• # question_answer You are given the electronic configuration of five neutral atoms - A, B, C, D and E. $A-1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{6}}\,3{{s}^{2}};$$B-1{{s}^{2}}2{{s}^{2}}\,2{{p}^{2}}\,3{{s}^{1}};$$C-1{{s}^{2}}2{{s}^{2}}2{{p}^{1}};$$D-1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{5}};$$E-1{{s}^{2}}\,2{{s}^{2}}\,2{{p}^{6}}$. Write the empirical formula for the substances containing (i) A and D (ii) B and D (iii) only D (iv) only E.

(i) The empirical formula for the compound containing A and D is $A{{D}_{2}}$. This is because atom A has two valence electrons while D has seven electrons. Therefore, A donates two electrons to two atoms of D and the compound has the formula $A{{D}_{2}}$. (ii) The empirical formula for the compound is BD because B loses one electron while D accept it. (iii) Two atoms of D can share one electron each to complete their octet. Therefore, the empirical formula of the compound is ${{D}_{2}}$. (iv) Since E has the configuration of a noble gas (Ne), it is, therefore, mono atomic and exists as E.