Question
Lanthanum, gadolinium and lutetium are extraordinarily stable in $+3$ oxidation state. Explain.
(Atomic number: $La = 57, Gd = 64, Lu = 71$)
(Atomic number: $La = 57, Gd = 64, Lu = 71$)
✓
Answer
$_{57}\text{La}\xrightarrow{\ \ \ \ \ \ }\ _{54}\text{La}^{3+}$
$[\text{Xe}]\ 5\text{d}1\ 6\text{s}2\ \ \ \ \ \ \ \ \ \ [\text{Xe}]\ 5\text{d}\ 0\ 6\text{s}0\\\ \ \ \ \ \ \ \ \ \ \ \ _{64}\text{Gd}\xrightarrow{\ \ \ \ \ \ \ \ \ }\ _{61}\text{Gd}^{3+}$
$[\text{Xe}]\ 4\text{f}^7\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^7\ 5\text{d}^0\ 6\text{s}^0\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{71}\text{Lu}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }{68}\text{Lu}^{3+}$
$[\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^0\ 6\text{s}^0$
From their electronic configurations, we find that $La^{3+}, Gd^{3+}$ and $Lu^{3+}$ have empty, exactly half-filled and fully filled valency shells, respectively, which make them extra stable.
$[\text{Xe}]\ 5\text{d}1\ 6\text{s}2\ \ \ \ \ \ \ \ \ \ [\text{Xe}]\ 5\text{d}\ 0\ 6\text{s}0\\\ \ \ \ \ \ \ \ \ \ \ \ _{64}\text{Gd}\xrightarrow{\ \ \ \ \ \ \ \ \ }\ _{61}\text{Gd}^{3+}$
$[\text{Xe}]\ 4\text{f}^7\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^7\ 5\text{d}^0\ 6\text{s}^0\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{71}\text{Lu}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }{68}\text{Lu}^{3+}$
$[\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^0\ 6\text{s}^0$
From their electronic configurations, we find that $La^{3+}, Gd^{3+}$ and $Lu^{3+}$ have empty, exactly half-filled and fully filled valency shells, respectively, which make them extra stable.
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