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Consider the followingelements: ._(4)Be, ._(19) F, ._(19)K, ._(20) C (a) Selectthe elementhavingone electron in theoutermostshell. (b) Selecttws element of thesamegroup . (c ) Writethe formula andnatureof the compoundformedwhen theelementK reacts with an elementX ofelectronic configuration2, 8, 7 |
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Answer» SOLUTION :(a) thefirstperiodhas onlytwoelements beginningwith Li (Z=3) andendingwith Ne (Z= 2) the secondperiodhasalso only8 elementsbeginning with Li (Z=3) and endingwith Ar (Z= 10) The thirdperiodalsohas 8 elementsbeginningwith Na (Z=11) and endingwithKr (Z=18) The Fourthperiodhas 18elementsbeginnings with K (Z=19) and endingwith Kr (Z=36) Since eachperiodbeginswith on AELEMENT havingonly oneelectron in thevalenceshell thereforeout ofthe fourelementslistedin the question(i.e.,._(4)Be, ._(9)F, ._(19)K, ._(20)Ca)` only._(19)K` has oneelectronin thevalence shell. (b) Sincesecondgrouphas twoelectrons in thevalenceshelltherefore`._(4)Be" and" ._(20)Ca` belongto thesamegroup. ( c) theelement Xwithelectronicconfiguration2,8.7 (i.e., , CI ,Z= 17) hasone electronlessthan thenearestinertgas i.e, Ar (Z=18)whileK withelectronicconfiguration (2,8,8,1)has ona electron morethan thenearest INERTGAS Ar (2,8,8) . Thereforeto achieve thethe nearestinertgas configuration, Kloses one electronto FORM `K^(+)`and CIgainsone electronto form `CI^(-)` .Thetwo ionsthencombineto formionic compound `K^(+) CI^(-)` or simplyKCI.
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