Question

The electronic configuration of an element is $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}4{s^1}$. This represents its:
A. Excited state
B. Ground state
C. Cationic form
D. Anionic form

Answer 1

Hint: Electronic configuration of an atom describes the distribution of number of electrons present inside the atom to the subshell, which is present in the orbitals around the nucleus. In the atom no. of electrons equals to the no. of protons inside the atom.

Step by step answer: Given that the electronic configuration of an element is $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}4{s^1}$. From the given configuration it is clear that the total no. of electrons present in that element is 24.
As we know that in an atom the total no. of electrons is equal to the total no. of protons present in that atom.
An atomic number of any atom is equal to the no. of protons in that atom.
According to the given configuration atomic number of the given element may be 24, because no. of electrons is 24 there which is equal to no. of protons.
If we assume that atomic number is 24 then element will be chromium $\left( {Cr} \right)$ whose electronic configuration is $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}4{s^1}$, it means it is the ground state configuration.
 If we assume that it is the cationic element from which 1 electron is escaped then again it follows the given configuration $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5}4{s^1}$ which is cationic form. So, the element is $M{n^ + }$ whose atomic number is 25.

Additional information: Anionic form of electronic configuration of an element occurs when there is one electron more than no. of protons present inside the atom. An excited state configuration occurs when electrons jump to a higher subshell by leaving their original shell.

Note: Here you may get confused in the filling of electrons in the subshell of the different orbitals, so for the perfect filling of electrons you have to know about Hund’s rule, Aufbau rule and Pauli's exclusion principle.