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For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Original topic: Predominance of III oxidation state for lanthanidesPredominance of III oxidation state for lanthanides

For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Original topic: Predominance of III oxidation state for lanthanides

For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Original topic: Predominance of III oxidation state for lanthanides

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Why don't we see these lanthanide species?

For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Original topic: Predominance of III oxidation state for lanthanides