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

Options

A. `CO_(2) lt SiO_(2) lt SnO_(2) lt PbO_(2)` : increasing oxidising power
B. `HF lt HCl lt HBr lt HI` : increasing acid strength
C. `NH_(3) lt PH_(3) lt AsH_(3) lt SbH_(3)` : increasing basic strength
D. `B lt C lt O lt N` : increasing first ionization enthalpy

Answer

To solve the question, we need to analyze each of the given options and determine if the sequences provided are strictly in accordance with the properties mentioned. Let's go through each option step by step:

### Step 1: Analyze Option A
**Sequence:** CO2 < SiO2 < SnO2 < PbO2
**Property:** Increasing oxidizing power

- All compounds are in the +4 oxidation state.
- Lead (Pb) in PbO2 is more stable in the +2 state due to the inert pair effect, meaning it has a greater tendency to reduce (and thus oxidize others).
- Therefore, the order of oxidizing power is correct.

**Conclusion:** This option is correct.

### Step 2: Analyze Option B
**Sequence:** HF < HCl < HBr < HI
**Property:** Increasing acid strength

- As we move down the group of hydrohalic acids, the bond length increases (HI > HBr > HCl > HF).
- A longer bond length means the bond is weaker and can dissociate more easily, leading to stronger acids.
- Thus, the order of acid strength is also correct.

**Conclusion:** This option is correct.

### Step 3: Analyze Option C
**Sequence:** NH3 < PH3 < AsH3 < SbH3
**Property:** Increasing basic strength

- Basic strength is determined by the ability to donate a lone pair of electrons.
- NH3 (ammonia) has a small size and a high tendency to donate its lone pair compared to PH3, AsH3, and SbH3.
- However, as we move down the group, the basicity decreases due to the increasing size and decreasing electronegativity of the central atom.
- Therefore, the order of basic strength should actually be NH3 > PH3 > AsH3 > SbH3, which contradicts the given sequence.

**Conclusion:** This option is incorrect.

### Step 4: Analyze Option D
**Sequence:** B < C < O < N
**Property:** Increasing first ionization enthalpy

- For elements in the second period, the ionization energy generally increases from left to right due to increasing nuclear charge.
- However, nitrogen has a half-filled p-orbital configuration, which is more stable, leading to a higher ionization energy than oxygen.
- Thus, the correct order should be B < C < N < O.

**Conclusion:** This option is also incorrect.

### Final Conclusion
The question asks for the arrangement that is not strictly according to the property written against it. We found that:

- Option A is correct.
- Option B is correct.
- Option C is incorrect (the order is not correct).
- Option D is also incorrect.

Thus, the answer to the question is **Option C**.

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