Question

State the electronic configuration for Nitrogen $\left[ \mathbf{p}=\mathbf{7},\text{ }\mathbf{n}=\mathbf{7} \right]$

Answer 1

As we know, atomic number is total number of a proton present in a nucleus of atom as well as atomic mass number is total number of a proton along with neutron. Nitrogen differs from other elements of the group $15$ due to its high electronegative character small size as well as high ionization enthalpy. Nitrogen could form multiple bonds with itself as well as other elements. It forms $pp$ multiple bonds.

Complete step-by-step answer: The atomic number of nitrogen is $7$ . This is the number of protons in a nuclei of nitrogen atoms. A neutral atom has the same number of electrons as protons. So electron configuration will include $7$ electron placed into appropriate $s$ along with $p$ orbitals in ground state, state of a lowest energy. The full electron configurations for nitrogen is $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$

So the nitrogen has an atomic number $7$ as well as mass number $7+7=14$ , $K$ shell will place $2$ electrons as well as rest will be accommodated into $L$ shell. Therefore configuration is $2,\text{ }5$ just adding power of configuration also we know that $s$ hold $2$ electron so the electron at $2{{s}^{2}}$ will be at second subshell since second subshell $p$ holds up to $6$ electrons so $2$ electrons from $2{{s}^{2}}$ and $3{{p}^{3}}$ which sums up and gives, $5$ electron in $p-orbital$

Therefore, the electronic configuration of Nitrogen is $2,\text{ }5$.

Note: Note that atomic number of nitrogen is $7$ . This is the number of protons in the nuclei of nitrogen atoms. A neutral atom has the same number of electrons as protons. So the electronic configurations will include $7$ electron placed into appropriate $s$ as well as $p$ orbital in ground state/state of lowest energy.