The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is
- C > N > O > F
- O > N > F > C
- O > F > N > C
- F > O > N > C
The Correct Option is C
Solution and Explanation
both $O^+ \, and \, F^+$ (periodic trend). Hence, overall order is 2nd IP : O > F > N > C.
Top Questions on Classification
- When the first electron gain enthalpy $( \Delta_{eg}H) $ of oxygen is $-141\, kJ/mol$, its second electron gain enthalpy is :
- JEE Main - 2019
- Chemistry
- Classification
- Choose the word which is least like the other words:
- In comparison to boron, berylium has :
- JEE Main - 2019
- Chemistry
- Classification
In general, the properties that decrease and increase down a group in the periodic table respectively are:
- JEE Main - 2019
- Chemistry
- Classification
- Which among the following elements is a metalloid?
- JKCET - 2019
- Chemistry
- Classification
Questions Asked in JEE Advanced exam
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
JEE Advanced Notification
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