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

The correct option that is not strictly arranged according to the indicated property is:
H₂O < H₂S < H₂Se < H₂Te : Increasing pK_a values

To understand why this is the incorrect arrangement, let us analyze the physical and chemical principles governing each of the given options:

1. Group 16 Hydrides (pK_a Values):
As we move down Group 16 (from O to Te), the size of the central atom increases (O < S < Se < Te). This increase in size leads to an increase in the H-E bond length (where E is the Group 16 element) and a corresponding decrease in the bond dissociation enthalpy. Since the H-E bond becomes weaker down the group, the ease of releasing hydrogen as proton ($H^{+}$) increases, which means the acidic strength increases:
$H_{2}O<H_{2}S<H_2\mathrm{Se}<H_2\mathrm{Te}$
Acidic strength is directly proportional to the acid dissociation constant ($K_a$) and inversely proportional to the $p{K}_a$ value:
$p{K}_a=-\log \left(K_a\right)$
Therefore, as the acidic strength increases down the group, the $p{K}_a$ values must decrease. Thus, the sequence H₂O < H₂S < H₂Se < H₂Te represents a sequence of decreasing pK_a values, making this sequence incorrect relative to its label.

2. Group 15 Hydrides (Acidic Character):
Down Group 15, the size of the central atom increases (N < P < As < Sb), causing the M-H bond length to increase and the bond dissociation enthalpy to decrease. This makes it easier to donate protons ($H^{+}$) down the group. Therefore, the acidic character increases as:
$N{H}_3<P{H}_3<\mathrm{As}{H}_3<\mathrm{Sb}{H}_3$
This sequence is correctly matched with its property.

3. Group 14 Dioxides (Oxidizing Power):
Down Group 14, due to the inert pair effect, the stability of the +4 oxidation state decreases, while the stability of the +2 oxidation state increases. Consequently, dioxides containing elements in the +4 state become increasingly stronger oxidizing agents down the group because they are more easily reduced to the +2 state. Thus, the oxidizing power increases as:
$C{O}_2<\mathrm{Si}{O}_2<\mathrm{Sn}{O}_2<\mathrm{Pb}{O}_2$
This sequence is correctly matched with its property.

4. Group 17 Hydrides (Acidic Strength):
As the size of the halogen atom increases down Group 17 (F < Cl < Br < I), the H-X bond length increases and the bond strength decreases. This facilitates the release of protons ($H^{+}$) in aqueous solution, thereby increasing the acidic strength:
$HF<H\mathrm{Cl}<H\mathrm{Br}<HI$
This sequence is correctly matched with its property.