Question:
Statement-I: An element ‘X’ of P-block forms a hydride H–X, which has the longest bond length, then element ‘X’ will have the shortest covalent radius.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.
Statement-I: An element ‘X’ of P-block forms a hydride H–X, which has the longest bond length, then element ‘X’ will have the shortest covalent radius.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.
Show Hint
Remember, the maximum covalency of Phosphorus is 5, and its B.P. is the least among Group 15 hydrides.
Updated On: Mar 27, 2026
- Both statements are correct.
- Statement-I is correct; statement-II is incorrect.
- Statement-I is incorrect; statement-II is correct.
- Both statements are incorrect.
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the statements.
- Statement-I is incorrect because a longer bond length typically corresponds to a larger atomic size, which is not necessarily related to a shorter covalent radius.
- Statement-II is incorrect because the least boiling point (B.P.) among the Group 15 hydrides is for PH$_3$ (Phosphine), and the maximum covalency of Phosphorus (P) is 5, not 4.
Step 2: Conclusion.
Both statements are incorrect. The correct answer is (4).
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