Answer:
\[{{H}_{2}}S\]is less acidic \[({{K}_{a}}=1.0\times
{{10}^{-7}})\] than \[{{H}_{2}}Te\,(2.3\times {{10}^{-7}})\] because the
cleavage of H-Te bond is easier than that of H-S bond due to its more bond
length. Therefore, release of \[{{H}^{+}}\] from \[{{H}_{2}}Te\] takes place
more readily than from \[{{H}_{2}}S\] molecule.
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