Question

Explain why $Na^+$ has completely filled the K and L shell.

Answer 1

Hint: The atomic number of Sodium is 11. So, the total number of electrons is 11 in the sodium atom. During the formation of the sodium ion, the sodium atom loses one electron from the atom. The electronic configuration of sodium ions becomes (2,8) on losing one electron. So, we can represent the electronic configuration as \[1\mathop s\nolimits^2 {\text{ }}2\mathop s\nolimits^2 {\text{ }}2\mathop p\nolimits^6 \] .

Complete step by step answer:
In this question we have been asked, $Na^+$ has completely filled the K and L shell and what does it mean. So, first of all we will learn what K and L shell.
An electron shell may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1st shell", followed by the "2nd shell", then the "3rd shell". Each shell can contain only a fixed number of electrons.
The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. These shells are denoted by K, L, M and so on.
An atom of Na has a total of 11 electrons. Its electronic configuration will be 2, 8, 1. This means it has 2 electrons in the K shell, 8 electrons in L and 1 electron in the M shell. By removing an electron from this atom we get a positively charged $Na^+$ ion that has a net charge of +1.
So, $Na^+$ ion has one electron less than the Na atom i.e. it has 10 electrons. When it gives away one electron, it is given away from the 3rd orbit. So, according to formula it has 2 electrons in K shell, 8 electrons in L shell, thereby having completely filled K and L shells.

Note:
The general formula is that the nth shell can in general hold up to $\mathop {2n}\nolimits^2 $ electrons, where, ‘n’ is the no. of shell. By putting the value of ‘n’ in this formula we can calculate the no. of maximum electrons acquired in a shell.