration of chromium. The element before chromium is vanadium and its electronic configuration is $[Ar]3{d^3}4{s^2}$ . So if we add one electron to the configuration of vanadium we should get $[Ar]3{d^4}4{s^2}$ . But this is not the case. We know partially filled and completely filled orbitals are more stable. So in this case, if we put the extra electron of chromium in the d-orbital of vanadium configuration and also one electron of the s-orbital, we would be left with five electrons in d orbital and one electron in s orbital. So we will have a half-filled d-orbital and half-filled s-orbital. Hence electronic configuration of chromium is $[Ar]3{d^5}4{s^1}$ .
So option D is the correct answer.

Note
Chromium is a constituent of steel as it does not undergo rusting. It is to be noted that it is the hardest metal. Because of its lustrous nature it is mainly used for chroming and finishing purposes.