Question

Consider two cations $X$ and $Y$ which have 8 and 18 electrons respectively in their outermost shell (both belong to $3d-$ series). Select the correct statement(s) among the following.

• A. Both the cations have 8 electrons in their penultimate shell.
• B. Difference between the total number of electrons of $X$ and $Y$ is 10.
• C. Cation $Y$ may be of element $Ni$.
• D. Cation $X$ may be of element $Sc$.

Answer 1

Answer: (A), (B), (D)

The correct options are
A Both the cations have 8 electrons in their penultimate shell.
B Difference between the total number of electrons of $X$ and $Y$ is 10.
D Cation $X$ may be of element $Sc$.

The electronic configuration of the elements of the $3d-$ series,
$\Rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^n$
where, ${}^{\prime }{n}^{\prime }$ varies from 1 to 10

Since both the cations ($X$ and $Y$) have 8 and 18 valence electrons respectively, the difference between the number electrons of $X$ and $Y=18-8=10(\because$ Both belong to same shell)

Penultimate shell $(n=2)$ has 8 electrons $\left(2s^{2}2p^6\right)$.

Electronic configuration of Scandium$\left(Sc\right)$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^1$
So, $S{c}^{3+}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^6\right)$ has 8 valence electrons and $X$ may be a cation of $Sc$.

Electronic configuration of Nickel$\left(Ni\right)$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^8$
Since Nickel can have a maximum of 16 valence electrons by losing the outermost two ${}^{\prime }4s^{\prime }$ electrons, cation $Y$ cannot be $Ni$.