Which condensed electronic configuration is correct for the [tex]$Cr^{4+}$[/tex] ion?

A. [tex][\operatorname{Ar}] 3d^2[/tex]
B. [tex][\operatorname{Ar}] 4s^2 3d^6[/tex]
C. [tex][\operatorname{Ar}] 4s^2 3d^4[/tex]
D. [tex][\operatorname{Ar}] 4s^2 3d^2[/tex]



Answer :

To determine the condensed electronic configuration for the \( Cr^{4+} \) ion, let's follow these steps step-by-step:

1. Find the ground-state electron configuration of neutral chromium (Cr):
- Chromium has an atomic number of 24, which means it has 24 electrons.
- The ground-state electron configuration of chromium is: \([Ar] 3d^5 4s^1\).

2. Determine the number of electrons to remove:
- For the \( Cr^{4+} \) ion, we need to remove 4 electrons from the neutral chromium atom.

3. Remove the electrons:
- Electrons are removed from the outermost orbitals first. In this case, we will remove electrons from the 4s orbital first and then from the 3d orbitals:
- First, remove 1 electron from the 4s orbital: \([Ar] 3d^5 4s^1 \rightarrow [Ar] 3d^5\).
- Then, remove 3 electrons from the 3d orbitals: \([Ar] 3d^5 \rightarrow [Ar] 3d^2\).

4. Write the resulting configuration:
- After removing the 4 electrons, the condensed electron configuration of the \( Cr^{4+} \) ion is: \([Ar] 3d^2\).

Thus, the correct condensed electronic configuration for the [tex]\( Cr^{4+} \)[/tex] ion is [tex]\([Ar] 3d^2\)[/tex].