The Ka values of H2Se and H2Te are respectively

2 .3 × 10 − 3 , 2 × 10 − 4

The Ka values of H2Se and H2Te are respectively

2 .3 × 10 − 3 , 2 × 10 − 4

1 .3 × 10 − 4 , 2 .3 × 10 − 3

2 × 10 − 14 , 2 .3 × 10 − 4

2 × 10 − 3 , 2 .3 × 10 − 3

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The Ka values of H₂Se and H₂Te are respectively

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$2 .3 \times 10^{- 3}, 2 \times 10^{- 4}$

$1 .3 \times 10^{- 4}, 2 .3 \times 10^{- 3}$

$2 \times 10^{- 14}, 2 .3 \times 10^{- 4}$

$2 \times 10^{- 3}, 2 .3 \times 10^{- 3}$

answer is A.

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Detailed Solution

Because of decrease in bond dissociation enthalpy, acidity of hydrides increases down the group.
Increasing order of acidity is (K_a) : H₂O<H₂S<H₂Se<H₂Te