MCQEasyJEE 2023Trends in Atomic Radius & Ionisation Energy

JEE Chemistry 2023 Question with Solution

For elements B, C, N, Li, Be, O and F, the correct order of first ionization enthalpy is:

  • A

    B < Li < Be < C < O < N < F

  • B

    Li < C < B < O < N < F

  • C

    Li < C < B < O < N < F

  • D

    Li < B < C < O < N < F

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The elements are Li, Be, B, C, N, O, F.

Find: The correct increasing order of first ionization enthalpy.

The first ionization enthalpy generally increases from left to right across a period because effective nuclear charge increases. However, there are known exceptions:

  1. B has lower first ionization enthalpy than Be because the electron removed from B is from the higher-energy 2p2p orbital, while in Be it is from the more stable 2s2s orbital.
  2. O has lower first ionization enthalpy than N because N has a stable half-filled 2p32p^3 configuration, whereas O has one paired electron in 2p2p, making removal easier.

Thus, the increasing order is

Li<B<Be<C<O<N<F\text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F}

the solution states option C, but its written sequence matches Li < B < C < O < N < F and omits Be. Among the listed options, the most defensible match to the extracted solution sequence is D.

Therefore, the correct option is D.

Using periodic trend exceptions

Given: First ionization enthalpy trend for second-period elements.

Find: The correct order among the given elements.

Across the second period, ionization enthalpy usually increases.

Important exceptions:

  • Be>B\text{Be} > \text{B} because removing a 2p2p electron from B is easier than removing a 2s2s electron from Be.
  • N>O\text{N} > \text{O} because N has extra stability due to half-filled configuration.

So the correct increasing order must place Li lowest, then B, then Be, then C, then O, then N, then F.

Since the options provided are inconsistent and duplicate one entry, the solution conclusion and the option list disagree. The closest supported listed option is D based on the written order in the solution.

Hence, the answer is D.

Common mistakes

  • Placing Be below B by assuming strict left-to-right increase is always valid is incorrect. B loses a higher-energy 2p2p electron more easily. Check subshell energies before finalizing the order.

  • Placing O above N by following only the periodic trend is incorrect. N has a stable half-filled 2p32p^3 configuration, so its first ionization enthalpy is higher. Account for electronic stability exceptions.

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