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Question
Atomic numbers of elements A, B, C, D, E, F are 8, 7, 11, 12, 13 and 9 respectively. State the type of ions they form.
Solution 1
The electronic configurations of the given elements are:
A = 8 (2, 6)
B = 7 (2, 5)
C = 11 (2, 8, 1)
D = 12 (2, 8, 2)
E = 13 (2, 8, 3)
F = 9 (2, 7)
Thus A, B, F will form anions and C, D, E will form cations.
Solution 2
| Element | Atomic no. | Electronic configuration | Difference from Octet | Ion formed |
| A | 8 | 2, 6 | 8 − 6 = 2 | −2 |
| B | 7 | 2, 5 | 8 − 5 = 3 | −3 |
| C | 11 | 2, 8, 1 | One electron more than octet | +1 |
| D | 12 | 2, 8, 2 | Two electrons more than octet | +2 |
| E | 13 | 2, 8, 3 | Three electrons more than octet | +3 |
| F | 9 | 2, 7 | 8 − 7 | −1 |
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