Periodic Classification And Periodic Properties Ques 20
20. The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is C < N < O < F
(1981, 1M)
(a) C>N>O>Cl
(b) O $>\mathrm{N}>\mathrm{F}>\mathrm{C}$
(c) O $>$ F $>$ N $>$ C
(d) F $>$ O $>$ N $>$ C
Objective Questions II
(One or more than one correct option)
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Answer:
Correct Answer: 20.(b)
Solution:
For second ionisation potential, electron will have to be removed from the valence shell of the following ions:
$$ \begin{alignedat} & \mathrm{C}^{+}\left(5 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad 2 p^{1} \ & \mathrm{N}^{+}\left(6 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \begin{array}{|l|l|l|} \hline 1 & 1 & 1 \ \hline & 2p \end{array} \ & \mathrm{O}^{+}\left(7 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad 2 p^{4} \ & \mathrm{F}^{+}\left(8 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \begin{array}{|l|l|l|} \hline 1 & 1 & 1 \ \hline 2p \end{array} \end{aligned} $$
In general, ionisation energy increases from left to right in a period. However, exceptions occur between adjacent atoms in a period, greater amount of energy is required for removal of electron from completely half-filled or completely filled orbital than the same for adjacent atom with either less than completely half-filled or less than completely filled orbital. Therefore, ionisation potential of $\mathrm{F}^{+}$is greater than that of $\mathrm{O}^{+}$. Also ionisation potential of $\mathrm{N}^{+}$is greater than $\mathrm{C}^{+}$but less than both $\mathrm{F}^{+}$and $\mathrm{O}^{+}$ (periodic trend). Hence, overall order is 2nd IP: $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}$.
BETA
