In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it ?

The correct answer is: H₂O < H₂S < H₂Se < H₂Te : Increasing pK_a values

This is the arrangement that is NOT strictly according to the property indicated. Let us verify each option one by one.

Option A: CO₂ < SiO₂ < SnO₂ < PbO₂ — Increasing oxidizing power

These are the dioxides (i.e., +4 oxidation state) of Group 14 elements. As we move down the group, the stability of the +4 oxidation state decreases due to the inert pair effect. This means the heavier dioxides (especially PbO₂ and SnO₂) have a strong tendency to get reduced to the +2 state, making them powerful oxidizing agents. CO₂ is extremely stable and shows virtually no oxidizing tendency. Therefore, the sequence of increasing oxidizing power is correct: CO₂ < SiO₂ < SnO₂ < PbO₂. ✔️

Option B: HF < HCl < HBr < HI — Increasing acidic strength

For binary hydracids of the same group, as we go down the group, the size of the halogen atom increases, the H–X bond length increases, and the bond dissociation enthalpy decreases. This makes it progressively easier to release H⁺ in aqueous solution. Hence, acidic strength increases in the order HF < HCl < HBr < HI. This sequence is correct. ✔️

Option C: H₂O < H₂S < H₂Se < H₂Te — Increasing pK_a values

These are the hydrides of Group 16 elements. As we move down the group from O to Te, the atomic size increases, the H–X bond becomes longer and weaker, and hence the bond dissociation enthalpy decreases. This makes it easier to release H⁺, so acidic strength increases in the order:

H₂O < H₂S < H₂Se < H₂Te

Now, pK_a is defined as:

${\mathrm{pK}}_a=-\log \left(K_a\right)$

Since pK_a is the negative logarithm of the acid dissociation constant (K_a), a stronger acid has a higher K_a but a lower pK_a. In other words, pK_a is inversely related to acid strength.

Therefore, since acid strength increases as H₂O < H₂S < H₂Se < H₂Te, the pK_a values must decrease in this order:

H₂O > H₂S > H₂Se > H₂Te (Decreasing pK_a)

The given sequence claims pK_a increases in this order, which is the exact opposite of the actual trend. Hence, this arrangement is NOT correct. ❌

Option D: NH₃ < PH₃ < AsH₃ < SbH₃ — Increasing acidic character

These are the hydrides of Group 15 elements. As we go down the group, the central atom becomes larger, the M–H bond becomes longer and weaker, and the bond dissociation enthalpy decreases. This facilitates the release of H⁺, so the acidic character increases from NH₃ to SbH₃. This sequence is correct. ✔️

Conclusion: The sequence in Option C is not strictly according to the property indicated, because pK_a values actually decrease (not increase) from H₂O to H₂Te as acid strength increases down the group.