Question

The electronic configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}3{{s}^{1}}$ shows:
A. Ground state of fluorine atom
B. Excited state of fluorine atom
C. Excited state of neon atom
D. Excited state of ion ${{O}_{2}}$

Answer 1

Hint: We are provided with the electronic configuration, we can figure out the atomic number of the element with the help of the electronic configuration and then with the help of the atomic number we can solve the given question by checking the options.

Complete step by step solution:
The electronic configuration given is: $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}3{{s}^{1}}$
Total number of electrons = 10
The atomic number of the given element is 10 but the 2p orbital can have 6 electrons but in the given configuration there are 5 electrons in the 2p orbital and 1 electron in the 3s orbital, which means that it is an excited electronic configuration of element having atomic number 10.
The first option is - ground state of fluorine atom
The atomic number of fluorine is 9 and there are 10 electrons in the given configuration hence it is not the correct answer.
The second option is – excited state of fluorine atom
The atomic number of fluorine is 9 and there are 10 electrons in the given configuration in excited state hence it is not the correct answer.
The third option is- excited state of neon atom
The atomic number of neon atoms is 10 and there are 10 electrons present in the given configuration. The electronic configuration of neon is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ and the excited electronic configuration of neon is $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}3{{s}^{1}}$.

Hence, the correct answer is option C.

Note: The ground state configuration is the lowest energy configuration which means that it is the most stable arrangement and an excited state configuration is the higher energy arrangement. The excited state configuration requires energy input to create the excited state.