Question

Correct order of ionisation energy are :
(l) Be > B (ll) Be > C
(lll) N > O (IV) N > F

• A. l, lV
• B. l, ll, lll, lV
• C. l, lll
• D. lI, lV

Answer 1

Answer: (C)

l, lll

$B1s^{2}2s^{2}2p^1$

$Be1s^{2}2s^2$

$C1s^{2}2s^{2}2p^2$ $\left[electronicconfiguration\right]$

$N1s^{2}2s^{2}2p^3$

$O1s^{2}2s^{2}2p^4$

$F1s^{2}2s^{2}2p^5$

From the above configuration we can predict

$IE$ of $Be>IE$ of $B$

and, $IE$ of $N>IE$ of $O$

$Be$ having fully filled electronic configuration will require more energy for removing an $e^{-}$.

And, in case of $N⟶$ it is having extra stability due to half-filled electronic configuration which makes it difficult for the removal of $e^{-}$. Hence answer is $C$.