An element ‘M’ with electronic configuration (2, 8, 2) combines separately with 2 3 4 (NO ) , (SO ) − − and 3 4 (PO ) − radicals. Write the formula of the three compounds so formed. To which group and period of the Modern Periodic Table does the elements ‘M’ belong? Will ‘M’ form covalent or ionic compounds? Give reason to justify your answer.
Answers
Answered by
224
The three compounds formed are
1. M(NO3) 2
2. MSO4
3. M3(PO4) 2
#M belongs to group 2 and period 3
M will form ionic compounds because it's valence electrons are 2 it can easily donate them and also it has a tendency to lose electrons
1. M(NO3) 2
2. MSO4
3. M3(PO4) 2
#M belongs to group 2 and period 3
M will form ionic compounds because it's valence electrons are 2 it can easily donate them and also it has a tendency to lose electrons
Answered by
6
Answer:
Electronic configuration of M= 2,8,2
Number of valence electron of M=2
valency of M = Number of valence electrons = 2
Valency of the nitrate ion (NO3) = 1
Valency of the sulphate ion (SO42) =2
Valency of phosphate ion (PO43) =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 such as (2,8,2), this element belongs to the group 3 and period 3 of the periodic table.
This element will tend to form an ionic compound by losing two valence electrons to achieve a noble gas electronic configuration, that is a stable octet in the valence shell.
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