Question

The number of occupied sub-shells in Cr is:
A.3
B.4
C.7
D.6

Answer 1

Hint: To answer this question, you must recall how to write the electronic configuration of an element. It is written using a number of rules namely, the Hund’s rule, Pauli’s exclusion principle and Aufbau principle.

Complete answer:
Chromium has the atomic number 24 and is present in the $d - $ block of the modern periodic table. The number of electrons present in a chromium atom is 24.
To write the electronic configuration of chromium atom, we must be familiar with the laws that determine the electronic configuration of an element.
The Hund’s rule of maximum multiplicity states that the pairing of electrons in an orbital does not start until all the orbitals in the sub- shell are singly filled.
The Pauli’s exclusion principle states that any two electrons in the same atom cannot have the exact same quantum number, i.e., each electron has a different set of quantum numbers associated to it.
The Aufbau principle states that atomic orbitals are filled in the increasing order of the energy level of the orbitals given by $\left( {n + l} \right)$
From these, we can write the electronic configuration of chromium as,
$Cr:1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^5},4{s^1}$
Thus, we can see that the number of subshells occupied in chromium atom are 7 $\left( {1s,2s,2p,3s,3p,3d,4s} \right)$

Hence, option C is correct.

Note:
It must be noted that chromium has an exceptional electronic configuration. Considering the rules and general trend, chromium atom is expected to have an electronic configuration as,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^2}{\text{3}}{{\text{d}}^4}$
But, one electron from the $4s$ orbital is shifted to the empty $3d$ orbital on account of the extra stability of a half- filled orbital due to symmetry and greater exchange energy. Thus, the electronic configuration of chromium atom is given as,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$.