$Na^{+}$ has completely filled $K$ and $L$ shells. Explain.
(N/A) The atomic number of sodium $(Na)$ is $11$.
In a neutral sodium atom,there are $11$ electrons,and its electronic configuration is $2, 8, 1$.
When a sodium atom loses one electron to form a sodium ion $(Na^{+})$,it is left with $10$ electrons.
The distribution of these $10$ electrons in the shells is as follows:
- $K$-shell: $2$ electrons
- $L$-shell: $8$ electrons
Since the maximum capacity of the $K$-shell is $2$ and the $L$-shell is $8$,both shells are completely filled in $Na^{+}$.
In a neutral sodium atom,there are $11$ electrons,and its electronic configuration is $2, 8, 1$.
When a sodium atom loses one electron to form a sodium ion $(Na^{+})$,it is left with $10$ electrons.
The distribution of these $10$ electrons in the shells is as follows:
- $K$-shell: $2$ electrons
- $L$-shell: $8$ electrons
Since the maximum capacity of the $K$-shell is $2$ and the $L$-shell is $8$,both shells are completely filled in $Na^{+}$.
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