JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

The correct order of C, N, O and F in terms of second ionisation potential is

A
F $<$ N $<$ C $<$ O
B
C $<$ O $<$ N $<$ F
C
C $<$ N $<$ F $<$ O
D
C $<$ F $<$ N $<$ O

Solution

Second ionisation removes an electron from the mono-cation:

$C^{+} : [He] 2 s^{2} 2 p^{1}$

$N^{+} : [He] 2 s^{2} 2 p^{2}$

$O^{+} : [He] 2 s^{2} 2 p^{3}$ (half-filled, extra stable)

$F^{+} : [He] 2 s^{2} 2 p^{4}$

Because $O^{+}$ has the extra-stable half-filled $2 p^{3}$ configuration, removing another electron is hardest $\Rightarrow$ O has the largest IE $_{2}$ . The chemical order is C $<$ N $<$ F $<$ O.

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