Question

The electronic configuration of $C{r^{3 + }}$ is
A.$\;\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{4}}}{\text{ 4}}{{\text{s}}^{\text{2}}}$
B. $\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{3}}}{\text{ 4}}{{\text{s}}^{\text{0}}}$
C. $\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{2}}}{\text{ 4}}{{\text{s}}^{\text{1}}}$
D. $\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{5}}}{\text{ 4}}{{\text{s}}^{\text{1}}}$

Answer 1

Hint - The element name of Cr is chromium. And the concept of this question is that the filling of electrons in the shells and subshells of Cr3+ takes place on the basis of Aufbau rule.

Complete step by step solution:
We have to remember that the chromium is a chemical element with the symbol ${\text{Cr}}$ and atomic number 24. It is the first element in group 6 and belongs to transition metal. Now, the question is on electron configuration, therefore, it is important to understand what is that? It basically describes how electrons are distributed in its atomic orbitals in an element. Electron configuration is a rule that follows certain rules to arrange the electrons in the main shell and subshell of atoms.
Now, the electronic configuration of chromium is discussed below

{{\text{Cr = 1}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{s}}^{\text{2}}}{\text{ 3}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{d}}^{\text{5}}}{\text{ 4}}{{\text{s}}^{\text{1}}}} \\\ {\text{where, Ar}} = {\text{1}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{s}}^{\text{2}}}{\text{ 3}}{{\text{p}}^{\text{6}}} \\\ {\text{Cr = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{5}}}{\text{ 4}}{{\text{s}}^{\text{1}}} \\\ \\\ {{\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ = }}\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{3}}}} \end{array}$$ Hence, we can conclude that the correct option for the given question is $$\left[ {{\text{Ar}}} \right]{\text{ 3}}{{\text{d}}^{\text{3}}}{\text{ 4}}{{\text{s}}^{\text{0}}}$$ For other things we need to be understood why this kind of filling is to be followed. According to the normal method, the electron configuration of $${\text{Cr}}$$ should be: $${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{\text{4}}}$$ or $$\left[ {{\text{Ar}}} \right]{\text{\;4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{\text{4}}}$$ But instead is: $${\text{1}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{s}}^{\text{2}}}{\text{ 3}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{d}}^{\text{5}}}{\text{ 4}}{{\text{s}}^{\text{1}}}$$ or $$\left[ {{\text{Ar}}} \right]{\text{\;4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$$as you can see above! This kind of filling actually violates the Aufbau Principle, which states that electrons orbiting one or more atoms fill the lowest available energy levels (subshells) before filling higher levels. To understand why this occurs, it is important to realize that a subshell which is exactly half filled is more stable than a partially filled subshell which is not half full. An electron therefore, moves from the 4s subshell to not completely half-fill the 3d subshell which gives the atom greater stability, so the change is favorable. It is important to understand that the minimized coulombic repulsion energy further stabilizes this configuration, i.e. the minimization comes from having all unpaired electrons in the 3d and 4s ($${\text{3}}{{\text{d}}^{\text{5}}}{\text{4}}{{\text{s}}^{\text{1}}}$$), rather than one electron pair in the 4s ($${\text{3}}{{\text{d}}^{\text{4}}}{\text{ 4}}{{\text{s}}^{\text{2}}}$$). Note: We must understand that the when$${\text{Cr}}$$get ionized to $${\text{C}}{{\text{r}}^{{\text{3 + }}}}$$ ion, first we must remove electrons from ‘s’ orbital then from‘d’ orbitals. This is because the outer shell is maximum shielded from the influence of the nucleus, and hence lower energy is required to extract electrons from the outermost shell than from an inner shell. Since 4s orbital lies in the fourth shell, and 3d in the third shell, electrons are removed first from 4s.