Consider the spectral line resulting from the transition $n=2\rightarrow n=1$ in the atoms and ions given below. The shortest wavelength is produced by
- hydrogen atom
- deuterium atom
- singly ionized helium
- doubly ionized lithium
The Correct Option is D
Solution and Explanation
$\therefore$ Correct option is (d).
Top Questions on Hydrogen bonding
- Preparation of KMnO\(_4\) from Mg(II):
- KEAM - 2025
- Chemistry
- Hydrogen bonding
- Intramolecular hydrogen bonding is found in
- BITSAT - 2024
- Chemistry
- Hydrogen bonding
- Which of the following statement is true?
- KEAM - 2024
- Chemistry
- Hydrogen bonding
- Para and ortho hydrogen differ in:
- COMEDK UGET - 2023
- Chemistry
- Hydrogen bonding
For hydrogen atom, $\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure The ratio of $\lambda_1$ and $\lambda_2$ is $\frac{x}{32}$ The value of $x$ is
- JEE Main - 2023
- Physics
- Hydrogen bonding
Questions Asked in JEE Advanced exam
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
- JEE Advanced - 2025
- Waves and Oscillations
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80- JEE Advanced - 2025
- Organic Chemistry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
Concepts Used:
Hydrogen Bonding
Hydrogen bonding implies the formation of hydrogen bonds which are an attractive intermolecular force. An example of hydrogen bonding is the bond between the H atom and the O atom in water.
A special type of intermolecular attractive force arises only in the compounds having Hydrogen atoms bonded to an electronegative atom. This force is known as the Hydrogen bond. For instance, in water molecules, the hydrogen atom is bonded to a highly electronegative Oxygen.
The conditions for hydrogen bonding are:
- The molecule must contain a strongly electronegative atom that is bound to the hydrogen atom. The higher the electronegativity, the more polarized is the molecule.
- The electronegative atom must be small. The smaller the size, the greater the electrostatic magnetism.
Effects of Hydrogen Bonding on Elements:
Association: The molecules of carboxylic acids exist as dimer because of the hydrogen bonding. The molecular masses of such compounds are found to be double than those calculated from their simple formula.
Dissociation: In aqueous solution, HF dissociates and gives the difluoride ion instead of fluoride ion. This is due to hydrogen bonding in HF. The molecules of HCl, HBr, HI do not form a hydrogen bond. This explains the non-existence of compounds like KHCl2, KHBr2, KHI2.
Types of Hydrogen bonding
- Intramolecular Hydrogen bonding: When hydrogen bonding takes place between different molecules of the same or different compounds, it is called intermolecular hydrogen bonding.
- Intermolecular hydrogen bonding: The hydrogen bonding which takes place within a molecule itself is called intramolecular hydrogen bonding.
- Symmetrical Hydrogen bonding: The symmetric hydrogen bond is a type of a three-centre four-electron bond.