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The number f-electrons in +3 oxidation state of gadolinium $(Z=64)$ is $x$ and in +2 oxidation state of Ytterbium $(\mathrm{Z}=70)$ is $\mathrm{y}$. The sum of $\mathrm{x}$ and $\mathrm{y}$ is
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The correct answer is:
$21$
The outer electronic configuration for $\mathrm{Gd}$ and $\mathrm{Yb}$ are:-
$\begin{aligned}
& \mathrm{Gd}=4 f^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2 \\
& \mathrm{Yb}=4 f^{14} 6 \mathrm{~s}^2
\end{aligned}$
Thus, $\mathrm{Gd}^{3+}=4 f^7$ and $\mathrm{yb}^{2+}=4 f^{14}$
Thus, $x=7$ and $y=14$
$\Rightarrow x+y=7+14=21$
$\begin{aligned}
& \mathrm{Gd}=4 f^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2 \\
& \mathrm{Yb}=4 f^{14} 6 \mathrm{~s}^2
\end{aligned}$
Thus, $\mathrm{Gd}^{3+}=4 f^7$ and $\mathrm{yb}^{2+}=4 f^{14}$
Thus, $x=7$ and $y=14$
$\Rightarrow x+y=7+14=21$
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