Given below are two statements:
Statement I: \( \mathrm{C<O<N<F} \) is the correct order in terms of first ionization enthalpy values.
Statement II: \( \mathrm{S>Se>Te>Po>O} \) is the correct order in terms of the magnitude of electron gain enthalpy values.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: \( \mathrm{C<O<N<F} \) is the correct order in terms of first ionization enthalpy values.
Statement II: \( \mathrm{S>Se>Te>Po>O} \) is the correct order in terms of the magnitude of electron gain enthalpy values.
In the light of the above statements, choose the correct answer from the options given below:
Show Hint
- Statement I is false but Statement II is true
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
The Correct Option is A
Solution and Explanation
Step 1: Analyze Statement I.
Across a period, first ionization enthalpy generally increases due to increasing nuclear charge. However, there is an exception between oxygen and nitrogen. Nitrogen has a half-filled \(2p^3\) configuration, which is more stable than oxygen’s \(2p^4\) configuration. Therefore, \[ \mathrm{C<N>O<F} \] is the correct trend, not \( \mathrm{C<O<N<F} \). Hence, Statement I is false.
Step 2: Analyze Statement II.
Electron gain enthalpy generally becomes less negative down a group due to increasing atomic size. However, oxygen has a much lower (less negative) electron gain enthalpy compared to sulfur because of strong inter-electronic repulsion in the compact \(2p\) orbitals. Thus, the correct order of magnitude is: \[ \mathrm{S>Se>Te>Po>O}. \] Hence, Statement II is true.
Step 3: Final conclusion.
Statement I is false but Statement II is true.
Final Answer: \[ \boxed{\text{Statement I is false but Statement II is true}} \]
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