In the long form of the periodic table, the valence shell electronic configuration of $5s^2 5p^4$ corresponds to the element present in :
- Group 16 and period 6
- Group 17 and period 5
- Group 16 and period 5
- Group 17 and period 6
The Correct Option is C
Solution and Explanation
$16^{\text {th }}$ group, $n=5$ (5th period)
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 Main exam
- Let \( [x] \) denote the greatest integer function, and let \( m \) and \( n \) respectively be the numbers of the points, where the function \( f(x) = [x] + |x - 2| \), \( -2<x<3 \), is not continuous and not differentiable. Then \( m + n \) is equal to:
- JEE Main - 2025
- Differential Equations
- Let \( f(x) = \frac{x - 5}{x^2 - 3x + 2} \). If the range of \( f(x) \) is \( (-\infty, \alpha) \cup (\beta, \infty) \), then \( \alpha^2 + \beta^2 \) equals to.
- JEE Main - 2025
- Functions
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
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