Periodic Classification and Periodic Properties - Result Question 28
####25. The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is
(1981, 1M)
(a) $C>N>O>F$
(b) O $>N>F>C$
(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: 25. (b)
Solution:
- For second ionisation potential, electron will have to be removed from valence shell of the following ions:
$$ \begin{aligned} & C^{+}\left(5 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \ & N^{+}\left(6 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \begin{array}{|l|l|l|} \hline 1 & 1 & \ \hline & 2 p \end{array} \ & O^{+}\left(7 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \ & F^{+}\left(8 e^{-}\right)=1 s^{2} \quad 2 s^{2} \quad \begin{array}{|l|l|l|} \hline 1 & 1 & 1 \ \hline 2 p \end{array} \end{aligned} $$
In general, ionisation energy increases from left to right in a period. However, exception occur between adjacent atoms in a period, greater amount 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 $O^{+}$is greater than that of $F^{+}$. Also ionisation potential of $N^{+}$is greater than $C^{+}$but less than both $O^{+}$and $F^{+}$ (periodic trend). Hence, overall order is 2nd IP: $O>F>N>C$.
