Electronic Configuration — Revision Notes

⚡ 30-Second Revision

General configuration: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$
Electrons removed first from $6s$, then $5d$, then $4f$.
Exceptions with $5d^1$ (ground state): — Ce ($4f^1 5d^1 6s^2$), Gd ($4f^7 5d^1 6s^2$), Lu ($4f^{14} 5d^1 6s^2$).
Stability: — Half-filled ($f^7$) and completely filled ($f^{14}$) $f$-orbitals are highly stable.
$+2$ Oxidation State: — Eu ($4f^7 6s^2$) and Yb ($4f^{14} 6s^2$) form stable $+2$ ions ($Eu^{2+}$ is $4f^7$, $Yb^{2+}$ is $4f^{14}$). They are good reducing agents.
$+4$ Oxidation State: — Ce ($4f^1 5d^1 6s^2$) forms stable $Ce^{4+}$ ($f^0$). It is a good oxidizing agent.
Diamagnetic ions: — $La^{3+}$ ($f^0$), $Lu^{3+}$ ($f^{14}$), $Ce^{4+}$ ($f^0$). All others are generally paramagnetic.

2-Minute Revision

The electronic configuration of lanthanoids (Z=58-71) is crucial for NEET. The general form is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$. Remember that $6s$ electrons are always present and are removed first during ionization.

The key challenge lies in the exceptions to the strict $4f$ filling. Cerium (Ce, Z=58), Gadolinium (Gd, Z=64), and Lutetium (Lu, Z=71) are the three lanthanoids that have a $5d^1$ electron in their ground state.

This $5d^1$ electron often contributes to achieving stable $f^0$ (for $Ce^{4+}$), $f^7$ (for Gd), or $f^{14}$ (for Lu) configurations. The stability of half-filled ($f^7$) and completely filled ($f^{14}$) $f$-orbitals is a major driving force.

This explains why Europium (Eu, Z=63, $4f^7 6s^2$) and Ytterbium (Yb, Z=70, $4f^{14} 6s^2$) readily form stable $+2$ ions ($Eu^{2+}$ is $4f^7$, $Yb^{2+}$ is $4f^{14}$), making them strong reducing agents.

Conversely, Cerium's tendency to form $Ce^{4+}$ ($f^0$) makes it an oxidizing agent. For magnetic properties, count unpaired $f$-electrons; $La^{3+}$ ($f^0$) and $Lu^{3+}$ ($f^{14}$) are diamagnetic.

5-Minute Revision

A thorough understanding of lanthanoid electronic configuration is vital for NEET. Start with the general configuration: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$. The Xenon core accounts for 54 electrons. The $6s^2$ electrons are the outermost and are always present, being the first to be lost during ionization, leading to the common $+3$ oxidation state for all lanthanoids. The distinguishing electrons enter the $4f$ subshell.

Key Exceptions and Stability:

1
Cerium (Ce, Z=58): — $[Xe] 4f^1 5d^1 6s^2$. The $5d^1$ electron is present. $Ce^{4+}$ ($[Xe]$ or $f^0$) is very stable, making $Ce^{3+}$ a good reducing agent and $Ce^{4+}$ a good oxidizing agent.
2
Europium (Eu, Z=63): — $[Xe] 4f^7 6s^2$. It achieves a stable half-filled $4f^7$ configuration. $Eu^{2+}$ ($[Xe] 4f^7$) is stable and a strong reducing agent.
3
Gadolinium (Gd, Z=64): — $[Xe] 4f^7 5d^1 6s^2$. The $5d^1$ electron allows the $4f$ subshell to remain half-filled ($f^7$), which is highly stable.
4
Ytterbium (Yb, Z=70): — $[Xe] 4f^{14} 6s^2$. It achieves a stable completely filled $4f^{14}$ configuration. $Yb^{2+}$ ($[Xe] 4f^{14}$) is stable and a strong reducing agent.
5
Lutetium (Lu, Z=71): — $[Xe] 4f^{14} 5d^1 6s^2$. The $5d^1$ electron is present after the $4f$ subshell is completely filled.

Ion Formation: When forming ions, electrons are removed in the order: $6s \rightarrow 5d \rightarrow 4f$. For example, for $Gd^{3+}$, you remove $6s^2$ and $5d^1$ from $[Xe] 4f^7 5d^1 6s^2$, resulting in $[Xe] 4f^7$.

Magnetic Properties: Paramagnetism arises from unpaired electrons. Diamagnetism occurs when all electrons are paired. Ions with $f^0$ ($La^{3+}$, $Ce^{4+}$) or $f^{14}$ ($Lu^{3+}$, $Yb^{2+}$) configurations are diamagnetic. All other lanthanoid ions are generally paramagnetic due to unpaired $f$-electrons. For example, $Gd^{3+}$ ($4f^7$) has 7 unpaired electrons and is highly paramagnetic.

Example: What is the configuration of $Sm^{3+}$ (Samarium, Z=62)? Neutral Sm: $[Xe] 4f^6 6s^2$. To form $Sm^{3+}$, remove two $6s$ electrons and one $4f$ electron. So, $Sm^{3+}$ is $[Xe] 4f^5$. This ion would be paramagnetic with 5 unpaired electrons.

Prelims Revision Notes

Lanthanoids: Electronic Configuration (NEET Revision)

1. General Configuration:

Lanthanoids (Z=58 to 71) are f-block elements.
General electronic configuration: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
$[Xe]$ represents the Xenon core ($1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6$).
$6s^2$ electrons are always present and are the first to be lost during ionization.

2. Key Exceptions (Ground State $5d^1$):

Cerium (Ce, Z=58): — $[Xe] 4f^1 5d^1 6s^2$. (Not $4f^2 6s^2$)
Gadolinium (Gd, Z=64): — $[Xe] 4f^7 5d^1 6s^2$. (Not $4f^8 6s^2$, due to stable half-filled $f^7$)
Lutetium (Lu, Z=71): — $[Xe] 4f^{14} 5d^1 6s^2$. (Not $4f^{15} 6s^2$, $4f$ is full, next electron goes to $5d$)

3. Other Important Configurations (Stability-Driven):

Europium (Eu, Z=63): — $[Xe] 4f^7 6s^2$. (Stable half-filled $f^7$)
Ytterbium (Yb, Z=70): — $[Xe] 4f^{14} 6s^2$. (Stable completely filled $f^{14}$)

4. Ionization Rules:

Electrons are removed from the outermost shell first (highest principal quantum number $n$).
Order of removal: $6s \rightarrow 5d \rightarrow 4f$.
Example: — For $Gd^{3+}$: From $[Xe] 4f^7 5d^1 6s^2$, remove $6s^2$ and $5d^1$. Result: $[Xe] 4f^7$.

5. Oxidation States & Stability:

Most Common: — $+3$ (loss of $6s^2$ and one $4f$/$5d$ electron).
$+2$ Oxidation State: — Stable for $Eu^{2+}$ ($[Xe] 4f^7$) and $Yb^{2+}$ ($[Xe] 4f^{14}$) due to $f^7$ and $f^{14}$ stability. These are good reducing agents.
$+4$ Oxidation State: — Stable for $Ce^{4+}$ ($[Xe]$ or $f^0$) due to noble gas configuration. $Ce^{4+}$ is a good oxidizing agent.

6. Magnetic Properties:

Paramagnetic: — Ions with unpaired electrons in $4f$ orbitals. Most lanthanoid ions are paramagnetic.
Diamagnetic: — Ions with zero unpaired electrons.

* $La^{3+}$ ($[Xe]$ or $4f^0$) * $Lu^{3+}$ ($[Xe] 4f^{14}$) * $Ce^{4+}$ ($[Xe]$ or $4f^0$)

**7. Mnemonic for $5d^1$ exceptions: Cute Girls Love U**s (Ce, Gd, Lu).

Vyyuha Quick Recall

To remember the lanthanoids with a $5d^1$ electron in their ground state: Cute Girls Love Us. (Ce, Gd, Lu)