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

Let us analyze each of the given sequences step-by-step to understand why the chosen arrangement is not strictly according to the properties indicated:

1. Sequence: HF < HCl < HBr < HI : Increasing acidic strength
As we move down the halogen group (Group 17), the size of the halogen atom increases in the order: F < Cl < Br < I. Consequently, the H-X bond length increases and the bond dissociation enthalpy decreases significantly. Since the H-I bond is the weakest, HI releases protons ($H^{+}$) most easily, making it the strongest acid. Thus, the acidic strength increases in the order of HF < HCl < HBr < HI. This arrangement is correct.

2. Sequence: NH₃ < PH₃ < AsH₃ < SbH₃ : Increasing acidic character
For hydrides of Group 15, as we go down the group from N to Sb, the atomic size of the element increases. This increases the E-H bond length (where E is the Group 15 element) and decreases the bond dissociation energy. As the bond becomes weaker down the group, the ease of releasing $H^{+}$ ions increases, which increases the acidic character. Thus, acidic character increases from NH₃ to SbH₃. This arrangement is correct.

3. Sequence: H₂O < H₂S < H₂Se < H₂Te : Increasing pK_a values
Let us evaluate the acid strength and ${\text{pK}}_{\text{a}}$ relationship for Group 16 hydrides:
As we go down Group 16 (O to Te), the atomic size increases, which decreases the H-E bond dissociation energy. Because the H-E bond becomes weaker, the ease of donation of $H^{+}$ increases, meaning the acidic strength increases in the order:
H₂O < H₂S < H₂Se < H₂Te
Recall the relationship between acid dissociation constant (${\text{K}}_{\text{a}}$), ${\text{pK}}_{\text{a}}$, and acidic strength:
Acidic strength is directly proportional to ${\text{K}}_{\text{a}}$ and inversely proportional to ${\text{pK}}_{\text{a}}$ because:
${\text{pK}}_{\text{a}}=-{\text{log}}_{10}{\text{K}}_{\text{a}}$
Therefore, a stronger acid has a higher ${\text{K}}_{\text{a}}$ but a lower ${\text{pK}}_{\text{a}}$ value.
Since H₂Te is the strongest acid in this series, it has the lowest ${\text{pK}}_{\text{a}}$ value, while H₂O (the weakest acid) has the highest ${\text{pK}}_{\text{a}}$ value.
Thus, the correct sequence for increasing ${\text{pK}}_{\text{a}}$ values (which corresponds to decreasing acidic strength) is:
H₂Te < H₂Se < H₂S < H₂O
Hence, the sequence H₂O < H₂S < H₂Se < H₂Te represents decreasing ${\text{pK}}_{\text{a}}$ values, making this arrangement incorrect (not strictly according to the property indicated).