The number f-electrons in +3 oxidation state of gadolinium $(Z=64)$ is $x$ and in +2 oxidation state of…

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
  1. $13$
  2. $20$
  3. $18$
  4. $21$

Solution

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 $y=14$ $\Rightarrow x+y=7+14=21$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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