The correct decreasing order of first ionisat | vedclass

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Question

(C) The general trend for first ionization enthalpy across a period is that it increases from left to right due to an increase in effective nuclear ch

Vedclass · Accepted Answer

$F > N > C > O > Be > B$

Vedclass · Answer

(C) The general trend for first ionization enthalpy across a period is that it increases from left to right due to an increase in effective nuclear charge and a decrease in atomic size. However,there are exceptions due to stable electronic configurations. The electronic configurations of the second-period elements are: $Li (1s^2 2s^1)$,$Be (1s^2 2s^2)$,$B (1s^2 2s^2 2p^1)$,$C (1s^2 2s^2 2p^2)$,$N (1s^2 2s^2 2p^3)$,$O (1s^2 2s^2 2p^4)$,$F (1s^2 2s^2 2p^5)$,$Ne (1s^2 2s^2 2p^6)$. $Be$ has a fully filled $2s$ orbital,which is more stable than the $2p^1$ configuration of $B$. $N$ has a half-filled $2p$ orbital,which is more stable than the $2p^4$ configuration of $O$. Thus,the correct decreasing order for the elements $Be, B, C, N, F$ is $F > N > C > Be > B$.

The correct decreasing order of first ionisation enthalpies of five elements of the second period is

Similar Questions

Which of the following orders for ionisation energy is correct?

How would you explain the fact that the first ionisation enthalpy of sodium is lower than that of magnesium,but its second ionisation enthalpy is higher than that of magnesium?

$A$ sudden large jump between the values of second and third ionisation energies of an element would be associated with the electronic configuration:

Which of the following represents the correct order of ionization enthalpy?
$(i) Be^{+} > Be$ $(ii) Be > Be^{+}$ $(iii) C > Be$ $(iv) B > Be$

In which of the following electronic configurations of an element is there a very large difference between the second and third ionization energy?

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