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
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
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.
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.