Question

Given,$H{{g}^{2+}}$, $N{{a}^{+}}$ have completely filled K and L shells. Explain?

Answer 1

Hint: The atomic number of an element is equal to the number of electrons present in an atom. So, the Atomic number of Na (Sodium) is 11 and that of Hg (mercury) is 80. Na loses one electron to become $N{{a}^{+}}$ and Hg loses two electrons to become $H{{g}^{2+}}$.

Complete answer:
The shells are namely- K, L, M and N in increasing order of the capacity. The first shell closer to the nucleus is K and so on.

Shell	K	L	M	N	O
Electron filling capacity	2	8	18	32	50

The general formula to find the electron filling capacity of a shell is $2({{n}^{2}})$, where n is the number of shells from the nucleus.
We know the atomic number of Na is 11 and Hg is 80.
Thus, the number of electrons will be:
No of electrons in Na: 11
No of electrons in Hg: 80
Na becomes $N{{a}^{+}}$and Hg becomes $H{{g}^{2+}}$when they lose one electron and two electrons respectively.
Therefore,
$N{{a}^{+}}$ will have 10 electrons and $H{{g}^{2+}}$ will have 78 electrons.
Now, let us see the electronic configuration of $N{{a}^{+}}$ and $H{{g}^{2+}}$.
Electronic configuration of $N{{a}^{+}}$:

K	L
2	8

Electronic configuration of $H{{g}^{2+}}$:

K	L	M	N	O
2	8	18	32	18

The electrons lost from the atom are from the last shells of the atom. In the case of Na, the electron was lost from the M shell and in Hg it was lost from the O shell.
So, we can observe that the K and L of $H{{g}^{2+}}$and $N{{a}^{+}}$are completely filled with electrons

Note:
When an atom loses electrons it gains a positive charge. The number of protons becomes more than the number of electrons. Also keep in mind the maximum accommodation capacity of the shells. But the electrons are filled as octets in shells due to the stability of the atom.