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Use Lewis symbols to show electron transfer between the following atoms to form cations and anions (a) K and S (b) Ca and O (c) Al nnd N |
Answer» Solution :K and S :  Alkali metal K lose one electron and S gain two electron and attain stable octet like Ar noble gas (Ar] containing `K^(+) and s^(2-)` ions and they are con1bine to FORM `K_(2)`S. (b) Ca and O : The electron configuration of Ca (Z = 20) and O (Z = 8) are `[Ar]4s^(2) and [He] 2s^(2) 2p^(4)`respectively Ca lose two electron and O gain two electron and form `Ca^(2+) and O^(2-)` . `underset([Ar] 4s^(2))overset(..)(Ca) rarr underset([A])(Ca^(2+)) + underset(Ca "lose " e^(-))(2e^(-))` In CaO, `Ca^(2+) and O^(2-)` POSITIVE and negative ions are combine by elector valence bond.  (c) Al and N : The configuration of Al (Z = 13) is (Ne]`3S^(2) 2p^(-1)`. This positive element lose 3 electron and attain stable octet like Ne noble gas and form `Al^(3+)` ION. `underset([Ne] 3s^(2)3p^(1))overset(.)(Al) rarr underset([Ne])(Al^(3+) + 3e^(-)` Nitrogen (Z = 7) is negative element. It accept 3 electron of Al and convert into `N^(3-)` Nytride ion like stable octet of Ne noble gas. In Al and N, the elector valence (ionic) bond between positive ion `AI^(3+)` and negative ion `N^(3-)` by electrostatic attraction. 
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