Answer :
Given the ion options and the requirement that the ion should have eight valence electrons, let's analyze each ion's electron configuration step by step.
1. Scandium (Sc) \(^{4+}\):
- Scandium has an atomic number of 21 with the electron configuration \([Ar] 3d^1 4s^2\).
- When Scandium loses 4 electrons (to form Sc\(^{4+}\)), it loses all electrons from the 3d and 4s orbitals.
- This means Sc\(^{4+}\) reverts to the electron configuration of \([Ar]\) (Argon).
- Argon has 8 valence electrons in its outer shell.
- However, these are core electrons now in the ion Sc\(^{4+}\), and no electrons remain in the outermost energy level.
2. Titanium (Ti) \(^{4+}\):
- Titanium has an atomic number of 22 with the electron configuration \([Ar] 3d^2 4s^2\).
- When Titanium loses 4 electrons (to form Ti\(^{4+}\)), it loses all electrons from the 3d and 4s orbitals.
- This means Ti\(^{4+}\) reverts to the electron configuration of \([Ar]\).
- Argon has 8 valence electrons in its outer shell, but again, none remain valence electrons in the ion Ti\(^{4+}\).
3. Chromium (Cr) \(^{4+}\):
- Chromium has an atomic number of 24 with the electron configuration \([Ar] 3d^5 4s^1\).
- When Chromium loses 4 electrons (to form Cr\(^{4+}\)), it will lose electrons from both 3d and 4s orbitals.
- This results in Cr\(^{4+}\) having the electron configuration \([Ar] 3d^1\).
- This leaves 1 electron in the outermost orbits, not satisfying the 8 valence electron condition.
4. Manganese (Mn) \(^{4+}\):
- Manganese has an atomic number of 25 with the electron configuration \([Ar] 3d^5 4s^2\).
- When Manganese loses 4 electrons (to form Mn\(^{4+}\)), it will lose electrons from both 3d and 4s orbitals.
- The configuration of Mn\(^{4+}\) becomes \([Ar] 3d^3\).
- This leaves 3 valence electrons, not satisfying the 8 valence electron requirement.
5. Vanadium (V) \(^{4+}\):
- Vanadium has an atomic number of 23 with the electron configuration \([Ar] 3d^3 4s^2\).
- When Vanadium loses 4 electrons (to form V\(^{4+}\)), it loses electrons from both 3d and 4s orbitals.
- This results in V\(^{4+}\) having the electron configuration \([Ar] 3d^1\).
- This leaves 1 valence electron, not satisfying the 8 valence electron condition.
Given all the information above, none of the given ions (Sc\(^{4+}\), Ti\(^{4+}\), Cr\(^{4+}\), Mn\(^{4+}\), and V\(^{4+}\)) satisfies the condition of having eight valence electrons. Therefore, the result is:
None.
1. Scandium (Sc) \(^{4+}\):
- Scandium has an atomic number of 21 with the electron configuration \([Ar] 3d^1 4s^2\).
- When Scandium loses 4 electrons (to form Sc\(^{4+}\)), it loses all electrons from the 3d and 4s orbitals.
- This means Sc\(^{4+}\) reverts to the electron configuration of \([Ar]\) (Argon).
- Argon has 8 valence electrons in its outer shell.
- However, these are core electrons now in the ion Sc\(^{4+}\), and no electrons remain in the outermost energy level.
2. Titanium (Ti) \(^{4+}\):
- Titanium has an atomic number of 22 with the electron configuration \([Ar] 3d^2 4s^2\).
- When Titanium loses 4 electrons (to form Ti\(^{4+}\)), it loses all electrons from the 3d and 4s orbitals.
- This means Ti\(^{4+}\) reverts to the electron configuration of \([Ar]\).
- Argon has 8 valence electrons in its outer shell, but again, none remain valence electrons in the ion Ti\(^{4+}\).
3. Chromium (Cr) \(^{4+}\):
- Chromium has an atomic number of 24 with the electron configuration \([Ar] 3d^5 4s^1\).
- When Chromium loses 4 electrons (to form Cr\(^{4+}\)), it will lose electrons from both 3d and 4s orbitals.
- This results in Cr\(^{4+}\) having the electron configuration \([Ar] 3d^1\).
- This leaves 1 electron in the outermost orbits, not satisfying the 8 valence electron condition.
4. Manganese (Mn) \(^{4+}\):
- Manganese has an atomic number of 25 with the electron configuration \([Ar] 3d^5 4s^2\).
- When Manganese loses 4 electrons (to form Mn\(^{4+}\)), it will lose electrons from both 3d and 4s orbitals.
- The configuration of Mn\(^{4+}\) becomes \([Ar] 3d^3\).
- This leaves 3 valence electrons, not satisfying the 8 valence electron requirement.
5. Vanadium (V) \(^{4+}\):
- Vanadium has an atomic number of 23 with the electron configuration \([Ar] 3d^3 4s^2\).
- When Vanadium loses 4 electrons (to form V\(^{4+}\)), it loses electrons from both 3d and 4s orbitals.
- This results in V\(^{4+}\) having the electron configuration \([Ar] 3d^1\).
- This leaves 1 valence electron, not satisfying the 8 valence electron condition.
Given all the information above, none of the given ions (Sc\(^{4+}\), Ti\(^{4+}\), Cr\(^{4+}\), Mn\(^{4+}\), and V\(^{4+}\)) satisfies the condition of having eight valence electrons. Therefore, the result is:
None.
