  Question

Classify the elements having atomic numbers as given below into three separate pairs on the basis of similar chemical properties. Give brief electronic explanation:$$9, 12, 16, 34, 53, 56$$.

Solution

The second period ends at atomic number $$10$$ while the third period ends at atomic number $$18$$. Therefore, $$9, 12$$and $$16$$ are the first elements in their respective groups. The atomic numbers of the other elements of the same group can be deduced by adding magic numbers of $$8, 18, 18$$ and $$32$$ to elements of $$2^{nd}$$ period and by adding magic numbers of $$18,18$$ and $$32$$ to the elements of $$3^{rd}$$ period. Thus,$$9 + 8 + 18 + 18 = 53$$$$12 + 8 +18+ 18= 56$$$$16 + 18 = 34$$elements with atomic numbers $$9 \,(F)$$ and $$53 \,(I)$$ belong to halogen family $$(group \,17)$$; elements with atomic numbers $$12 \,(Mg)$$ and $$56 \,(Ba)$$ belong to alkaline earth metals $$(Group \,2)$$ while elements with atomic numbers $$16 \,(S)$$ and $$34 \,(Se)$$ belong to oxygen family $$(Group \,16)$$.Chemistry

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