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NEET CHEMISTRYClassification of Elements and Periodicity in PropertiesEasy

Question

The value of electron gain enthalpy of Na+Na^+, if IE1IE_1 of Na = 5.1 eV, is:

A

+10.2 eV

B

–5.1 eV

C

–10.2 eV

D

+2.55 eV

Step-by-Step Solution

The first ionization enthalpy (IE1IE_1) corresponds to the energy required to remove an electron from a neutral gaseous atom: Na(g)Na+(g)+e;ΔH=+5.1 eVNa(g) \rightarrow Na^+(g) + e^-; \quad \Delta H = +5.1 \text{ eV} The electron gain enthalpy of the cation (Na+Na^+) corresponds to the energy change when an electron is added to the ion to form the neutral atom. This is the exact reverse of the ionization process: Na+(g)+eNa(g)Na^+(g) + e^- \rightarrow Na(g) According to thermodynamic principles, reversing a chemical reaction reverses the sign of the enthalpy change (ΔH\Delta H) while the magnitude remains the same. Therefore, the electron gain enthalpy of Na+Na^+ is 5.1 eV-5.1 \text{ eV}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Classification of Elements and Periodicity in Properties. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYClassification of Elements and Periodicity in Propertieselectronenthalpy

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