Two elements ‘A’ and ‘B’ belong to the 3rd period of Modern periodic table and are in group 2 and 13, respectively. Compare their following characteristics in tabular form:-
(a) Number of electrons in their atoms
(b) Size of their atoms
(c) Their tendencies to lose electrons
(d) The formula of their oxides
(e) Their metallic character
(f) The formula of their chlorides
Solution
Since elements ‘A’ and ‘B’ belong to the 3rd period of the modern periodic table and are in groups 2 and 13, respectively,
Electronic configuration of A = 2, 8, 2
Electronic configuration of B = 2, 8, 3
(a) Number of electrons in an atom of A = 12
Number of electrons in an atom of B = 13
(b) Along a period, the atomic size of elements decreases. Thus, element A has a higher atomic size than element B.
(c) Along a period, due to increase in effective nuclear charge, the tendency to lose electrons decreases. Thus, element A has a higher tendency to lose electrons than element B.
(d) Valency of an O atom is −2.
Number of valence electrons in an atom of A = 2
Thus, formula of the oxide of A is A2O2 or AO.
Number of valence electrons in an atom of B = 3
Thus, formula of the oxide of B is B2O3.
(e) Along a period, the metallic character of elements decreases. Thus, element A has a higher metallic character than element B.
(f) Valency of a Cl atom is −1.
Number of valence electrons in an atom of A = 2
Thus, formula of the chloride of A is ACl2.
Number of valence electrons in an atom of B = 3
Thus, formula of the chloride of B is BCl3.
