An element 'M' with electronic configuration (2, 8, 2) combines separately with (NO3)–, (SO4)2– and (PO4)3– radicals. Write the formula of the three compounds so formed. To which group and period of the Modern Periodic Table does the element 'M' belong? Will 'M' form covalent or ionic compounds? Give reason to justify your answer.
Solution
Electronic configuration of M = 2, 8, 2
Number of valence electrons of M = 2
Valency of M = Number of valence electrons = 2
Valency of the nitrate ion `(NO_3^-)` = 1
Valency of the sulphate ion `(SO_4^-2)` = 2
Valency of the phosphate ion `(PO_4^-3)` = 3
The formulae of nitrate, sulphate and phosphate formed by M are M(NO3)2, MSO4 and M3(PO4)2, respectively.
Since the valency of M is 2 and the atomic number of M is 12 (i.e. 2 + 8 + 2), this element belongs to the group 2 and period 3 of the periodic table.
This element will tend to form ionic compounds by losing two valence electrons to achieve a noble gas electronic configuration, that is, a stable octet in the valence shell.
