$(a)$ Which of the following electronic configurations are not possible? Give reasons.
$(i)$ $X: 2, 8, 4$
$(ii)$ $Y: 3, 8, 2$
$(iii)$ $Z: 2, 8, 9$
$(b)$ Write electronic configurations of the following elements and predict their valencies:
Fluorine: $9$,Aluminium: $13$,Argon: $18$
$(i)$ $X: 2, 8, 4$
$(ii)$ $Y: 3, 8, 2$
$(iii)$ $Z: 2, 8, 9$
$(b)$ Write electronic configurations of the following elements and predict their valencies:
Fluorine: $9$,Aluminium: $13$,Argon: $18$
(A) Electronic configurations $Y: 3, 8, 2$ and $Z: 2, 8, 9$ are not possible. According to the Bohr-Bury scheme,the first shell ($K$-shell) can hold a maximum of $2$ electrons,and the outermost shell of an atom can hold a maximum of $8$ electrons.
$(b)$ The electronic configurations and valencies are as follows:
$(b)$ The electronic configurations and valencies are as follows:
| Element | Electronic Configuration $(K, L, M)$ | Valency |
|---|---|---|
| Fluorine $(9)$ | $2, 7$ | $1$ |
| Aluminium $(13)$ | $2, 8, 3$ | $3$ |
| Argon $(18)$ | $2, 8, 8$ | $0$ |
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