Question

The electronic configuration of four elements are given in brackets $L\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{1}} \right)$; $M\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{5}} \right)$; $Q\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}} \right)$; $R\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{2}} \right)$. The element that would most readily form a diatomic molecule is:
A) $Q$
B) $M$
C) $R$
D) $L$

Answer 1

Hint: The elements with seven electrons in their valence shell (Group 17 elements) readily combine with the other element of the same group through a nonpolar covalent bond resulting in the formation of a diatomic molecule.


Complete step by step solution:As per the given electronic configuration in the question, the element which require only one electron to complete its octet is $M\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{5}} \right)$ which belongs to group 17 group i.e., it is a part of halogen family in which all the elements shows a tendency to readily share their electrons to form a nonpolar covalent bond and results in formation of diatomic molecule as shown in the figure below:


Also, diatomic molecules of alkali metals with electronic configuration $L\left( 1{{s}^{2}},2{{s}^{2}}2{{p}^{1}} \right)$ are detected in their gas phase but the bond between the elements is very weak because the outer orbitals of the alkali metals are very diffuse and thus, the possibility to exist as a diatomic molecule is very less. Due to this reason, the alkali metals mostly form ionic bonds with the elements.
Hence, the element that would most readily form a diatomic molecule is $M$.


Option ‘B’ is correct

Note: It is important to note that the only element in group 17 which does not exist as a diatomic molecule is Astatine because it is a short lived radioactive element which does not occur in nature. Thus, only fluorine, chlorine, bromine and iodine elements in group 17 exist as diatomic molecule.