NEET 2021 Chemistry Question 91 Solution

For Group 16 hydrides ($H_2O, H_2S, H_2Se, H_2Te$), the size of the central element $E$ increases down the group ($O < S < Se < Te$), leading to longer $E-H$ bonds and decreasing $E-H$ bond dissociation enthalpies: $$H_2O (463\text{ kJ/mol}) > H_2S (347\text{ kJ/mol}) > H_2Se (276\text{ kJ/mol}) > H_2Te (238\text{ kJ/mol})$$ Consequently, ease of proton release increases down the group, so acidic strength increases: $$\text{Acidic Strength: } H_2O < H_2S < H_2Se < H_2Te \implies K_a(H_2O) < K_a(H_2S) < K_a(H_2Se) < K_a(H_2Te)$$

Because $pK_a = -\log_{10} K_a$, an increase in acid dissociation constant $K_a$ strictly corresponds to a decrease in $pK_a$: - $pK_a(H_2O) \approx 14.0$ (or $15.74$ auto-ionization) - $pK_a(H_2S) \approx 7.0$ - $pK_a(H_2Se) \approx 3.89$ - $pK_a(H_2Te) \approx 2.6$ Thus, the sequence $H_2O < H_2S < H_2Se < H_2Te$ is the order of DECREASING pKa, making the label "Increasing $pK_a$ values" false.

- Option (1): $HF < HCl < HBr < HI$ correctly follows decreasing $H-X$ bond dissociation enthalpy, increasing acidic strength. - Option (3): $NH_3 < PH_3 < AsH_3 < SbH_3$ correctly reflects decreasing $E-H$ bond stability, increasing acidic character. - Option (4): In Group 14, due to the inert pair effect, the stability of the $+2$ state increases relative to $+4$ down the column ($Pb^{2+} > Pb^{4+}$). Thus, $PbO_2$ readily accepts electrons to reduce to $Pb^{2+}$, functioning as a powerful oxidizing agent.