(a) H2O2 structure in gas phase, dihedral angle is 111.5°. (b) H2O2 structure in solid phase at 110K, dihedral angle is 90.2° Hence given statement (A) is not correct.
But statement (B) is correct.
(a) H2O2 structure in gas phase, dihedral angle is 111.5°. (b) H2O2 structure in solid phase at 110K, dihedral angle is 90.2° Hence given statement (A) is not correct.
But statement (B) is correct.
Assertion A : Isotopes of hydrogen (protium, deuterium, and tritium) have almost the same chemical properties because they have the same number of protons and electrons, which dictate chemical behavior.
However, they differ in their rates of reaction due to differences in their atomic masses, which affect reaction kinetics.
Reason R : The isotopes of hydrogen have different enthalpy of bond dissociation due to their differences in atomic mass.
The stronger bond in deuterium compared to protium, for example, leads to higher bond dissociation enthalpy.
This, in turn, influences the rate of reactions involving these isotopes, as breaking the bonds requires different amounts of energy.
Therefore, the correct answer is Both A and R are correct, and R is the correct explanation of A.
NOTE : In one electron species, such as
-atom, the energy of orbital depends only on the principal quantum number,
Hence answer
is
H2O2 act as both oxidising and reducing agent.
H2O2 is miscible with water due to open book like structure both -OH group are not in the same plane H2O2 used in the treatment of effluents.
Total number of isotopes of hydrogen is 3
or
or
and only
or
is an Radioactive element.
Since, H2O2 gains electron to convert into H2O. Hence, H2O2 acts as an oxidizing agent.
There is extensive inter-molecular hydrogen bonding in the condensed phase instead of intra-molecular
-bonding.
Enthalpy of bond dissociation (kJ/mole) at 298.2 K For, Hydrogen = 435.88 For, Deuterium = 443.35
Heavy water is used as moderator in nuclear reactors to control the speed of neutrons.
In this reaction is acting as a reducing agent as is undergoing a change in oxidation state from to .