When the first electron gain enthalpy $( \Delta_{eg}H) $ of
oxygen is $-141\, kJ/mol$, its second electron gain enthalpy is :
- almost the same as that of the first
- negative , but less negative than the first
- a positive value
- a more negative value than the first
The Correct Option is C
Solution and Explanation
${O^{-} \, _{(g)} + e^{-} -> O^{-2} \, _{(g)} ; \Delta H = positive}$
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In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
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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