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JEE Chemistry 2026 Question with Solution

The correct order of CC, NN, OO and FF in terms of second ionisation potential is

  • A

    C<N<F<OC < N < F < O

  • B

    F<N<C<OF < N < C < O

  • C

    C<O<N<FC < O < N < F

  • D

    C<F<N<OC < F < N < O

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: We need the correct order of CC, NN, OO and FF in terms of second ionisation potential.

Find: The increasing order of second ionisation potential.

Second ionisation potential is the energy required to remove an electron from a singly charged positive ion. The stability of the resulting ion is important.

After first ionisation, the configurations are:

C+:1s2 2s2 2p1N+:1s2 2s2 2p2O+:1s2 2s2 2p3F+:1s2 2s2 2p4\begin{aligned} C^+ &: 1s^2\,2s^2\,2p^1 \\ N^+ &: 1s^2\,2s^2\,2p^2 \\ O^+ &: 1s^2\,2s^2\,2p^3 \\ F^+ &: 1s^2\,2s^2\,2p^4 \end{aligned}

Stability Comparison

The ion O+O^+ has the half-filled configuration 2p32p^3, which is extra stable. Removing one more electron from this ion is therefore especially difficult.

Hence, oxygen has the highest second ionisation potential among the given elements. Using the stability comparison given in the solution, the increasing order is:

C<N<F<OC < N < F < O

Therefore, the correct option is A.

Common mistakes

  • Confusing first ionisation potential with second ionisation potential. For second ionisation potential, compare the stability of C+C^+, N+N^+, O+O^+ and F+F^+, not the neutral atoms.

  • Ignoring the special stability of a half-filled subshell. O+O^+ has 2p32p^3, which is half-filled and extra stable, so removing the next electron requires more energy.

  • Assuming the trend follows atomic number directly. Here, subshell stability alters the simple trend, so electronic configuration must be checked before ordering the values.

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