Question

Which of the following elements can form bonds with $s{p^3}$ hybridization?
A. Sodium
B. Nitrogen
C. Carbon
D. Oxygen
E. Fluorine

Answer 1

Hybridization is defined as the mixing of two or more atomic orbital having the same energy level to form a new hybrid orbital. In $s{p^3}$ hybridization, one s- orbital and three p orbital of the same electronic configuration of the atom mix with each other to form four new hybrid orbital.

Complete step by step answer:
The atomic weight of sodium Na is 11. The electronic configuration of sodium is $[Ne]3{s^1}$. The valence shell of sodium is 3s which only contains 1 electron, it does not have any p-orbital in the valence shell. Therefore, it will not form a bond with $s{p^3}$ hybridization.
The atomic weight of nitrogen N is 7. The electronic configuration of nitrogen is $[He]2{s^2}2{p^3}$. In the ground state of valence shell, the nitrogen contains two electrons in 2s orbital and three electrons is 2p orbital. The electrons move to the excited state and one electron from 2s orbital will move to 2p orbital. Thus 2s will contain one electron and 2p will contain four electrons. Thus in hybridization 1 s orbital and 3 p orbital take part forming $s{p^3}$ hybridization.
The atomic weight of carbon C is 6. The electronic configuration of carbon is $[He]2{s^2}2{p^2}$. In the ground state of valence shell, the carbon contains two electrons in 2s orbital and two electrons is 2p orbital. The electrons move to the excited state and one electron from 2s orbital will move to 2p orbital. Thus 2s will contain one electron and 2p will contain three electrons. Thus in hybridization 1 s orbital and 3 p orbital take part forming $s{p^3}$ hybridization.
The atomic weight of oxygen O is 8. The electronic configuration of carbon is $[He]2{s^2}2{p^4}$. In the ground state of valence shell, the oxygen contains two electrons in 2s orbital and four electrons is 2p orbital. The electrons move to the excited state and one electron from 2s orbital will move to 2p orbital. Thus 2s will contain one electron and 2p will contain five electrons. Thus in hybridization 1 s orbital and 3 p orbital take part forming $s{p^3}$ hybridization.
The atomic weight of fluorine F is 9. The electronic configuration of fluorine is $[He]2{s^2}2{p^5}$. In the ground state of valence shell, the oxygen contains two electrons in 2s orbital and five electrons is 2p orbital. The electrons move to the excited state and one electron from 2s orbital will move to 2p orbital. Thus 2s will contain one electron and 2p will contain six electrons. As all the orbital in the p subshell contains paired electrons, it cannot take part in hybridization.
Thus, nitrogen, carbon and oxygen can form bonds with $s{p^3}$ hybridization.
Therefore, the correct options are B,C and D.

Note: In fluorine atoms the valence shell contains s orbital and p orbital but it cannot form bond with $s{p^3}$ hybridization. In $s{p^3}$ hybrid orbital, 25 % is of s-character and 75% is of p-character.