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

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

This sequence is incorrect because basic strength actually decreases down the group. 

The correct order is:

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

[A] $$\text{CO}_2 < \text{SiO}_2 < \text{SnO}_2 < \text{PbO}_2$$ : increasing oxidising power

• Correct Trend: Down Group 14, the inert pair effect becomes more pronounced. This means the $$+2$$ oxidation state becomes more stable than the $+4$ oxidation state.

[B] $$\text{HF} < \text{HCl} < \text{HBr} < \text{HI}$$ : increasing acid strength

• Correct Trend: Down Group 17, the size of the halogen atom increases, which increases the $$\text{H-X}$$ bond length. A longer bond is weaker and breaks more easily to release $$\text{H}^+$$ ions.
  

[D] $$\text{B} < \text{C} < \text{O} < \text{N}$$ : increasing first ionization enthalpy

• Correct Trend: Generally, ionization energy increases across a period.

Hence, Option C is correct.