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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:
(i) Number of electrons in their atoms
(ii) Size of their atoms
(iii) Their tendencies to lose electrons
(iv) The formula of their oxides
(v) Their metallic character
(vi) The formula of their chlorides.
Answer
A and B both belong to third period which means they have 3 shells
|
Characteristic |
A |
B |
|
Electronic Configuration |
2, 8, 2 |
2, 8, 3 |
|
Valency |
2 |
3 |
|
Number of electrons |
12 |
13 |
|
Size of atoms |
Bigger |
Smaller |
|
Tendency to lose electrons |
More |
Less |
|
Formula of their oxides |
AO |
B 2 O 3 |
|
Metallic Character |
More metallic |
Less Metallic |
|
Formula of chlorides |
ACl 2 |
BCl 3 |
Explanation:
1. Number of electrons in
- A = 12
- B = 13
2. The size of A is bigger than that of B because they belong to the same period and across the period size decreases.
3. As the size increases, the tendency to lose electrons increases. So, it is more for A.
4. The valency of A is 2 and the valency of O is also 2 . So, the formula of its oxide will be AO.
The valency of B is 3 and that of O is 2. So, the formula of its oxide will be B 2 O 3 .
5. Metallic character depends on the tendency to lose electrons . So, it will be more for A.
6. A has a valency of 2 and that of chloride is 1 . So the formula of its chloride is ACl 2
B has a valency of 3 and that of chloride is 1 . So, the formula of its chloride is BCl 3
