Why is \[{{\text{H}}_2}{\text{S}}\] is less acidic than \[{{\text{H}}_2}{\text{Te}}\]?
Answer
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HINT-Proceed with the solution of this issue by comparing bond dissociation enthalpy of the compounds in this question.
Complete answer:
As we know that
\[{{\text{H}}_2}{\text{S}}\] known as hydrogen sulfide do not contain carbon in this chemical compound and are therefore not an organic compound, because organic compounds contain carbon-hydrogen bonds
Therefore we may assume that hydrogen sulfide is a gas that is an inorganic compound.
\[{{\text{H}}_2}{\text{S}}\]and \[{{\text{H}}_2}{\text{Te}}\] are inorganic acids, and mineral acids are also recognized. Once dissolved in water, all inorganic acids contain hydrogen ions ( H+) and the base ions of the conjugate. These are corrosive and water soluble but organic solvents are insoluble.
\[{{\text{H}}_2}{\text{S}}\] is less acidic than \[{{\text{H}}_2}{\text{Te}}\] as the size of the atom increases and enthalpy of ionization decreases when going down the chain. That helps to easily dissociate the bond. So taking the case of \[{{\text{H}}_2}{\text{Te}}\] and thereby extracting a hydrogen ion from these compounds is simple and because of this, acidity is higher. But the dissociation of bonds is less in the case of \[{{\text{H}}_2}{\text{S}}\] the bond dissociation is less and hence acidity is lower. So \[{{\text{H}}_2}{\text{S}}\] is less acidic than \[{{\text{H}}_2}{\text{Te}}\].
Note- In this particular question, we should be aware that Hydrogen sulfide \[{{\text{H}}_2}{\text{S}}\] is a colorless chalcogen hydride gas with the characteristic foul odor of rotten eggs and also very poisonous, corrosive and flammable, used to separate deuterium oxide or heavy water from normal water through the Girdler sulfide process.
Although \[{{\text{H}}_2}{\text{Te}}\] is a simplest tellurium hydride, which is a colorless, unstable gas in ambient air, the gas can exist at very low concentrations long enough to be readily detectable at very low concentrations by the odour of rotting garlic. \[{{\text{H}}_2}{\text{Te}}\] is similar in both chemical and structural terms to hydrogen sulfide.
Complete answer:
As we know that
\[{{\text{H}}_2}{\text{S}}\] known as hydrogen sulfide do not contain carbon in this chemical compound and are therefore not an organic compound, because organic compounds contain carbon-hydrogen bonds
Therefore we may assume that hydrogen sulfide is a gas that is an inorganic compound.
\[{{\text{H}}_2}{\text{S}}\]and \[{{\text{H}}_2}{\text{Te}}\] are inorganic acids, and mineral acids are also recognized. Once dissolved in water, all inorganic acids contain hydrogen ions ( H+) and the base ions of the conjugate. These are corrosive and water soluble but organic solvents are insoluble.
\[{{\text{H}}_2}{\text{S}}\] is less acidic than \[{{\text{H}}_2}{\text{Te}}\] as the size of the atom increases and enthalpy of ionization decreases when going down the chain. That helps to easily dissociate the bond. So taking the case of \[{{\text{H}}_2}{\text{Te}}\] and thereby extracting a hydrogen ion from these compounds is simple and because of this, acidity is higher. But the dissociation of bonds is less in the case of \[{{\text{H}}_2}{\text{S}}\] the bond dissociation is less and hence acidity is lower. So \[{{\text{H}}_2}{\text{S}}\] is less acidic than \[{{\text{H}}_2}{\text{Te}}\].
Note- In this particular question, we should be aware that Hydrogen sulfide \[{{\text{H}}_2}{\text{S}}\] is a colorless chalcogen hydride gas with the characteristic foul odor of rotten eggs and also very poisonous, corrosive and flammable, used to separate deuterium oxide or heavy water from normal water through the Girdler sulfide process.
Although \[{{\text{H}}_2}{\text{Te}}\] is a simplest tellurium hydride, which is a colorless, unstable gas in ambient air, the gas can exist at very low concentrations long enough to be readily detectable at very low concentrations by the odour of rotting garlic. \[{{\text{H}}_2}{\text{Te}}\] is similar in both chemical and structural terms to hydrogen sulfide.
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