question_answer

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                                                                                 [NCERT 1983]

A)                 Q            

B)                 M

C)                 R            

D)                 L

Correct Answer: B

Solution :

           Non-metals readily form diatomic molecules by sharing of electrons. Element \[M\,(1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{5}})\] has seven electrons in its valence shell and thus needs one more electron to complete its octet. Therefore, two atoms share one electron each to form a diatomic molecule \[({{M}_{2}})\] \[:\underset{.\,\,.}{\mathop{\overset{.\,\,.}{\mathop{M}}\,}}\,.+.\overset{.\,\,.}{\mathop{\underset{.\,\,.}{\mathop{M}}\,}}\,:\] \[\to \] \[:\underset{.\,\,.}{\mathop{\overset{.\,\,.}{\mathop{M}}\,}}\,:\overset{.\,\,.}{\mathop{\underset{.\,\,.}{\mathop{M}}\,}}\,:\]