The first ionisation potential of $Na$ is $5.1\, eV$. The value of electron gain enthalpy of $Na^+$ will be
- $- 2.55\, eV$
- $- 5.1\, eV$
- $- 10.2\, eV$
- $+ 2.55\, eV$
The Correct Option is B
Solution and Explanation
$Na^+ + e^- \longrightarrow Na$
Electron gain enthalpy of Na+ is reverse of (IE) Because reaction is reverse so
$ \, \, \, \, \, ? H (eq) = - 5.1 \, eV$
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Concepts Used:
Haloalkanes and Haloarenes - Classification
Classification of Haloalkanes and Haloarenes are as follows:
Haloalkanes and Haloarenes are Classified on the Basis of the Number of Halogen Atoms:
These are further classified as mono, di, or polyhalogen (tri-,tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures. For example;
- Monohaloalkane
- Dihaloalkane
- Trihaloalkane
Haloalkanes and Haloarenes are Classified on Compounds Containing sp3 C—X Bond (X= F, Cl, Br, I)
This class includes:
- Alkyl halides or haloalkanes (R—X)
- Allylic halides
- Benzylic halides
Haloalkanes and Haloarenes are Classified on Compounds Containing sp2 C—X Bond
This class includes:
- Vinylic halides
- Aryl halides