In which of the following arrangements, the sequence is not strictly according to the property written against it?

We have to check, for each option, whether the order shown really follows the stated periodic/property trend. Whichever order is wrong will be the answer.

Case A : $$\text{CO}_2 \lt \text{SiO}_2 \lt \text{SnO}_2 \lt \text{PbO}_2$$ - increasing oxidising power

Down group 14 the oxidation state $$+4$$ becomes less stable while $$+2$$ becomes more stable (inert‐pair effect). Hence the higher oxide $$\text{PbO}_2$$ ($$+4$$) very readily gets reduced to $$\text{Pb}^{2+}$$, making it a strong oxidising agent. $$\text{SnO}_2$$ is moderately oxidising, whereas $$\text{SiO}_2$$ is almost chemically inert and $$\text{CO}_2$$ is not an oxidising agent at all. Thus oxidising power indeed increases downwards as written. Option A is correct with respect to the property.

Case B : $$\text{NH}_3 \lt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3$$ - increasing basic strength

Basicity of group 15 hydrides depends on the availability of the lone-pair on the central atom. Down the group, atomic size increases, the lone-pair becomes more diffused, and its donating tendency falls. Therefore basic strength follows the order

$$\text{SbH}_3 \lt \text{AsH}_3 \lt \text{PH}_3 \lt \text{NH}_3$$

This is exactly the reverse of that given in the option. Hence the sequence supplied is NOT in accordance with increasing basic strength. So Option B is incorrect with respect to the stated property.

Case C : $$\text{HF} \lt \text{HCl} \lt \text{HBr} \lt \text{HI}$$ - increasing acid strength

For hydrogen halides, acid strength mainly depends on H-X bond dissociation energy, which decreases sharply from $$\text{HF}$$ to $$\text{HI}$$. Therefore $$\text{HF}$$ is the weakest and $$\text{HI}$$ is the strongest acid in water. The given order is correct, so Option C matches the trend.

Case D : $$\text{B} \lt \text{C} \lt \text{O} \lt \text{N}$$ - increasing first ionisation enthalpy

Across the second period, first ionisation enthalpy generally increases, but there are two well-known exceptions:

(i) $$\text{B}$$ is lower than $$\text{Be}$$, and (ii) $$\text{O}$$ is lower than $$\text{N}$$ (half-filled $$2p^3$$ stability for $$\text{N}$$).

The numerical values (kJ mol-1) are $$\text{B}=801 \lt \text{C}=1086 \lt \text{O}=1314 \lt \text{N}=1402$$, precisely as listed. Therefore Option D is also correct.

Only Option B fails to follow the stated trend. Hence,

Option B which is: $$\text{NH}_3 \lt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3$$ : increasing basic strength