Question

A neutral atom of an element has $2K,8L,9M{\text{ and }}2N$ electrons, the total number of s-electrons are?
Options-
A.$8$
B.$6$
C.$4$
D.$10$

Answer 1

The structure of an atom is defined in such a manner that the electrons are present in specific energy levels called shells which further contain subshells and orbitals in which the electrons are placed. The total number of electrons in the shells can be used to calculate the atomic number of the unknown element.

Complete answer:
The electronic configuration of any element is preferably written in the form of energy levels and orbitals as it contains information about the positions at which electrons are placed and can be used to determine the various periodic trends of the modern periodic table.
The given configuration in the form of shells can be used to find out the total number of electrons present in the atom which is equal to its atomic number. Therefore a neutral atom with configuration $2K,8L,9M{\text{ and }}2N$ has a total of $2 + 8 + 9 + 2 = 21$ electrons.
The s-orbitals are always filled first, followed by the p and d orbitals. The configuration is written in a manner so as to ensure that the electrons are filled sequentially from a lower energy position to a higher energy position.
Therefore the configuration is:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^1}4{s^2}$
There are two electrons in the $1s$ orbital, two electrons in the $2s$ orbital, two electrons in the $3s$ orbital and two of them in the $4s$ orbital.

$\Rightarrow$ Thus, there are a total of eight s-electrons and option (a) is correct.

Note:
The electrons in each shell were filled according to the given configuration. The $M{\text{ }}$ shell stands for the third energy level and only had nine electrons therefore no more electrons were added to the d-orbital in the electronic configuration.