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].
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].
