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Which given order is wrong according to ...

Which given order is wrong according to given property :-

A

`K_(2)OgtCaO` basic nature

B

`NH_(3)gtH_(2)O` basic nature

C

`BH_(3)gtNH_(3)` Acidic nature

D

`KHgtNaH` acidic nature

Text Solution

AI Generated Solution

The correct Answer is:
To determine which order is wrong according to the given property, we need to analyze each option based on the basicity and acidity of the compounds mentioned. ### Step-by-Step Solution: 1. **Understanding Basicity and Acidity**: - Basicity refers to the ability of a substance to donate electrons or accept protons. - Acidity refers to the ability of a substance to accept electrons or donate protons. 2. **Analyzing Option A: Potassium Oxide vs. Calcium Oxide**: - Potassium (K) is in Group 1 and Calcium (Ca) is in Group 2 of the periodic table. - Basicity increases down the group. Therefore, K2O (Potassium Oxide) should be more basic than CaO (Calcium Oxide). - **Conclusion**: Option A is correct. 3. **Analyzing Option B: Ammonia vs. Water**: - Ammonia (NH3) contains nitrogen from Group 15, while water (H2O) contains oxygen from Group 16. - Ammonia is more basic than water because nitrogen has a lower electronegativity than oxygen, allowing it to donate its lone pair more readily. - **Conclusion**: Option B is correct. 4. **Analyzing Option C: BH3 vs. NH3**: - Boron (B) is in Group 13 and nitrogen (N) is in Group 15. - As we move across the period, the acidic character decreases. Therefore, NH3 (Ammonia) is less acidic than BH3 (Boron Hydride). - **Conclusion**: Option C is correct. 5. **Analyzing Option D: KH vs. NaH**: - Potassium (K) and Sodium (Na) are both in Group 1. - As we move down the group, the acidic character decreases. Therefore, KH (Potassium Hydride) should be less acidic than NaH (Sodium Hydride). - **Conclusion**: Option D is incorrect because it states that KH is more acidic than NaH, which contradicts the trend. ### Final Answer: The wrong order according to the given property is **Option D: KH is more acidic than NaH**. ---
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