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Question Number 22883 by Tinkutara last updated on 23/Oct/17

The correct order of second ionisation  energy in the following :  (a) F > O > N > C  (b) O > F > N > C  (c) O > N > F > C  (d) C > N > O > F

$$\mathrm{The}\:\mathrm{correct}\:\mathrm{order}\:\mathrm{of}\:\mathrm{second}\:\mathrm{ionisation} \\ $$ $$\mathrm{energy}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:: \\ $$ $$\left({a}\right)\:\mathrm{F}\:>\:\mathrm{O}\:>\:\mathrm{N}\:>\:\mathrm{C} \\ $$ $$\left({b}\right)\:\mathrm{O}\:>\:\mathrm{F}\:>\:\mathrm{N}\:>\:\mathrm{C} \\ $$ $$\left({c}\right)\:\mathrm{O}\:>\:\mathrm{N}\:>\:\mathrm{F}\:>\:\mathrm{C} \\ $$ $$\left({d}\right)\:\mathrm{C}\:>\:\mathrm{N}\:>\:\mathrm{O}\:>\:\mathrm{F} \\ $$

Commented bymath solver last updated on 23/Oct/17

B ?

$${B}\:? \\ $$

Answered by math solver last updated on 23/Oct/17

draw electronic configuration of each...  2nd I.E means removing electron  from monovalent cation....  2p^(3  )  ⟩ 2p^4  { in case of stability }  becaouse 2 p^3  is half filled and the  rest you know.....

$${draw}\:{electronic}\:{configuration}\:{of}\:{each}... \\ $$ $$\mathrm{2}{nd}\:{I}.{E}\:{means}\:{removing}\:{electron} \\ $$ $${from}\:{monovalent}\:{cation}.... \\ $$ $$\mathrm{2}{p}^{\mathrm{3}\:\:} \:\rangle\:\mathrm{2}{p}^{\mathrm{4}} \:\left\{\:{in}\:{case}\:{of}\:{stability}\:\right\} \\ $$ $${becaouse}\:\mathrm{2}\:{p}^{\mathrm{3}} \:{is}\:{half}\:{filled}\:{and}\:{the} \\ $$ $${rest}\:{you}\:{know}..... \\ $$

Commented byTinkutara last updated on 24/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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