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

Options

A. `NH_3ltPH_3ltAsH_3ltSbH_3` : Increasing acidic character
B. `CO_2ltSiO_2ltSnO_2ltPbO_2` : Increasing oxidizing power
C. `HFltHClltHBrltHI` : Increasing acidic strength
D. `H_2OltH_2SltH_2SeltH_2Te` : Increasing `pK_a` values

Answer

To solve the question, we need to analyze the given sequences based on the properties indicated and identify which one does not follow the expected trend. Let's break it down step by step.

### Step 1: Analyze Acidic Strength
1. **Given Sequence**: HF, HCl, HBr, HI
2. **Trend**: As we move down the group (from HF to HI), the acidic strength increases due to the increasing atomic size which leads to longer bond lengths. This results in weaker bonds that break more easily, releasing H⁺ ions.
3. **Conclusion**: The sequence is correct as HI is the strongest acid and HF is the weakest.

**Hint**: Remember that bond strength decreases as atomic size increases, leading to stronger acids.

### Step 2: Analyze pKa Values
1. **Given Sequence**: H2O, H2S, H2Se, H2Te
2. **Trend**: Similar to acidic strength, as we move down the group, the acidic strength increases, which means the Ka value increases and the pKa value decreases. Therefore, H2Te should have the lowest pKa and H2O the highest.
3. **Conclusion**: The sequence is incorrect because it states that H2O has the lowest pKa, which contradicts the expected trend.

**Hint**: pKa is inversely related to acid strength; lower pKa means stronger acid.

### Step 3: Analyze Acidic Strength in Group 15
1. **Given Sequence**: NH3, PH3, AsH3, SbH3
2. **Trend**: As we move down the group, the bond length increases, leading to weaker bonds and thus stronger acids. Hence, the order should be SbH3 > AsH3 > PH3 > NH3.
3. **Conclusion**: This sequence is correct.

**Hint**: Similar to previous steps, larger atoms lead to weaker bonds and stronger acids.

### Step 4: Analyze Oxidizing Power
1. **Given Sequence**: CO2, SiO2, SnO2, PbO2
2. **Trend**: As we go down the group, the inert pair effect stabilizes the lower oxidation states. Thus, the oxidizing power decreases down the group, with PbO2 being the strongest oxidizer and CO2 being the weakest.
3. **Conclusion**: This sequence is also correct.

**Hint**: The inert pair effect leads to stability in lower oxidation states, affecting oxidizing power.

### Final Conclusion
The only sequence that does not follow the expected trend is the one related to pKa values, where H2O is incorrectly stated to have the lowest pKa.

### Answer: The incorrect arrangement is the pKa values: H2O, H2S, H2Se, H2Te.

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